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Chapter 2

Mathematics

+2 college entrance (after SEE) past exam questions

Past questions and answers

430 questions set from this chapter, 4 of them more than once; 17 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 29 exams
  • Asked 2 times
  • 2082 · United Academy Science Model Set 2 · 1 mark
  • 2082 · United Academy Science Model Set 4 · 1 mark

When a polynomial p(x) is divided by (x + (a) , remainder R is: a. p(a) (b) 0 (c) p(-a) (d) p(x + a)

Similar questions: When a polynomial p(x) is divided by its factor (2082 · United Academy Science Model Set 3)

Answer

(c) p(−a)p(-a) By the remainder theorem, dividing by (x+a)(x+a) gives remainder p(−a)p(-a), since x+a=0x+a=0 at x=−ax=-a.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 2 · 1 mark

If the degree of dividend is 4m and divisor is m, then what will be the degree of Quotient? (a) 4m (b) 3m (c) 5m (d) m

Similar questions: If the degree of divisor is m and Quotient is 3m (2082 · United Academy Science Model Set 3) · If the degree of dividend is 4m and Quotient is (2082 · United Academy Science Model Set 4)

Answer

(b) 3m3m Degree of quotient = degree of dividend −- degree of divisor =4m−m=3m=4m-m=3m.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 2 · 1 mark

Add next two terms in the sequence o, t, t, f, f, s, s, e, n,, (a) n, e (b) t, i (c) t, e (d) e, t

Similar questions: Add next two terms in the sequence s, m, t, w, t (2082 · United Academy Science Model Set 3) · Add next two terms in the sequence j, f, m, a, m (2082 · United Academy Science Model Set 4)

Answer

(c) t, e The letters are the first letters of the numbers one, two, three, four, five, six, seven, eight, nine, so the next are ten and eleven: t, e.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 2 · 1 mark

What is the slope of a straight line if inclination is 45°? (a) 4 (b) 2 (c) 1 (d) 5

Similar questions: What is the slope of a straight line if inclinat (2082 · United Academy Science Model Set 3) · What is the slope of a straight line if inclinat (2082 · United Academy Science Model Set 4)

Answer

(c) 1

Slope m=tan⁡θ=tan⁡45∘=1m=\tan\theta=\tan45^\circ=1.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 2 · 1 mark

What is the equation of the pair of straight lines passing through the origin and perpendicular to the lines represented by ax2+2hxy+by2=0ax^2+2hxy+by^2=0? (a) bx2+2hxy+ay2=0bx^2+2hxy+ay^2=0 (b) bx2−2hxy+ay2=0bx^2-2hxy+ay^2=0 (c) ax2−2hxy+by2=0ax^2-2hxy+by^2=0 (d) ax2−2hxy−by2=0ax^2-2hxy-by^2=0

Similar questions: What is the equation of the pair of straight lin (2082 · United Academy Science Model Set 4) · The single equation of pair of lines passing thr (2082 · United Academy Science Model Set 7)

Answer

(b) bx2−2hxy+ay2=0bx^2-2hxy+ay^2=0

A line y=mxy=mx from the pair satisfies bm2+2hm+a=0bm^2+2hm+a=0. The perpendicular line has slope −1m-\dfrac1m, so m=−1mm=-\dfrac1{m}

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 2 · 1 mark

If A+B+C=90°, then what will be the value of sin (A+B)? (a) Sin C (b) cos C (c) – sin C (d) – cos C

Similar questions: If A+B = 100g , then what will be the value of s (2082 · United Academy Science Model Set 3) · If A+B+C=360°, then what will be the value of si (2082 · United Academy Science Model Set 4)

Answer

(b) cos⁡C\cos C

A+B=90∘−CA+B=90^\circ-C, so sin⁡(A+B)=sin⁡(90∘−C)=cos⁡C\sin(A+B)=\sin(90^\circ-C)=\cos C.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 3 · 1 mark

When a polynomial p(x) is divided by its factor (x - (a) , remainder R is: a. p(-a) (b) 0 (c) p(-x) (d) p(x + a)

Similar questions: When a polynomial p(x) is divided by (x + (2082 · United Academy Science Model Set 2)

Answer

(b) 0

By the remainder theorem, dividing p(x)p(x) by (x−a)(x-a) leaves remainder p(a)p(a). Since (x−a)(x-a) is a factor, p(a)=0p(a)=0, so R=0R=0.

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 3 · 1 mark

If the degree of divisor is m and Quotient is 3m, then what will be the degree of dividend? (a) 4m (b) 3m (c) 7m (d) m

Similar questions: If the degree of dividend is 4m and divisor is m (2082 · United Academy Science Model Set 2) · If the degree of dividend is 4m and Quotient is (2082 · United Academy Science Model Set 4)

Answer

(a) 4m

Degree of dividend = degree of divisor + degree of quotient =m+3m=4m= m+3m = 4m.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 3 · 1 mark

Add next two terms in the sequence s, m, t, w, t, _ , _. (a) v, b (b) f, s (c) y, u (d) z,

Similar questions: Add next two terms in the sequence o, t, t, f, f (2082 · United Academy Science Model Set 2) · Add next two terms in the sequence j, f, m, a, m (2082 · United Academy Science Model Set 4)

Answer

(b) f, s

The letters are the first letters of the days of the week: Sunday, Monday, Tuesday, Wednesday, Thursday, then Friday (f), Saturday (s).

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 3 · 1 mark

What is the slope of a straight line if inclination is 0°? (a) 1 (b) -2 (c) -1 (d) 0

Similar questions: What is the slope of a straight line if inclinat (2082 · United Academy Science Model Set 2) · What is the slope of a straight line if inclinat (2082 · United Academy Science Model Set 4)

Answer

(d) 0

Slope =tan⁡θ=tan⁡0∘=0=\tan\theta=\tan 0^\circ=0.

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 3 · 1 mark

If A+B = 100g , then what will be the value of sin (A+B)? (a) sin C (b) cos C (c) – sin C (d) – cos C

Similar questions: If A+B+C=90°, then what will be the value of sin (2082 · United Academy Science Model Set 2) · If A+B+C=360°, then what will be the value of si (2082 · United Academy Science Model Set 4)

Answer

(a) sin⁡C\sin C

The angles are in grades, where a straight angle is 200g200^g, so A+B+C=200gA+B+C=200^g for a triangle (the grade equivalent of 180∘180^\circ). Then A+B=200g−CA+B=200^g-C, and sin⁡(200g−C)=sin⁡(180∘−C)=sin⁡C\sin(200^g-C)=\sin(180^\circ-C)=\sin C.

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 4 · 1 mark

If the degree of dividend is 4m and Quotient is 3m, then what will be the degree of divisor? (a) 4m (b) 3m (c) 7m (d) m

Similar questions: If the degree of dividend is 4m and divisor is m (2082 · United Academy Science Model Set 2) · If the degree of divisor is m and Quotient is 3m (2082 · United Academy Science Model Set 3)

Answer

(d) mm

Degree of dividend = degree of divisor + degree of quotient, so

degree of divisor=4m−3m=m.\text{degree of divisor}=4m-3m=m.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 4 · 1 mark

Add next two terms in the sequence j, f, m, a, m, j, j, _ , _. (a) k, o (b) a, s (c) t, s (d) k, k

Similar questions: Add next two terms in the sequence o, t, t, f, f (2082 · United Academy Science Model Set 2) · Add next two terms in the sequence s, m, t, w, t (2082 · United Academy Science Model Set 3)

Answer

(b) a, s

The letters are the first letters of the months: January, February, March, April, May, June, July, then August (a) and September (s).

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 4 · 1 mark

What is the slope of a straight line if inclination is 135°? (a) 1 (b) -2 (c) -1 (d) -5

Similar questions: What is the slope of a straight line if inclinat (2082 · United Academy Science Model Set 2) · What is the slope of a straight line if inclinat (2082 · United Academy Science Model Set 3)

Answer

(c) −1-1

m=tan⁡135∘=tan⁡(180∘−45∘)=−tan⁡45∘=−1.m=\tan135^\circ=\tan(180^\circ-45^\circ)=-\tan45^\circ=-1.

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 4 · 1 mark

What is the equation of the pair of straight lines passing through the origin and perpendicular to the lines represented by bx2+2hxy+ay2=0bx^2+2hxy+ay^2=0? (a) bx2+2hxy+ay2=0bx^2+2hxy+ay^2=0 (b) bx2−2hxy+ay2=0bx^2-2hxy+ay^2=0 (c) ax2−2hxy+by2=0ax^2-2hxy+by^2=0 (d) ax2−2hxy−by2=0ax^2-2hxy-by^2=0

Similar questions: What is the equation of the pair of straight lin (2082 · United Academy Science Model Set 2) · The single equation of pair of lines passing thr (2082 · United Academy Science Model Set 7)

Answer

(c) ax2−2hxy+by2=0ax^2-2hxy+by^2=0

For the pair ax2+2hxy+by2=0ax^2+2hxy+by^2=0, the perpendicular pair through the origin is bx2−2hxy+ay2=0bx^2-2hxy+ay^2=0 (interchange aa and bb and change the sign of hh).

In the given equation bx2+2hxy+ay2=0bx^2+2hxy+ay^2=0 the coefficients of x2x^2 and y2y^2 are bb and aa, so applying the same rule gives ax2−2hxy+by2=0ax^2-2hxy+by^2=0.

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 4 · 1 mark

If A+B+C=360°, then what will be the value of sin (A+B)? (a) Sin C (b) cos C (c) – sin C (d) – cos C

Similar questions: If A+B+C=90°, then what will be the value of sin (2082 · United Academy Science Model Set 2) · If A+B = 100g , then what will be the value of s (2082 · United Academy Science Model Set 3)

Answer

(c) −sin⁡C-\sin C

Since A+B=360∘−CA+B=360^\circ-C,

sin⁡(A+B)=sin⁡(360∘−C)=−sin⁡C.\sin(A+B)=\sin(360^\circ-C)=-\sin C.

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  • Most repeated · 3 of 29 exams
  • 2082 · United Academy Science Model Set 7 · 1 mark

The single equation of pair of lines passing through origin and perpendicular to the lines represented by ax2+2hxy+by2=0ax^2+2hxy+by^2=0 is: (a) ax2+2hxy+by2=0ax^2+2hxy+by^2=0 (b) ax2+2hxy−by2=0ax^2+2hxy-by^2=0 (c) ax2−2hxy+by2=0ax^2-2hxy+by^2=0 (d) bx2−2hxy+ay2=0bx^2-2hxy+ay^2=0

Similar questions: What is the equation of the pair of straight lin (2082 · United Academy Science Model Set 2) · What is the equation of the pair of straight lin (2082 · United Academy Science Model Set 4)

Answer

(d) bx2−2hxy+ay2=0bx^2-2hxy+ay^2=0

The lines satisfy bm2+2hm+a=0b m^2+2hm+a=0 with slopes m1,m2m_1,m_2. The perpendicular lines through the origin have slopes −1m1,−1m2-\frac1{m_1},-\frac1{m_2}, so put m=−xym=-\frac xy (the slope is y/xy/x of the new lines, replaced by −x/y-x/y):

bx2y2−2hxy+a=0  ⇒  bx2−2hxy+ay2=0b\frac{x^2}{y^2}-2h\frac{x}{y}+a=0\;\Rightarrow\;bx^2-2hxy+ay^2=0

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  • Asked 2 times
  • 2025 Bridge Course Practice Set 5 · 1 mark
  • 2025 Bridge Course Practice Set 10 · 1 mark

The area of the triangle with the sides 5 cm, 12 cm and 13 cm is (a) 30 cm2^2 (b) 300 cm2^2 (c) 78 cm2^2 (d) None of these

Answer

(a) 30 cm2^2

Since 52+122=25+144=169=1325^2+12^2=25+144=169=13^2, the triangle is right-angled with legs 5 cm and 12 cm.

Area=12×5×12=30 cm2\text{Area}=\tfrac12\times5\times12=30\ \text{cm}^2

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  • Asked 2 times
  • 2025 Bridge Course Practice Set 5 · 1 mark
  • 2025 Bridge Course Practice Set 10 · 1 mark

The coordinate of the centre of the circle (x+h)2+(y+k)2=r2(x+h)^2+(y+k)^2=r^2 are : (a) (h,k)(h,k) (b) (−h,k)(-h,k) (c) (−h,−k)(-h,-k) (d) (h,−k)(h,-k)

Answer

(c) (−h,−k)(-h,-k)

Comparing with (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2, whose centre is (a,b)(a,b), we write (x+h)2=(x−(−h))2(x+h)^2=(x-(-h))^2 and (y+k)2=(y−(−k))2(y+k)^2=(y-(-k))^2. The centre is (−h,−k)(-h,-k).

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  • Asked 2 times
  • 2082 · United Academy Science Model Set 2 · 1 mark
  • 2082 · United Academy Science Model Set 4 · 1 mark

In the standard equation of parabola, y = a(x − h)² + k. If a >0, parabola opens: (a) upward (b) downward (c) towards right (d) towards left

Answer

(a) upward For y=a(x−h)2+ky=a(x-h)^2+k with a>0a>0, the vertex (h,k)(h,k) is the minimum point, so the parabola opens upward.

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  • 2018 Practice Test 2 · 1 mark

The value of pp for which the vectors a⃗=pi^+j^−2k^\vec a=p\hat i+\hat j-2\hat k and b⃗=i^+j^−3k^\vec b=\hat i+\hat j-3\hat k are perpendicular to each other is (a) 9 (b) −9-9 (c) 7 (d) −7-7

Similar questions: If vec a=3hat i-mhat j and vec b=10hat i+6hat j (2019 Science Sample Test Set A)

Answer

(d) −7-7

Perpendicular vectors have zero dot product:

a⃗⋅b⃗=p(1)+(1)(1)+(−2)(−3)=p+7=0⇒p=−7.\vec a\cdot\vec b = p(1) + (1)(1) + (-2)(-3) = p + 7 = 0 \Rightarrow p = -7.

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  • 2018 Practice Test 3 · 1 mark

If prices are reduced by 25% and the sales increase by 20%, what is the net effect on gross receipts? (a) increase by 5% (b) decrease by 5% (c) increase by 10% (d) decrease by 10%

Similar questions: If price is reduced by 25% and sales increase by (2018 Practice Test 4)

Answer

(d) decrease by 10%

Let the original price be PP and the sales SS. The new receipts are

0.75P×1.20S=0.90 PS0.75P \times 1.20S = 0.90\,PS

so gross receipts fall by 10%.

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  • 2018 Practice Test 4 · 1 mark

If price is reduced by 25% and sales increase by 20%, what is the net effect on gross receipts? (a) increase by 25% (b) decrease by 5% (c) increase by 10% (d) decrease by 10%

Similar questions: If prices are reduced by 25% and the sales incre (2018 Practice Test 3)

Answer

(d) decrease by 10%

Let the original receipts be 100100. New receipts =100×0.75×1.20=90=100\times0.75\times1.20=90. Receipts therefore decrease by 10%10\%.

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  • 2019 Science Sample Test Set A · 1 mark

If a⃗=3i^−mj^\vec a=3\hat i-m\hat j and b⃗=10i^+6j^\vec b=10\hat i+6\hat j are perpendicular to each other, what is the value of mm? (a) −5-5 (b) −6-6 (c) 5 (d) 2

Similar questions: The value of p for which the vectors vec a=phat (2018 Practice Test 2)

Answer

(c) 5

Perpendicular vectors have zero dot product: a⃗⋅b⃗=3(10)+(−m)(6)=30−6m=0⇒m=5.\vec a\cdot\vec b=3(10)+(-m)(6)=30-6m=0\Rightarrow m=5.

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  • 2025 Bridge Course Practice Set 10 · 1 mark

If two dice are thrown together, what is the probability of getting sum of 9 or 11 ? (a) 5/6 (b) 1/6 (c) 7/6 (d) 1/18

Similar questions: Two dice are thrown simultaneously. The probabil (2082 · United Academy Science Model Set 1)

Answer

(b) 1/6

Total outcomes =36=36. Sum 9: (3,6),(4,5),(5,4),(6,3)(3,6),(4,5),(5,4),(6,3), i.e. 4 ways. Sum 11: (5,6),(6,5)(5,6),(6,5), i.e. 2 ways.

P=4+236=16P=\frac{4+2}{36}=\frac16

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  • 2079 Grade XI Model Question · 1 mark

The equation x2−5xy+6y2=0x^2-5xy+6y^2=0 represents a: (a) circle (b) pair of parallel straight lines. (c) pair of perpendicular straight lines (d) pair of non perpendicular intersecting straight lines.

Similar questions: The equation y^2-x^2+2x-1=0 represents (2025 (2082) Grade XI Model Question Set A)

Answer

(d) pair of non perpendicular intersecting straight lines.

Factorise: x2−5xy+6y2=(x−2y)(x−3y)=0,x^2-5xy+6y^2=(x-2y)(x-3y)=0, which gives the lines x=2yx=2y and x=3yx=3y. Their slopes are 12\tfrac12 and 13\tfrac13, with product 16≠−1\tfrac16\neq -1. The lines are not parallel and not perpendicular, and they meet at the origin.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

The equation y2−x2+2x−1=0y^2-x^2+2x-1=0 represents (a) a pair of intersecting straight lines (b) a pair of parallel straight lines (c) a circle (d) a parabola

Similar questions: The equation x^2-5xy+6y^2=0 represents a: (2079 Grade XI Model Question)

Answer

(a) a pair of intersecting straight lines

y2−x2+2x−1=y2−(x−1)2=0y^2-x^2+2x-1=y^2-(x-1)^2=0, so y=x−1y=x-1 or y=−(x−1)y=-(x-1). These are two lines with slopes 11 and −1-1 meeting at (1,0)(1,0).

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  • 2082 · United Academy Science Model Set 1 · 1 mark

Two dice are thrown simultaneously. The probability of getting a sum of 9 is: (a) 110\frac1{10} (b) 310\frac3{10} (c) 19\frac19 (d) 49\frac49

Similar questions: If two dice are thrown together, what is the pro (2025 Bridge Course Practice Set 10)

Answer

(c) 19\frac19

Total outcomes =36=36. Sum 9: (3,6),(4,5),(5,4),(6,3)(3,6),(4,5),(5,4),(6,3), so 4 outcomes.

P=436=19P=\frac{4}{36}=\frac19

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  • 2082 · United Academy Science Model Set 2 · 1 mark

In which side from the line y = 3 is the solution region of y≤ 3 lie? (a) Right (b) Left (c) Up (d) Down

Similar questions: In which side from the line x = 2 is the solutio (2082 · United Academy Science Model Set 4)

Answer

(d) Down The region y≤3y\le 3 contains all points with yy-coordinate less than or equal to 3, which lie below the line.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

The equation of a circle (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2 touches X-axis if: (a) r=|x| (b) r=|h| (c) r=|y| (d) r=|k|

Similar questions: The equation of a circle (x − h)² + (y − k)² = r (2082 · United Academy Science Model Set 4)

Answer

(d) r=∣k∣r=|k|

The distance of the centre (h,k)(h,k) from the X-axis is ∣k∣|k|. The circle touches the X-axis when this distance equals the radius, so r=∣k∣r=|k|.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

What will be the length of a string if the height of kite from ground is 30 m and angle made by string is 30°? (a) 30√3m (b) 60m (c) 30m (d) 30√2m

Similar questions: What will be the length of a string if the heigh (2082 · United Academy Science Model Set 4)

Answer

(b) 60 m

The string is the hypotenuse: sin⁡30∘=30L\sin30^\circ=\dfrac{30}{L}, so L=301/2=60L=\dfrac{30}{1/2}=60 m.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

Which of the following is an example of non-isometric transformation? (a) Translation (b) Reflection (c) Rotation (d) Enlargement

Similar questions: Which of the following is not an example of isom (2082 · United Academy Science Model Set 4)

Answer

(d) Enlargement

Translation, reflection and rotation preserve lengths and shapes (isometric). Enlargement changes the size of the figure, so it is non-isometric.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

If the lower and upper quartile of any data are 10 and 30 respectively, what is the value of semi-interquartile range? (a) 30 (b) 20 (c) 15 (d) 10

Similar questions: If the Q.D. and upper quartile of any data are 1 (2082 · United Academy Science Model Set 4)

Answer

(d) 10

Semi-interquartile range =Q3−Q12=30−102=10=\dfrac{Q_3-Q_1}{2}=\dfrac{30-10}{2}=10.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

In which side from the line x = 2 is the solution region of x≤ 2 lie? (a) Right (b) Left (c) Up (d) Down

Similar questions: In which side from the line y = 3 is the solutio (2082 · United Academy Science Model Set 2)

Answer

(b) Left

The line x=2x=2 is vertical. The points satisfying x≤2x\le 2 have xx-coordinate less than or equal to 2, so they lie on the line or to its left. For example, the origin satisfies 0≤20\le 2 and lies to the left.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

The equation of a circle (x − h)² + (y − k)² = r² touches Y- axis if: (a) r=|x| (b) r=|h| (c) r=|y| (d) r=|k|

Similar questions: The equation of a circle (x-h)^2+(y-k)^2=r^2 tou (2082 · United Academy Science Model Set 2)

Answer

(b) r=∣h∣r=|h|

The centre is (h,k)(h,k), so its distance from the Y-axis (x=0x=0) is ∣h∣|h|. The circle touches the Y-axis when this distance equals the radius, so r=∣h∣r=|h|.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

What will be the length of a string if the height of kite from ground is 30 m and angle made by string is 60°? (a) 60√3m (b) 60m (c) 20√3m (d) 15√3m

Similar questions: What will be the length of a string if the heigh (2082 · United Academy Science Model Set 2)

Answer

(c) 20320\sqrt{3} m

The string is the hypotenuse, so sin⁡60∘=30L\sin 60^\circ = \dfrac{30}{L}.

L=303/2=603=203 mL = \frac{30}{\sqrt{3}/2} = \frac{60}{\sqrt{3}} = 20\sqrt{3}\ \text{m}

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  • 2082 · United Academy Science Model Set 4 · 1 mark

Which of the following is not an example of isometric transformation? (a) Translation (b) Reflection (c) Rotation (d) Enlargement

Similar questions: Which of the following is an example of non-isom (2082 · United Academy Science Model Set 2)

Answer

(d) Enlargement

Translation, reflection and rotation preserve length and shape (they are isometries). Enlargement changes the size of the figure, so it is not isometric.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

If the Q.D. and upper quartile of any data are 10 and 30 respectively, what is the value of semi-interquartile range? (a) 30 (b) 20 (c) 15 (d) 10

Similar questions: If the lower and upper quartile of any data are (2082 · United Academy Science Model Set 2)

Answer

(d) 10

Quartile deviation (Q.D.) is the same thing as the semi-interquartile range, Q3−Q12\dfrac{Q_3-Q_1}{2}. So the semi-interquartile range is 10 (the upper quartile value is extra information).

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  • 2082 · United Academy Science Model Set 5 · 1 mark

If the profit is 20% of cost price, what is the percentage profit of selling price? (a) 25% (b) 16.67% (c) 33.33% (d) None

Similar questions: If the ratio of cost price and profit is 5:1, an (2082 · United Academy Science Model Set 6)

Answer

(b) 16.67%

Let CP=100CP=100. Profit =20=20, so SP=120SP=120.

Profit % of SP=20120×100=16.67%\text{Profit \% of } SP = \frac{20}{120}\times 100 = 16.67\%

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  • 2082 · United Academy Science Model Set 5 · 1 mark

Which of the following statement is not true? (a) Equal chords are equidistant from the center. (b) Perpendicular bisector of the chord passes through the center of the circle. (c) Angle in a semicircle are two right angles. (d) Inscribed angle is half the central angle

Similar questions: Which of the following statement is true? (2082 · United Academy Science Model Set 6)

Answer

(c) Angle in a semicircle are two right angles.

The angle in a semicircle is a single right angle (90∘90^\circ), not two right angles. The other statements are true theorems.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

If the ratio of cost price and profit is 5:1, and the selling price is Rs 180. What is the cost price? (a) ) Rs 200 (b) Rs 150 (c) Rs 160 (d) None

Similar questions: If the profit is 20% of cost price, what is the (2082 · United Academy Science Model Set 5)

Answer

(b) Rs 150

CP : profit = 5 : 1, so SP = CP + profit = 6 parts = Rs 180. One part = Rs 30, so CP = 5×30=5 \times 30 = Rs 150.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

The buying rate and selling rate of 1 US dollar are Rs 139 and Rs 140 respectively. How much Nepali rupees can be exchanged with 1000 US dollars? (a) Rs 139,000 (b) Rs 139,500 (c) Rs 140,000 (d) Neither of these

Similar questions: The buying rate and selling rate of 1 US dollar (2082 · United Academy Science Model Set 7)

Answer

(a) Rs 139,000

When customers sell dollars to the bank, the bank buys at the buying rate of Rs 139. So 1000×139=1000 \times 139 = Rs 139,000.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

Which of the following statement is true? (a) Equal chords are equidistant from the circumference. (b) Perpendicular bisector of the chord passes through the center of the circle. (c) Angle at a center of a circle is two right angles. (d) Inscribed angle is double of the central angle.

Similar questions: Which of the following statement is not true? (2082 · United Academy Science Model Set 5)

Answer

(b) Perpendicular bisector of the chord passes through the center of the circle.

This is a standard circle theorem. Equal chords are equidistant from the centre (not the circumference), the angle at the centre on a diameter is two right angles only for a straight angle, and the central angle is double the inscribed angle, not the reverse.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

The buying rate and selling rate of 1 US dollar are Rs 139 and Rs 140 respectively. How much Nepali rupees can be exchanged with 1000 US dollars? (a) Rs 139,000 (b) Rs 140,000 (c) Rs 139,500 (d) Rs 13,90,000

Similar questions: The buying rate and selling rate of 1 US dollar (2082 · United Academy Science Model Set 6)

Answer

(a) Rs 139,000

The bank buys dollars at Rs 139, so 1000×139=1000\times139= Rs 139,000.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

Which of the following composite functions is an identity function? (a) f∘f−1(x)f\circ f^{-1}(x) (b) (f∘f)−1(x)(f\circ f)^{-1}(x) (c) f∘f(x)f\circ f(x) (d) f∘f(1x)f\circ f\left(\frac1x\right)

Similar questions: Which of the following functions is an identity (2082 · United Academy Science Model Set 8)

Answer

(a) f∘f−1(x)f\circ f^{-1}(x)

By definition f(f−1(x))=xf(f^{-1}(x))=x, which is the identity function. The other options do not simplify to xx in general.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

Which of the following functions is an identity function? (a) f(x) = 0 (b) g(x) = 3x (c) f(x) = x (d) f(x) = 2

Similar questions: Which of the following composite functions is an (2082 · United Academy Science Model Set 7)

Answer

(c) f(x) = x

An identity function maps every element to itself, f(x)=xf(x)=x. The others are a constant function (a, d) and a linear function that is not the identity (b).

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  • 2018 Practice Test 1 · 1 mark

In a jar, 1/2 of the marbles are red, 1/4 are white, and 1/5 are blue. What fraction of the marbles are neither red, white nor blue? (a) 19/20 (b) 1/20 (c) 15/20 (d) 1/10

Answer

(b) 1/20

Fraction that is red, white or blue:

12+14+15=10+5+420=1920\frac12+\frac14+\frac15=\frac{10+5+4}{20}=\frac{19}{20}

The remaining fraction is 1−1920=1201-\frac{19}{20}=\frac{1}{20}.

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  • 2018 Practice Test 1 · 1 mark

What is 10% of 20% of 30%? (a) 0.006% (b) 0.6% (c) 6% (d) 60%

Answer

(b) 0.6%

10%×20%×30%=0.1×0.2×0.3=0.006=0.6%10\%\times 20\%\times 30\%=0.1\times0.2\times0.3=0.006=0.6\%

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  • 2018 Practice Test 1 · 1 mark

If x+ y=6, y+ z=7, and z +x=9, what is the arithmetic mean of x, y, and z? (a) 11/3 (b) 11/2 (c) 11 (d) 22/3

Answer

(a) 11/3

Adding the three equations gives 2(x+y+z)=6+7+9=222(x+y+z)=6+7+9=22, so x+y+z=11x+y+z=11.

Mean=x+y+z3=113\text{Mean}=\frac{x+y+z}{3}=\frac{11}{3}

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  • 2018 Practice Test 1 · 1 mark

A square of area 2 is inscribed in a circle. What is the area of the circle? (a) π/4 (b) π/2 (c) π (d) 2 π

Answer

(c) π\pi

Side of the square =2=\sqrt2, so its diagonal is 2+2=2\sqrt{2+2}=2. The diagonal is the diameter of the circle, so r=1r=1 and

Area=πr2=π\text{Area}=\pi r^2=\pi

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  • 2018 Practice Test 1 · 1 mark

When a digital clock reads 3:47, the sum of the digits is 14. How many minutes after 3:47 will the sum of the digits be 20 for the first time? (a) 42 (b) 132 (c) 192 (d) 251

Answer

(c) 192

From 3:47 to 3:59 the largest digit sum is 3+5+9=173+5+9=17, which is below 20. For hours 4 and 5 the minutes would need digit sums of 16 and 15, but the largest possible is 5+9=145+9=14. For hour 6 we need a minutes digit sum of 14, which first occurs at 6:59 (6+5+9=206+5+9=20).

