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Chapter 2

Mathematics

+2 college entrance (after SEE) past exam questions

Past questions and answers

300 questions set from this chapter, 36 of them more than once; 11 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 4 of 18 exams
  • Asked 2 times
  • 2083 · Science Model Set B · 1 mark
  • 2083 · Science Model Set C · 1 mark

The graphs of the equations 3x+6y=93x + 6y = 9 and 9x+18y=279x + 18y = 27 are (a) intersecting (b) parallel (c) coincident (d) perpendicular

Similar questions: The graphs of the equations 3x + 6y = 9 and 9x + (2080 · Science Model) · The graphs of the equations 3x + 2y = 5 and 2x - (2083 · Science Model Set D)

Answer

(c) coincident

a1a2=39=13,b1b2=618=13,c1c2=927=13.\frac{a_1}{a_2} = \frac{3}{9} = \frac13,\quad \frac{b_1}{b_2} = \frac{6}{18} = \frac13,\quad \frac{c_1}{c_2} = \frac{9}{27} = \frac13. All three ratios are equal (the second equation is 3 times the first), so the lines coincide.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 4 of 18 exams
  • 2083 · Science Model Set D · 1 mark

The graphs of the equations 3x+2y=53x + 2y = 5 and 2x−y=112x - y = 11 are (a) intersecting (b) parallel (c) coincident (d) None

Similar questions: The graphs of the equations 3x + 6y = 9 and 9x + (2083 · Science Model Set B) · The graphs of the equations 3x + 6y = 9 and 9x + (2080 · Science Model)

Answer

(a) intersecting

The slopes are −32-\frac{3}{2} and 22, which are different, so the lines are neither parallel nor coincident. (Equivalently 32≠2−1\frac{3}{2} \neq \frac{2}{-1}.) They meet at x=277x = \frac{27}{7}, y=−297y = -\frac{29}{7}.

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  • Most repeated · 4 of 18 exams
  • 2080 · Science Model · 1 mark

The graphs of the equations 3x+6y=93x + 6y = 9 and 9x+18y=279x + 18y = 27 are (a) coincident (b) parallel (c) intersecting (d) perpendicular

Similar questions: The graphs of the equations 3x + 6y = 9 and 9x + (2083 · Science Model Set B) · The graphs of the equations 3x + 2y = 5 and 2x - (2083 · Science Model Set D)

Answer

(a) coincident

Dividing 9x+18y=279x+18y=27 by 3 gives 3x+6y=93x+6y=9, so both equations represent the same line. Equivalently 39=618=927\frac{3}{9}=\frac{6}{18}=\frac{9}{27}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 18 exams
  • Asked 3 times
  • 2083 · Management and Law Model Set B
  • 2079 · Science Model · 1 mark
  • 2080 · Management and Law Model Set I

The factors of 6x2+x−126x^2 + x - 12 are (a) (3x+4)(2x−3)(3x+4)(2x-3) (b) (3x−4)(2x+3)(3x-4)(2x+3) (c) (3x−4)(2x−4)(3x-4)(2x-4) (d) (3x+4)(2x+4)(3x+4)(2x+4)

Answer

(b) (3x−4)(2x+3)(3x-4)(2x+3)

(3x−4)(2x+3)=6x2+9x−8x−12=6x2+x−12.(3x-4)(2x+3) = 6x^2 + 9x - 8x - 12 = 6x^2 + x - 12.

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  • Most repeated · 3 of 18 exams
  • Asked 2 times
  • 2083 · Science Model Set C · 1 mark
  • 2083 · Management and Law Model Set C

HCF of two numbers is 21 and their product is 2646. How many pairs of such numbers are possible? (a) 3 (b) 1 (c) 2 (d) None

Similar questions: HCF of two numbers is 101 and their sum is 1212. (2083 · Science Model Set B)

Answer

(c) 2

Let the numbers be 21a21a and 21b21b with a,ba, b co-prime. Then 21a⋅21b=2646⇒ab=2646441=6.21a \cdot 21b = 2646 \Rightarrow ab = \frac{2646}{441} = 6. The co-prime pairs are (1,6)(1,6) and (2,3)(2,3), giving the number pairs (21,126)(21,126) and (42,63)(42,63). So there are 2 pairs.

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  • Most repeated · 3 of 18 exams
  • Asked 2 times
  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

The number of distinct prime factors of 210 is (a) 1 (b) 2 (c) 3 (d) 4

Similar questions: The number of distinct prime factors of 27 is (2083 · Science Model Set B)

Answer

(d) 4

210=2×3×5×7210 = 2 \times 3 \times 5 \times 7, which has 4 distinct prime factors.

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  • Most repeated · 3 of 18 exams
  • 2083 · Science Model Set B · 1 mark

HCF of two numbers is 101 and their sum is 1212. How many pairs of such numbers are possible? (a) 3 (b) 2 (c) 1 (d) None

Similar questions: HCF of two numbers is 21 and their product is (2083 · Science Model Set C)

Answer

(b) 2

Let the numbers be 101a101a and 101b101b with a,ba, b co-prime. Then 101(a+b)=1212101(a+b) = 1212, so a+b=12a+b = 12. The co-prime pairs with sum 12 are (1,11)(1,11) and (5,7)(5,7) ((2,10)(2,10), (3,9)(3,9), (4,8)(4,8), (6,6)(6,6) share factors). So there are 2 pairs: (101,1111)(101, 1111) and (505,707)(505, 707).

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  • Most repeated · 3 of 18 exams
  • 2083 · Science Model Set B · 1 mark

The number of distinct prime factors of 27 is (a) 1 (b) 2 (c) 3 (d) 4

Similar questions: The number of distinct prime factors of 210 is (2083 · Management and Law Model Set B)

Answer

(a) 1

27=3327 = 3^3, so 3 is the only prime factor.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 18 exams
  • 2083 · Science Model Set D · 1 mark

xn−ynx^n - y^n is divisible by x+yx + y for (a) All values of nn (b) Only even values of nn (c) Only odd values of nn (d) No value of nn

Similar questions: a^n + b^n is divisible by a + b (2083 · Science Model Set E) · x^n + y^n is divisible by x - y (2079 · Science Model)

Answer

(b) Only even values of nn

xn−ynx^n - y^n is divisible by x+yx + y iff it vanishes at x=−yx = -y: (−y)n−yn=0(-y)^n - y^n = 0 only when nn is even.

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  • Most repeated · 3 of 18 exams
  • 2083 · Science Model Set E · 1 mark

an+bna^n + b^n is divisible by a+ba + b (a) For all values of nn (b) Only for even values of nn (c) Only for odd values of nn (d) For no value of nn

Similar questions: x^n - y^n is divisible by x + y for (2083 · Science Model Set D) · x^n + y^n is divisible by x - y (2079 · Science Model)

Answer

(c) Only for odd values of nn

Put a=−ba=-b in an+bna^n+b^n: it gives (−b)n+bn(-b)^n+b^n, which is 00 only when nn is odd. By the factor theorem, a+ba+b divides an+bna^n+b^n only for odd nn. For example, a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2).

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  • Most repeated · 3 of 18 exams
  • 2079 · Science Model · 1 mark

xn+ynx^n + y^n is divisible by x−yx - y (a) For all values of nn (b) For even values of nn (c) For odd values of nn (d) For no value of nn

Similar questions: x^n - y^n is divisible by x + y for (2083 · Science Model Set D) · a^n + b^n is divisible by a + b (2083 · Science Model Set E)

Answer

(d) For no value of nn

By the factor theorem, x−yx-y divides xn+ynx^n+y^n only if the expression vanishes at x=yx=y. But xn+ynx^n+y^n at x=yx=y equals 2yn≠02y^n\ne0. So it is never divisible by x−yx-y (for odd nn it is divisible by x+yx+y instead).

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  • Asked 2 times
  • 2083 · Science Model Set A · 1 mark
  • 2083 · Management and Law Model Set D

If £1 = Rs 135 and USD 1 = Rs 105, then convert £63 into dollar. (a) USD 63 (b) USD 30 (c) USD 81 (d) USD 36

Answer

(c) USD 81

£63 =63×135==63\times135= Rs 8505, and Rs 8505 ÷ 105=81\div\,105=81 dollars.