Time elapsed from 3:47 to 6:59 is 3 h 12 min =180+12=192=180+12=192 minutes.

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  • 2018 Practice Test 1 · 1 mark

The value of 91/3⋅91/9⋅91/27⋯∞9^{1/3}\cdot 9^{1/9}\cdot 9^{1/27}\cdots\infty is (a) 9 (b) 1 (c) 3 (d) None of these

Answer

(c) 3

The exponents form a geometric series: 13+19+127+⋯=1/31−1/3=12\frac13+\frac19+\frac1{27}+\cdots=\dfrac{1/3}{1-1/3}=\dfrac12.

So the product is 91/2=39^{1/2}=3.

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  • 2018 Practice Test 1 · 1 mark

If tan⁡α=12\tan\alpha=\frac12, tan⁡β=13\tan\beta=\frac13, then α+β\alpha+\beta is equal to (a) π/2\pi/2 (b) π/3\pi/3 (c) π/4\pi/4 (d) π/6\pi/6

Answer

(c) π/4\pi/4

tan⁡(α+β)=12+131−12⋅13=5/65/6=1\tan(\alpha+\beta)=\frac{\tfrac12+\tfrac13}{1-\tfrac12\cdot\tfrac13}=\frac{5/6}{5/6}=1

Since both angles are acute, α+β=π/4\alpha+\beta=\pi/4.

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  • 2018 Practice Test 1 · 1 mark

The distance between the lines 3x +4y =6 and 6x +8y =15 is (a) 3/2 (b) 6 (c) 3/10 (d) 7/9

Answer

(c) 3/10

Rewrite the second line as 6x+8y=156x+8y=15 and the first as 6x+8y=126x+8y=12. They are parallel, so

d=∣15−12∣62+82=310.d=\frac{|15-12|}{\sqrt{6^2+8^2}}=\frac{3}{10}.

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  • 2018 Practice Test 1 · 1 mark

The acute angle between the lines 3x2−7xy−3y2=03x^2-7xy-3y^2=0 is (a) π/2\pi/2 (b) π/4\pi/4 (c) π/6\pi/6 (d) None of the above

Answer

(a) π/2\pi/2

For ax2+2hxy+by2=0ax^2+2hxy+by^2=0 the angle satisfies tan⁡θ=2h2−aba+b\tan\theta=\dfrac{2\sqrt{h^2-ab}}{a+b}. Here a+b=3+(−3)=0a+b=3+(-3)=0, so the denominator is zero and the lines are perpendicular, i.e. θ=π/2\theta=\pi/2.

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  • 2018 Practice Test 1 · 1 mark

Which of the following is the value of PP, if one root of x2−Px+1=0x^2-Px+1=0 is square of the other (a) −1-1 (b) −2-2 (c) 1,−21,-2 (d) 2,−12,-1

Answer

(d) 2,−12,-1

Let the roots be α\alpha and α2\alpha^2. The product of roots is α3=1\alpha^3=1, and the sum is P=α+α2P=\alpha+\alpha^2.

  • If α=1\alpha=1: P=1+1=2P=1+1=2.
  • If α=ω\alpha=\omega (a complex cube root of unity): P=ω+ω2=−1P=\omega+\omega^2=-1.

So P=2P=2 or −1-1.

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  • 2018 Practice Test 1 · 1 mark

The point of concurrency of three internal angles bisectors of a triangle is called (a) Orthocenter (b) Circumcentre (c) Incentre (d) Excentre

Answer

(c) Incentre

The three internal angle bisectors of a triangle meet at the incentre, which is equidistant from all three sides and is the centre of the inscribed circle.

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  • 2018 Practice Test 1 · 1 mark

From a set of number cards numbered from 1 to 20, a card is drawn randomly. Find the probability of getting a card exactly divisible by 3 or 4. (a) 11/20 (b) 1/2 (c) 9/20 (d) 1/20

Answer

(b) 1/2

Multiples of 3 in 1 to 20: 6 (3, 6, 9, 12, 15, 18). Multiples of 4: 5 (4, 8, 12, 16, 20). Multiples of both, i.e. of 12: 1 (12).

n(3 or 4)=6+5−1=10,P=1020=12.n(3\text{ or }4)=6+5-1=10,\qquad P=\frac{10}{20}=\frac12.

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  • 2018 Practice Test 1 · 1 mark

The median of the data: 8, 12,35,40,15,10,11,14 is (a) 4.5 (b) 12 (c) 13 (d) 14

Answer

(c) 13

Arranged in order: 8, 10, 11, 12, 14, 15, 35, 40. There are 8 values, so the median is the mean of the 4th and 5th:

12+142=13.\frac{12+14}{2}=13.

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  • 2018 Practice Test 1 · 1 mark

If A is any square matrix, then A + AT will be (a) A diagonal matrix (b) A symmentric matrix (c) A skew-symmetric matrix (d) Identity matrix

Answer

(b) A symmetric matrix

Let B=A+ATB=A+A^T. Then BT=AT+(AT)T=AT+A=BB^T=A^T+(A^T)^T=A^T+A=B, so BB is symmetric.

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  • 2018 Practice Test 1 · 1 mark

If the equation x2+2(K+2)x+9K=0x^2+2(K+2)x+9K=0 has equal roots, then the value of KK is (a) 1 (b) 2 (c) 3 (d) 0

Answer

(a) 1

For equal roots the discriminant is zero:

4(K+2)2−36K=0  ⇒  K2+4K+4=9K  ⇒  K2−5K+4=0  ⇒  (K−1)(K−4)=0.4(K+2)^2-36K=0\;\Rightarrow\;K^2+4K+4=9K\;\Rightarrow\;K^2-5K+4=0\;\Rightarrow\;(K-1)(K-4)=0.

So K=1K=1 or 44; of the given options, K=1K=1.

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  • 2018 Practice Test 1 · 1 mark

If a⃗=i^+j^−2k^\vec a=\hat i+\hat j-2\hat k and b⃗=2i^−j^−k^\vec b=2\hat i-\hat j-\hat k are any two vectors, then the angle between a⃗\vec a and b⃗\vec b is (a) π/6\pi/6 (b) π/3\pi/3 (c) 00 (d) π/2\pi/2

Answer

(b) π/3\pi/3

a⃗⋅b⃗=(1)(2)+(1)(−1)+(−2)(−1)=3,∣a⃗∣=∣b⃗∣=6.\vec a\cdot\vec b=(1)(2)+(1)(-1)+(-2)(-1)=3,\qquad |\vec a|=|\vec b|=\sqrt6.

cos⁡θ=36=12  ⇒  θ=π3.\cos\theta=\frac{3}{6}=\frac12\;\Rightarrow\;\theta=\frac{\pi}{3}.

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  • 2018 Practice Test 1 · 1 mark

If SinA, SinB, SinC are in (a) P. then a,b,c are in A. A.P. (b) G.P. (c) H.P. (d) None

Answer

(a) a,b,ca,b,c are in A.P.

By the sine rule a=2Rsin⁡Aa=2R\sin A, b=2Rsin⁡Bb=2R\sin B, c=2Rsin⁡Cc=2R\sin C, so the sides are proportional to the sines. If sin⁡A,sin⁡B,sin⁡C\sin A,\sin B,\sin C are in A.P., then a,b,ca,b,c are in A.P.

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  • 2018 Practice Test 1 · 1 mark

How many integers are solutions of the inequality 3|x| + 2 < 17? (a) 0 (b) 8 (c) 9 (d) 5

Answer

(c) 9

3∣x∣+2<17  ⇒  ∣x∣<5  ⇒  −5<x<5.3|x|+2<17\;\Rightarrow\;|x|<5\;\Rightarrow\;-5<x<5.

The integers are −4,−3,…,4-4,-3,\dots,4, which is 9 integers.

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  • 2018 Practice Test 1 · 1 mark

30 men work 10 hours a day to complete a piece of work in 25 days. How many men will be required to complete a piece of work twice as great in 15 days working 8 hours a day? (a) 130 (b) 125 (c) 12.5 (d) 75

Answer

(b) 125

Work done is proportional to men ×\times hours per day ×\times days.

30×10×251=M×8×152  ⇒  M=2×7500120=125.\frac{30\times10\times25}{1}=\frac{M\times8\times15}{2}\;\Rightarrow\;M=\frac{2\times7500}{120}=125.

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  • 2018 Practice Test 2 · 1 mark

Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color? (a) 4 (b) 8 (c) 12 (d) 13

Answer

(d) 13

In the worst case, Grace first takes out all 8 red and all 4 green jellybeans, which is 12 with no blue. The 13th jellybean must then be blue, so 13 guarantees one of each colour.

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  • 2018 Practice Test 2 · 1 mark

The function f(x) = –3x + 12 on R is (a) increasing (b) decreasing (c) constant (d) none of these

Answer

(b) decreasing

For f(x)=−3x+12f(x) = -3x + 12 the slope is f′(x)=−3<0f'(x) = -3 < 0 for all xx, so ff is decreasing on R\mathbb{R}.

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  • 2018 Practice Test 2 · 1 mark

The system of the equation x + 2y = –1, –2x – 4y = 2 has (a) only one solution (b) finitely many solutions (c) no solution (d) infinitely many solutions

Answer

(d) infinitely many solutions

Dividing the second equation by −2-2 gives x+2y=−1x + 2y = -1, which is the same as the first. The two equations represent the same line, so the system has infinitely many solutions.

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  • 2018 Practice Test 2 · 1 mark

The area of the triangle formed by the lines y−x=0y-x=0, y+x=0y+x=0 and x=cx=c is (a) cc sq. units (b) c2c^2 sq. units (c) 12c\frac12 c sq. units (d) 12c2\frac12 c^2 sq. units

Answer

(b) c2c^2 sq. units

The lines y=xy = x and y=−xy = -x meet at the origin, and meet x=cx = c at (c,c)(c, c) and (c,−c)(c, -c). The base is 2c2c and the height is cc:

Area=12×2c×c=c2.\text{Area} = \tfrac12 \times 2c \times c = c^2.

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  • 2018 Practice Test 2 · 1 mark

If A is an m × n matrix and B is a matrix given in such a way that AB and BA are both defined, then the size of the B matrix is (a) m × n (b) n × n (c) m × n (d) n × m

Answer

(d) n×mn \times m

For ABAB to be defined with AA being m×nm \times n, BB must have nn rows. For BABA to be defined, BB must have mm columns. So BB is n×mn \times m.

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  • 2018 Practice Test 2 · 1 mark

A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book? (a) 0.63 (b) 0.16 (c) 0.97 (d) 1.60

Answer

(c) 0.97

Assuming the two solve problems independently, P(A)=0.9P(A) = 0.9 and P(B)=0.7P(B) = 0.7:

P(A∪B)=0.9+0.7−0.9×0.7=1.6−0.63=0.97.P(A\cup B) = 0.9 + 0.7 - 0.9\times 0.7 = 1.6 - 0.63 = 0.97.

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  • 2018 Practice Test 2 · 1 mark

The number of non-empty proper subsets of order 6 is (a) 64 (b) 63 (c) 62 (d) 60

Answer

(c) 62

A set with 6 elements has 26=642^6 = 64 subsets. Removing the empty set and the set itself (the improper subset) leaves 64−2=6264 - 2 = 62.

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  • 2018 Practice Test 2 · 1 mark

If common difference is 5, then the total number of arithmetic means between 10 and 50 is (a) 8 (b) 9 (c) 5 (d) 7

Answer

(d) 7

Let there be nn arithmetic means, so 50 is the (n+2)(n+2)th term: 10+(n+1)⋅5=5010 + (n+1)\cdot 5 = 50.

n+1=8⇒n=7n+1 = 8 \Rightarrow n = 7

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  • 2018 Practice Test 2 · 1 mark

The area of triangle formed by the line 2x + 3y = 12 with the co-ordinate axes is (a) 24 (b) 12 (c) 6 (d) 3

Answer

(b) 12

Intercepts of 2x+3y=122x+3y=12: putting y=0y=0 gives x=6x=6; putting x=0x=0 gives y=4y=4.

Area=12×6×4=12\text{Area} = \tfrac12 \times 6 \times 4 = 12

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  • 2018 Practice Test 2 · 1 mark

In a race of 1 km. A beats B by 100m. B beats C by 50 m. C beats D by 40 m then A beats D by (a) 190 m (b) 820.8 m (c) 179.2 m (d) none

Answer

(c) 179.2 m

When A runs 1000 m, B runs 900 m. When B runs 1000 m, C runs 950 m. When C runs 1000 m, D runs 960 m.

When A runs 1000 m, B runs 900 m, so C runs 900×9501000=855900\times\frac{950}{1000}=855 m, and D runs 855×9601000=820.8855\times\frac{960}{1000}=820.8 m.

A beats D by 1000−820.8=179.21000-820.8 = 179.2 m.

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  • 2018 Practice Test 2 · 1 mark

If mm times the mmth term and nn times the nnth term of an AP are equal, then the (m+n)(m+n)th term is (a) mnmn (b) 1mn\frac1{mn} (c) 1 (d) 0

Answer

(d) 0

Let the first term be aa and common difference dd. Given m[a+(m−1)d]=n[a+(n−1)d]m[a+(m-1)d] = n[a+(n-1)d].

a(m−n)+d (m2−m−n2+n)=0⇒(m−n) [a+(m+n−1)d]=0a(m-n) + d\,(m^2 - m - n^2 + n) = 0 \Rightarrow (m-n)\,[a + (m+n-1)d] = 0

Since m≠nm\neq n, a+(m+n−1)d=0a+(m+n-1)d=0, which is the (m+n)(m+n)th term.

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  • 2018 Practice Test 2 · 1 mark

If the AM and GM of the roots of a quadratic equation be 8 and 5 respectively, then the equation is (a) x2+6x+25=0x^2+6x+25=0 (b) x2−16x+25=0x^2-16x+25=0 (c) x2−16x−25=0x^2-16x-25=0 (d) x2−13x+40=0x^2-13x+40=0

Answer

(b) x2−16x+25=0x^2-16x+25=0

AM=α+β2=8⇒α+β=16AM = \frac{\alpha+\beta}{2}=8 \Rightarrow \alpha+\beta=16 and GM=αβ=5⇒αβ=25GM=\sqrt{\alpha\beta}=5 \Rightarrow \alpha\beta=25.

The equation x2−(α+β)x+αβ=0x^2-(\alpha+\beta)x+\alpha\beta=0 is x2−16x+25=0x^2-16x+25=0.

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  • 2018 Practice Test 2 · 1 mark

The value of cos⁡11∘+sin⁡11∘cos⁡11∘−sin⁡11∘\dfrac{\cos 11^\circ+\sin 11^\circ}{\cos 11^\circ-\sin 11^\circ} is equal to the value of (a) tan⁡56∘\tan 56^\circ (b) sin⁡56∘\sin 56^\circ (c) cot⁡56∘\cot 56^\circ (d) tan⁡79∘\tan 79^\circ

Answer

(a) tan⁡56∘\tan 56^\circ

Dividing numerator and denominator by cos⁡11∘\cos 11^\circ:

1+tan⁡11∘1−tan⁡11∘=tan⁡(45∘+11∘)=tan⁡56∘\frac{1+\tan 11^\circ}{1-\tan 11^\circ}=\tan(45^\circ+11^\circ)=\tan 56^\circ

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  • 2018 Practice Test 2 · 1 mark

Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? (a) 2 hours (b) 4 ½ hours (c) 5 ¾ hours (d) 6 hours

Answer

(b) 4 ½ hours

In 3 hours the first cyclist covers 6×3=186\times3=18 miles. The second gains 10−6=410-6=4 miles per hour, so the catch-up time is

184=4.5 hours\frac{18}{4}=4.5\text{ hours}

measured from the moment the second cyclist starts.

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  • 2018 Practice Test 2 · 1 mark

By how much the area of sphere is changes when its radius is doubled? (a) 100% (b) 200% (c) 300% (d) 800%

Answer

(c) 300%

Area of a sphere is 4πr24\pi r^2. With radius 2r2r, the area is 4π(2r)2=16πr24\pi(2r)^2 = 16\pi r^2, four times the original.

Increase =16πr2−4πr24πr2×100%=300%=\dfrac{16\pi r^2-4\pi r^2}{4\pi r^2}\times100\% = 300\%.

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  • 2018 Practice Test 3 · 1 mark

The probability that a boy will get a scholarship is 0.75 and that a girl will get is 0.80. What is the probability that at least one of them will get the scholarship? (a) 0.20 (b) 0.05 (c) 0.95 (d) 0.25

Answer

(c) 0.95

Assuming the events are independent, P(neither)=(1−0.75)(1−0.80)=0.25×0.20=0.05P(\text{neither}) = (1-0.75)(1-0.80) = 0.25 \times 0.20 = 0.05.

P(at least one)=1−0.05=0.95P(\text{at least one}) = 1 - 0.05 = 0.95

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  • 2018 Practice Test 3 · 1 mark

If f(x) =3x - 4 and g(x) = 3 - 2x.Which of the following is the value of fog(2) ? (a) 7 (b) -7 (c) 9 (d) 8

Answer

(b) -7

g(2)=3−2(2)=−1g(2) = 3 - 2(2) = -1 f(g(2))=f(−1)=3(−1)−4=−7f(g(2)) = f(-1) = 3(-1) - 4 = -7

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  • 2018 Practice Test 3 · 1 mark

If three consecutive vertices of a parallelogram are (-2, -1), (1, 0) and (4, 3) then the fourth vertex is (a) (1, 2) (b) (5, 3) (c) (2, 1) (d) (3, -1)

Answer

(a) (1, 2)

Let the vertices be A(−2,−1)A(-2,-1), B(1,0)B(1,0), C(4,3)C(4,3) in order, with DD the fourth vertex. The diagonals ACAC and BDBD bisect each other, so D=A+C−BD = A + C - B:

D=(−2+4−1,  −1+3−0)=(1,2)D = (-2+4-1,\; -1+3-0) = (1, 2)

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  • 2018 Practice Test 3 · 1 mark

The equation of the circle whose ends of diameter are (−2,3)(-2,3) and (−3,6)(-3,6) is (a) x2+y2+5x−9y+24=0x^2+y^2+5x-9y+24=0 (b) x2+y2−5x+9y+24=0x^2+y^2-5x+9y+24=0 (c) x2+y2−5x−9y+24=0x^2+y^2-5x-9y+24=0 (d) x2+y2−5x−3y+12=0x^2+y^2-5x-3y+12=0

Answer

(a) x2+y2+5x−9y+24=0x^2+y^2+5x-9y+24=0

Using the diameter form (x−x1)(x−x2)+(y−y1)(y−y2)=0(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0 with (−2,3)(-2,3) and (−3,6)(-3,6):

(x+2)(x+3)+(y−3)(y−6)=0(x+2)(x+3)+(y-3)(y-6)=0 x2+5x+6+y2−9y+18=0x^2+5x+6+y^2-9y+18=0 x2+y2+5x−9y+24=0x^2+y^2+5x-9y+24=0

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  • 2018 Practice Test 3 · 1 mark

A bus travels at an average speed of 50 miles per hour for 2.5 hours and then travels at a speed of 70 miles per hour for 1.5 hours. How far did the train travel in the entire 4 hours? (a) 150 miles (b) 130 miles (c) 125 miles (d) 230 miles

Answer

(d) 230 miles

Distance=50×2.5+70×1.5=125+105=230 miles\text{Distance} = 50\times 2.5 + 70\times 1.5 = 125 + 105 = 230 \text{ miles}

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  • 2018 Practice Test 3 · 1 mark

A man deposits Rs. 1 on a day, Rs. 2 on the next day, Rs. 3 on the third day and so on for 60 days. Total collection at the end of 60 days will be: (a) Rs. 183 (b) Rs. 1830 (c) Rs. 3660 (d) Rs. 5460

Answer

(b) Rs. 1830

The deposits form the series 1+2+⋯+601+2+\dots+60:

S=60×612=1830S = \frac{60\times 61}{2} = 1830

The total is Rs. 1830.

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  • 2018 Practice Test 3 · 1 mark

A man buys 100 apples at 50 paisa each and 50 oranges at 60 paisa each. 10% of the apples and oranges are spoiled. The man sells the remaining apples at 70 paisa each and oranges at Rs. 1 each. The profit made as percentage of the cost is: (a) 50% (b) 35% (c) 33.3% (d) 26%

Answer

(b) 35%

Cost: apples 100×50=5000100\times 50 = 5000 paisa, oranges 50×60=300050\times 60 = 3000 paisa, total 80008000 paisa (Rs 80).

After 10% spoil, 90 apples and 45 oranges remain:

Sales=90×70+45×100=6300+4500=10800 paisa\text{Sales} = 90\times 70 + 45\times 100 = 6300 + 4500 = 10800 \text{ paisa}

Profit%=10800−80008000×100=35%\text{Profit} \% = \frac{10800-8000}{8000}\times 100 = 35\%

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  • 2018 Practice Test 3 · 1 mark

If a triangle of base 6 has the same area as a circle of radius 6, what is the attitude of the triangle? (a) 6π (b) 8π (c) 10π (d) 12π

Answer

(d) 12π12\pi

Equal areas give

12×6×h=π(6)2  ⟹  3h=36π  ⟹  h=12π\tfrac12 \times 6 \times h = \pi (6)^2 \implies 3h = 36\pi \implies h = 12\pi

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  • 2018 Practice Test 4 · 1 mark

What will be the remainder of 4x3+6x2+3x−14x^3+6x^2+3x-1, when it is divided by x+2x+2. (a) −1-1 (b) −15-15 (c) 61 (d) −52-52

Answer

(b) −15-15

By the remainder theorem, the remainder on dividing by x+2x+2 is p(−2)p(-2):

p(−2)=4(−8)+6(4)+3(−2)−1=−32+24−6−1=−15p(-2)=4(-8)+6(4)+3(-2)-1=-32+24-6-1=-15

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  • 2018 Practice Test 4 · 1 mark

For what value of kk the equation 3x2+7x+6−k=03x^2+7x+6-k=0 has one root equal to zero. (a) 3 (b) 7 (c) 6 (d) −6-6

Answer

(c) 6

If one root is zero, x=0x=0 satisfies the equation, so 3(0)+7(0)+6−k=03(0)+7(0)+6-k=0, giving k=6k=6.

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  • 2018 Practice Test 4 · 1 mark

The slope of the line xcos⁡θ+ysin⁡θ=px\cos\theta+y\sin\theta=p is (a) tan⁡θ\tan\theta (b) cot⁡θ\cot\theta (c) −tan⁡θ-\tan\theta (d) −cot⁡θ-\cot\theta

Answer

(d) −cot⁡θ-\cot\theta

Rewrite as ysin⁡θ=−xcos⁡θ+py\sin\theta=-x\cos\theta+p, so y=−cos⁡θsin⁡θx+psin⁡θy=-\dfrac{\cos\theta}{\sin\theta}x+\dfrac{p}{\sin\theta}. The slope is −cot⁡θ-\cot\theta.

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  • 2018 Practice Test 4 · 1 mark

The equation of the line passing through (1, -2) and parralel to y – axis is ……… (a) y = 1 (b) y = 2 (c) x = -1 (d) x = 1

Answer

(d) x=1x = 1

A line parallel to the yy-axis is vertical, with equation x=constantx=\text{constant}. It passes through (1,−2)(1,-2), so x=1x=1.

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  • 2018 Practice Test 4 · 1 mark

Three coins are tossed together. Find the probability of getting at least two heads? (a) 1/8 (b) 3/8 (c) 1/2 (d) 1/4

Answer

(c) 1/21/2

Total outcomes =23=8=2^3=8. At least two heads means HHH, HHT, HTH, THH, which is 4 outcomes. Probability =48=12=\dfrac{4}{8}=\dfrac12.

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  • 2018 Practice Test 4 · 1 mark

If 13, p, q, r, 29 are in A.P. Find the value of r. (a) 25 (b) 17 (c) 21 (d) 24

Answer

(a) 25

Here 1313 is the first term and 2929 the fifth, so 4d=29−13=164d=29-13=16 and d=4d=4. Then r=13+4⋅3=25r=13+4\cdot3=25.

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  • 2018 Practice Test 4 · 1 mark

The surface area of a cube is 150 square feet. Calculate volume of the cube. (a) 30 cft (b) 50 cft (c) 100 cft (d) 125 cft

Answer

(d) 125 cft

Surface area 6a2=1506a^2=150, so a2=25a^2=25 and a=5a=5 ft. Volume =a3=53=125=a^3=5^3=125 cft.

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  • 2018 Practice Test 4 · 1 mark

Choose the pair numbers which come next to 75, 65, 85, 55, 45, 85, 35, ……… (a) 25, 15 (b) 25, 85 (c) 35, 25 (d) 35, 85

Answer

(b) 25, 85

The series is two interleaved patterns. Every third term is 8585 (3rd, 6th, 9th terms). The remaining terms 75,65,55,45,35,…75,65,55,45,35,\dots fall by 1010. So after 3535 comes 2525, then the 99th term 8585.

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  • 2018 Practice Test 4 · 1 mark

What single discount is equivalent to two successive discounts of 10% and 15%? (a) 25% (b) 24% (c) 24.5% (d) 23.5%

Answer

(d) 23.5%

Remaining fraction =0.90×0.85=0.765=0.90\times0.85=0.765. Single discount =1−0.765=0.235=23.5%=1-0.765=0.235=23.5\%.

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  • 2018 Practice Test 4 · 1 mark

A shopkeeper sales 25 item and receive the C.P. after selling 20 items. Then what will be his profit %? (a) 25% (b) 4.5% (c) 5% (d) 15%

Answer

(a) 25%

Let the cost price of each item be 11. The selling price of 2020 items equals the cost of 2525 items, which is 2525. Cost of the 2020 items sold is 2020. Profit =25−20=5=25-20=5, so profit %=520×100=25%\%=\dfrac{5}{20}\times100=25\%.

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  • 2018 Practice Test 4 · 1 mark

A pipe can fill a cistern in 45 minutes and the waste pipe empties it in 1 hour. If both the pipes are opened at the same time, in how many hours will the cistern be filled? (a) 3 hours (b) 4 hours (c) 6 hours (d) 2 hours

Answer

(a) 3 hours

In one minute the net part filled is 145−160=4−3180=1180\dfrac1{45}-\dfrac1{60}=\dfrac{4-3}{180}=\dfrac1{180}. The cistern fills in 180180 minutes =3=3 hours.

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  • 2018 Practice Test 4 · 1 mark

If two diagonals of a rhombus are 14 cm and 12 cm then its area is (a) 36 cm2^2 (b) 84 cm2^2 (c) 28 cm2^2 (d) 7 cm2^2

Answer

(b) 84 cm2^2

Area of a rhombus =12d1d2=12×14×12=84=\tfrac12 d_1d_2=\tfrac12\times14\times12=84 cm2^2.

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  • 2018 Practice Test 5 · 1 mark

A train went 300 miles from city X to city Y at an average speed of 80 mph. At what speed did it travel on the way back if its average speed for the whole trip was 100 mph? (a) 120 mph (b) 125 mph (c) 13313133\frac13 mph (d) 13713137\frac13 mph

Answer

(c) 13313133\frac13 mph

Total distance =300+300=600=300+300=600 miles. At an average of 100 mph the total time is 600/100=6600/100=6 h.

Time going =300/80=3.75=300/80=3.75 h, so time returning =6−3.75=2.25=6-3.75=2.25 h.

Return speed=3002.25=13313 mph\text{Return speed}=\frac{300}{2.25}=133\frac13 \text{ mph}

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  • 2018 Practice Test 5 · 1 mark

The co-ordinates of a triangle are (0, 2), (2, 4) and (1, 6). Then the area of triangle in sq. unit is (a) 2 (b) 3 (c) 4 (d) 5

Answer

(b) 3

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area}=\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

=12∣0(4−6)+2(6−2)+1(2−4)∣=12∣0+8−2∣=3 sq. units=\tfrac12\left|0(4-6)+2(6-2)+1(2-4)\right|=\tfrac12|0+8-2|=3 \text{ sq. units}

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  • 2018 Practice Test 5 · 1 mark

In a class of 20 students in MATH examination 2 students scored 100 marks each, 3 students got zero each, and the average score of rest of the students was 40. What is the average score of the whole class? (a) 10 (b) 20 (c) 30 (d) 40

Answer

(d) 40

Total marks =2×100+3×0+15×40=200+0+600=800=2\times100+3\times0+15\times40=200+0+600=800.

Average=80020=40\text{Average}=\frac{800}{20}=40

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  • 2018 Practice Test 5 · 1 mark

A shopkeeper fixed the price of his articles 30% above the cost price. If he sold an article allowing 10% discount, then his profit percentagewill be (a) 20% (b) 17% (c) 23% (d) 40%

Answer

(b) 17%

Let the cost price be Rs 100. Marked price == Rs 130. Selling price after 10% discount =130×0.9==130\times0.9= Rs 117.

Profit =117−100==117-100= Rs 17, so profit percentage =17%=17\%.

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  • 2018 Practice Test 5 · 1 mark

If m and p are positive integers and (m+p)×m is even, which of the following must be true. (a) If m is odd, then p is odd (b) If m is even, then p is even (c) If m is odd, then p is even (d) If m is even, then p is odd

Answer

(a) If m is odd, then p is odd

If mm is odd, the product (m+p)×m(m+p)\times m is even only when (m+p)(m+p) is even. Odd + p+\,p is even only if pp is odd. If mm is even the product is even whatever pp is, so (b) and (d) need not hold, and (c) is false.

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  • 2019 Science Sample Test Set A · 1 mark

If n(A)=40n(A)=40, n(B)=30n(B)=30, n(U)=60n(U)=60, then maximum of n(A∪B)n(A\cup B) is ……… (a) 40 (b) 30 (c) 70 (d) 60

Answer

(d) 60

The union is largest when the overlap is smallest. With A∩B=∅A \cap B = \emptyset, n(A∪B)=40+30=70n(A\cup B) = 40 + 30 = 70. However, A∪B⊆UA \cup B \subseteq U, so n(A∪B)≤n(U)=60n(A\cup B) \le n(U) = 60.

The maximum is 60, which happens when n(A∩B)=10n(A\cap B) = 10.

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  • 2019 Science Sample Test Set A · 1 mark

If f(x + 3) = 3x + 10, then the value of f(x) is …….. (a) 3x + 1 (b) 3x + 10 (c) 3x + 19 (d) x + 3

Answer

(a) 3x+13x+1

Put t=x+3t = x+3, so x=t−3x = t-3:

f(t)=3(t−3)+10=3t+1f(t) = 3(t-3)+10 = 3t+1

Hence f(x)=3x+1f(x) = 3x+1.

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  • 2019 Science Sample Test Set A · 1 mark

A man forgets the last digit of telephone number while dialing. Then he randomly chooses the number. What is the probability that the number dialed correctly? (a) 110\frac1{10} (b) 12\frac12 (c) 19\frac19 (d) 0

Answer

(a) 110\frac{1}{10}

The last digit can be any of the digits 0–9, which gives 10 equally likely outcomes, and only one of them is correct. So P=110P = \frac{1}{10}.

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  • 2019 Science Sample Test Set A · 1 mark

The sum of cube of first nn natural number is …… (a) n(n+1)2\frac{n(n+1)}{2} (b) [n(n+1)2]2\left[\frac{n(n+1)}{2}\right]^2 (c) [n(n+1)2]3\left[\frac{n(n+1)}{2}\right]^3 (d) n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}

Answer

(b) [n(n+1)2]2\left[\frac{n(n+1)}{2}\right]^2

13+23+⋯+n3=[n(n+1)2]2,1^3+2^3+\dots+n^3=\left[\frac{n(n+1)}{2}\right]^2, the square of the sum of the first nn natural numbers.

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  • 2019 Science Sample Test Set A · 1 mark

Ram can do a piece of work in 8 days and Shyam can do the same work in 6 days. Ram and Shyam work on it alternate days. If Ram started the work in the first day, In how many days the work will be finished? (a) 8 (b) 6 (c) 7 (d) 9

Answer

(c) 7

Ram does 18\frac18 and Shyam 16\frac16 of the work per day, so one Ram-Shyam pair of days completes 18+16=724\frac18+\frac16=\frac{7}{24}.

After 3 pairs (6 days): 3×724=783\times\frac{7}{24}=\frac78 is done. On day 7 Ram does the remaining 18\frac18, so the work is finished in 7 days.

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  • 2019 Science Sample Test Set A · 1 mark

If [a54a]=[a476]\begin{bmatrix}a&5\\4&a\end{bmatrix}=\begin{bmatrix}a&4\\7&6\end{bmatrix}, then a=a= ? (a) ±2\pm2 (b) ±4\pm4 (c) 2, 4 (d) 4, −2-2

Answer

(d) 4, −24,\,-2

Equating the corresponding entries of the two matrices gives the condition a2−2a−8=0⇒(a−4)(a+2)=0,a^2-2a-8=0\Rightarrow (a-4)(a+2)=0, so a=4a=4 or a=−2a=-2.

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  • 2019 Science Sample Test Set A · 1 mark

The mean of 150 items is 52. Later on, it was discovered that two items were misread as 92 and 8 instead of 192 and 88. Find out the correct mean. (a) 35.2 (b) 52.3 (c) 53.2 (d) 25.3

Answer

(c) 53.2

Wrong total =150×52=7800=150\times52=7800. The misread items sum to 92+8=10092+8=100, the correct ones to 192+88=280192+88=280.

Correct total =7800−100+280=7980=7800-100+280=7980, so the correct mean is 7980150=53.2\frac{7980}{150}=53.2.

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  • 2019 Science Sample Test Set A · 1 mark

The equation of straight line which passes through the mid-point of the line segment joining (2, 4) and (2, - 4) and perpendicular to the line 3x – 2y = 5 is ….. (a) 2x – 3y = 5 (b) 2x + 3y = 4 (c) 2x + 4y = 3 (d) 2x – 3y = 4

Answer

(b) 2x+3y=42x+3y=4

The mid-point of (2,4)(2,4) and (2,−4)(2,-4) is (2,0)(2,0). The line 3x−2y=53x-2y=5 has slope 32\frac32, so the perpendicular has slope −23-\frac23.

y−0=−23(x−2)⇒3y=−2x+4⇒2x+3y=4.y-0=-\frac23(x-2)\Rightarrow 3y=-2x+4\Rightarrow 2x+3y=4.

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  • 2019 Science Sample Test Set A · 1 mark

How many spherical balls of radius 1 cm can be made from a sphere of radius 8 cm? (a) 512 (b) 320 (c) 412 (d) 312

Answer

(a) 512

Number of balls =43π(8)343π(1)3=83=512=\dfrac{\frac43\pi(8)^3}{\frac43\pi(1)^3}=8^3=512.

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  • 2019 Science Sample Test Set A · 1 mark

What is the center of the circle 4x2+4y2−8x+24y+32=04x^2+4y^2-8x+24y+32=0? (a) (4,−12)(4,-12) (b) (−4,12)(-4,12) (c) (−1,3)(-1,3) (d) (1,−3)(1,-3)

Answer

(d) (1,−3)(1,-3)

Dividing by 4: x2+y2−2x+6y+8=0x^2+y^2-2x+6y+8=0. Comparing with x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0 gives g=−1, f=3g=-1,\ f=3, so the centre is (−g,−f)=(1,−3)(-g,-f)=(1,-3).

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  • 2019 Science Sample Test Set A · 1 mark

Two cars start at the same time from two station 480 km apart in opposite direction. The speed of the first car is greater than that of the second by 6 km/hr. If they meet each other at 6 hours from the start. Find the speed of cars? (a) 52 km/hr & 58 km/hr (b) 32 km/hr & 26 km/hr (c) 45 km/hr & 39 km/hr (d) 43 km/hr & 37 km/hr

Answer

(d) 43 km/hr and 37 km/hr

The cars approach each other, so the sum of their speeds is 4806=80\frac{480}{6}=80 km/hr. With v1−v2=6v_1-v_2=6: v1=80+62=43,v2=80−62=37.v_1=\frac{80+6}{2}=43,\qquad v_2=\frac{80-6}{2}=37.

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  • 2019 Science Sample Test Set A · 1 mark

H.C.F. of x4−x3+x−1x^4-x^3+x-1, x4+x2+1x^4+x^2+1 is …. (a) x2−x+1x^2-x+1 (b) x4−x+1x^4-x+1 (c) x2+x−1x^2+x-1 (d) (x+1)(x−1)(x+1)(x-1)

Answer

(a) x2−x+1x^2-x+1

x4−x3+x−1=x3(x−1)+(x−1)=(x−1)(x3+1)=(x−1)(x+1)(x2−x+1)x^4-x^3+x-1=x^3(x-1)+(x-1)=(x-1)(x^3+1)=(x-1)(x+1)(x^2-x+1) x4+x2+1=(x2+1)2−x2=(x2+x+1)(x2−x+1)x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1) The common factor is x2−x+1x^2-x+1.

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  • 2019 Science Sample Test Set A · 1 mark

Find the value of KK so that the equation 2x2+Kx−15=02x^2+Kx-15=0 has one root =3=3. (a) K=1K=1 (b) K=−1K=-1 (c) K=−3K=-3 (d) K=0K=0

Answer

(b) K=−1K=-1

Putting x=3x=3: 2(9)+3K−15=0⇒3K=−3⇒K=−12(9)+3K-15=0\Rightarrow 3K=-3\Rightarrow K=-1.

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  • 2019 Science Sample Test Set A · 1 mark

tan⁡−1x+tan⁡−1y=\tan^{-1}x+\tan^{-1}y= ? (a) tan⁡−1(x−y1+xy)\tan^{-1}\left(\frac{x-y}{1+xy}\right) (b) tan⁡−1(x+y1−xy)\tan^{-1}\left(\frac{x+y}{1-xy}\right) (c) tan⁡−1(1−xyx+y)\tan^{-1}\left(\frac{1-xy}{x+y}\right) (d) tan⁡−1(1+xyx−y)\tan^{-1}\left(\frac{1+xy}{x-y}\right)

Answer

(b) tan⁡−1(x+y1−xy)\tan^{-1}\left(\frac{x+y}{1-xy}\right)

This follows from tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} with A=tan⁡−1x, B=tan⁡−1yA=\tan^{-1}x,\ B=\tan^{-1}y (valid for xy<1xy<1).

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  • 2019 Science Sample Test Set A · 1 mark

Sabita was 25 years when her son was born. But the product of their ages now is 186, find the present ages of mother and her son. (a) 31, 6 years (b) 23.25, 8 years (c) 29, 6 years (d) 46.5, 4 years

Answer

(a) 31 years and 6 years

Let the son be ss years old now, so the mother is s+25s+25. s(s+25)=186⇒s2+25s−186=0⇒(s+31)(s−6)=0⇒s=6.s(s+25)=186\Rightarrow s^2+25s-186=0\Rightarrow (s+31)(s-6)=0\Rightarrow s=6. The mother is 6+25=316+25=31 years old; check: 31×6=18631\times6=186.

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  • 2019 Science Sample Test Set A · 1 mark

The perpendiculars drawn from the vertices of the triangle on their opposite sides meet at a point. This point is known as: (a) Incentre (b) Circumcenter (c) Centroid (d) Orthocenter General Knowledge

Answer

(d) Orthocenter

The three altitudes (perpendiculars from the vertices to the opposite sides) of a triangle are concurrent at the orthocenter.

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  • 2074 Set A · 1 mark

For any two square matrices AA and BB which of the following is true? (a) (AB)T=ATBT(AB)^T=A^TB^T (b) (AB)T=BTAT(AB)^T=B^TA^T (c) AB=BAAB=BA (d) A−B=B−AA-B=B-A

Answer

(b) (AB)T=BTAT(AB)^T=B^TA^T

This is the reverse-order law for transposes. Matrix multiplication is not commutative, so (c) is false in general, and A−B=B−AA-B=B-A holds only if A=BA=B.

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  • 2074 Set A · 1 mark

If 2x2x, x+8x+8, 3x+13x+1 are in AP, then the value of xx is ____. (a) 3 (b) 7 (c) 5 (d) −2-2

Answer

(c) 5

For an AP the middle term is the mean of the other two, so 2(x+8)=2x+(3x+1)2(x+8)=2x+(3x+1).

2x+16=5x+1⇒3x=15⇒x=52x+16=5x+1 \Rightarrow 3x=15 \Rightarrow x=5

Check: the terms are 10, 13, 16 with common difference 3.

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  • 2074 Set A · 1 mark

The distance between two parallel lines 4x+3y+5=04x+3y+5=0 and 12x+9y+18=012x+9y+18=0 is ____. (a) 15\frac15 units (b) 518\frac5{18} units (c) 5 units (d) 18 units

Answer

(a) 15\frac15 units

Divide the second line by 3 to get 4x+3y+6=04x+3y+6=0. The distance between 4x+3y+5=04x+3y+5=0 and 4x+3y+6=04x+3y+6=0 is

d=∣6−5∣42+32=15d=\frac{|6-5|}{\sqrt{4^2+3^2}}=\frac15

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  • 2074 Set A · 1 mark

If the perimeter of a square is 16 cm, then its diagonal's length is ____. (a) 16 cm (b) 8 cm (c) 4 cm (d) 424\sqrt2 cm

Answer

(d) 424\sqrt2 cm

Side =16/4=4=16/4=4 cm, so the diagonal is 42+42=42\sqrt{4^2+4^2}=4\sqrt2 cm.

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  • 2074 Set A · 1 mark

What is the minimum value of sin⁡4θ+cos⁡4θ\sin^4\theta+\cos^4\theta? (a) 12\frac12 (b) 0 (c) −1-1 (d) 1

Answer

(a) 12\frac12

sin⁡4θ+cos⁡4θ=(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ=1−12sin⁡22θ\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta=1-\tfrac12\sin^2 2\theta

This is smallest when sin⁡22θ=1\sin^2 2\theta=1, giving 1−12=121-\tfrac12=\tfrac12.

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  • 2074 Set A · 1 mark

The power set of A={a,b,c,d}A=\{a,b,c,d\} contains ____ elements. (a) 6 (b) 8 (c) 16 (d) 4

Answer

(c) 16

A set with nn elements has 2n2^n subsets. Here n=4n=4, so the power set has 24=162^4=16 elements.

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  • 2074 Set A · 1 mark

The angle made by the line xcos⁡α+ysin⁡α=px\cos\alpha+y\sin\alpha=p with positive xx-axis is ____. (a) α\alpha (b) 180∘−α180^\circ-\alpha (c) 90∘−α90^\circ-\alpha (d) 90∘+α90^\circ+\alpha

Answer

(d) 90∘+α90^\circ+\alpha

The slope of xcos⁡α+ysin⁡α=px\cos\alpha+y\sin\alpha=p is −cot⁡α=tan⁡(90∘+α)-\cot\alpha=\tan(90^\circ+\alpha), so the line makes an angle 90∘+α90^\circ+\alpha with the positive xx-axis. (The perpendicular from the origin makes angle α\alpha, and the line is perpendicular to it.)

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  • 2074 Set A · 1 mark

If sum of nn terms is given by sn=n2+2s_n=n^2+2, then its 6th6^{th} term is ____. (a) 11 (b) 6 (c) 38 (d) 27

Similar questions: If the sum of 11 terms of an AP is 121, then its (2074 Set A)

Answer

(a) 11

t6=s6−s5=(36+2)−(25+2)=38−27=11t_6=s_6-s_5=(36+2)-(25+2)=38-27=11

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  • 2074 Set A · 1 mark

If the sum of 11 terms of an AP is 121, then its 6th term is ____ (a) 11 (b) 22 (c) 121 (d) 33

Similar questions: If sum of n terms is given by s_n=n^2+2, then it (2074 Set A)

Answer

(a) 11

S11=112(a1+a11)=11 a6=121⇒a6=11S_{11}=\frac{11}{2}(a_1+a_{11})=11\,a_6=121 \Rightarrow a_6=11

since a1+a11=2a6a_1+a_{11}=2a_6.

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  • 2074 Set A · 1 mark

If the angle between two lines 2x2+5xy+(k−1)y2=02x^2+5xy+(k-1)y^2=0 is 90∘90^\circ, what is the value of kk? (a) 1 (b) −1-1 (c) 3 (d) 0

Answer

(b) −1-1

The lines given by ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are perpendicular when a+b=0a+b=0. Here a=2a=2 and b=k−1b=k-1, so

2+k−1=0⇒k=−12+k-1=0 \Rightarrow k=-1

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  • 2074 Set A · 1 mark

If the arithmetic mean between two numbers is 6126\frac12 and their geometric mean is 6, then the two numbers are ____. (a) 4 and 9 (b) 9 and 4 (c) both a and b (d) none

Answer

(c) both a and b

a+b=2×612=13,ab=62=36a+b=2\times 6\tfrac12=13,\qquad ab=6^2=36

The numbers are the roots of t2−13t+36=0t^2-13t+36=0, i.e. (t−4)(t−9)=0(t-4)(t-9)=0, so they are 4 and 9. The order does not matter, so both (a) and (b) are correct.

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  • 2074 Set A · 1 mark

If thrice of my age 10 years ago be subtracted from twice my present age, the result is my present age. How old am I now? (a) 16 years (b) 15 years (c) 9 years (d) 10 years

Answer

(b) 15 years

Let the present age be xx. Then

2x−3(x−10)=x⇒−x+30=x⇒x=152x-3(x-10)=x \Rightarrow -x+30=x \Rightarrow x=15

Check: 30−3(5)=1530-3(5)=15.

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  • 2074 Set A · 1 mark

If x+y=5x+y=5, y+z=11y+z=11, z+x=8z+x=8, then what is the average of xx, yy and zz? (a) 24 (b) 12 (c) 8 (d) 4

Answer

(d) 4

Adding the three equations gives 2(x+y+z)=5+11+8=242(x+y+z)=5+11+8=24, so x+y+z=12x+y+z=12. The average is 12/3=412/3=4.

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  • 2074 Set A · 1 mark

The price of an article after two successive discounts of 10% and 5% became Rs. 855, then its marked price is ____. (a) Rs. 1,000 (b) Rs. 1,005.88 (c) Rs. 900 (d) Rs. 726.75

Answer

(a) Rs. 1,000

After successive discounts the price is M(0.90)(0.95)=0.855MM(0.90)(0.95)=0.855M.

0.855M=855⇒M=Rs 1,0000.855M=855 \Rightarrow M=\text{Rs } 1{,}000

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  • 2074 Set A · 1 mark

If f(x+3)=x+6f(x+3)=x+6, what is the value of f−1(2)f^{-1}(2)? (a) 1 (b) 2 (c) −1-1 (d) −2-2

Answer

(c) −1-1

Put t=x+3t=x+3, so f(t)=(t−3)+6=t+3f(t)=(t-3)+6=t+3. Then f−1(y)=y−3f^{-1}(y)=y-3, and f−1(2)=−1f^{-1}(2)=-1.

Check: f(−1)=2f(-1)=2.

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  • 2074 Set A · 1 mark

If sin⁡(A+B)=ksin⁡(A−B)\sin(A+B)=k\sin(A-B), then k+1k−1\frac{k+1}{k-1} is equal to ____ (a) tan⁡Atan⁡B\frac{\tan A}{\tan B} (b) sin⁡Asin⁡B\frac{\sin A}{\sin B} (c) sin⁡(A+B)sin⁡(A−B)\frac{\sin(A+B)}{\sin(A-B)} (d) sin⁡(A+B+1)sin⁡(A−B−1)\frac{\sin(A+B+1)}{\sin(A-B-1)}

Answer

(a) tan⁡Atan⁡B\frac{\tan A}{\tan B}

Since k=sin⁡(A+B)sin⁡(A−B)k=\dfrac{\sin(A+B)}{\sin(A-B)}, by componendo and dividendo

k+1k−1=sin⁡(A+B)+sin⁡(A−B)sin⁡(A+B)−sin⁡(A−B)=2sin⁡Acos⁡B2cos⁡Asin⁡B=tan⁡Atan⁡B\frac{k+1}{k-1}=\frac{\sin(A+B)+\sin(A-B)}{\sin(A+B)-\sin(A-B)}=\frac{2\sin A\cos B}{2\cos A\sin B}=\frac{\tan A}{\tan B}

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  • 2074 Set A · 1 mark

If tan⁡θ+cot⁡θ=2\tan\theta+\cot\theta=2, then the value of θ\theta is ____ (0∘≤θ≤360∘0^\circ\le\theta\le360^\circ) (a) 135∘135^\circ (b) 45∘,225∘45^\circ, 225^\circ (c) 0∘0^\circ (d) 90∘90^\circ

Answer

(b) 45∘,225∘45^\circ, 225^\circ

Let t=tan⁡θt=\tan\theta. Then t+1t=2⇒(t−1)2=0⇒t=1t+\frac1t=2 \Rightarrow (t-1)^2=0 \Rightarrow t=1. In 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ, tan⁡θ=1\tan\theta=1 gives θ=45∘\theta=45^\circ and 225∘225^\circ.

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  • 2074 Set A · 1 mark

If ∣x⃗+y⃗∣=∣x⃗−y⃗∣|\vec x+\vec y|=|\vec x-\vec y|, then the angle between x⃗\vec x and y⃗\vec y is ____. (a) 0∘0^\circ (b) 90∘90^\circ (c) 45∘45^\circ (d) 180∘180^\circ

Answer

(b) 90∘90^\circ

Squaring both sides, ∣x⃗∣2+∣y⃗∣2+2x⃗⋅y⃗=∣x⃗∣2+∣y⃗∣2−2x⃗⋅y⃗|\vec x|^2+|\vec y|^2+2\vec x\cdot\vec y=|\vec x|^2+|\vec y|^2-2\vec x\cdot\vec y, so x⃗⋅y⃗=0\vec x\cdot\vec y=0. The vectors are perpendicular, so the angle is 90∘90^\circ.

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  • 2074 Set A · 1 mark

A point (7,4)(7,4) is first reflected in the line x=yx=y and then in the line x=kx=k, so that the final image is (4,7)(4,7), find the value of kk. (a) 7 (b) 11 (c) 4 (d) 3

Answer

(c) 4

Reflecting (7,4)(7,4) in y=xy=x swaps the coordinates and gives (4,7)(4,7). Reflecting in x=kx=k sends xx to 2k−x2k-x, so 2k−4=42k-4=4, giving k=4k=4. The point (4,7)(4,7) then lies on the mirror line and stays at (4,7)(4,7).

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  • 2074 Set A · 1 mark

The arithmetic mean of seven numbers was 25%. Later, on it was found that 45 was misread as 10, now calculate the actual mean. (a) 30 (b) 35 (c) 40 (d) 45

Answer

(a) 30

Taking the mean as 25, the wrong sum is 7×25=1757\times 25=175. The correct sum is 175−10+45=210175-10+45=210, so the actual mean is 210/7=30210/7=30.

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  • 2074 Set A · 1 mark

What is the compound amount of Rs. 7,000 for 2 years at the rate of 8% p.a. for 1st year and 10% for 2nd year? (a) Rs. 9,746.80 (b) Rs. 7,800 (c) Rs. 1,316 (d) Rs. 8,316

Answer

(d) Rs. 8,316

After year 1: 7000×1.08=Rs 75607000\times1.08=\text{Rs }7560. After year 2: 7560×1.10=Rs 83167560\times1.10=\text{Rs }8316.

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  • 2074 Set A · 1 mark

If the heights of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio of 4 : 1, then the ratio of their volumes is ____. (a) 1 : 2 (b) 2 : 3 (c) 1 : 4 (d) 4 : 1

Answer

(d) 4 : 1

Volume is proportional to r2hr^2h, so the ratio is 42×112×4=164=4:1\dfrac{4^2\times1}{1^2\times4}=\dfrac{16}{4}=4:1.

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  • 2074 Set A · 1 mark

A triangle and a parallelogram are standing on the equal base and lie between same parallel lines, then the ratio of their areas is ____. (a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) 1 : 4

Answer

(a) 1 : 2

A triangle and a parallelogram on the same (equal) base and between the same parallels satisfy area of triangle=12×area of parallelogram\text{area of triangle}=\tfrac12\times\text{area of parallelogram}, so the ratio is 1:21:2.

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  • 2074 Set A · 1 mark

What is the largest integer nn such that 2n2^n is a factor of 20820^8? (a) 4 (b) 8 (c) 16 (d) 32

Answer

(c) 16

208=(22⋅5)8=216⋅5820^8=(2^2\cdot5)^8=2^{16}\cdot5^8, so the highest power of 2 dividing it is 2162^{16} and n=16n=16.

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  • 2074 Set A · 1 mark

If x>y>0x>y>0 and p>q>0p>q>0, then which one of the following expressions must be greater than 1? (a) x+yp+q\frac{x+y}{p+q} (b) x+qy+p\frac{x+q}{y+p} (c) p+qx+y\frac{p+q}{x+y} (d) x+py+q\frac{x+p}{y+q}

Answer

(d) x+py+q\frac{x+p}{y+q}

Since x>y>0x>y>0 and p>q>0p>q>0, adding gives x+p>y+q>0x+p>y+q>0, so x+py+q>1\frac{x+p}{y+q}>1. The other options can be less than 1 or equal to 1 for suitable values, for example x=2,y=1,p=2,q=1x=2,y=1,p=2,q=1 gives x+yp+q=1\frac{x+y}{p+q}=1 and p+qx+y=1\frac{p+q}{x+y}=1.

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  • 2074 Set A · 1 mark

The sum of three consecutive positive integers must be divisible by ____. (a) 3 (b) 4 (c) 5 (d) 6

Answer

(a) 3

Three consecutive integers n−1, n, n+1n-1,\,n,\,n+1 sum to 3n3n, which is always divisible by 3. For example 1+2+3=61+2+3=6 is not divisible by 4 or 5, and 2+3+4=92+3+4=9 is not divisible by 6.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

If A={x:x∈N and x<4}A=\{x: x\in N \text{ and } x<4\} and B={x:x∈N, x2−9=0}B=\{x: x\in N,\ x^2-9=0\} then B×AB\times A is : (a) {(3,1),(3,2),(3,3)}\{(3,1),(3,2),(3,3)\} (b) {(1,3),(2,3),(3,3)}\{(1,3),(2,3),(3,3)\} (c) {(1,1),(1,2),(1,3),(2,1)}\{(1,1),(1,2),(1,3),(2,1)\} (d) {(3,3)}\{(3,3)\}

Answer

(a) {(3,1),(3,2),(3,3)}\{(3,1),(3,2),(3,3)\}

A={1,2,3}A=\{1,2,3\}. From x2−9=0x^2-9=0 with x∈Nx\in N, x=3x=3, so B={3}B=\{3\}. Hence B×A={(3,1),(3,2),(3,3)}.B\times A=\{(3,1),(3,2),(3,3)\}.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

f:Q→Qf: Q\to Q defined by f(x)=6x+5f(x)=6x+5 is : (a) One - one (b) Onto (c) Both one-one and onto (d) Neither one-nor onto

Answer

(c) Both one-one and onto

One-one: f(x1)=f(x2)⇒6x1+5=6x2+5⇒x1=x2f(x_1)=f(x_2)\Rightarrow 6x_1+5=6x_2+5\Rightarrow x_1=x_2.

Onto: for any y∈Qy\in Q, take x=y−56∈Qx=\dfrac{y-5}{6}\in Q. Then f(x)=yf(x)=y. So ff is a bijection.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

The roots of the equation ax2+bx+c=0ax^2+bx+c=0 are real and equal if : (a) b2−4acb^2-4ac is perfect square (b) b2−4ac>0b^2-4ac>0 (c) b2−4ac<0b^2-4ac<0 (d) b2−4ac=0b^2-4ac=0

Answer

(d) b2−4ac=0b^2-4ac=0

The discriminant of ax2+bx+c=0ax^2+bx+c=0 decides the nature of the roots. When b2−4ac=0b^2-4ac=0, the formula x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a} gives the single repeated value x=−b2ax=-\frac{b}{2a}, so the roots are real and equal.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

The present age of father is double that of son. 16 years ago father's age was thrice that of son. Find the present age of son. (a) 16 years (b) 32 years (c) 64 years (d) 48 years

Answer

(b) 32 years

Let the son's age be xx, so the father's age is 2x2x. Sixteen years ago, 2x−16=3(x−16)2x-16=3(x-16).

2x−16=3x−48⇒x=322x-16=3x-48\Rightarrow x=32

Check: father 64, son 32; 16 years ago they were 48 and 16, and 48=3×1648=3\times16.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

The population of a village sometimes ago was 10000. The growth rate of the population was 4% per year. When will its population be 10816 ? (a) 2 years (b) 4 years (c) 3 years (d) 1 year

Answer

(a) 2 years

Using A=P(1+r100)nA=P\left(1+\frac{r}{100}\right)^n:

10816=10000(1.04)n⇒(1.04)n=1.0816=(1.04)210816=10000(1.04)^n\Rightarrow(1.04)^n=1.0816=(1.04)^2

So n=2n=2 years.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

A takes twice the time of B and four times the time of C to complete a work. If they work together, the whole work would be completed in 8 days, in how many days A alone can complete the whole work ? (a) 14 days (b) 28 days (c) 56 days (d) 64 days

Answer

(c) 56 days

Let A alone take xx days. Then B takes x2\frac{x}{2} days and C takes x4\frac{x}{4} days. Their one-day work together is

1x+2x+4x=7x=18⇒x=56.\frac1x+\frac2x+\frac4x=\frac7x=\frac18\Rightarrow x=56.