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  • Asked 2 times
  • 2083 · Science Model Set A · 1 mark
  • 2079 · Science Model · 1 mark

Find the probability of occurring a squared number or a cubed number while drawing a flash card from the set of cards numbered from 3 to 32. (a) 14\frac{1}{4} (b) 15\frac{1}{5} (c) 12\frac{1}{2} (d) 13\frac{1}{3}

Answer

(b) 15\frac{1}{5}

Cards 3 to 32 give n(S)=30n(S)=30. Squares: 4, 9, 16, 25 (4 cards). Cubes: 8, 27 (2 cards). No number is both, so favourable cases =6=6. P=630=15P=\frac{6}{30}=\frac15

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  • Asked 2 times
  • 2083 · Science Model Set B · 1 mark
  • 2083 · Science Model Set D · 1 mark

The mean of the first nn natural numbers is (a) n+12\frac{n+1}{2} (b) n(n+1)2\frac{n(n+1)}{2} (c) n2\frac{n}{2} (d) n2+12\frac{n^2+1}{2}

Answer

(a) n+12\frac{n+1}{2}

Mean=1+2+⋯+nn=n(n+1)/2n=n+12.\text{Mean} = \frac{1+2+\dots+n}{n} = \frac{n(n+1)/2}{n} = \frac{n+1}{2}.

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  • Asked 2 times
  • 2083 · Science Model Set B · 1 mark
  • 2083 · Science Model Set C · 1 mark

The value of cc for which the equation 2x2−9x+c=02x^2 - 9x + c = 0 has equal roots is (a) 818\frac{81}{8} (b) 881\frac{8}{81} (c) 9 (d) 3

Answer

(a) 818\frac{81}{8}

Equal roots need D=b2−4ac=0D = b^2 - 4ac = 0: 81−8c=0⇒c=81881 - 8c = 0 \Rightarrow c = \frac{81}{8}.

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  • Asked 2 times
  • 2083 · Science Model Set B · 1 mark
  • 2083 · Management and Law Model Set E

The value of probability ranges from (a) 0 to 1 (b) −1-1 to 1 (c) 0.5 to 1 (d) 0.1 to 1

Answer

(a) 0 to 1

Probability of an event is 0≤P≤10\le P\le 1: 0 for an impossible event and 1 for a certain event.

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  • Asked 2 times
  • 2083 · Science Model Set C · 1 mark
  • 2083 · Management and Law Model Set C

Which of the following pairs is co-prime? (a) (3, 9) (b) (3, 7) (c) (6, 2) (d) (5, 25)

Answer

(b) (3, 7)

Co-prime numbers have HCF 1. HCF(3,9)=3\text{HCF}(3,9)=3, HCF(3,7)=1\text{HCF}(3,7)=1, HCF(6,2)=2\text{HCF}(6,2)=2, HCF(5,25)=5\text{HCF}(5,25)=5. Only (3,7)(3,7) is co-prime.

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  • Asked 2 times
  • 2083 · Science Model Set C · 1 mark
  • 2083 · Management and Law Model Set E

A sector of a circle has a central angle of 270∘270^\circ. If the circle has a radius of 4, what is the area of the sector? (a) 4π4\pi (b) 8π8\pi (c) 12π12\pi (d) 16π16\pi

Answer

(c) 12π12\pi

Area=270360πr2=34π(16)=12π.\text{Area} = \frac{270}{360}\pi r^2 = \frac{3}{4}\pi(16) = 12\pi.

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  • Asked 2 times
  • 2083 · Science Model Set C · 1 mark
  • 2080 · Science Model · 1 mark

If 1,x,y,z,161, x, y, z, 16 are in G.P., what is the value of x+y+zx + y + z? (a) 8 (b) 12 (c) 14 (d) 16

Answer

(c) 14

The fifth term is ar4=r4=16ar^4 = r^4 = 16, so r=2r = 2 (taking the real positive ratio). Then x=2x=2, y=4y=4, z=8z=8 and x+y+z=14x+y+z = 14.

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  • 2083 · Science Model Set D · 1 mark
  • 2080 · Science Model · 1 mark

What is the area of a circle whose circumference is π\pi? (a) π4\frac{\pi}{4} (b) π2\frac{\pi}{2} (c) π\pi (d) 2π2\pi

Answer

(a) π4\frac{\pi}{4}

2πr=π2\pi r = \pi gives r=12r = \frac{1}{2}, so the area is πr2=π4\pi r^2 = \frac{\pi}{4}.

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  • 2083 · Science Model Set E · 1 mark
  • 2079 · Science Model · 1 mark

If y2=4y^2 = 4 and x2y=18x^2 y = 18, x+yx + y could equal which of the following values? (a) −5-5 (b) −1-1 (c) 1 (d) 6

Answer

(b) −1-1

From y2=4y^2=4, y=±2y=\pm2.