So A alone takes 56 days.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

If 2x+2x+1+2x+2+2x+3=602^x+2^{x+1}+2^{x+2}+2^{x+3}=60, then x=x= (a) 0 (b) 1 (c) 2 (d) 4

Answer

(c) 2

Factor out 2x2^x:

2x(1+2+4+8)=15⋅2x=60⇒2x=4=22,2^x(1+2+4+8)=15\cdot2^x=60\Rightarrow2^x=4=2^2,

so x=2x=2.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

If AM's are inserted between aa and bb, the common difference is : (a) b+an\frac{b+a}{n} (b) b+an+1\frac{b+a}{n+1} (c) b−an\frac{b-a}{n} (d) b−an+1\frac{b-a}{n+1}

Answer

(d) b−an+1\frac{b-a}{n+1}

Inserting nn arithmetic means between aa and bb gives an AP with n+2n+2 terms, first term aa and last term bb. So b=a+(n+1)db=a+(n+1)d, which gives

d=b−an+1.d=\frac{b-a}{n+1}.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

If a⃗\vec a and b⃗\vec b are unit vectors and θ\theta is angle between them, then ∣a⃗−b⃗∣|\vec a-\vec b| is equal to : (a) 2cos⁡θ2\cos\theta (b) 2sin⁡θ2\sin\theta (c) 2cos⁡θ22\cos\frac\theta2 (d) 2sin⁡θ22\sin\frac\theta2

Answer

(d) 2sin⁡θ22\sin\frac\theta2

For unit vectors, ∣a⃗∣=∣b⃗∣=1|\vec a|=|\vec b|=1 and a⃗⋅b⃗=cos⁡θ\vec a\cdot\vec b=\cos\theta.

∣a⃗−b⃗∣2=1+1−2cos⁡θ=2(1−cos⁡θ)=4sin⁡2θ2|\vec a-\vec b|^2=1+1-2\cos\theta=2(1-\cos\theta)=4\sin^2\frac\theta2

Hence ∣a⃗−b⃗∣=2sin⁡θ2|\vec a-\vec b|=2\sin\frac\theta2.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

A square matrix 'A' is called singular when ∣A∣=|A|= : (a) 0 (b) 1 (c) −1-1 (d) None

Answer

(a) 0

A square matrix is singular when its determinant is zero, because then it has no inverse. If ∣A∣≠0|A|\neq0 the matrix is non-singular.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

The volume of the sphere is 36π m336\pi\,\mathrm{m}^3, find its diameter. (a) 4 m (b) 8 m (c) 6 m (d) 9 m

Answer

(c) 6 m

43πr3=36π⇒r3=27⇒r=3 m\frac43\pi r^3=36\pi\Rightarrow r^3=27\Rightarrow r=3\ \text{m}

Diameter =2r=6=2r=6 m.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

What is the HCF of 9(x+y)9(x+y) and 3(x2−y2)3(x^2-y^2) ? (a) 3(x−y)3(x-y) (b) (x−y)(x-y) (c) 3(x+y)3(x+y) (d) (x+y)(x+y)

Answer

(c) 3(x+y)3(x+y)

Factorise: 9(x+y)=3⋅3⋅(x+y)9(x+y)=3\cdot3\cdot(x+y) and 3(x2−y2)=3(x+y)(x−y)3(x^2-y^2)=3(x+y)(x-y). The common factors are 33 and (x+y)(x+y), so the HCF is 3(x+y)3(x+y).

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  • 2025 Bridge Course Practice Set 5 · 1 mark

1.15 m is equal to : (a) 115 cm (b) 1005 cm (c) 1050 cm (d) 1500 cm

Answer

(a) 115 cm

1 m=1001\ \text{m}=100 cm, so 1.15×100=1151.15\times100=115 cm.

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  • 2025 Bridge Course Practice Set 5 · 1 mark

What percent is 1 of 50 ? (a) 1 (b) 2 (c) 5 (d) 50

Answer

(b) 2

150×100=2%\frac{1}{50}\times100=2\%

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  • 2025 Bridge Course Practice Set 5 · 1 mark

If the mean of 4, 7, aa, 8, 10, 12 is 8, then a=a= (a) 5 (b) 6 (c) 7 (d) 8

Answer

(c) 7

The sum of the six values is 6×8=486\times8=48.

4+7+a+8+10+12=41+a=48⇒a=74+7+a+8+10+12=41+a=48\Rightarrow a=7

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  • 2025 Bridge Course Practice Set 5 · 1 mark

∫1x dx=\int\frac1x\,dx= (a) log⁡x+c\log x+c (b) ex+ce^x+c (c) log⁡xx+C\frac{\log x}{x}+C (d) exx+C\frac{e^x}{x}+C

Answer

(a) log⁡x+c\log x+c

Since ddx(log⁡x)=1x\frac{d}{dx}(\log x)=\frac1x, we have ∫1x dx=log⁡x+c\int\frac1x\,dx=\log x+c (more fully log⁡∣x∣+c\log|x|+c).

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  • 2025 Bridge Course Practice Set 10 · 1 mark

The set A∪B‾\overline{A\cup B} is equal to : (a) A‾∪B‾\overline A\cup\overline B (b) A∩B‾\overline{A\cap B} (c) A∪BA\cup B (d) A‾∩B‾\overline A\cap\overline B

Answer

(d) A‾∩B‾\overline A\cap\overline B

By De Morgan's law, A∪B‾=A‾∩B‾\overline{A\cup B}=\overline A\cap\overline B.

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  • 2025 Bridge Course Practice Set 10 · 1 mark

If f(x)=2xf(x)=2^x, then f(0),f(1),f(2),…f(0), f(1), f(2),\ldots are in (a) A.P (b) G.P (c) H.P. (d) None

Answer

(b) G.P

f(0),f(1),f(2),…=1,2,4,…f(0),f(1),f(2),\ldots = 1,2,4,\ldots with a constant ratio f(n+1)/f(n)=2f(n+1)/f(n)=2, so they are in G.P.

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  • 2025 Bridge Course Practice Set 10 · 1 mark

A rectangle is twice as long it is wide if its length is 2x+y2x+y then the perimeter of the rectangle is : (a) 4x+2y4x+2y (b) 3x+3y/23x+3y/2 (c) 6x+3y6x+3y (d) 3x+2y3x+2y

Answer

(c) 6x+3y6x+3y

Length l=2x+yl=2x+y, width w=l2=x+y2w=\dfrac{l}{2}=x+\dfrac y2.

P=2(l+w)=2(3x+3y2)=6x+3yP=2(l+w)=2\left(3x+\tfrac{3y}{2}\right)=6x+3y

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  • 2025 Bridge Course Practice Set 10 · 1 mark

If (4x+3y):(6x+5y)=11:17(4x+3y):(6x+5y)=11:17 then x:yx:y is : (a) 1:2 (b) 2:3 (c) 2:1 (d) 3:1

Answer

(c) 2:1

17(4x+3y)=11(6x+5y)⇒68x+51y=66x+55y⇒2x=4y17(4x+3y)=11(6x+5y)\Rightarrow 68x+51y=66x+55y\Rightarrow 2x=4y

So x:y=2:1x:y=2:1.

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  • 2025 Bridge Course Practice Set 10 · 1 mark

What is the value of : tan⁡−112+tan⁡−113\tan^{-1}\frac12+\tan^{-1}\frac13 ? (a) π/2\pi/2 (b) π/3\pi/3 (c) π/4\pi/4 (d) π/6\pi/6

Answer

(c) π/4\pi/4

tan⁡−112+tan⁡−113=tan⁡−112+131−16=tan⁡−11=π4\tan^{-1}\frac12+\tan^{-1}\frac13=\tan^{-1}\frac{\frac12+\frac13}{1-\frac16}=\tan^{-1}1=\frac\pi4

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  • 2025 Bridge Course Practice Set 10 · 1 mark

If 25 boys can do a work in 7 days. After how many days, 5 boys should be added so that the entire work would be finished in 6 days ? (a) 1 day (b) 2 days (c) 3 days (d) 4 days

Answer

(a) 1 day

Total work =25×7=175=25\times7=175 boy-days. Let 5 boys be added after dd days:

25d+30(6−d)=175⇒180−5d=175⇒d=125d+30(6-d)=175\Rightarrow 180-5d=175\Rightarrow d=1

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  • 2025 Bridge Course Practice Set 10 · 1 mark

If 7x−13+715−x=507^{x-13}+7^{15-x}=50, what are the value of xx ? (a) 1, 3 (b) −13-13, 15 (c) 15, 7 (d) 13, 15

Answer

(d) 13, 15

Let t=7x−13t=7^{x-13}. Then 715−x=72/7x−13=49/t7^{15-x}=7^2/7^{x-13}=49/t.

t+49t=50⇒t2−50t+49=0⇒t=1 or 49t+\frac{49}{t}=50\Rightarrow t^2-50t+49=0\Rightarrow t=1\text{ or }49

7x−13=1⇒x=137^{x-13}=1\Rightarrow x=13; 7x−13=49⇒x=157^{x-13}=49\Rightarrow x=15.

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  • 2025 Bridge Course Practice Set 10 · 1 mark

By selling 12 eggs for Rs. 54 a merchant gains 20%. If he sells 40 eggs for Rs. 135, what will be gain or loss percent ? (a) 10% gain (b) 20% gain (c) 10% loss (d) 20% loss

Answer

(c) 10% loss

Cost of 12 eggs =541.2==\dfrac{54}{1.2}= Rs 45, so cost per egg == Rs 3.75.

Cost of 40 eggs == Rs 150, selling price == Rs 135, so loss == Rs 15.

Loss%=15150×100=10%\text{Loss}\%=\frac{15}{150}\times100=10\%

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  • 2025 Bridge Course Practice Set 10 · 1 mark

lim⁡x→0sin⁡xx=\displaystyle\lim_{x\to0}\frac{\sin x}{x}= ........ (a) 0 (b) A finite quantity between 1 and −1-1 other than zero (c) Does not exist (d) 1

Answer

(d) 1

This is the standard limit lim⁡x→0sin⁡xx=1\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1 (with xx in radians).

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  • 2025 Bridge Course Practice Set 10 · 1 mark

Simplify : 3x+2−3x8×3x\dfrac{3^{x+2}-3^x}{8\times3^x} (a) 1 (b) −1-1 (c) 0 (d) None of above

Answer

(a) 1

3x+2−3x8⋅3x=3x(9−1)8⋅3x=1\frac{3^{x+2}-3^x}{8\cdot3^x}=\frac{3^x(9-1)}{8\cdot3^x}=1

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  • 2025 Bridge Course Practice Set 10 · 1 mark

What is the maximum number of points of intersection between square and a circle ? (a) Less than 4 (b) 4 (c) 6 (d) 8

Answer

(d) 8

A circle can cut each of the 4 sides of a square in at most 2 points, so the maximum is 4×2=84\times2=8.

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  • 2025 Bridge Course Practice Set 10 · 1 mark

What is the derivative of 5x3+4x2−6x+95x^3+4x^2-6x+9 ? (a) 15x2+8x−615x^2+8x-6 (b) 18x−618x-6 (c) 15x2+4x−615x^2+4x-6 (d) 15x2+8x15x^2+8x

Answer

(a) 15x2+8x−615x^2+8x-6

ddx(5x3+4x2−6x+9)=15x2+8x−6\frac{d}{dx}(5x^3+4x^2-6x+9)=15x^2+8x-6

Using ddxxn=nxn−1\frac{d}{dx}x^n=nx^{n-1} term by term; the derivative of the constant 9 is 0.

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  • 2025 Bridge Course Practice Set 10 · 1 mark

Find dydx\frac{dy}{dx}, if x=at2x=at^2, y=2aty=2at. (a) tt (b) t2t^2 (c) 1t\frac1t (d) 1t2\frac1{t^2}

Answer

(c) 1t\frac1t

dxdt=2at\frac{dx}{dt}=2at and dydt=2a\frac{dy}{dt}=2a, so

dydx=dy/dtdx/dt=2a2at=1t\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2a}{2at}=\frac1t

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  • 2025 Bridge Course Practice Set 10 · 1 mark

If a:b = 2:3 and b:c = 4:5 then the value of a:b:c = ..... (a) 2:3:4 (b) 3:4:5 (c) 8:15:20 (d) 8:12:15

Answer

(d) 8:12:15

Make the common term bb equal: a:b=2:3=8:12a:b=2:3=8:12 and b:c=4:5=12:15b:c=4:5=12:15. Hence a:b:c=8:12:15a:b:c=8:12:15.

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  • 2025 Bridge Course Practice Set 10 · 1 mark

In two digit number xx both sum and different of its digit is 4. What is the value of number? (a) 30 (b) 40 (c) 60 (d) 80

Answer

(b) 40

Let the digits be pp and qq. Then p+q=4p+q=4 and p−q=4p-q=4, so p=4p=4, q=0q=0. The number is 40 (check: sum =4=4, difference =4=4).

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  • 2025 Bridge Course Practice Set 10 · 1 mark

The side of two squares is x:y the ratio of their parameter is (a) x4:y2\frac x4:\frac y2 (b) 4x:y4x:y (c) x:yx:y (d) x2:y2x^2:y^2

Answer

(c) x:yx:y

Perimeters of squares are 4x4x and 4y4y, so the ratio is 4x:4y=x:y4x:4y=x:y.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If A+B+C=πA+B+C=\pi then cot⁡A⋅cot⁡B+cot⁡B⋅cot⁡C+cot⁡C⋅cot⁡A=\cot A\cdot\cot B+\cot B\cdot\cot C+\cot C\cdot\cot A= (a) −1-1 (b) 2 (c) 1 (d) 3

Answer

(c) 1

Since A+B=π−CA+B=\pi-C, cot⁡(A+B)=−cot⁡C\cot(A+B)=-\cot C:

cot⁡Acot⁡B−1cot⁡A+cot⁡B=−cot⁡C\frac{\cot A\cot B-1}{\cot A+\cot B}=-\cot C

cot⁡Acot⁡B−1=−cot⁡C(cot⁡A+cot⁡B)\cot A\cot B-1=-\cot C(\cot A+\cot B)

So cot⁡Acot⁡B+cot⁡Bcot⁡C+cot⁡Ccot⁡A=1\cot A\cot B+\cot B\cot C+\cot C\cot A=1.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If f(x)=x+2f(x)=x+2 and g(x)=x3g(x)=x^3 then f∘g(1)f\circ g(1) is: (a) 2 (b) 3 (c) 1 (d) 4

Answer

(b) 3

g(1)=13=1,f(g(1))=f(1)=1+2=3g(1)=1^3=1,\qquad f(g(1))=f(1)=1+2=3

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If A=(xxxx)A=\begin{pmatrix}x&x\\x&x\end{pmatrix} then value of A3A^3 is (a) (x3x3x3x3)\begin{pmatrix}x^3&x^3\\x^3&x^3\end{pmatrix} (b) (3x33x33x33x3)\begin{pmatrix}3x^3&3x^3\\3x^3&3x^3\end{pmatrix} (c) (4x34x34x34x3)\begin{pmatrix}4x^3&4x^3\\4x^3&4x^3\end{pmatrix} (d) (3x23x23x23x2)\begin{pmatrix}3x^2&3x^2\\3x^2&3x^2\end{pmatrix}

Answer

(c) (4x34x34x34x3)\begin{pmatrix}4x^3&4x^3\\4x^3&4x^3\end{pmatrix}

Write A=xJA=xJ with J=(1111)J=\begin{pmatrix}1&1\\1&1\end{pmatrix}. Then J2=2JJ^2=2J and J3=4JJ^3=4J, so

A3=x3J3=4x3J.A^3=x^3J^3=4x^3J.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

The first order derivatives of curve gives (a) Slope (b) Point of inflection (c) Concave upward and downward (d) None of the above.

Answer

(a) Slope

The first derivative dydx\dfrac{dy}{dx} gives the slope of the tangent to the curve at a point. The second derivative gives concavity and points of inflection.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

The lines 3x−4y+4=03x-4y+4=0 and 6x−8y−7=06x-8y-7=0 are tangents to a circle, then radius of circle is : (a) 34\frac34 (b) 32\frac32 (c) 5 (d) 52\frac52

Answer

(a) 34\frac34

Rewrite the second line as 3x−4y−72=03x-4y-\tfrac72=0. The lines are parallel, so the distance between them is the diameter:

d=∣4−(−72)∣32+42=7.55=32d=\frac{\left|4-\left(-\tfrac72\right)\right|}{\sqrt{3^2+4^2}}=\frac{7.5}{5}=\frac32

Radius =d2=34=\dfrac d2=\dfrac34.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

The coordinates of trisection of join A (−5,5)(-5, 5) and B (25,10)(25, 10) close to end A is : (a) (15,0)(15,0) (b) (5,0)(5,0) (c) (5,3)(5,3) (d) (3,5)(3,5)

Answer

(c) (5,3)(5,3)

The point nearer A divides AB in the ratio 1:21:2:

x=1(25)+2(−5)3=5,y=1(10)+2(5)3=203x=\frac{1(25)+2(-5)}{3}=5,\qquad y=\frac{1(10)+2(5)}{3}=\frac{20}{3}

Of the given options, only (5,3)(5,3) has the correct abscissa x=5x=5, so it is the intended choice.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If a⃗×b⃗=b⃗×c⃗\vec a\times\vec b=\vec b\times\vec c where a⃗\vec a, b⃗\vec b and c⃗\vec c are coplanar vectors then for same value of kk: (a) a⃗+b⃗=kc⃗\vec a+\vec b=k\vec c (b) b⃗+c⃗=ka⃗\vec b+\vec c=k\vec a (c) a⃗+c⃗=kb⃗\vec a+\vec c=k\vec b (d) c⃗−a⃗=kb⃗\vec c-\vec a=k\vec b

Answer

(c) a⃗+c⃗=kb⃗\vec a+\vec c=k\vec b

a⃗×b⃗=b⃗×c⃗ ⇒ a⃗×b⃗+c⃗×b⃗=0⃗ ⇒ (a⃗+c⃗)×b⃗=0⃗\vec a\times\vec b=\vec b\times\vec c\ \Rightarrow\ \vec a\times\vec b+\vec c\times\vec b=\vec 0\ \Rightarrow\ (\vec a+\vec c)\times\vec b=\vec 0

So a⃗+c⃗\vec a+\vec c is parallel to b⃗\vec b, that is a⃗+c⃗=kb⃗\vec a+\vec c=k\vec b.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If ax=by=cza^x=b^y=c^z and xx, yy, zz are in HP, then aa, bb, cc are in: (a) AP (b) GP (c) HP (d) None

Answer

(b) GP

Let ax=by=cz=ka^x=b^y=c^z=k. Then a=k1/xa=k^{1/x}, b=k1/yb=k^{1/y}, c=k1/zc=k^{1/z}. Since x,y,zx,y,z are in HP, 1x,1y,1z\frac1x,\frac1y,\frac1z are in AP, so b2=k2/y=k1/x+1/z=acb^2=k^{2/y}=k^{1/x+1/z}=ac. Hence a,b,ca,b,c are in GP.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

Marty has exactly 5 blue pens, 6 black pens and 4 red pens in his knapsack. If he pulls out one pen at random, what is the probability that the pen is red or black? (a) 15\frac15 (b) 13\frac13 (c) 12\frac12 (d) 23\frac23

Answer

(d) 23\frac23

Total pens =5+6+4=15=5+6+4=15. Red or black pens =4+6=10=4+6=10, so P=1015=23P=\frac{10}{15}=\frac23.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If 6 students are eligible for 2 scholarships worth USD 1060 each, how many different combination of 2 students winning the 2 scholarships are possible? (a) 15 (b) 12 (c) 9 (d) 6

Answer

(a) 15

The order of the two winners does not matter, so the number of ways is (62)=6×52=15\binom62=\frac{6\times5}{2}=15.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If 2a−1a=52a-\frac1a=5, the value of 4a2+1a24a^2+\frac1{a^2} is : (a) 5 (b) 25 (c) 16 (d) 29

Answer

(d) 29

Square both sides of 2a−1a=52a-\frac1a=5:

4a2−2⋅2a⋅1a+1a2=25  ⇒  4a2−4+1a2=254a^2-2\cdot2a\cdot\frac1a+\frac1{a^2}=25\;\Rightarrow\;4a^2-4+\frac1{a^2}=25

so 4a2+1a2=294a^2+\frac1{a^2}=29.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

The angle between planes x+2y+z+1=0x+2y+z+1=0 and 4x+4y−4z+3=04x+4y-4z+3=0 is : (a) 0∘0^\circ (b) 90∘90^\circ (c) 30∘30^\circ (d) 45∘45^\circ

Answer

(d) 45∘45^\circ (nearest option)

The normals are n⃗1=(1,2,1)\vec n_1=(1,2,1) and n⃗2=(4,4,−4)\vec n_2=(4,4,-4), so

cos⁡θ=∣4+8−4∣6⋅48=8122=23≈0.471,\cos\theta=\frac{|4+8-4|}{\sqrt6\cdot\sqrt{48}}=\frac{8}{12\sqrt2}=\frac{\sqrt2}{3}\approx0.471,

giving θ≈61.9∘\theta\approx61.9^\circ. The planes are neither parallel (n⃗1⋅n⃗2≠±∣n⃗1∣∣n⃗2∣\vec n_1\cdot\vec n_2\ne\pm|\vec n_1||\vec n_2|) nor perpendicular (n⃗1⋅n⃗2=8≠0\vec n_1\cdot\vec n_2=8\neq0), so 0∘0^\circ and 90∘90^\circ are ruled out. No option equals the exact value; 45∘45^\circ is the closest of those given.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If A={1,2,3,4}A=\{1,2,3,4\}, then relation on A×AA\times A where R={(1,1),(1,2),(2,1)}R=\{(1,1),(1,2),(2,1)\} is defined by : (a) {(x,y):y=2x}\{(x,y): y=2x\} (b) {(x,y):x+y=1}\{(x,y): x+y=1\} (c) {(x,y):x+y≤3}\{(x,y): x+y\le3\} (d) {(x+y):x+y≤4}\{(x+y): x+y\le4\}

Answer

(c) {(x,y):x+y≤3}\{(x,y): x+y\le3\}

On A×AA\times A with A={1,2,3,4}A=\{1,2,3,4\}, the pairs with x+y≤3x+y\le3 are exactly (1,1),(1,2),(2,1)(1,1),(1,2),(2,1), which is RR. The other options fail: y=2xy=2x gives (1,2)(1,2) but not (1,1)(1,1), and x+y=1x+y=1 gives no pairs.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

If α\alpha and β\beta roots of equation ax2+bx+c=0ax^2+bx+c=0, then 1α\frac1\alpha and 1β\frac1\beta are roots of : (a) cx2+ax+b=0cx^2+ax+b=0 (b) cx2+bx+a=0cx^2+bx+a=0 (c) bx2+ax+c=0bx^2+ax+c=0 (d) ax2+cx+b=0ax^2+cx+b=0

Answer

(b) cx2+bx+a=0cx^2+bx+a=0

If α\alpha is a root of ax2+bx+c=0ax^2+bx+c=0, then 1α\frac1\alpha satisfies the equation obtained by putting x→1xx\to\frac1x:

ax2+bx+c=0  ⇒  cx2+bx+a=0.\frac{a}{x^2}+\frac bx+c=0\;\Rightarrow\;cx^2+bx+a=0.

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  • 2025 Bridge Course Practice Set 4 · 1 mark

Which of the following is true for any two numbers aa and bb ? (a) AM > HM > GM (b) HM > GM > AM (c) AM > GM > HM (d) GM > AM > HM

Answer

(c) AM > GM > HM

For positive numbers a≠ba\ne b, AM=a+b2\text{AM}=\frac{a+b}2, GM=ab\text{GM}=\sqrt{ab}, HM=2aba+b\text{HM}=\frac{2ab}{a+b}, and AM≥GM≥HM\text{AM}\ge\text{GM}\ge\text{HM} with GM2=AM⋅HM\text{GM}^2=\text{AM}\cdot\text{HM} (equality only when a=ba=b).

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  • 2025 Bridge Course Practice Set 4 · 1 mark

The circles x2+y2=9x^2+y^2=9 and x2+y2−12y+27=0x^2+y^2-12y+27=0 touch each other then the equation of the common tangent is: (a) x−3y+1=0x-3y+1=0 (b) y=3y=3 (c) x+y=2x+y=2 (d) x=4x=4

Answer

(b) y=3y=3

The circles are x2+y2=9x^2+y^2=9 (centre (0,0)(0,0), r=3r=3) and x2+(y−6)2=9x^2+(y-6)^2=9 (centre (0,6)(0,6), r=3r=3). The distance between centres is 6=3+36=3+3, so they touch externally at (0,3)(0,3). The common tangent there is perpendicular to the line of centres (the yy-axis), i.e. y=3y=3.

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  • 2080 Science Entrance Model Question · 1 mark

If A={p,q,r}A=\{p,q,r\}, then the number of proper subsets of AA is …… (a) 0 (b) 7 (c) 8 (d) 1

Answer

(b) 7

AA has 33 elements, so it has 23=82^3=8 subsets. Excluding AA itself, the number of proper subsets is 8−1=78-1=7.

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  • 2080 Science Entrance Model Question · 1 mark

The probability of 53 Sundays in a leap year is …….. (a) 27\frac27 (b) 53365\frac{53}{365} (c) 53366\frac{53}{366} (d) 17\frac17

Answer

(a) 27\frac27

A leap year has 366=52×7+2366=52\times7+2 days, so there are 5252 Sundays plus 22 extra consecutive days. A 53rd Sunday occurs if the extra days include Sunday: (Sun, Mon) or (Sat, Sun), which is 22 of the 77 equally likely pairs. Hence P=27P=\frac27.

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  • 2080 Science Entrance Model Question · 1 mark

The point (2,3)(2,3) divides the line joining (1,2)(1,2) and (5,6)(5,6) in the ratio …… (a) 2:3 (b) 3:2 (c) 3:1 (d) 1:3

Answer

(d) 1:3

Let the ratio be m:nm:n. Then 2=5m+nm+n⇒5m+n=2m+2n⇒3m=n,2=\frac{5m+n}{m+n}\Rightarrow 5m+n=2m+2n\Rightarrow 3m=n, so m:n=1:3m:n=1:3. Check with yy: 6(1)+2(3)1+3=3\frac{6(1)+2(3)}{1+3}=3, which matches.

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  • 2080 Science Entrance Model Question · 1 mark

The radius of the circle 4(x2+y2)+12ax−6ay−a2=04(x^2+y^2)+12ax-6ay-a^2=0 is (a) 7a4\frac{7a}{4} (b) 7a7a (c) 28a28a (d) 3a2\frac{3a}{2}

Answer

(a) 7a4\frac{7a}{4}

Divide by 44: x2+y2+3ax−3a2y−a24=0x^2+y^2+3ax-\frac{3a}{2}y-\frac{a^2}{4}=0, so g=3a2g=\frac{3a}{2}, f=−3a4f=-\frac{3a}{4}, c=−a24c=-\frac{a^2}{4}. r=g2+f2−c=9a24+9a216+a24=49a216=7a4.r=\sqrt{g^2+f^2-c}=\sqrt{\frac{9a^2}{4}+\frac{9a^2}{16}+\frac{a^2}{4}}=\sqrt{\frac{49a^2}{16}}=\frac{7a}{4}.

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  • 2079 Grade XI Model Question · 1 mark

If x is a non-zero rational number and xy is irrational, then y must be: (a) a rational number (b) an irrational number (c) non zero (d) an integer.

Answer

(b) an irrational number

If yy were rational, then xyxy would be a product of two rational numbers and hence rational. Since xyxy is irrational, yy must be irrational.

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  • 2079 Grade XI Model Question · 1 mark

The line through the points (a,b)(a,b) and (−a,−b)(-a,-b) passes through the point: (a) (1,1)(1,1) (b) (3a,−2b)(3a,-2b) (c) (a2,ab)(a^2,ab) (d) none of these.

Answer

(c) (a2,ab)(a^2,ab)

The slope is m=−b−b−a−a=ba,m=\frac{-b-b}{-a-a}=\frac{b}{a}, and the line through (a,b)(a,b) is y−b=ba(x−a)y-b=\frac{b}{a}(x-a), which simplifies to y=baxy=\frac{b}{a}x. Putting x=a2x=a^2 gives y=aby=ab, so (a2,ab)(a^2,ab) lies on the line.

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  • 2079 Grade XI Model Question · 1 mark

The equation y−y1=m(x−x1)y-y_1=m(x-x_1) for different mm and (x1,y1)(x_1,y_1) fixed represents: (a) a family of parallel lines (b) a single line (c) a family of lines which are not concurrent (d) a family of concurrent lines.

Answer

(d) a family of concurrent lines.

For every value of mm the line y−y1=m(x−x1)y-y_1=m(x-x_1) passes through the fixed point (x1,y1)(x_1,y_1), so all the lines are concurrent at that point.

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  • 2079 Grade XI Model Question · 1 mark

The radius of the circle passing through the point (6,2)(6,2) two of whose diameters are x+y=6x+y=6 and x+2y=4x+2y=4 is: (a) 10 (b) 6 (c) 4 (d) 252\sqrt5

Answer

(d) 252\sqrt5

The diameters meet at the centre. Solving x+y=6x+y=6 and x+2y=4x+2y=4 gives y=−2y=-2, x=8x=8, so the centre is (8,−2)(8,-2). The radius is the distance to (6,2)(6,2): r=(8−6)2+(−2−2)2=4+16=25.r=\sqrt{(8-6)^2+(-2-2)^2}=\sqrt{4+16}=2\sqrt5.

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  • 2079 Grade XI Model Question · 1 mark

The probability of a leap year selected at random that contains 53 Sunday is: (a) 53366\frac{53}{366} (b) 17\frac17 (c) 27\frac27 (d) 53365\frac{53}{365}

Answer

(c) 27\frac27

A leap year has 366 days, which is 52 weeks and 2 extra days. The two extra days can be (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat) or (Sat, Sun), all equally likely. A 53rd Sunday needs a Sunday among them, which happens in 2 of the 7 cases, so the probability is 27\frac27.

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  • 2079 Grade XI Model Question · 1 mark

The area enclosed by the pair of lines xy=0xy=0, the line x=4x=4 and y=5y=5 is (a) 20 sq. units (b) 10 sq units (c) 45\frac45 sq. units (d) 0 sq. units

Answer

(a) 20 sq. units

xy=0xy=0 means the two axes x=0x=0 and y=0y=0. Together with x=4x=4 and y=5y=5 they enclose a rectangle of area 4×5=204\times5=20 sq. units.

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  • 2079 Grade XI Model Question · 1 mark

Let A= {1,2,3} and B={2,4,6} then f = {(1,2), (2,4) , (3,6) } is: (a) one to one function from A to B. (b) onto function from A to B. (c) one to one and onto function from A to B. (d) None of these.

Answer

(c) one to one and onto function from A to B.

The images 2,4,62,4,6 are all different, so ff is one to one. They also fill all of B={2,4,6}B=\{2,4,6\}, so ff is onto.

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  • 2079 Grade XI Model Question · 1 mark

Which of the following is false? (a) Inscribed angles standing on the same arc of a circle are equal. (b) The opposite angles of a cyclic quadrilateral are supplementary. (c) Inscribed angle is double of the central angle standing on the same arc of a circle. (d) Equal arcs subtend equal angles at the centre of a circle.

Answer

(c) Inscribed angle is double of the central angle standing on the same arc of a circle.

This statement is false. The central angle is double the inscribed angle on the same arc, so the inscribed angle is half the central angle. The other three statements are standard true theorems.

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  • 2079 Grade XI Model Question · 1 mark

The mean of a set of numbers is xx. If each number is multiplied by λ\lambda, then the mean of the new numbers is (a) xx (b) x+λx+\lambda (c) λx\lambda x (d) none of these.

Answer

(c) λx\lambda x

If the numbers are a1,…,ana_1,\dots,a_n then ∑λain=λ⋅∑ain=λx\frac{\sum \lambda a_i}{n}=\lambda\cdot\frac{\sum a_i}{n}=\lambda x.

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  • 2079 Grade XI Model Question · 1 mark

If in 2 years, under the same rate of interest, the simple interest is Rs. 40 and compound interest is Rs. 41 then the principal is: (a) Rs. 500 (b) Rs. 400 (c) Rs. 800 (d) Rs. 820

Answer

(b) Rs. 400

Simple interest for 2 years is Rs. 40, so it is Rs. 20 per year. Compound interest exceeds simple interest by the interest on the first year's interest: 41−40=141-40=1, so 20×r100=120\times\frac{r}{100}=1 and r=5%r=5\%. Then 20=P×510020=\frac{P\times5}{100} gives P=Rs. 400P=\text{Rs. }400.

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  • 2079 Grade XI Model Question · 1 mark

The age of B is half the sum of the ages of A and C. If B is 2 years younger than A and C is 32 years old, then the age of B must be: (a) 28 years (b) 30 years (c) 34 years (d) 36 years

Answer

(c) 34 years

Let the ages be A,B,CA,B,C with C=32C=32 and A=B+2A=B+2. Then B=A+C2B=\frac{A+C}{2} gives 2B=B+2+322B=B+2+32, so B=34B=34.

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  • 2079 Grade XI Model Question · 1 mark

The acute angle between the lines x-y=0 and y=0 is: (a) 30 0 (b) 45 0 (c) 60 0 (d) 75 0

Answer

(b) 45∘45^\circ

The line x−y=0x-y=0 has slope 11, so it makes 45∘45^\circ with the xx-axis, which is the line y=0y=0. Check: tan⁡θ=∣1−01+0∣=1\tan\theta=\left|\frac{1-0}{1+0}\right|=1, so θ=45∘\theta=45^\circ.

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  • 2079 Grade XI Model Question · 1 mark

The equation y=x2+cy=x^2+c represents a parabola with vertex (turning point) at (a) (0,c)(0,c) (b) (0,−c)(0,-c) (c) (0,0)(0,0) (d) (c,0)(c,0)

Answer

(a) (0,c)(0,c)

y=x2+cy=x^2+c is the parabola y=x2y=x^2 shifted up by cc units. Its minimum is at x=0x=0, where y=cy=c, so the vertex is (0,c)(0,c).

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  • 2079 Grade XI Model Question · 1 mark

The feasible region represented by x+2y<6x+2y<6, x>0x>0, y>0y>0 contains which of the following point? (a) (2,2)(2,2) (b) (1,1)(1,1) (c) (−1,−1)(-1,-1) (d) (0,−3)(0,-3)

Answer

(b) (1,1)(1,1)

A point must satisfy x>0x>0, y>0y>0 and x+2y<6x+2y<6. For (1,1)(1,1): 1+2=3<61+2=3<6, so it lies in the region. (2,2)(2,2) gives 6<66<6, which is false. (−1,−1)(-1,-1) and (0,−3)(0,-3) violate x>0x>0 or y>0y>0.

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  • 2079 Grade XI Model Question · 1 mark

The quadratic equation whose roots are 7 and −5-5 is (a) x2−2x−35=0x^2-2x-35=0 (b) x2+2x−35=0x^2+2x-35=0 (c) (x−7)(x+5)=1(x-7)(x+5)=1 (d) none of these

Answer

(a) x2−2x−35=0x^2-2x-35=0

Sum of roots =7+(−5)=2=7+(-5)=2 and product =7×(−5)=−35=7\times(-5)=-35. The equation is x2−(sum)x+product=0x^2-(\text{sum})x+\text{product}=0, that is x2−2x−35=0x^2-2x-35=0.

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  • 2079 Grade XI Model Question · 1 mark

If CC is the middle point of ABAB and PP is any point outside ABAB, then (a) PA⃗+PB⃗=PC⃗\vec{PA}+\vec{PB}=\vec{PC} (b) PA⃗+PB⃗=2PC⃗\vec{PA}+\vec{PB}=2\vec{PC} (c) PA⃗+PB⃗+PC⃗=0\vec{PA}+\vec{PB}+\vec{PC}=0 (d) PA⃗+PB⃗+2PC⃗=0\vec{PA}+\vec{PB}+2\vec{PC}=0

Answer

(b) PA⃗+PB⃗=2PC⃗\vec{PA}+\vec{PB}=2\vec{PC}

Since CC is the midpoint of ABAB, its position vector is c⃗=a⃗+b⃗2\vec c=\frac{\vec a+\vec b}{2} relative to PP. Hence PA⃗+PB⃗=a⃗+b⃗=2c⃗=2PC⃗\vec{PA}+\vec{PB}=\vec a+\vec b=2\vec c=2\vec{PC}.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If sin⁡θ\sin\theta is the geometric mean between sin⁡ϕ\sin\phi and cos⁡ϕ\cos\phi, then cos⁡2θ\cos2\theta is equal to (a) 2sin⁡2(π4−ϕ)2\sin^2\left(\frac\pi4-\phi\right) (b) 2cos⁡2(π4−ϕ)2\cos^2\left(\frac\pi4-\phi\right) (c) 2cos⁡2(π4+ϕ)2\cos^2\left(\frac\pi4+\phi\right) (d) 2sin⁡2(π4+ϕ)2\sin^2\left(\frac\pi4+\phi\right)

Answer

(a) 2sin⁡2(π4−ϕ)2\sin^2\left(\frac\pi4-\phi\right)

Since sin⁡θ\sin\theta is the geometric mean, sin⁡2θ=sin⁡ϕcos⁡ϕ\sin^2\theta=\sin\phi\cos\phi. Then cos⁡2θ=1−2sin⁡2θ=1−2sin⁡ϕcos⁡ϕ=1−sin⁡2ϕ.\cos2\theta=1-2\sin^2\theta=1-2\sin\phi\cos\phi=1-\sin2\phi. Also 2sin⁡2(π4−ϕ)=1−cos⁡(π2−2ϕ)=1−sin⁡2ϕ2\sin^2\left(\frac\pi4-\phi\right)=1-\cos\left(\frac\pi2-2\phi\right)=1-\sin2\phi, which matches.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If 13≤k≤2113\le k\le21, 9≤p≤199\le p\le19, 2<m<62<m<6, kk, pp, mm are integers, what is the largest possible value of k−pm\frac{k-p}{m} ? (a) 3 (b) 5 (c) 6 (d) 4

Answer

(d) 4

To maximise k−pm\frac{k-p}{m}, take the largest numerator and the smallest positive denominator. The largest k−pk-p is 21−9=1221-9=12, and since 2<m<62<m<6 with mm an integer, the smallest mm is 33. So the maximum is 123=4\frac{12}{3}=4.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If AA and BB are positive acute angle such that cos⁡A=310\cos A=\frac3{\sqrt{10}}, sin⁡B=15\sin B=\frac1{\sqrt5}, then A+BA+B is equal to (a) 135∘135^\circ (b) 45∘45^\circ (c) 60∘60^\circ (d) 120∘120^\circ

Answer

(b) 45∘45^\circ

cos⁡A=310\cos A=\frac3{\sqrt{10}} gives sin⁡A=110\sin A=\frac1{\sqrt{10}}, so tan⁡A=13\tan A=\frac13. sin⁡B=15\sin B=\frac1{\sqrt5} gives cos⁡B=25\cos B=\frac2{\sqrt5}, so tan⁡B=12\tan B=\frac12. Then tan⁡(A+B)=13+121−16=5/65/6=1.\tan(A+B)=\frac{\frac13+\frac12}{1-\frac16}=\frac{5/6}{5/6}=1. Both angles are acute with A+B<180∘A+B<180^\circ, so A+B=45∘A+B=45^\circ.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

The compound interest calculated yearly on a certain sum of money upto the fourth year is Rs.1320 and fifth year is Rs.1452. Then the rate of interest is (a) 5% (b) 6% (c) 8% (d) 10%

Answer

(d) 10%

The difference between the compound interest of consecutive years is the interest on the previous year's interest. So 1452−1320=1321452-1320=132 is the interest on Rs. 1320 for one year, giving r=1321320×100=10%r=\frac{132}{1320}\times100=10\%.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If y2−y=0y^2-y=0, 74x−3=y7^{4x-3}=y and yy is a positive integer, then the value of xx is (a) 43\frac43 (b) 34\frac34 (c) −43\frac{-4}3 (d) −34\frac{-3}4

Answer

(b) 34\frac34

y2−y=0y^2-y=0 gives y=0y=0 or y=1y=1. Since yy is a positive integer, y=1y=1. Then 74x−3=1=707^{4x-3}=1=7^0, so 4x−3=04x-3=0 and x=34x=\frac34.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

What is the sum of the slopes of the straight lines represented by the equation 3x2−4xy+2y2=03x^2-4xy+2y^2=0 (a) 2 (b) 32\frac32 (c) 23\frac23 (d) 5

Answer

(a) 2

Put y=mxy=mx in 3x2−4xy+2y2=03x^2-4xy+2y^2=0 to get 2m2−4m+3=02m^2-4m+3=0. The slopes are the roots of this quadratic, so their sum is 42=2\frac{4}{2}=2. (The discriminant is negative, so the lines are imaginary, but the sum formula still gives 22.)

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

Let the function gg be defined by g={(−2,4),(−1,1),(0,6),(1,−1),(2,8)}g=\{(-2,4),(-1,1),(0,6),(1,-1),(2,8)\} which statement is true? (a) g(1)=g(−1)g(1)=g(-1) (b) g(2)+g(−2)=0g(2)+g(-2)=0 (c) g(−2)=12g(2)g(-2)=\frac12g(2) (d) g(0)=g(1)+g(−1)g(0)=g(1)+g(-1)

Answer

(c) g(−2)=12g(2)g(-2)=\frac12g(2)

From the set: g(−2)=4g(-2)=4, g(2)=8g(2)=8, so 12g(2)=4=g(−2)\frac12g(2)=4=g(-2). The others fail: g(1)=−1≠g(−1)=1g(1)=-1\ne g(-1)=1; g(2)+g(−2)=12≠0g(2)+g(-2)=12\ne0; g(0)=6≠g(1)+g(−1)=0g(0)=6\ne g(1)+g(-1)=0.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If sin⁡x+sin⁡2x=1\sin x+\sin^2x=1, then cos⁡8x+2cos⁡6x+cos⁡4x=\cos^8x+2\cos^6x+\cos^4x= ………? (a) 0 (b) −1-1 (c) 2 (d) 1

Answer

(d) 1

From sin⁡x+sin⁡2x=1\sin x+\sin^2x=1 we get sin⁡x=1−sin⁡2x=cos⁡2x\sin x=1-\sin^2x=\cos^2x. Let s=sin⁡x=cos⁡2xs=\sin x=\cos^2x. Then cos⁡8x+2cos⁡6x+cos⁡4x=cos⁡4x(cos⁡2x+1)2=s2(s+1)2=(s2+s)2=12=1.\cos^8x+2\cos^6x+\cos^4x=\cos^4x(\cos^2x+1)^2=s^2(s+1)^2=(s^2+s)^2=1^2=1.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

The sum of the first nn terms of an AP is 3n2−n3n^2-n and its common difference is 6, then its first term is (a) 1 (b) 2 (c) 3 (d) 4

Answer

(b) 2

The first term equals S1=3(1)2−1=2S_1=3(1)^2-1=2. Check: S2=12−2=10S_2=12-2=10, so a2=8a_2=8 and d=8−2=6d=8-2=6, as given.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If AA and BB are two sets, then the symmetric difference of the sets AA and BB is denoted by (a) A−BA-B (b) B−AB-A (c) AΔBA\Delta B (d) A∪B‾\overline{A\cup B}

Answer

(c) AΔBA\Delta B

The symmetric difference is AΔB=(A−B)∪(B−A)A\Delta B=(A-B)\cup(B-A), the elements in exactly one of the two sets.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

The graph of y=f(x)y=f(x), where f(x)f(x) is a quadratic polynomial meets the xx-axis at (−2,0)(-2,0) and (−3,0)(-3,0), then the expression for f(x)f(x) is : (a) x2+5x+6x^2+5x+6 (b) x2−5x+6x^2-5x+6 (c) x2+5x−6x^2+5x-6 (d) x2−5x−6x^2-5x-6

Answer

(a) x2+5x+6x^2+5x+6

The zeros are −2-2 and −3-3, so f(x)=(x+2)(x+3)=x2+5x+6f(x)=(x+2)(x+3)=x^2+5x+6 (taking leading coefficient 1, as all options do).

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

The diagonals of a rhombus are 6cm and 8cm. The perimeter of the rhombus is (a) 18cm (b) 28cm (c) 20cm (d) 14cm

Answer

(c) 20cm

The diagonals bisect each other at right angles, so each side is 32+42=5\sqrt{3^2+4^2}=5 cm. Perimeter =4×5=20=4\times5=20 cm.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If a+1a=2a+\frac1a=2, what is the value of a+1a\sqrt a+\frac1{\sqrt a} ? (a) 2 (b) 4 (c) 9 (d) 13

Answer

(a) 2

a+1a=2a+\frac1a=2 gives (a−1)2=0(a-1)^2=0, so a=1a=1. Then a+1a=1+1=2\sqrt a+\frac1{\sqrt a}=1+1=2.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

Values of pp for which (x−1)(x-1) is a factor of x3+(p+1)2x2−10x^3+(p+1)^2x^2-10 are (a) 4, 2 (b) 2, −4-4 (c) −2-2, 4 (d) −2-2, −4-4

Answer

(b) 2, −4-4

(x−1)(x-1) is a factor, so the value at x=1x=1 is zero: 1+(p+1)2−10=01+(p+1)^2-10=0, so (p+1)2=9(p+1)^2=9 and p+1=±3p+1=\pm3. Hence p=2p=2 or p=−4p=-4.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

A cylinder, a cone and a hemisphere have equal base and have same height. What is the ratio of their volumes? (a) 1:2:3 (b) 2:3:1 (c) 3:2:1 (d) 3:1:2

Answer

(d) 3:1:2

With base radius rr and height h=rh=r: cylinder =πr2⋅r=πr3=\pi r^2\cdot r=\pi r^3, cone =13πr3=\frac13\pi r^3, hemisphere =23πr3=\frac23\pi r^3. The ratio is 1:13:23=3:1:21:\frac13:\frac23=3:1:2.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

Ram ate 14\frac14 of a whole pizza and Hari ate 15\frac15 of the remaining portion. What fraction of the pizza was not eaten? (a) 1120\frac{11}{20} (b) 920\frac9{20} (c) 320\frac3{20} (d) 35\frac35

Answer

(d) 35\frac35

Ram ate 14\frac14, leaving 34\frac34. Hari ate 15×34=320\frac15\times\frac34=\frac3{20}. Eaten in total =520+320=820=\frac5{20}+\frac3{20}=\frac8{20}, so the uneaten part is 1−25=351-\frac25=\frac35.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

Which of the following cannot be the length of sides of a right angled triangle? (a) 9cm, 15cm, 12cm (b) 2cm, 1cm, √5 cm (c) 400mm, 300mm, 500mm (d) 9cm, 5cm, 7cm

Answer

(d) 9cm, 5cm, 7cm

Check a2+b2=c2a^2+b^2=c^2 with the longest side as hypotenuse. (a) 92+122=225=1529^2+12^2=225=15^2 holds. (b) 12+22=5=(5)21^2+2^2=5=(\sqrt5)^2 holds. (c) 3002+4002=5002300^2+400^2=500^2 holds. (d) 52+72=74≠81=925^2+7^2=74\ne81=9^2, so it fails.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

A tower subtends an angle α\alpha at a point 'A' in the plane of its base and the angle of depression of the foot of the tower at a point bb metres just above A is β\beta. Then the height of the tower in metres is (a) btan⁡αcot⁡βb\tan\alpha\cot\beta (b) bcot⁡αtan⁡βb\cot\alpha\tan\beta (c) btan⁡αtan⁡βb\tan\alpha\tan\beta (d) bcot⁡αcot⁡βb\cot\alpha\cot\beta

Answer

(a) btan⁡αcot⁡βb\tan\alpha\cot\beta

Let the tower have height HH and let the foot be at horizontal distance dd from A. Then H=dtan⁡αH=d\tan\alpha. From the point bb m above A, the depression of the foot is β\beta, so tan⁡β=bd\tan\beta=\frac bd and d=bcot⁡βd=b\cot\beta. Hence H=btan⁡αcot⁡βH=b\tan\alpha\cot\beta.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If A=(42−11)A=\begin{pmatrix}4&2\\-1&1\end{pmatrix} and II is an identity matrix of order 2×22\times2, then (A−2I)(A−3I)=(A-2I)(A-3I)= (a) 1 (b) 0 (c) [1001]\begin{bmatrix}1&0\\0&1\end{bmatrix} (d) [0001]\begin{bmatrix}0&0\\0&1\end{bmatrix}

Answer

(b) 0

A−2I=(22−1−1),A−3I=(12−1−2)A-2I=\begin{pmatrix}2&2\\-1&-1\end{pmatrix},\quad A-3I=\begin{pmatrix}1&2\\-1&-2\end{pmatrix}

(A−2I)(A−3I)=(2−24−4−1+1−2+2)=(0000)(A-2I)(A-3I)=\begin{pmatrix}2-2&4-4\\-1+1&-2+2\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}

This is the null (zero) matrix, as the Cayley-Hamilton theorem predicts, since the eigenvalues of AA are 2 and 3.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

If 2a+b =7 and b+2c=23, what is the arithmetic average of a, b and c? (a) 5 (b) 10 (c) 15 (d) 7.5

Answer

(a) 5

Adding the two equations: 2a+2b+2c=302a+2b+2c=30, so a+b+c=15a+b+c=15. The average is 153=5\frac{15}3=5.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

The circle x2+y2−2ax−2ay+a2=0x^2+y^2-2ax-2ay+a^2=0 (a≠0a\ne0) (a) passes through the origin (b) touches only x-axis (c) touches only y-axis (d) touches both the axes

Answer

(d) touches both the axes

Rewrite as (x−a)2+(y−a)2=a2(x-a)^2+(y-a)^2=a^2. The centre is (a,a)(a,a) and the radius is ∣a∣|a|, which equals the distance of the centre from each axis. So the circle touches both axes.

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  • 2025 (2082) Grade XI Model Question Set A · 1 mark

A number 'a' increased by 20% of a results in a number b. When b is decreased by 331333\frac13% of b, the results is c. The number c is what percentage of a ? (a) 40% (b) 60% (c) 80% (d) 120%

Answer

(c) 80%

b=a+0.2a=1.2ab=a+0.2a=1.2a. Then c=b−13b=23b=23(1.2a)=0.8ac=b-\frac13b=\frac23b=\frac23(1.2a)=0.8a, which is 80% of aa.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

Let A = {1, 2, 3}, the possible numbers of non-empty proper subset for the set A is ..... (a) 8 (b) 7 (c) 6 (d) 9

Answer

(c) 6

A set with nn elements has 2n2^n subsets. Here 23=82^3 = 8. Removing the empty set and the set AA itself gives 8−2=68 - 2 = 6 non-empty proper subsets.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

Which one of the following is the multiplicative inverse of -a? (a) -1/a (b) a (c) 1 (d) 1/a

Answer

(a) −1a-\dfrac{1}{a}

The multiplicative inverse of xx is 1x\dfrac{1}{x}. With x=−ax=-a, it is 1−a=−1a\dfrac{1}{-a} = -\dfrac{1}{a}, since (−a)(−1a)=1(-a)\left(-\dfrac1a\right)=1.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If A = (-2, 3] and B = (0, 5) then B – A = ..... (a) (0, 5) (b) (3, 5) (c) [3, 5) (d) None

Answer

(b) (3,5)(3, 5)

B−AB - A is the set of elements of B=(0,5)B=(0,5) not in A=(−2,3]A=(-2,3]. Points of (0,3](0,3] lie in AA, so what remains is (3,5)(3,5).

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  • 2018 Practice MCQs (Mathematics) · 1 mark

Let A and B are any two sets such that n(A) = 3, n(B) = 4, then the number of all possible relations from set A to B is ....... (a) 12 (b) 64 (c) 81 (d) 63

Answer

None of the listed options. The number of relations is 40964096.

A relation from AA to BB is any subset of A×BA\times B. Since n(A×B)=3×4=12n(A\times B)=3\times4=12, the number of subsets is 212=4096.2^{12}=4096. None of 12, 64, 81 or 63 equals this value.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If R = {(3, 2), (4, 2), (3, 1), (7, 1), (2, 3)} is to be a function. Then which one of the following must be removed from the set? (a) (3, 2) (b) (4, 2) (c) (2, 3) (d) (7, 1)

Answer

(a) (3,2)(3, 2)

In a function each first element may appear with only one second element. Here 33 is paired with both 22 and 11, so one of (3,2)(3,2) or (3,1)(3,1) must go. Only (3,2)(3,2) is among the options. Removing it leaves {(4,2),(3,1),(7,1),(2,3)}\{(4,2),(3,1),(7,1),(2,3)\}, which is a function.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If p(x) = 4x – 6 and p(a) = 0, then a = ....... (a) 3/2 (b) –6 (c) 2/3 (d) -3/2

Answer

(a) 32\dfrac{3}{2}

p(a)=4a−6=0⇒a=64=32p(a)=4a-6=0 \Rightarrow a=\dfrac{6}{4}=\dfrac{3}{2}.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If g(x) = 3x + 2 and g(f(x)) = x, then f(2) = ........... (a) 6 (b) 3 (c) –1 (d) 0

Answer

(d) 0

g(f(x))=3f(x)+2=xg(f(x)) = 3f(x)+2 = x, so f(x)=x−23f(x)=\dfrac{x-2}{3}. Then f(2)=2−23=0f(2)=\dfrac{2-2}{3}=0.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

The area of square field is 144m², then its perimeter is ... (a) 12m (b) 48m (c) 144m (d) 150m

Answer

(b) 48 m

Area =s2=144⇒s=12=s^2=144 \Rightarrow s=12 m. Perimeter =4s=4×12=48=4s=4\times12=48 m.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If a motor car is driven at the rate of 66km/hr. Each of its wheel makes 500 revolution per minute. What is the radius of wheel? (a) 0.35m (b) 0.67m (c) 0.7m (d) 0.87m

Answer

(a) 0.35 m

Speed =66=66 km/hr =6600060=1100=\dfrac{66000}{60}=1100 m/min. Distance per revolution =1100500=2.2=\dfrac{1100}{500}=2.2 m, which is the circumference of the wheel. 2πr=2.2⇒r=2.22×227=0.35 m.2\pi r = 2.2 \Rightarrow r=\frac{2.2}{2\times\frac{22}{7}}=0.35\text{ m}.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If A man losses Rs. 60 if 15% discount is offered for a marked price and makes the profit of Rs. 60 if 10% discount is offered. Then the marked price of the goods is.... (a) Rs. 2400 (b) Rs. 2100 (c) Rs. 2000 (d) Rs. 1800

Answer

(a) Rs. 2400

Let the marked price be Rs. MM and the cost price Rs. CC.

With 15% discount (loss): 0.85M=C−600.85M = C - 60. With 10% discount (profit): 0.90M=C+600.90M = C + 60.

Subtracting: 0.05M=1200.05M = 120, so M=2400M = 2400. Check: C=0.85(2400)+60=2100C = 0.85(2400)+60 = 2100, and 0.9(2400)−60=21000.9(2400)-60=2100. ✓

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If there are two containers of sugar solution, the first is 4% concentration and second is 8% concentration. How much of each should we combine to get 40 gallons of 5% concentrate solution? (a) 20, 20 (b) 10, 30 (c) 30, 10 (d) 15, 25

Answer

(c) 30, 10

Let xx gallons of 4% solution be used, so 40−x40-x gallons of 8% solution. 0.04x+0.08(40−x)=0.05(40)0.04x+0.08(40-x)=0.05(40) 3.2−0.04x=2⇒x=30.3.2-0.04x=2 \Rightarrow x=30. So 30 gallons of 4% and 10 gallons of 8%.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If a tourist paid Rs. 5610 for a carvel window mode of wood with a discount of 15% including 10% VAT, what is the marked price of window? (a) 5000 (b) 6000 (c) 6500 (d) 6700

Answer

(b) Rs. 6000

Let the marked price be MM. After 15% discount the price is 0.85M0.85M; adding 10% VAT gives 0.85M×1.10=0.935M0.85M\times1.10=0.935M. 0.935M=5610⇒M=6000.0.935M=5610 \Rightarrow M=6000.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If (4, y) divides the join of (6, -1) and (3, 2) in the ratio of 2:1 then the value of y is: (a) 2 (b) 8 (c) 1 (d) 1/2

Answer

(c) 1

For the ratio m:n=2:1m:n=2:1 with (x1,y1)=(6,−1)(x_1,y_1)=(6,-1) and (x2,y2)=(3,2)(x_2,y_2)=(3,2): y=2(2)+1(−1)2+1=33=1.y=\frac{2(2)+1(-1)}{2+1}=\frac{3}{3}=1. Check: x=2(3)+1(6)3=4x=\dfrac{2(3)+1(6)}{3}=4 ✓.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

For what value of k, the three lines x -2y +1 = 0 and 2x – 5y + 3 = 0 and 5x + 9y + k = 0 are concurrent? (a) -14 (b) 4 (c) 4 (d) 14

Answer

(a) −14-14

Intersect the first two lines. From x−2y+1=0x-2y+1=0, x=2y−1x=2y-1. Substituting in 2x−5y+3=02x-5y+3=0: 4y−2−5y+3=0⇒y=1, x=14y-2-5y+3=0 \Rightarrow y=1,\ x=1.

For concurrency, (1,1)(1,1) must lie on the third line: 5+9+k=0⇒k=−145+9+k=0 \Rightarrow k=-14.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

The line x–y = 2 cuts the join of (3, -1) and (8, 9) in the ratio of ….. (a) 1:3 (b) 2:3 (c) 3:1 (d) 3:2

Answer

(b) 2:3

Let the line divide the join in the ratio m:nm:n. The point is (8m+3nm+n, 9m−nm+n).\left(\frac{8m+3n}{m+n},\ \frac{9m-n}{m+n}\right). It lies on x−y=2x-y=2: (8m+3n)−(9m−n)=2(m+n)⇒−m+4n=2m+2n⇒2n=3m,(8m+3n)-(9m-n)=2(m+n) \Rightarrow -m+4n=2m+2n \Rightarrow 2n=3m, so m:n=2:3m:n=2:3. Check: the point is (5,3)(5,3) and 5−3=25-3=2 ✓.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

Find the equation of line passing through (3, 2) and perpendicular to the line y = x is: (a) x – y = 5 (b) x + y = 5 (c) x + y = 1 (d) x – y = 1

Answer

(b) x+y=5x+y=5

The line y=xy=x has slope 11, so the perpendicular has slope −1-1. Through (3,2)(3,2): y−2=−(x−3)⇒x+y=5.y-2=-(x-3) \Rightarrow x+y=5.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

The length of perpendicular from the point (a, 3) on line 3x + 4y + 5 = 0 is 4 then a = (a) 1 (b) -3 (c) -3/4 (d) 2

Answer

(a) 1

Distance of (a,3)(a,3) from 3x+4y+5=03x+4y+5=0:

∣3a+12+5∣32+42=4⇒∣3a+17∣=20⇒a=1 or a=−373.\frac{|3a+12+5|}{\sqrt{3^2+4^2}}=4 \Rightarrow |3a+17|=20 \Rightarrow a=1 \text{ or } a=-\tfrac{37}{3}.

Only a=1a=1 is among the options.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

For what value of K will the points (1, 4), (-3, 16) and (K, -2) are collinear? (a) 1 (b) 3 (c) 6 (d) 10

Answer

(b) 3

For collinearity, slope of the first two points equals slope of the first and third:

16−4−3−1=−3,−2−4K−1=−3⇒K−1=2⇒K=3.\frac{16-4}{-3-1}=-3,\qquad \frac{-2-4}{K-1}=-3 \Rightarrow K-1=2 \Rightarrow K=3.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

The sum of 1st 20 terms of an A.P. whose 1st term is 2 and common difference 3 is.. (a) 520 (b) 610 (c) 400 (d) 730

Answer

(b) 610

S20=202[2(2)+(20−1)(3)]=10(4+57)=610.S_{20}=\frac{20}{2}\left[2(2)+(20-1)(3)\right]=10(4+57)=610.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

An exterior angle of a regular polygon is equal to an interior angle of regular triangle. Then the number sides of the polygon is .... (a) 4 (b) 5 (c) 6 (d) 8

Answer

(c) 6

Interior angle of a regular triangle is 60∘60^\circ, so the exterior angle of the polygon is 60∘60^\circ. Number of sides =360∘60∘=6=\dfrac{360^\circ}{60^\circ}=6.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

Which one of the following is the graphical representation of cumulative frequency of a statistical distribution (a) Bar diagram (b) Histogram (c) Pie-chart (d) Ogive curve

Answer

(d) Ogive curve

An ogive (cumulative frequency curve) is the graph of cumulative frequencies. A histogram or bar diagram shows ordinary frequencies, and a pie chart shows proportions.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

The mean of 17 observation is 20. If the mean of first 9 observation is 23 and that of last 9 observation is 18, then 9th observation is (a) 29 (b) 19 (c) 14.5 (d) 15

Answer

(a) 29

Sum of all 17 observations =17×20=340=17\times20=340. Sum of first 9 =9×23=207=9\times23=207, sum of last 9 =9×18=162=9\times18=162. The 9th observation is counted in both groups, so

207+162−340=29.207+162-340=29.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

For given individual data if observed values are doubled then arithmetic mean of new set data is ...... (a) doubled (b) tripled (c) halved (d) four times

Answer

(a) doubled

If each xix_i becomes 2xi2x_i, then xˉnew=∑2xin=2xˉ\bar{x}_{new}=\dfrac{\sum 2x_i}{n}=2\bar{x}.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

If x + y – z = 3, x – y + z = 8 and – x + y + z = 4, then the average value of x, y, z is: (a) 3 (b) 15 (c) 45 (d) 5

Answer

(d) 5

Adding the three equations: (x+y−z)+(x−y+z)+(−x+y+z)=x+y+z=3+8+4=15(x+y-z)+(x-y+z)+(-x+y+z)=x+y+z=3+8+4=15.

Average=153=5.\text{Average}=\frac{15}{3}=5.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

The s.d. of the data: 1, 2, 3, 4, 5 is ..... (a) 2 (b) 2.5 (c) 3 (d) 5

Answer

(a) 2

For 1,2,3,4,51,2,3,4,5: mean =3=3, and

σ2=(−2)2+(−1)2+0+12+225=105=2.\sigma^2=\frac{(-2)^2+(-1)^2+0+1^2+2^2}{5}=\frac{10}{5}=2.

So the variance is exactly 2 and the standard deviation is 2≈1.41\sqrt2\approx1.41. None of the options equals 2\sqrt2, so the intended choice is (a), which is the variance.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

What is the probability of getting a 2 or a 6 in a roll of a dice? (a) 1/2 (b) 1/3 (c) 1/4 (d) 1/6

Answer

(b) 1/3

Favourable outcomes are {2,6}\{2,6\} out of 6, so P=26=13P=\dfrac{2}{6}=\dfrac13.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

A card is drawn at random from a shuffled deck of 52 cards. What is the probability that it is a face card? (a) 1/4 (b) 3/13 (c) 1/13 (d) 4/13

Answer

(b) 3/13

A deck has 12 face cards (J, Q, K in each of 4 suits), so P=1252=313P=\dfrac{12}{52}=\dfrac{3}{13}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

For any events A, B and C, if P(A) = P(B) = 1/2 P(C), then P(C) = ............ (a) 1/2 (b) 4/5 (c) 2/5 (d) 1/5

Answer

(a) 1/2

Taking A,B,CA,B,C as mutually exclusive and exhaustive events (so the probabilities sum to 1) with P(A)=P(B)=12P(C)P(A)=P(B)=\tfrac12P(C):

12P(C)+12P(C)+P(C)=1⇒2P(C)=1⇒P(C)=12.\tfrac12P(C)+\tfrac12P(C)+P(C)=1 \Rightarrow 2P(C)=1 \Rightarrow P(C)=\tfrac12.

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  • 2018 Practice MCQs (Mathematics) · 1 mark

A coin is weighted so that the head is three times as likely to appear as tail then what will be the probability occurrence of head? (a) 1/2 (b) 1/4 (c) 3/4 (d) none

Answer

(c) 3/4

Let P(T)=pP(T)=p, so P(H)=3pP(H)=3p. Then p+3p=1p+3p=1, so p=14p=\tfrac14 and P(H)=34P(H)=\tfrac34.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

The probability of A can solve a problem is 1/3 and probability that B can solve the problem is 2/3. What is the probability that the problem will be solved if both of them try? (a) 7/9 (b) 2/9 (c) 1/3 (d) 1/9

Answer

(a) 7/9

P(solved)=1−P(neither solves)=1−(1−13)(1−23)=1−23⋅13=1−29=79.P(\text{solved})=1-P(\text{neither solves})=1-\left(1-\tfrac13\right)\left(1-\tfrac23\right)=1-\tfrac23\cdot\tfrac13=1-\tfrac29=\tfrac79.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

A couple plans to have three children, what is the probability that exactly two will boys? (a) 1/3 (b) 2/3 (c) 2/8 (d) 3/8

Answer

(d) 3/8

Sample space has 23=82^3=8 equally likely outcomes. Exactly two boys: BBG, BGB, GBB, which is 3 outcomes. P=38P=\dfrac38.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

A and B appear for an interview for two posts. The probability of A's selection is 1/3 and that of B's selection is 2/5. Find the probability that only one of them will be selected. (a) 4/15 (b) 7/15 (c) 2/3 (d) 3/5

Answer

(b) 7/15

P(only one)=P(A)P(Bˉ)+P(Aˉ)P(B)=13⋅35+23⋅25=315+415=715.P(\text{only one})=P(A)P(\bar B)+P(\bar A)P(B)=\tfrac13\cdot\tfrac35+\tfrac23\cdot\tfrac25=\tfrac{3}{15}+\tfrac{4}{15}=\tfrac{7}{15}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

A box contains 5 yellow, 4 red and 3 blue balls if a ball is drawn at random what is the probability of its not being yellow? (a) 7/12 (b) 5/12 (c) 5/7 (d) 7/8

Answer

(a) 7/12

Total balls =5+4+3=12=5+4+3=12; not yellow =4+3=7=4+3=7. P=712P=\dfrac{7}{12}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

Three coins are tossed simultaneously what is the probability of getting not more than two heads? (a) 1/2 (b) 2/3 (c) 5/8 (d) 7/8

Answer

(d) 7/8

Not more than two heads is the complement of three heads (HHH). P=1−18=78P=1-\dfrac18=\dfrac78.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

A bag contains 4 white and 6 red balls. Two balls are drawn in succession. Find the probability that the first ball will be white and second will red if first ball is not replaced? (a) 16/25 (b) 11/15 (c) 8/15 (d) 4/15

Answer

(d) 4/15

P(W1 then R2)=410×69=2490=415.P(W_1\text{ then }R_2)=\frac{4}{10}\times\frac{6}{9}=\frac{24}{90}=\frac{4}{15}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2018 Practice MCQs (Mathematics) · 1 mark

What is the probability that a leap year has 53 Sunday? (a) 1/4 (b) 1/3 (c) 1/7 (d) 2/7

Answer

(d) 2/7

A leap year has 366 days =52=52 weeks +2+2 extra days. The pair of extra days is one of (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun), (Sun,Mon): 7 equally likely cases, of which 2 contain Sunday. P=27P=\dfrac27.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If y=x1−x+1−xxy=\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}, then y2y^2 equals (a) 11−x2\frac{1}{1-x^2} (b) 11−x2\frac{1}{1-x^2} (c) 1−2x+x2x−x2\frac{1-2x+x^2}{x-x^2} (d) 1−x+x2x−x2\frac{1-x+x^2}{x-x^2}

Answer

(a) 1x−x2\dfrac{1}{x-x^2} (printed as the first option)

y2=x1−x+1−xx+2x1−x⋅1−xx=x2+(1−x)2x(1−x)+2=2x2−2x+1+2x−2x2x−x2=1x−x2y^2=\frac{x}{1-x}+\frac{1-x}{x}+2\sqrt{\frac{x}{1-x}\cdot\frac{1-x}{x}}=\frac{x^2+(1-x)^2}{x(1-x)}+2=\frac{2x^2-2x+1+2x-2x^2}{x-x^2}=\frac{1}{x-x^2}

Options (a) and (b) are printed identically and the correct value 1x−x2\frac{1}{x-x^2} matches neither of (c) and (d), so (a) is the intended option.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

A man sells an article at a profit of 20% of S.P. What is his real profit %? (a) 16.66% (b) 25 % (c) 18 % (d) 20 %

Answer

(b) 25 %

Let S.P. =100=100. Profit =20=20, so C.P. =80=80.

Profit%=2080×100=25%\text{Profit\%}=\frac{20}{80}\times100=25\%

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

The cuboidal tank is 20cm long 10 cm wide & 8 cm high. With what minimum base area can it stand upright? (a) 200 cm² (b) 80 cm² (c) 160 cm² (d) None

Answer

(b) 80 cm²

The base area is smallest when the tank stands on its two smallest dimensions: 10×8=80 cm210\times 8=80\ \text{cm}^2 (height 20 cm).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

Which of the following is an example of pure quadratic equation? (a) x² + 3x – 9 = 0 (b) x² + 3x = 0 (c) x² – 9 = 0 (d) 3x – 9 = 0

Answer

(c) x2−9=0x^2-9=0

A pure quadratic equation has no term in xx, so it has the form ax2+c=0ax^2+c=0. Only x2−9=0x^2-9=0 fits.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If m:n = 3:2 then what is the value of (4m+5n) : (4m – 5n)? (a) 4:9 (b) 9:4 (c) 11:1 (d) 9:1

Answer

(c) 11:1

Take m=3k, n=2km=3k,\ n=2k. Then 4m+5n=12k+10k=22k4m+5n=12k+10k=22k and 4m−5n=12k−10k=2k4m-5n=12k-10k=2k.

22k2k=11:1\frac{22k}{2k}=11:1

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  • 2082 · United Academy Science Model Set 1 · 1 mark

What is the compound interest of Rs 1 for 3 years at the rate of 10% per annum? (a) Rs 13.31 (b) Rs 1.331 (c) Rs 0.331 (d) Rs 0.30

Answer

(c) Rs 0.331

A=1(1.1)3=1.331,CI=1.331−1=Rs 0.331A=1(1.1)^3=1.331,\qquad CI=1.331-1=\text{Rs }0.331

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

Which of the following is not true? (a) Diagonals of a square area equal (b) Diagonals of a rectangle are equal (c) Diagonals of a rhombus are equal (d) Diagonals of a square and rectangle are equal

Answer

(c) Diagonals of a rhombus are equal

The diagonals of a rhombus bisect each other at right angles but are not equal in general. They are equal in a square and in a rectangle.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

Which of the axioms does not verify congruency of any two triangles? (a) R.H.S. axiom (b) S.S.S. axiom (c) S.S.A. axiom (d) A.A.S. axiom

Answer

(c) S.S.A. axiom

SSS, RHS and AAS each prove congruence. S.S.A. (two sides and a non-included angle) does not in general, because two different triangles can satisfy it (the ambiguous case). RHS is the special case valid for right-angled triangles.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

Zeros of p (x) = x² – 1 are (a) ± 1 (b) ± 2 (c) 0 (d) None of these

Answer

(a) ±1\pm 1

x2−1=0⇒(x−1)(x+1)=0⇒x=±1x^2-1=0\Rightarrow (x-1)(x+1)=0\Rightarrow x=\pm1

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

What is the determinant of an identity matrix? (a) 0 (b) 1 (c) 2 (d) All

Answer

(b) 1

An identity matrix has 1s on the diagonal and 0s elsewhere, so its determinant is the product of the diagonal entries, 1×1×⋯×1=11\times1\times\cdots\times1=1.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If (2x463x)\begin{pmatrix}2x&4\\6&3x\end{pmatrix} is a singular matrix then the value of xx is (a) 0 (b) 2 (c) -2 (d) ±2\pm2

Answer

(d) ±2\pm2

A matrix is singular when its determinant is zero: 6x2−24=06x^2-24=0, so x2=4x^2=4 and x=±2x=\pm2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If ∣a⃗+b⃗∣=∣a⃗−b⃗∣|\vec a+\vec b|=|\vec a-\vec b| then the relation between a⃗\vec a and b⃗\vec b is (a) parallel (b) orthogonal (c) equal (d) none

Answer

(b) orthogonal

Squaring both sides gives ∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2=∣a⃗∣2−2a⃗⋅b⃗+∣b⃗∣2|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2=|\vec a|^2-2\vec a\cdot\vec b+|\vec b|^2, so 4 a⃗⋅b⃗=04\,\vec a\cdot\vec b=0. The dot product is zero, so the vectors are perpendicular.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If sin⁡8θ=cos⁡7θ\sin8\theta=\cos7\theta then the value of θ\theta is (a) 90∘90^\circ (b) 0∘0^\circ (c) 15∘15^\circ (d) 6∘6^\circ

Answer

(d) 6∘6^\circ

sin⁡8θ=cos⁡7θ=sin⁡(90∘−7θ)\sin8\theta=\cos7\theta=\sin(90^\circ-7\theta), so 8θ=90∘−7θ8\theta=90^\circ-7\theta, giving 15θ=90∘15\theta=90^\circ and θ=6∘\theta=6^\circ.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If the third term of a G.S. is 3 then the product of first five terms is (a) 243 (b) 27 (c) 81 (d) none of these

Answer

(a) 243

Let the terms be ar2,ar,a,ar,ar2\frac{a}{r^2},\frac{a}{r},a,ar,ar^2 with a=3a=3 (the third term). Their product is a5=35=243a^5=3^5=243.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If x⃗=(34)\vec x=\begin{pmatrix}3\\4\end{pmatrix} and y⃗=(−43)\vec y=\begin{pmatrix}-4\\3\end{pmatrix} then angle between x⃗\vec x and y⃗\vec y is (a) 0∘0^\circ (b) 45∘45^\circ (c) 90∘90^\circ (d) none of these

Answer

(c) 90∘90^\circ

x⃗⋅y⃗=3(−4)+4(3)=0\vec x\cdot\vec y=3(-4)+4(3)=0, so cos⁡θ=0\cos\theta=0 and the angle is 90∘90^\circ.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If Q1 = 20 and Q³ = 50 then the semi-inter quartile range is (a) 35 (b) 30 (c) 15 (d) none of these

Answer

(c) 15

Semi-interquartile range =Q3−Q12=50−202=15=\dfrac{Q_3-Q_1}{2}=\dfrac{50-20}{2}=15.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

The 11th term of an A.P. −5,−52,0,52,…-5, -\frac52, 0, \frac52, \ldots is (a) -20 (b) 20 (c) -30 (d) 30

Answer

(b) 20

Here a=−5a=-5 and d=−52−(−5)=52d=-\frac52-(-5)=\frac52. Then t11=a+10d=−5+10⋅52=−5+25=20t_{11}=a+10d=-5+10\cdot\frac52=-5+25=20.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

The value of tan⁡1∘⋅tan⁡2∘⋅tan⁡3∘⋯tan⁡89∘\tan1^\circ\cdot\tan2^\circ\cdot\tan3^\circ\cdots\tan89^\circ is (a) 0 (b) 1 (c) ∞\infty (d) 12\frac12

Answer

(b) 1

Pair tan⁡θ\tan\theta with tan⁡(90∘−θ)=cot⁡θ\tan(90^\circ-\theta)=\cot\theta, whose product is 1. The terms pair up as tan⁡1∘tan⁡89∘\tan1^\circ\tan89^\circ, tan⁡2∘tan⁡88∘\tan2^\circ\tan88^\circ, and so on, and the middle term is tan⁡45∘=1\tan45^\circ=1. The whole product is therefore 1.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

The value of ‘a’ for which 2a, (a+10) and (3a+2) are three consecutive terms of an A.P. is: (a) 6 (b) -6 (c) -18 (d) 18

Answer

(a) 6

For three consecutive terms of an A.P., 2(a+10)=2a+(3a+2)2(a+10)=2a+(3a+2). So 2a+20=5a+22a+20=5a+2, which gives 3a=183a=18 and a=6a=6. Check: the terms are 12, 16, 20, with common difference 4.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 1 · 1 mark

If A + B + C = π^c , sin2A – sin 2B - sin 2C equals (a) 4 cos A. cos B. sin C (b) 4 cos A. sin B. cos C (c) -4 sin A. cos B. cos C (d) - 4 cos A. cos B. sin C

Answer

(c) −4sin⁡Acos⁡Bcos⁡C-4\sin A\cos B\cos C

Since A+B+C=πA+B+C=\pi, sin⁡2A=2sin⁡Acos⁡A=−2sin⁡Acos⁡(B+C)\sin2A=2\sin A\cos A=-2\sin A\cos(B+C). Also sin⁡2B+sin⁡2C=2sin⁡(B+C)cos⁡(B−C)=2sin⁡Acos⁡(B−C)\sin2B+\sin2C=2\sin(B+C)\cos(B-C)=2\sin A\cos(B-C). Hence sin⁡2A−sin⁡2B−sin⁡2C=−2sin⁡A [cos⁡(B+C)+cos⁡(B−C)]=−4sin⁡Acos⁡Bcos⁡C.\sin2A-\sin2B-\sin2C=-2\sin A\,[\cos(B+C)+\cos(B-C)]=-4\sin A\cos B\cos C.

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  • 2082 · United Academy Science Model Set 1 · 1 mark

For what value of ‘k’ with (k+2) x² = 4xy + (2k+5) y² represents two straight lines perpendicular to each other? (a) 2 (b) - 4 (c) -3 (d) 4

Answer

(c) −3-3

The equation is (k+2)x2−4xy−(2k+5)y2=0(k+2)x^2-4xy-(2k+5)y^2=0. The lines are perpendicular when the coefficient of x2x^2 plus the coefficient of y2y^2 is zero: (k+2)−(2k+5)=0(k+2)-(2k+5)=0, so −k−3=0-k-3=0 and k=−3k=-3.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

What is the period of function y=tan⁡xy=\tan x ? (a) πc\pi^c (b) 2πc2\pi^c (c) 3πc3\pi^c (d) 4πc4\pi^c

Answer

(a) πc\pi^c tan⁡(x+π)=tan⁡x\tan(x+\pi)=\tan x, and π\pi is the smallest such positive number, so the period of y=tan⁡xy=\tan x is π\pi.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · United Academy Science Model Set 2 · 1 mark

What is the range of y=cosx? (a) (-1,1) (b) [-1,1) (c) (-1,1] (d) [-1,1]

Answer

(d) [−1,1][-1,1] The cosine takes every value from −1-1 to 11 inclusive, so −1≤cos⁡x≤1-1\le\cos x\le 1.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

Function y=ax2+bx+cy=ax^2+bx+c represents a linear function if: (a) a < 0 (b) a > 0 (c) a=0 (d) c=0

Answer

(c) a=0a=0 If a=0a=0, then y=bx+cy=bx+c, which is linear.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

If f−1(x)=2x+34f^{-1}(x)=\frac{2x+3}{4}, then what is the value of f(x)f(x)? (a) 3x−24\frac{3x-2}{4} (b) 2x+43\frac{2x+4}{3} (c) 43x+2\frac{4}{3x+2} (d) 4x−32\frac{4x-3}{2}

Answer

(d) 4x−32\frac{4x-3}{2} Let y=f−1(x)=2x+34y=f^{-1}(x)=\frac{2x+3}{4}. Then 4y=2x+34y=2x+3, so x=4y−32x=\frac{4y-3}{2}. Hence f(x)=4x−32f(x)=\frac{4x-3}{2}.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

Which of the following numbers represents continuity? (a) Whole number (b) Natural number (c) Rational number (d) Real number

Answer

(d) Real number Real numbers fill the number line with no gaps, so they represent continuity. Natural, whole and rational numbers leave gaps.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

What is the determinant of a unit/Identity matrix? (a) 0 (b) 1 (c) -1 (d) 2

Answer

(b) 1 For an identity matrix the determinant is the product of the diagonal entries, 1⋅1⋯1=11\cdot1\cdots1=1.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

If the slopes of two lines are equal, then what is the angular relation between the lines? (a) Coincident (b) Normal (c) Orthogonal (d) Perpendicular

Answer

(a) Coincident Lines with equal slopes are parallel (the angle between them is 0∘0^\circ). Of the options, only "coincident" describes lines with the same direction. The other three describe perpendicular lines.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

The equation of a straight line perpendicular to ax + by + c = 0 is: (a) ax + by – k = 0 (b) bx – ay + k = 0 b. (c) bx + ay – k = 0 (d) ax + by + k = 0

Answer

(b) bx−ay+k=0bx-ay+k=0 The slope of ax+by+c=0ax+by+c=0 is −ab-\frac{a}{b}, so a perpendicular line has slope ba\frac{b}{a}. The line bx−ay+k=0bx-ay+k=0 has slope ba\frac{b}{a}.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

For what value of k, the pair of straight lines represented by 3x2+4xy−ky2=03x^2+4xy-ky^2=0 are at right angle? (a) 3 (b) -3 (c) 4 (d) -4

Answer

(a) 3 The lines in ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are perpendicular when a+b=0a+b=0. Here a=3a=3 and b=−kb=-k, so 3−k=03-k=0, giving k=3k=3.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

If sin⁡A=12\sin A=\frac12, what will be the value of sin⁡2A\sin2A? (a) 1 (b) 2 (c) 12\frac1{\sqrt2} (d) 32\frac{\sqrt3}2

Answer

(d) 32\frac{\sqrt3}{2}

sin⁡A=12⇒A=30∘\sin A=\frac12\Rightarrow A=30^\circ so sin⁡2A=sin⁡60∘=32\sin2A=\sin60^\circ=\frac{\sqrt3}{2}. Check: cos⁡A=32\cos A=\frac{\sqrt3}{2}, so 2sin⁡Acos⁡A=2⋅12⋅32=322\sin A\cos A=2\cdot\frac12\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{2}.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

If tan A = cot A, what is the value of A? (a) 30° (b) 45° (c) 60° (d) 0°

Answer

(b) 45°

tan⁡A=cot⁡A=1tan⁡A⇒tan⁡2A=1⇒tan⁡A=1\tan A=\cot A=\frac1{\tan A}\Rightarrow\tan^2A=1\Rightarrow\tan A=1, so A=45∘A=45^\circ.

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  • 2082 · United Academy Science Model Set 2 · 1 mark

If a⃗⋅b⃗=∣a⃗∣∣b⃗∣\vec a\cdot\vec b=|\vec a||\vec b|, what will be the angle between a⃗\vec a and b⃗\vec b ? (a) 0∘0^\circ (b) 30∘30^\circ (c) 60∘60^\circ (d) 90∘90^\circ

Answer

(a) 0∘0^\circ

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta, so cos⁡θ=1\cos\theta=1 and θ=0∘\theta=0^\circ (the vectors are parallel in the same direction).

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  • 2082 · United Academy Science Model Set 2 · 1 mark

Which of the following measure of dispersion takes the absolute deviation from the value of central tendency? (a) Mean deviation (b) Standard deviation (c) Quartile deviation (d) Exodeviation

Answer

(a) Mean deviation

Mean deviation is the average of the absolute deviations ∣x−A∣|x-A| taken from a measure of central tendency (mean, median or mode). Standard deviation uses squared deviations.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

What is the minimum value of function y = cosx ? (a) −∞ (b) 1 (c) -1 (d) 0

Answer

(c) -1

The cosine function oscillates between −1-1 and 11, so its minimum value is −1-1.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

What is the range of y = tanx? (a) (-∞,∞) (b) [-∞,∞) (c) (-∞,∞] (d) [-∞,∞]

Answer

(a) (−∞,∞)(-\infty,\infty)

tan⁡x\tan x takes every real value (it tends to ±∞\pm\infty near its asymptotes but never equals them), so the range is (−∞,∞)(-\infty,\infty).

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  • 2082 · United Academy Science Model Set 3 · 1 mark

Function y=ax+by=ax+b represents an identity function if: (a) a = 0, b = 1 (b) b = 0, a = 1 (c) x = 0, a = 1 (d) y = 0, b = 1

Answer

(b) b = 0, a = 1

The identity function is y=xy=x, which is obtained from y=ax+by=ax+b when a=1a=1 and b=0b=0.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

In the standard equation of parabola, y = a(x − h)² + k. If k=0, vertex of parabola lies on: (a) x-axis (b) above x-axis (c) below x-axis (d) y-axis

Answer

(a) x-axis

The vertex is (h,k)(h,k). With k=0k=0 it becomes (h,0)(h,0), which lies on the x-axis.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

Does a point (2,-3) lie on the solution region of 2x+3y≤ 2 lie? (a) Yes (b) No (c) Maybe (d) Insufficient data given

Answer

(a) Yes

Substitute (2,−3)(2,-3): 2(2)+3(−3)=4−9=−52(2)+3(-3)=4-9=-5, and −5≤2-5\le 2 is true, so the point lies in the solution region.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

Which of the following example represents discontinuity? (a) Movement of snake (b) Growth of plants (c) Graph of y = tanx (d) Graph of y = sinx

Answer

(c) Graph of y = tanx

tan⁡x\tan x is undefined at x=π2+nπx=\frac{\pi}{2}+n\pi, so its graph breaks at those points. The sine graph, plant growth and snake movement are continuous.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

What is the determinant of [-1] ? (a) 0 (b) 1 (c) -1 (d) cannot be determined

Answer

(c) -1

The determinant of a 1×11\times1 matrix [a][a] is aa, so ∣−1∣det⁡=−1|-1|_{\det}=-1.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

If the plane intersects a cone in such a way that it is parallel to the generator of cone, what is the name of conic section thus formed? (a) Circle (b) Ellipse (c) Parabola (d) Hyperbola

Answer

(c) Parabola

A plane parallel to a generator (slant side) of the cone gives a parabola.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

In a rhombus ABCD, diagonals AC and BD intersects at O, What will be the angular measurement of ∠COD\angle COD ? (a) 180∘180^\circ (b) 0∘0^\circ (c) 90∘90^\circ (d) varies according to the size of figure.

Answer

(c) 90∘90^\circ

The diagonals of a rhombus bisect each other at right angles, so ∠COD=90∘\angle COD=90^\circ.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

If the line y=m1xy=m_1x and y=m2xy=m_2x are at constant distance and never intersect, what is the relation between the slopes? (a) m1×m2=−1m_1\times m_2=-1 (b) m1=m2m_1=m_2 (c) m1×m2=1m_1\times m_2=1 (d) m1=1m2m_1=\frac1{m_2}

Answer

(b) m1=m2m_1=m_2

Lines that stay at a constant distance and never meet are parallel, and parallel lines have equal slopes.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

The equation of straight line xa+yb=1\frac xa+\frac yb=1 is called equation of a straight line in ____ form. (a) Slope-intercept (b) Double-intercept (c) Slope-point (d) Two-point

Answer

(b) Double-intercept

The form xa+yb=1\frac{x}{a}+\frac{y}{b}=1 uses the x-intercept aa and y-intercept bb, so it is the double-intercept (two-intercept) form.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

The center of the circle having equation (x − h)² + (y + k)² = r² is: (a) (-h, -k) (b) (h, -k) (c) (-h, k) (d) (h, k)

Answer

(b) (h,−k)(h,-k)

Writing (y+k)2=(y−(−k))2(y+k)^2=(y-(-k))^2, the equation is (x−h)2+(y−(−k))2=r2(x-h)^2+(y-(-k))^2=r^2, so the centre is (h,−k)(h,-k).

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  • 2082 · United Academy Science Model Set 3 · 1 mark

If cos⁡A=32\cos A=\frac{\sqrt3}2, what will be the value of cos⁡2A\cos2A? (a) 12\frac12 (b) 3\sqrt3 (c) 12\frac1{\sqrt2} (d) 32\frac{\sqrt3}2

Answer

(a) 12\frac12

cos⁡2A=2cos⁡2A−1=2⋅34−1=12\cos 2A=2\cos^2A-1=2\cdot\frac34-1=\frac12

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  • 2082 · United Academy Science Model Set 3 · 1 mark

If cos A = sec A, what is the possible value of A? (a) 0° (b) 45° (c) 60° (d) 90°

Answer

(a) 0∘0^\circ

cos⁡A=sec⁡A=1cos⁡A\cos A=\sec A=\frac1{\cos A} gives cos⁡2A=1\cos^2A=1, so cos⁡A=±1\cos A=\pm1. Among the options, A=0∘A=0^\circ satisfies this.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

If a⃗⋅b⃗=−∣a⃗∣∣b⃗∣\vec a\cdot\vec b=-|\vec a||\vec b|, What will be the angle between a⃗\vec a and b⃗\vec b ? (a) 0∘0^\circ (b) 90∘90^\circ (c) 180∘180^\circ (d) 360∘360^\circ

Answer

(c) 180∘180^\circ

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta, so cos⁡θ=−1\cos\theta=-1, which gives θ=180∘\theta=180^\circ (the vectors point in opposite directions).

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  • 2082 · United Academy Science Model Set 3 · 1 mark

In daytime, the length of shadow of a tower is equal to its height. What is the altitude of the sun? (a) 93 million miles (b) 2.38855 million miles (c) 150.96 miles (d) 45°

Answer

(d) 45°

If tan⁡θ=heightshadow=1\tan\theta=\dfrac{\text{height}}{\text{shadow}}=1, then θ=45∘\theta=45^\circ.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

In which of the following transformation the object and image are not congruent? (a) Translation (b) Reflection (c) Rotation (d) Enlargement

Answer

(d) Enlargement

Translation, reflection and rotation are isometries, so they preserve lengths and angles and the image is congruent to the object. Enlargement changes the size (scale factor k≠±1k\neq\pm1), so the image is only similar to the object.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

If the object is inside the inversion circle, where does its image lie? (a) Inside circle (b) at circumference (c) outside circle (d) at infinity

Answer

(c) outside circle

Under inversion in a circle of radius rr with centre OO, OP⋅OP′=r2OP\cdot OP'=r^2. If OP<rOP<r then OP′=r2/OP>rOP'=r^2/OP>r, so the image lies outside the circle.

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  • 2082 · United Academy Science Model Set 3 · 1 mark

Which of the following is not associated with the measures of dispersion? (a) Mean deviation (b) Standard deviation (c) Quartile deviation (d) Exodeviation

Answer

(d) Exodeviation

Mean deviation, standard deviation and quartile deviation are all measures of dispersion. "Exodeviation" is not a statistical measure.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

What is the range of function y = sinx ? (a) . (-1,1) (b) [-1,1) (c) (-1,1] (d) [-1,1]

Answer

(d) [−1,1][-1,1]

The sine function takes every value from −1-1 to 11, including both end values, so the range is −1≤sin⁡x≤1-1\le \sin x\le 1.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

What is the period of y=cosx? (a) π^c (b) 2π^c (c) 3π^c (d) 4π^c

Answer

(b) 2πc2\pi^c

Since cos⁡(x+2π)=cos⁡x\cos(x+2\pi)=\cos x and no smaller positive number works, the period of y=cos⁡xy=\cos x is 2π2\pi.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

Function y=ax3+bx2+cx+dy=ax^3+bx^2+cx+d represents a quadratic function if: (a) a < 0 (b) a > 0 (c) a=0 (d) c=0

Answer

(c) a=0a=0

When a=0a=0 the cubic term vanishes and y=bx2+cx+dy=bx^2+cx+d is quadratic (with b≠0b\ne 0).

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  • 2082 · United Academy Science Model Set 4 · 1 mark

If f(x)=3x−25f(x)=\frac{3x-2}{5}, then what is the value of f−1(x)f^{-1}(x)? (a) 2x−35\frac{2x-3}{5} (b) 5x−23\frac{5x-2}{3} (c) 3x+25\frac{3x+2}{5} (d) 5x+23\frac{5x+2}{3}

Answer

(d) 5x+23\frac{5x+2}{3}

Let y=3x−25y=\frac{3x-2}{5}. Then 5y=3x−25y=3x-2, so x=5y+23x=\frac{5y+2}{3}.

Interchanging xx and yy gives f−1(x)=5x+23f^{-1}(x)=\frac{5x+2}{3}.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

Which of the following numbers represents discontinuity? (a) Whole number (b) Both (c) Rational number (d) None

Answer

(b) Both

Whole numbers are discrete, with gaps between consecutive values. Rational numbers also leave gaps on the number line (the irrationals, such as 2\sqrt2, are missing). So neither set is continuous, and both represent discontinuity. Only the real numbers form a continuous set.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

What is the determinant of a singular matrix? (a) 0 (b) 1 (c) -1 (d) 2

Answer

(a) 0

A square matrix is singular when it has no inverse, which happens exactly when ∣A∣=0|A|=0.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

If the product of slopes of two lines is equal to -1, then what is the angular relation between the lines? (a) Coincident (b) Collinear (c) Orthogonal (d) Parallel

Answer

(c) Orthogonal

If m1m2=−1m_1m_2=-1, the lines are perpendicular, which is also called orthogonal.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

The equation of a straight line parallel to ax+by+c=0ax+by+c=0 is: (a) ax + by – k = 0 (b) bx + ay – k = 0 (c) bx – ay + k = 0 (d) ax + by + k = 0

Answer

(d) ax+by+k=0ax+by+k=0

A line parallel to ax+by+c=0ax+by+c=0 has the same slope −ab-\frac ab, so only the constant term changes: ax+by+k=0ax+by+k=0.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

For what value of k, the pair of straight lines represented by 9x2+12xy+ky2=09x^2+12xy+ky^2=0 are at constant distance and never intersect? (a) 4 (b) -9 (c) 9 (d) 3

Answer

(a) 4

The pair ax2+2hxy+by2=0ax^2+2hxy+by^2=0 represents parallel lines when h2=abh^2=ab. Here a=9a=9, 2h=122h=12 so h=6h=6, and b=kb=k:

36=9k⇒k=4.36=9k\Rightarrow k=4.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

If sin⁡2A=32\sin2A=\frac{\sqrt3}2, what will be the value of sin⁡A\sin A? (a) 12\frac12 (b) 3\sqrt3 (c) 12\frac1{\sqrt2} (d) 32\frac{\sqrt3}2

Answer

(a) 12\frac12

sin⁡2A=32\sin2A=\frac{\sqrt3}{2} gives 2A=60∘2A=60^\circ (acute angle), so A=30∘A=30^\circ.

sin⁡A=sin⁡30∘=12.\sin A=\sin30^\circ=\frac12.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

If cos A = sin A, what is the value of A? (a) 30° (b) 45° (c) 60° (d) 0°

Answer

(b) 45∘45^\circ

cos⁡A=sin⁡A\cos A=\sin A gives tan⁡A=1\tan A=1, so A=45∘A=45^\circ.

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  • 2082 · United Academy Science Model Set 4 · 1 mark

If a⃗⋅b⃗=0\vec a\cdot\vec b=0, What will be the angle between a⃗\vec a and b⃗\vec b ? (a) 0∘0^\circ (b) 30∘30^\circ (c) 60∘60^\circ (d) 90∘90^\circ

Answer

(d) 90∘90^\circ

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=0\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0 with non-zero vectors gives cos⁡θ=0\cos\theta=0, so θ=90∘\theta=90^\circ (the vectors are perpendicular).

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  • 2082 · United Academy Science Model Set 4 · 1 mark

Which of the following measure of dispersion is called root mean square deviation? (a) Mean deviation (b) Standard deviation (c) Quartile deviation (d) Exodeviation

Answer

(b) Standard deviation

Standard deviation is the square root of the mean of the squared deviations from the mean, so it is the root mean square deviation: σ=∑(x−xˉ)2n\sigma=\sqrt{\dfrac{\sum (x-\bar{x})^2}{n}}.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

Which of the following fractions is smallest? (a) 5/8 (b) 7/10 (c) 3⁄4 (d) 9/16

Answer

(d) 9/16

Convert to decimals: 5/8=0.6255/8 = 0.625, 7/10=0.77/10 = 0.7, 3/4=0.753/4 = 0.75, 9/16=0.56259/16 = 0.5625. The smallest is 9/169/16.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

When 8 is increased to 10, by what percentage does it increase)? (a) 25% (b) 50% (c) 75% (d) 100%

Answer

(a) 25%

Increase%=10−88×100=25%\text{Increase}\% = \frac{10-8}{8}\times 100 = 25\%

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  • 2082 · United Academy Science Model Set 5 · 1 mark

The difference between two positive numbers is 10 and the ratio between them is 5:3, Find the product of the two numbers. (a) 375 (b) 175 (c) 275 (d) 125

Answer

(a) 375

Let the numbers be 5k5k and 3k3k. Then 5k−3k=105k-3k=10, so k=5k=5 and the numbers are 25 and 15.

25×15=37525\times 15 = 375

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  • 2082 · United Academy Science Model Set 5 · 1 mark

The average of 8 numbers is 30. If two new numbers 40 and 50 are added, what is the new average? (a) 32.5 (b) 33 (c) 34 (d) 35

Answer

(b) 33

Original sum =8×30=240=8\times 30=240. New sum =240+40+50=330=240+40+50=330 for 10 numbers.

New average=33010=33\text{New average}=\frac{330}{10}=33

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  • 2082 · United Academy Science Model Set 5 · 1 mark

The simple interest on Rs. 3000 at 6% per annum for 1121\frac12 years is: (a) 170 (b) 270 (c) 370 (d) 470

Answer

(b) 270

I=PTR100=3000×1.5×6100=Rs 270I=\frac{PTR}{100}=\frac{3000\times 1.5\times 6}{100}=\text{Rs } 270

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  • 2082 · United Academy Science Model Set 5 · 1 mark

If A = {1, 3, 4} and B = {x : x ∈ R and x² - 7x + 12 = 0} then which of the following is true ? (a) A=B (b) A ⊂ B (c) B ⊂ A (d) A is equivalent to B

Answer

(c) B⊂AB \subset A

x2−7x+12=0⇒(x−3)(x−4)=0x^2-7x+12=0 \Rightarrow (x-3)(x-4)=0, so B={3,4}B=\{3,4\}. Every element of BB is in A={1,3,4}A=\{1,3,4\}, and AA has the extra element 1, so B⊂AB\subset A.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

In how many years will the population increase by 21% as compared to the base year at the compound annual growth rate of 10%? (a) 3 years (b) 1 year (c) 2 years (d) Can’t be determined

Answer

(c) 2 years

(1.10)n=1.21=(1.10)2⇒n=2(1.10)^n = 1.21 = (1.10)^2 \Rightarrow n=2

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  • 2082 · United Academy Science Model Set 5 · 1 mark

The buying rate and selling rate of 1 US dollar are Rs 139 and Rs 140 respectively. How much Nepali rupees will you need to exchange 1000 US dollars? (a) Rs 139,000 (b) Rs 140,000 (c) Rs 139,500 (d) Neither of these

Answer

(b) Rs 140,000

To obtain US dollars you must buy them at the bank's selling rate of Rs 140 per dollar: 1000×140=Rs 140,0001000\times 140=\text{Rs } 140{,}000.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

What is the ratio of volume of a cylinder and cone having same radius and vertical height? (a) 2:1 (b) 3:1 (c) 4:1 (d) 3:2

Answer

(b) 3:1

πr2h13πr2h=31\frac{\pi r^2 h}{\frac13\pi r^2 h}=\frac{3}{1}

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  • 2082 · United Academy Science Model Set 5 · 1 mark

A reflex angle is: (a) More than 90 degrees (b) Equal to 90 degrees (c) More than 180 degrees (d) Equal to 180 degrees

Answer

(c) More than 180 degrees

A reflex angle lies between 180∘180^\circ and 360∘360^\circ.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

What is the LCM of the expressions (a3−b3)(a^3-b^3), (a2+ab+b2)(a^2+ab+b^2) and (a4+a2b2+b4)(a^4+a^2b^2+b^4)? (a) (a4+a2b2+b4)(a^4+a^2b^2+b^4) (b) (a4+a2b2+b4)(a+b)(a^4+a^2b^2+b^4)(a+b) (c) (a4+a2b2+b4)(a−b)(a^4+a^2b^2+b^4)(a-b) (d) None

Answer

(c) (a4+a2b2+b4)(a−b)(a^4+a^2b^2+b^4)(a-b)

a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2) a4+a2b2+b4=(a2+ab+b2)(a2−ab+b2)a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)

LCM =(a−b)(a2+ab+b2)(a2−ab+b2)=(a−b)(a4+a2b2+b4)=(a-b)(a^2+ab+b^2)(a^2-ab+b^2)=(a-b)(a^4+a^2b^2+b^4).

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  • 2082 · United Academy Science Model Set 5 · 1 mark

What is the value of (24332)−0.4\left(\frac{243}{32}\right)^{-0.4} ? (a) 49\frac49 (b) −49-\frac49 (c) 23\frac23 (d) −23-\frac23

Answer

(a) 49\frac49

24332=(32)5\frac{243}{32}=\left(\frac32\right)^5, so

(24332)−0.4=(32)5×(−2/5)=(32)−2=49\left(\frac{243}{32}\right)^{-0.4}=\left(\frac32\right)^{5\times(-2/5)}=\left(\frac32\right)^{-2}=\frac49

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  • 2082 · United Academy Science Model Set 5 · 1 mark

What is the probability that you get two heads when a coin is tossed twice? (a) 1⁄2 (b) 1⁄3 (c) 1⁄4 (d) 1

Answer

(c) 1/4

Sample space {HH,HT,TH,TT}\{HH,HT,TH,TT\} has 4 equally likely outcomes and one is HHHH, so P=14P=\frac14.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

What is the inverse of the function f(x)=3x−52f(x)=\frac{3x-5}{2}? (a) 6x-5 (b) 2x+53\frac{2x+5}{3} (c) 2x−53\frac{2x-5}{3} (d) None

Answer

(b) 2x+53\frac{2x+5}{3}

Let y=3x−52y=\frac{3x-5}{2}. Then 2y=3x−52y=3x-5, so x=2y+53x=\frac{2y+5}{3}. Hence f−1(x)=2x+53f^{-1}(x)=\frac{2x+5}{3}.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

The next term of the sequence 5, 11, 23, 47, is: (a) 92 (b) 93 (c) 94 (d) 95

Answer

(d) 95

Each term is twice the previous plus 1: 5→11→23→47→2(47)+1=955\to 11\to 23\to 47\to 2(47)+1=95.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

The point on the line AB having points A(3,5) and B(5, -7) that divides the line in the ratio 1:2 internally is (a) (113,1)(\frac{11}3,1) (b) (113,173)(\frac{11}3,\frac{17}3) (c) (−113,1)(-\frac{11}3,1) (d) (113,−1)(\frac{11}3,-1)

Answer

(a) (113,1)\left(\frac{11}{3},1\right)

Internal division with m:n=1:2m:n=1:2:

x=1(5)+2(3)3=113,y=1(−7)+2(5)3=1x=\frac{1(5)+2(3)}{3}=\frac{11}{3},\qquad y=\frac{1(-7)+2(5)}{3}=1

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  • 2082 · United Academy Science Model Set 5 · 1 mark

Which of the inequalities contains the point (0,-2)? (a) 2x − 3y ≤ 5 (b) 2x − 3y ≥ 5 (c) 2x + 3y ≤ 5 (d) 2x + 3y ≥ 5

Answer

(b) 2x−3y≥52x-3y\ge 5

At (0,−2)(0,-2): 2x−3y=0+6=6≥52x-3y=0+6=6\ge 5, so (b) holds. (Also 2x+3y=−6≤52x+3y=-6\le 5 satisfies (c), but (a) and (d) are false.)

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  • 2082 · United Academy Science Model Set 5 · 1 mark

Which of the formulae does not represent the scalar product of vectors? (a) Work = Force × Displacement (b) Pressure = Force Area (c) Velocity = Displacement Time (d) Power = Current ×Volatge

Answer

(d) Power = Current ×\times Voltage

Work =F⃗⋅d⃗=\vec F\cdot\vec d is a scalar product of two vectors. Current and voltage are not vectors, so their product is not a scalar (dot) product of vectors.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

If SinA=1 /2, what is the balue of sin3A (a) 0 (b) 1 (c) 1⁄2 (d) None

Answer

(b) 1

sin⁡A=12\sin A=\frac12, so A=30∘A=30^\circ and sin⁡3A=sin⁡90∘=1\sin 3A=\sin 90^\circ=1. Check: 3sin⁡A−4sin⁡3A=32−12=13\sin A-4\sin^3A=\frac32-\frac12=1.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

If A+B+C=90∘A+B+C=90^\circ, what is the value of sin⁡(B+C)\sin(B+C)? (a) Cos A (b) - Cos A (c) sin A (d) –Sin A

Answer

(a) cos⁡A\cos A

B+C=90∘−AB+C=90^\circ-A, so sin⁡(B+C)=sin⁡(90∘−A)=cos⁡A\sin(B+C)=\sin(90^\circ-A)=\cos A.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

A pair of lines is represented by an equation mx2−6xy+9y2mx^2-6xy+9y^2. If the lines are co-incident, what is the value of m? (a) 0 (b) 1 (c) -1 (d) 4

Answer

(b) 1

For a pair of lines ax2+2hxy+by2=0ax^2+2hxy+by^2=0 to be coincident, h2=abh^2=ab. Here a=ma=m, 2h=−62h=-6 so h=−3h=-3, b=9b=9:

9=9m⇒m=19=9m \Rightarrow m=1

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  • 2082 · United Academy Science Model Set 5 · 1 mark

What is the general equation of circle? (a) ax2+by2+2gx+2fy−c=0ax^2+by^2+2gx+2fy-c=0 (b) ax2+by2+2gx+2fy+c=0ax^2+by^2+2gx+2fy+c=0 (c) x2+y2+gx+fy+c=0x^2+y^2+gx+fy+c=0 (d) x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0

Answer

(d) x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0

The general circle has equal coefficients of x2x^2 and y2y^2 (taken as 1) and no xyxy term, with centre (−g,−f)(-g,-f) and radius g2+f2−c\sqrt{g^2+f^2-c}.

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  • 2082 · United Academy Science Model Set 5 · 1 mark

Which of the following is the inverse of the matrix [2513]\begin{bmatrix}2&5\\1&3\end{bmatrix}? (a) [2−5−13]\begin{bmatrix}2&-5\\-1&3\end{bmatrix} (b) [3−5−12]\begin{bmatrix}3&-5\\-1&2\end{bmatrix} (c) [−215−3]\begin{bmatrix}-2&1\\5&-3\end{bmatrix} (d) [3−512]\begin{bmatrix}3&-5\\1&2\end{bmatrix}

Answer

(b) [3−5−12]\begin{bmatrix}3&-5\\-1&2\end{bmatrix}

det⁡=2⋅3−5⋅1=1\det = 2\cdot3-5\cdot1=1. Swap the diagonal entries and change the signs of the off-diagonal ones:

A−1=11[3−5−12]A^{-1}=\frac{1}{1}\begin{bmatrix}3&-5\\-1&2\end{bmatrix}

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  • 2082 · United Academy Science Model Set 6 · 1 mark

If 154 ×18 = 2772, then 27. 72÷1.8 =? (a) 1.54 (b) 1.44 (c) 15.4 (d) 1.45

Answer

(c) 15.4

27.72÷1.8=277.2÷18=2772÷180=277218×110=154×110=15.427.72 \div 1.8 = 277.2 \div 18 = 2772 \div 180 = \dfrac{2772}{18} \times \dfrac{1}{10} = 154 \times \dfrac{1}{10} = 15.4.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

What number's 15% equals Rs. 60) ? (a) Rs 200 (b) Rs 300 (c) Rs 400 (d) Rs 500

Answer

(c) Rs 400

15%15\% of x=60⇒x=60×10015=400x = 60 \Rightarrow x = \dfrac{60 \times 100}{15} = 400.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

The ratio of water and milk in a 30 liter mixture is 7:3. Find the quantity of water to be added to the mixture in order to make this ratio 6:1. (a) 30 (b) 32 (c) 33 (d) 35

Answer

(c) 33

Water =710×30=21= \dfrac{7}{10}\times 30 = 21 L, milk =9= 9 L. Adding xx L of water:

21+x9=61⇒x=54−21=33.\frac{21+x}{9} = \frac{6}{1} \Rightarrow x = 54 - 21 = 33.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

The average of 5 consecutive even numbers is 30. What is the largest number? (a) 32 (b) 34 (c) 36 (d) 38

Answer

(b) 34

The middle (third) number is the average, 30, so the numbers are 26, 28, 30, 32, 34. The largest is 34.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

The sum of money at the rate of 5% per annum will yield an interest of Rs. 120 in 3 years is (a) 1200 (b) 900 (c) 1000 (d) 800

Answer

(d) 800

I=PTR100⇒120=P×3×5100⇒P=1200015=I = \dfrac{PTR}{100} \Rightarrow 120 = \dfrac{P \times 3 \times 5}{100} \Rightarrow P = \dfrac{12000}{15} = Rs 800.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

How many proper subsets can be formed from a set with 3 elements? (a) 3 (b) 6 (c) 7 (d) 8

Answer

(c) 7

A set with nn elements has 2n2^n subsets, of which 2n−12^n - 1 are proper. For n=3n=3: 23−1=72^3 - 1 = 7.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

The price of a computer is Rs 100(say). If it depreciates by 10% per annum compounded annually, what will be it’s price after 2 years? (a) Rs 80 (b) Rs 81 (c) Rs 79 (d) Rs 82

Answer

(b) Rs 81

A=100(1−10100)2=100×0.92=100×0.81=Rs 81.A = 100\left(1 - \frac{10}{100}\right)^2 = 100 \times 0.9^2 = 100 \times 0.81 = \text{Rs } 81.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

What is the ratio of volume of a pyramid and cube having same base area and vertical height? (a) 1:2 (b) 1:3 (c) 1:4 (d) 3:4

Answer

(b) 1:3

Pyramid volume =13Bh= \tfrac13 Bh and the cube (prism) volume =Bh= Bh for the same base area BB and height hh. The ratio is 13:1=1:3\tfrac13 : 1 = 1:3.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is: (a) 37 1⁄2° (b) 72 1⁄2° (c) 75° (d) 52 1⁄2°

Answer

(d) 5212∘52\tfrac12^\circ

An exterior angle equals the sum of the two opposite interior angles, so 2x=105∘2x = 105^\circ and x=52.5∘=5212∘x = 52.5^\circ = 52\tfrac12^\circ.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

What is the HCF of the expressions (a3−b3)(a^3-b^3), (a2+ab+b2)(a^2+ab+b^2) and (a4+a2b2+b4)(a^4+a^2b^2+b^4)? (a) (a2−ab+b2)(a^2-ab+b^2) (b) (a4+a2b2+b4)(a^4+a^2b^2+b^4) (c) (a−b)(a-b) (d) (a2+ab+b2)(a^2+ab+b^2)

Answer

(d) (a2+ab+b2)(a^2+ab+b^2)

a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2+ab+b^2), and a4+a2b2+b4=(a2+ab+b2)(a2−ab+b2)a^4 + a^2b^2 + b^4 = (a^2+ab+b^2)(a^2-ab+b^2). The factor a2+ab+b2a^2+ab+b^2 is itself the second expression, and it is common to all three, so it is the HCF.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

What is the value of 2564\sqrt[4]{\sqrt{256}} ? (a) 2 (b) 4 (c) 6 (d) 8

Answer

(a) 2

256=16\sqrt{256} = 16, and 164=2\sqrt[4]{16} = 2 since 24=162^4 = 16.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

What is the probability of getting a faced card or an Ace when a card is drawn from a well-shuffled deck of 52 cards? (a) 1/13 (b) 2/13 (c) 3/13 (d) 4/13

Answer

(d) 4/13

There are 12 face cards (J, Q, K) and 4 aces, which do not overlap, giving 16 favourable cards:

P=1652=413.P = \frac{16}{52} = \frac{4}{13}.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

What is the necessary condition for a function to have it’s inverse? (a) One to one function (b) Many to one function (c) One to one and onto function (d) One to one and into function

Answer

(c) One to one and onto function

A function has an inverse exactly when it is bijective: one-to-one so each output has a single input, and onto so every element of the codomain is mapped.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

Which term of the sequence 18,14,12,…\frac18,\frac14,\frac12,\ldots is 32? (a) 7th term (b) 8th term (c) 10th term (d) 9th term

Answer

(d) 9th term

This is a G.P. with a=18a = \tfrac18 and r=2r = 2. Then arn−1=32⇒2n−1=256=28ar^{n-1} = 32 \Rightarrow 2^{n-1} = 256 = 2^8, so n−1=8n - 1 = 8 and n=9n = 9.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

Which of the given lines passes through the point (-3,2)? (a) 2x = 3y (b) 2x − 3y = 5 (c) 2x − 3y + 12 = 0 (d) None

Answer

(c) 2x−3y+12=02x - 3y + 12 = 0

Substitute (−3,2)(-3, 2): 2(−3)−3(2)+12=−6−6+12=02(-3) - 3(2) + 12 = -6 - 6 + 12 = 0. For (a), −6≠6-6 \ne 6; for (b), −12≠5-12 \ne 5.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

The length of a rectangle is four times the breadth. If the minimum perimeter of the rectangle is 160 cm, then (a) breadth > 16 cm (b) length < 20 cm (c) breadth ≥ 16 cm (d) length ≤ 20 cm

Answer

(c) breadth ≥\ge 16 cm

Let breadth =b= b, length =4b= 4b. Perimeter =2(4b+b)=10b≥160⇒b≥16= 2(4b + b) = 10b \ge 160 \Rightarrow b \ge 16 cm.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

Two vectors (3i^+2j^)(3\hat i+2\hat j) and (ai^−3j^)(a\hat i-3\hat j) are orthogonal. What should be the value of aa? (a) 2 (b) -3 (c) -2 (d) 3

Answer

(a) 2

Orthogonal vectors have zero dot product: 3a+2(−3)=0⇒3a=6⇒a=23a + 2(-3) = 0 \Rightarrow 3a = 6 \Rightarrow a = 2.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

In a 90°, 60°, 30° triangle, what is the ratio of length of perpendicular and hypotenuse with reference to the angle of 30°? (a) 1:2 (b) 1:3 (c) 2:1 (d) 3:1

Answer

(a) 1:2

The side opposite 30∘30^\circ is half the hypotenuse, so sin⁡30∘=perpendicularhypotenuse=12\sin 30^\circ = \dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac12, i.e. 1:21:2.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

Which of the following formulae is not correct? (a) Sin2a = 2SinaCosa (b) Cos2a=1+2Sin²a (c) Sin(A+B) = SinA.CosB+CosA.SinB (d) Sin (A-B) = SinA.CosB-CosA.SinB

Answer

(b) cos⁡2a=1+2sin⁡2a\cos 2a = 1 + 2\sin^2 a

This one is incorrect. The correct identity is cos⁡2a=1−2sin⁡2a\cos 2a = 1 - 2\sin^2 a. The other three are standard formulae.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

What is the acute angle between the lines having slopes 2 and −3-3? (a) 30∘30^\circ (b) 45∘45^\circ (c) 60∘60^\circ (d) 90∘90^\circ

Answer

(b) 45∘45^\circ

tan⁡θ=∣m1−m21+m1m2∣=∣2−(−3)1+(2)(−3)∣=∣5−5∣=1⇒θ=45∘.\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \left|\frac{2-(-3)}{1 + (2)(-3)}\right| = \left|\frac{5}{-5}\right| = 1 \Rightarrow \theta = 45^\circ.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

What is the equation of the circle having center at (1,0) and having radius 3 units? (a) x² + 2x + y² + 8 = 0 (b) x² + 2x + y² − 8 = 0 (c) x² − 2x + y² + 8 = 0 (d) x² − 2x + y² − 8 = 0

Answer

(d) x2−2x+y2−8=0x^2 - 2x + y^2 - 8 = 0

The circle is (x−1)2+(y−0)2=32(x-1)^2 + (y-0)^2 = 3^2. Expanding gives x2−2x+1+y2=9x^2 - 2x + 1 + y^2 = 9, so x2−2x+y2−8=0.x^2 - 2x + y^2 - 8 = 0.

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  • 2082 · United Academy Science Model Set 6 · 1 mark

Identify the transformation represented by the matrix [100−1]\begin{bmatrix}1&0\\0&-1\end{bmatrix} (a) Reflection by x- axis (b) Reflection by y-axis (c) Reflection by the line y=x (d) Reflection by the line y=-x

Answer

(a) Reflection by x-axis

The matrix sends (x,y)(x, y) to (x,−y)(x, -y): the x-coordinate is unchanged and the y-coordinate changes sign, which is reflection in the x-axis.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

Out of 120 employees in a company, one-third are female. 1/5 of the females are married and half of the married females have a baby. How many married females have babies? (a) 4 (b) 5 (c) 3 (d) 6

Answer

(a) 4

Females =13×120=40=\frac13\times120=40. Married =15×40=8=\frac15\times40=8. With babies =12×8=4=\frac12\times8=4.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

If an item is sold at Rs 960 after 4% loss, what should be the selling price with 20% profit? (a) Rs 1180 (b) Rs 1200 (c) Rs 1320 (d) Rs 1080

Answer

(b) Rs 1200

Cost price =9600.96==\frac{960}{0.96}= Rs 1000. Selling price for 20% profit =1000×1.2==1000\times1.2= Rs 1200.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

The ratio of current age of father and son is 5:2. After 5 years, the ratio of their age will be 11:5. What is the present age of son? (a) 15 years (b) 18 years (c) 20 years (d) 25 years

Answer

(c) 20 years

Let the ages be 5x5x and 2x2x. Then 5x+52x+5=115\frac{5x+5}{2x+5}=\frac{11}{5}, so 25x+25=22x+5525x+25=22x+55 and x=10x=10. Son =2x=20=2x=20 years.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

If 6 lemons are bought for Rs 1, how many lemons should be sold for Rs 1 to gain a profit of 20%? (a) 4 (b) 5 (c) 7 (d) 8

Answer

(b) 5

Cost of one lemon =16=\frac16 Rs. For a 20% profit, selling price of one lemon =16×1.2=15=\frac16\times1.2=\frac15 Rs. So 5 lemons are sold for Rs 1.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

What is the average of first 25 odd numbers? (a) 21 (b) 23 (c) 25 (d) 26

Answer

(c) 25

The sum of the first nn odd numbers is n2n^2, so the average is 25225=25\frac{25^2}{25}=25.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

The time and rate of simple interest of Rs 2000 are same. If Rs 980 is paid as interest, find the rate of interest. (a) 6% (b) 7% (c) 8% (d) 9%

Answer

(b) 7%

Time == rate =r=r. Then I=PTR100=2000 r2100=980I=\frac{PTR}{100}=\frac{2000\,r^2}{100}=980, so r2=49r^2=49 and r=7%r=7\%.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

40 people like coffee while 60 people like tea only. 20 people like both drinks while 10 people prefer none. How many people like at most one drink? (a) 100 (b) 90 (c) 110 (d) 130

Answer

(b) 90

Taking 40 as the total who like coffee, coffee only =40−20=20=40-20=20. Tea only is 60 and none is 10. At most one drink means everyone except those who like both: 20+60+10=9020+60+10=90.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

The ratio of initial price of car and its price after 2 years is 16:9. Calculate the rate of depreciation? (a) 10% (b) 15% (c) 25% (d) 30%

Answer

(c) 25%

(1−r100)2=916\left(1-\frac{r}{100}\right)^2=\frac{9}{16}, so 1−r100=341-\frac{r}{100}=\frac34 and r=25%r=25\%.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

The perimeter of base of a pyramid is 40 cm and vertical height is 12 cm. What is the area of triangular faces of pyramid? (a) 160 sq. cm (b) 260 sq. cm (c) 360 sq. cm (d) 480 sq. cm

Answer

(b) 260 sq. cm

Take a square base (perimeter 40 cm), so the side is 10 cm and the apothem is 5 cm. Slant height l=122+52=13l=\sqrt{12^2+5^2}=13 cm. Area of triangular faces =12×40×13=260=\frac12\times40\times13=260 sq. cm.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

In the figure below, lines l, k and are parallel. What is the value of y – x? (a) 45 (b) 60 (c) 90 (d) 135

Answer

(c) 90

Using corresponding and alternate angles between the parallel lines, the two angles combine so that y−x=90∘y-x=90^\circ.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

At 3:00 A.M. the temperature was 13° below zero. By noon it had risen to 32°. What was the average hourly increase in temperature? (a) 19/9 degrees (b) 19/6 degrees (c) 5 degrees (d) 7.5 degrees

Answer

(c) 5 degrees

From 3:00 A.M. to noon is 9 hours. Total rise =32−(−13)=45∘=32-(-13)=45^\circ. Average hourly rise =459=5∘=\frac{45}{9}=5^\circ.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

If (7a)(7b)=7c7d(7^a)(7^b)=\frac{7^c}{7^d}, what is dd in terms of aa, bb, and cc? (a) cab\frac{c}{ab} (b) c−a−bc-a-b (c) a+b−ca+b-c (d) c−abc-ab

Answer

(b) c−a−bc-a-b

7a+b=7c−d7^{a+b}=7^{c-d}, so a+b=c−da+b=c-d and d=c−a−bd=c-a-b.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

If the radius of a circle is increased by 10%, by what percent will it’s area increase? (a) 10% (b) 20% (c) 10π% (d) 21%

Answer

(d) 21%

New radius =1.1r=1.1r, so new area =π(1.1r)2=1.21πr2=\pi(1.1r)^2=1.21\pi r^2. The increase is 21%.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

A dice is rolled and a card is drawn from a deck of 52 cards. What is the probability of getting 3 on a dice and a red card? (a) 1/8 (b) 1/12 (c) 2/3 (d) 1/3

Answer

(b) 1/12

The events are independent: P=16×2652=16×12=112P=\frac16\times\frac{26}{52}=\frac16\times\frac12=\frac1{12}.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

Which of the following does not exhibit the property of geometric sequence? (a) Population growth (b) Compound amount for successive years (c) Radioactive decay (d) Filling water from the tap

Answer

(d) Filling water from the tap

Population growth, compound amount and radioactive decay change by a constant ratio each period. Filling water from a tap adds a constant amount per unit time, which is arithmetic.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

By what ratio does the centroid divide the median drawn from the vertex to the base of the triangle? (a) 2:1 (b) 1:2 (c) 1:3 (d) 3:1

Answer

(a) 2:1

The centroid divides each median in the ratio 2:12:1, measured from the vertex.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

If the half plane with the boundary 2x-3y =6 contains origin, how is the inequality represented? (a) 2x − 3y ≤ −6 (b) 2x − 3y ≥ −6 (c) 2x − 3y ≤ 6 (d) 2x − 3y ≥ 6

Answer

(c) 2x−3y≤62x-3y\le6

At the origin 2(0)−3(0)=0<62(0)-3(0)=0<6, so the half plane containing the origin satisfies 2x−3y≤62x-3y\le6.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

If the two vectors (a,m)(a,m) and (p,m)(p,m) are perpendicular to each other, which of the following is true? (a) ap−m2=0ap-m^2=0 (b) am+pm=0am+pm=0 (c) ap+m2=0ap+m^2=0 (d) am−pm=0am-pm=0

Answer

(c) ap+m2=0ap+m^2=0

Perpendicular vectors have zero dot product: ap+m⋅m=ap+m2=0ap+m\cdot m=ap+m^2=0.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

If 1+cos⁡2A=01+\cos2A=0, what is the value of sin⁡2A3\sin\frac{2A}{3} ? (a) 0 (b) 32\frac{\sqrt3}2 (c) −12-\frac12 (d) None

Answer

(b) 32\frac{\sqrt3}{2}

1+cos⁡2A=01+\cos2A=0 gives cos⁡2A=−1\cos2A=-1, so 2A=180∘2A=180^\circ. Then sin⁡2A3=sin⁡60∘=32\sin\frac{2A}{3}=\sin60^\circ=\frac{\sqrt3}{2}.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

What is the value of sin⁡165∘\sin165^\circ? (a) 12\frac12 (b) 12\frac1{\sqrt2} (c) −12-\frac12 (d) −12-\frac1{\sqrt2}

Answer

(a) 12\frac12

Exactly, sin⁡165∘=sin⁡(180∘−15∘)=sin⁡15∘=6−24≈0.259\sin165^\circ=\sin(180^\circ-15^\circ)=\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}\approx0.259, which is positive. Of the listed options only the positive ones can fit, and 12\frac12 is the nearest.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

The radius of a circle is 5 units and touches the both-axis. What should be the center of the circle? (a) (5,2) (b) (-5,2) (c) (5,5) (d) (2,5)

Answer

(c) (5,5)

A circle touching both axes has its centre at distance equal to the radius from each axis, so the centre is (±5,±5)(\pm5,\pm5); in the options only (5,5)(5,5) fits.

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  • 2082 · United Academy Science Model Set 7 · 1 mark

If the matrix [13x227]\begin{bmatrix}1&3\\x^2&27\end{bmatrix} is a singular matrix, what is the value of x? (a) 3 (b) -3 (c) Neither of the two (d) Both a and b

Answer

(d) Both a and b

A matrix is singular when its determinant is zero:

1⋅27−3x2=0⇒x2=9⇒x=±3.1\cdot27-3x^2=0\Rightarrow x^2=9\Rightarrow x=\pm3.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

A man saves Rs. 27000 monthly which is three-fifth of his total income. What amount does he saves? (a) Rs. 15000 (b) Rs. 18000 (c) Rs. 25000 (d) Rs. 27000

Answer

(d) Rs. 27000

The amount saved is given directly as Rs. 27000. It is three-fifths of the income, so the total income is 27000×53=27000\times\frac53= Rs. 45000, but the question asks for the savings.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

40% of the greater number is equal to 60% of the smaller number. If their sum is 50, what is the smaller number? (a) 30 (b) 16 (c) 20 (d) 40

Answer

(c) 20

Let the greater number be GG and the smaller SS. From 0.4G=0.6S0.4G=0.6S, G:S=3:2G:S=3:2.

S=25×50=20S=\frac{2}{5}\times 50=20

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  • 2082 · United Academy Science Model Set 8 · 1 mark

If a:b = 5:7, and c:d=2a:3b, then ad:bc is? (a) 1:3 (b) 2:3 (c) 3:2 (d) 4:5

Answer

(c) 3:2

ab=57\frac ab=\frac57 and cd=2a3b=23⋅57=1021\frac cd=\frac{2a}{3b}=\frac23\cdot\frac57=\frac{10}{21}, so dc=2110\frac dc=\frac{21}{10}.

adbc=ab⋅dc=57⋅2110=32\frac{ad}{bc}=\frac ab\cdot\frac dc=\frac57\cdot\frac{21}{10}=\frac32

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  • 2082 · United Academy Science Model Set 8 · 1 mark

If Ram sold an article for Rs. 1243 including 13% VAT, he had 10% Profit. What is his profit amount? (a) Rs. 100 (b) Rs. 130 (c) Rs. 90 (d) Rs. 60

Answer

(a) Rs. 100

Selling price without VAT =12431.13==\dfrac{1243}{1.13}= Rs. 1100.

Since the profit is 10%, 1100=1.1×CP1100=1.1\times CP, so CP=CP= Rs. 1000 and profit =1100−1000==1100-1000= Rs. 100.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

The average age of 5 sisters is 10 years. If the age of 4 sisters are 6, 9, 12, 13, what is the age of fifth one? (a) 11 (b) 14 (c) 10 (d) 7

Answer

(c) 10

Total age =5×10=50=5\times10=50. The four known ages sum to 6+9+12+13=406+9+12+13=40, so the fifth sister is 50−40=1050-40=10 years old.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

If Akshay wants to double his investment at the rate of 10% p.a. annual compound interest, for how many years at least he needs to invest his money? (a) 6 (b) 7 (c) 8 (d) 9

Answer

(c) 8

We need (1.1)n≥2(1.1)^n\ge 2. Now 1.17≈1.949<21.1^7\approx1.949<2 and 1.18≈2.144>21.1^8\approx2.144>2, so the least whole number of years is 8.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

In an election of a local level, 2500 votes were secured by candidate A, 3000 votes were secured by candidate B and 50 votes were invalid. If there were 6000 names in voters’ list, how many people did not cast the vote? (a) 500 (b) 450 (c) 400 (d) 350

Answer

(b) 450

Votes cast =2500+3000+50=5550=2500+3000+50=5550 (the invalid votes were still cast). People who did not vote =6000−5550=450=6000-5550=450.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

A motorbike is bought for Rs. 2,00,000 and its price depreciates each year by 10%. Its value after 2 years will be: (a) .Rs. 1,80,000 (b) Rs. 1,62,000 (c) .Rs. 1,45,800 (d) None of above

Answer

(b) Rs. 1,62,000

V=200000(1−10100)2=200000×0.81=162000V=200000\left(1-\frac{10}{100}\right)^2=200000\times0.81=162000

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  • 2082 · United Academy Science Model Set 8 · 1 mark

The buying rate and selling rate of 1 US dollar are Rs 139 and Rs 140 respectively. How much loss will be there while exchanging NRs. 1,40,000 into US dollar and again exchanging the dollars back into Nepali rupees? (a) Rs 1,000 (b) Rs 10,000 (c) Rs 10,500 (d) Rs 14,000

Answer

(a) Rs 1,000

Rs 1,40,000 buys 140000140=1000\frac{140000}{140}=1000 dollars (bank selling rate Rs 140). Selling them back at the buying rate gives 1000×139=1000\times139= Rs 1,39,000.

Loss =140000−139000==140000-139000= Rs 1,000.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

If the volume of a cube is 600m³ , what will be the volume of pyramid of same height and base? (a) 600 m³ (b) 60 m³ (c) 200 m³ (d) 300 m³

Answer

(c) 200 m³

The volume of a pyramid is one third of the volume of a prism (or cube) with the same base and height: 13×600=200 m3\frac13\times600=200\ \text{m}^3.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

What is the sum of interior angles of a hexagon? (a) 360° (b) 450° (c) 540° (d) 720°

Answer

(d) 720°

(n−2)×180∘=(6−2)×180∘=720∘(n-2)\times180^\circ=(6-2)\times180^\circ=720^\circ

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  • 2082 · United Academy Science Model Set 8 · 1 mark

10 years ago the younger sister was half the age of the older sister. At present, older sister is 28 years old, what is the age of older sister? (a) 19 (b) 14 (c) 25 (d) 16

Answer

(a) 19

The question has a misprint: the older sister is given as 28, so the age asked for must be that of the younger sister.

Ten years ago the older sister was 28−10=1828-10=18, so the younger sister was 99. Her present age is 9+10=199+10=19.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

If x0=1x^0=1, then what is 0x0^x ? (a) 1 (b) 0 (c) x (d) none of above

Answer

(b) 0

The rule x0=1x^0=1 holds for x≠0x\ne0. For the positive values of xx intended here, 0x=0×0×⋯=00^x=0\times0\times\cdots=0.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

If the radius of a circle is doubled, what will be its new area? (consider the area of original circle A) (a) 2A (b) 4A (c) A (d) 6A

Answer

(b) 4A

A=πr2A=\pi r^2. With radius 2r2r, the new area is π(2r)2=4πr2=4A\pi(2r)^2=4\pi r^2=4A.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

A dice is rolled 100 times. The results were : Event 1 2 3 4 5 6 No. of occurrence 15 30 10 20 12 13 What is the empirical probability of getting 3? (a) 0.5 (b) 0.2 (c) 0.1 (d) 0.12

Answer

(c) 0.1

P(3)=number of 3stotal throws=10100=0.1P(3)=\frac{\text{number of 3s}}{\text{total throws}}=\frac{10}{100}=0.1

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  • 2082 · United Academy Science Model Set 8 · 1 mark

A child started saving money in piggy bank. He started from Rs. 1 in the first day and he doubled the amount each day. How much money would be there in his piggy bank at 10th day? (a) 100 (b) 1023 (c) 1020 (d) 2000

Answer

(b) 1023

The daily deposits are 1,2,4,…,291,2,4,\dots,2^9, a geometric series with a=1a=1, r=2r=2, n=10n=10.

S10=210−12−1=1023S_{10}=\frac{2^{10}-1}{2-1}=1023

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  • 2082 · United Academy Science Model Set 8 · 1 mark

What is the angle between the lines represented by 2x – 3y = 5 and 3x + 2y = 5? (a) 45° (b) 60° (c) 0° (d) 90°

Answer

(d) 90°

The slope of 2x−3y=52x-3y=5 is m1=23m_1=\frac23 and the slope of 3x+2y=53x+2y=5 is m2=−32m_2=-\frac32. Since m1m2=−1m_1m_2=-1, the lines are perpendicular.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

The highest value of a⃗⋅b⃗\vec a\cdot\vec b can be obtained while angle between vectors a⃗\vec a and b⃗\vec b is (a) 0∘0^\circ (b) 90∘90^\circ (c) 180∘180^\circ (d) 45∘45^\circ

Answer

(a) 0∘0^\circ

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta is greatest when cos⁡θ=1\cos\theta=1, that is θ=0∘\theta=0^\circ.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

If sin⁡A=12\sin A=\frac12, then what is the value of cos⁡2A\cos2A if A is acute angle? (a) 12\frac12 (b) 32\frac{\sqrt3}2 (c) −12-\frac12 (d) None

Answer

(a) 12\frac12

Since sin⁡A=12\sin A=\frac12 and AA is acute, A=30∘A=30^\circ. Then cos⁡2A=cos⁡60∘=12\cos2A=\cos60^\circ=\frac12.

Check: 1−2sin⁡2A=1−2⋅14=121-2\sin^2A=1-2\cdot\frac14=\frac12.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

What is the value of sin⁡(270∘−2A)\sin(270^\circ-2A)? (a) sin⁡2A\sin2A (b) −sin⁡2A-\sin2A (c) −cos⁡2A-\cos2A (d) cos⁡2A\cos2A

Answer

(c) −cos⁡2A-\cos2A

sin⁡(270∘−θ)=−cos⁡θ\sin(270^\circ-\theta)=-\cos\theta

With θ=2A\theta=2A, this gives −cos⁡2A-\cos2A.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

The pair of lines represented by ax2+2hxy+by2=0ax^2+2hxy+by^2=0 represent perpendicular lines if: (a) a−b=0a-b=0 (b) h2+ab=0h^2+ab=0 (c) a+b=0a+b=0 (d) h2−ab=0h^2-ab=0

Answer

(c) a+b=0a+b=0

For ax2+2hxy+by2=0ax^2+2hxy+by^2=0, the lines are perpendicular when the sum of the coefficients of x2x^2 and y2y^2 is zero, since m1m2=ab=−1m_1m_2=\frac ab=-1.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

A circle is represented by the equation 3x2+3y2−12x+18y−3=03x^2+3y^2-12x+18y-3=0, its center is: (a) (−2,3)(-2,3) (b) (2,3)(2,3) (c) (−2,−3)(-2,-3) (d) (2,−3)(2,-3)

Answer

(d) (2,−3)(2,-3)

Divide by 3: x2+y2−4x+6y−1=0x^2+y^2-4x+6y-1=0. Comparing with x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0, g=−2g=-2 and f=3f=3. The centre is (−g,−f)=(2,−3)(-g,-f)=(2,-3).

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  • 2082 · United Academy Science Model Set 8 · 1 mark

The combined transformation of reflection through x axis followed by reflection through y axis is equivalent to which of the following transformations? (a) Reflection through line x + y = 0 (b) Reflection through line x – y = 0 (c) Rotation +90° about origin (d) Rotation 180° about origin

Answer

(d) Rotation 180° about origin

Reflection in the x-axis sends (x,y)→(x,−y)(x,y)\to(x,-y), and then reflection in the y-axis sends this to (−x,−y)(-x,-y). That is a half-turn about the origin.

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  • 2082 · United Academy Science Model Set 8 · 1 mark

In interval notation, 3 ≤ x < 5, x ∈ R can be written as: (a) [3,5] (b) (3,5] (c) [3,5) (d) (3,5)

Answer

(c) [3,5)[3,5)

The square bracket at 3 includes 3 (3≤x3 \le x), and the round bracket at 5 excludes 5 (x<5x < 5).

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Questions from KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Practice Test 1 (Science) (College-published practice test (100 MCQs: English, Mathematics, Science, GK/IQ) in the April 2018 entrance preparation book; model set, not a recalled paper), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Practice Test 2 (Science) (College-published practice test (100 MCQs: English, Mathematics, Science, GK/IQ) in the April 2018 entrance preparation book; model set, not a recalled paper), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Practice Test 3 (Science) (College-published practice test (100 MCQs: English, Mathematics, Science, GK/IQ) in the April 2018 entrance preparation book; model set, not a recalled paper), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Practice Test 4 (Science) (College-published practice test (100 MCQs: English, Mathematics, Science, GK/IQ) in the April 2018 entrance preparation book; model set, not a recalled paper), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Practice Test 5 (Science) (College-published practice test (100 MCQs: English, Mathematics, Science, GK/IQ) in the April 2018 entrance preparation book; model set, not a recalled paper), KIST College & SS, +2 Science Entrance Sample Test Set A (2019) (College-published model set (100 MCQs, 1 hr 30 min: English 20, Science 50, Mathematics 20, GK 10); year inferred from its 2019 references, file hosted on the KIST site), Motherland Secondary School, Pokhara, +2 Science Entrance Examination 2074 (Set A) (College entrance paper for Class 11 Science (100 marks, 2 hrs: English, Physics, Chemistry, Biology, Mathematics, Nepali), 2074 BS (2017 AD); collected on the Gaurab Subedi +2 entrance blog), Study Notes Nepal: SEE Entrance Model Question for top ten colleges (2023), Set 1 (Unattributed practice set compiled by a study-notes site (June 2023) for +2 Science entrance preparation (English, Maths, Physics, Chemistry, Biology); not an official college paper), Study Notes Nepal: SEE Entrance Model Question for top ten colleges (2023), Set 2 (Unattributed practice set compiled by a study-notes site (June 2023) for +2 Science entrance preparation (English, Maths, Physics, Chemistry, Biology); not an official college paper), Colleges Nepal, Bridge Course Model Questions 2025, Practice Set 5 (Science) (Bridge-course practice set (100 MCQs for +2 Science entrance: English, Physics, Chemistry, Biology, Mathematics) published on the Colleges Nepal bridge course page, 2025; model set, not a recalled paper), Colleges Nepal, Bridge Course Model Questions 2025, Practice Set 10 (Science) (Bridge-course practice set (100 MCQs for +2 Science entrance: English, Physics, Chemistry, Biology, Mathematics) published on the Colleges Nepal bridge course page, 2025; model set, not a recalled paper), Colleges Nepal, Bridge Course Model Questions 2025, Practice Set 4 (Science) (Bridge-course practice set (100 MCQs for +2 Science entrance: English, Physics, Chemistry, Biology, Mathematics) published on the Colleges Nepal bridge course page, 2025; model set, not a recalled paper), United Academy, Lalitpur, Faculty of Science Model Question for Entrance Examination 2080 (School-published sample questions (selection from a 100-question paper: Maths, Physics, Chemistry, Biology, English), 2080 BS; collected on the Gaurab Subedi +2 entrance blog), SOS Hermann Gmeiner School, Sanothimi, Grade XI Entrance Exam Model Question 2079 (School-published model question (100 MCQs, 2 hours, Science stream: Physics, Biology, Chemistry, English/GK, Mathematics), 2079 BS (2022 AD)), SOS Hermann Gmeiner School, Sanothimi, Grade XI Entrance Exam 2025 (2082 BS) Model Question, Set A (School-published model question (100 MCQs, 2 hours, Science stream: Mathematics, Physics, Chemistry, Biology, English, GK), 2025 AD), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), English practice MCQs (Practice MCQs at the end of topics in the April 2018 entrance preparation book published by KIST College (for +2 Science/Management entrance); practice items, not recalled papers), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Mathematics practice MCQs (Practice MCQs at the end of topics in the April 2018 entrance preparation book published by KIST College (for +2 Science/Management entrance); practice items, not recalled papers), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Physics practice MCQs (Practice MCQs at the end of topics in the April 2018 entrance preparation book published by KIST College (for +2 Science/Management entrance); practice items, not recalled papers), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Chemistry practice MCQs (Practice MCQs at the end of topics in the April 2018 entrance preparation book published by KIST College (for +2 Science/Management entrance); practice items, not recalled papers), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Zoology practice MCQs (Practice MCQs at the end of topics in the April 2018 entrance preparation book published by KIST College (for +2 Science/Management entrance); practice items, not recalled papers), KIST College & SS, Post-SEE Students Gearing Up for +2 Entrance Exams (2018 book), Botany practice MCQs (Practice MCQs at the end of topics in the April 2018 entrance preparation book published by KIST College (for +2 Science/Management entrance); practice items, not recalled papers), United Academy, Lalitpur, Entrance Model Questions 2082 (2025 booklet), Science Model Set 1 (College-published model set for the Grade XI entrance (2 hours, 100 MCQs: Mathematics 25, Physics 20, Chemistry 20, Biology 20, English 15), booklet dated 2025 (2082 BS); model/practice set, not a recalled paper), United Academy, Lalitpur, Entrance Model Questions 2082 (2025 booklet), Science Model Set 2 (College-published model set for the Grade XI entrance (2 hours, 100 MCQs: Mathematics 25, Physics 20, Chemistry 20, Biology 20, English 15), booklet dated 2025 (2082 BS); model/practice set, not a recalled paper), United Academy, Lalitpur, Entrance Model Questions 2082 (2025 booklet), Science Model Set 3 (College-published model set for the Grade XI entrance (2 hours, 100 MCQs: Mathematics 25, Physics 20, Chemistry 20, Biology 20, English 15), booklet dated 2025 (2082 BS); model/practice set, not a recalled paper), United Academy, Lalitpur, Entrance Model Questions 2082 (2025 booklet), Science Model Set 4 (College-published model set for the Grade XI entrance (2 hours, 100 MCQs: Mathematics 25, Physics 20, Chemistry 20, Biology 20, English 15), booklet dated 2025 (2082 BS); model/practice set, not a recalled paper), United Academy, Lalitpur, Entrance Model Questions 2082 (2025 booklet), Science Model Set 5 (College-published model set for the Grade XI entrance (2 hours, 100 MCQs: Mathematics 25, Physics 20, Chemistry 20, Biology 20, English 15), booklet dated 2025 (2082 BS); model/practice set, not a recalled paper), United Academy, Lalitpur, Entrance Model Questions 2082 (2025 booklet), Science Model Set 6 (College-published model set for the Grade XI entrance (2 hours, 100 MCQs: Mathematics 25, Physics 20, Chemistry 20, Biology 20, English 15), booklet dated 2025 (2082 BS); model/practice set, not a recalled paper), United Academy, Lalitpur, Entrance Model Questions 2082 (2025 booklet), Science Model Set 7 (College-published model set for the Grade XI entrance (2 hours, 100 MCQs: Mathematics 25, Physics 20, Chemistry 20, Biology 20, English 15), booklet dated 2025 (2082 BS); model/practice set, not a recalled paper) and United Academy, Lalitpur, Entrance Model Questions 2082 (2025 booklet), Science Model Set 8 (College-published model set for the Grade XI entrance (2 hours, 100 MCQs: Mathematics 25, Physics 20, Chemistry 20, Biology 20, English 15), booklet dated 2025 (2082 BS); model/practice set, not a recalled paper). 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