If y=2y=2, then x2=9x^2=9, so x=±3x=\pm3 and x+y=5x+y=5 or −1-1.

If y=−2y=-2, then x2=−9x^2=-9, which has no real solution.

Of the options, only x+y=−1x+y=-1 is possible.

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  • 2083 · Science Model Set E · 1 mark
  • 2079 · Science Model · 1 mark

The average of a set of 18 consecutive integers is 22.5. What is the smallest integer in the set? (a) 14 (b) 16 (c) 12 (d) 28

Answer

(a) 14

Let the smallest integer be nn. The integers are n,n+1,…,n+17n, n+1, \dots, n+17, with average n+172=22.5⇒n=14n+\frac{17}{2}=22.5 \Rightarrow n=14

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  • 2083 · Science Model Set E · 1 mark
  • 2080 · Science Model · 1 mark

58\frac{5}{8} of 24 is equal to 157\frac{15}{7} of what number? (a) 7 (b) 8 (c) 15 (d) 7225\frac{7}{225}

Answer

(a) 7

58×24=15\frac{5}{8} \times 24 = 15. Let 157x=15\frac{15}{7} x = 15; then x=7x = 7.

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  • 2083 · Science Model Set E · 1 mark
  • 2080 · Science Model · 1 mark

If there are 24 white marbles and 36 blue marbles in a bag, what is the ratio of blue to white marbles? (a) 23\frac{2}{3} (b) 32\frac{3}{2} (c) 49\frac{4}{9} (d) 94\frac{9}{4}

Answer

(b) 32\frac{3}{2}

Blue : white =36:24=3:2= 36 : 24 = 3 : 2, i.e. 32\frac{3}{2}.

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  • 2083 · Science Model Set E · 1 mark
  • 2080 · Science Model · 1 mark

What is the probability of getting 53 Sundays in a non-leap year? (a) 17\frac{1}{7} (b) 27\frac{2}{7} (c) 37\frac{3}{7} (d) 47\frac{4}{7}

Answer

(a) 17\frac{1}{7}

A non-leap year has 365=52×7+1365=52\times7+1 days, so it has 52 full weeks plus 1 extra day. A 53rd Sunday occurs only if the extra day is a Sunday, which is 1 of 7 equally likely days. So P=17P=\frac17.

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  • 2083 · Management and Law Model Set A
  • 2079 · Management Model

In a group of 100 people, 70 can speak Nepali and 45 speak English, then the number of people who can speak English only. (a) 30 (b) 40 (c) 50 (d) 60

Answer

(a) 30

Assuming everyone speaks at least one language, n(N∪E)=100n(N \cup E) = 100.

n(N∩E)=70+45−100=15n(N \cap E) = 70 + 45 - 100 = 15

English only =45−15=30= 45 - 15 = 30.

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  • 2083 · Management and Law Model Set A
  • 2079 · Management Model

Two identical pies are cut into a total of 16 equal parts. If each part is then split equally among three people, what fraction of a pie does each person receive? (a) 148\frac{1}{48} (b) 124\frac{1}{24} (c) 116\frac{1}{16} (d) 316\frac{3}{16}

Answer

(b) 124\frac{1}{24}

Each part is 216=18\frac{2}{16} = \frac{1}{8} of a pie. Split among three people, each gets 18×13=124\frac{1}{8} \times \frac{1}{3} = \frac{1}{24}.

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  • 2083 · Management and Law Model Set A
  • 2079 · Management Model

Two pipes X and Y can fill a cistern in 24 minutes and 32 minutes respectively. If both the pipes are opened together, then after how much time should Y be closed so that the tank is full in 18 minutes? (a) 8 min. (b) 10 min. (c) 18 min. (d) 20 min.

Answer

(a) 8 min.

X works the whole 18 minutes and fills 1824=34\frac{18}{24} = \frac{3}{4} of the tank. The remaining 14\frac{1}{4} is filled by Y at 132\frac{1}{32} per minute:

t=1/41/32=8 mint = \frac{1/4}{1/32} = 8 \text{ min}

So Y is closed after 8 minutes.

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  • 2083 · Management and Law Model Set A
  • 2079 · Management Model

A trade man charges 20% above the cost price but allowed a discount of 10% for cash payment. What is the gain percent? (a) 8% (b) 10% (c) 20% (d) 30%

Answer

(a) 8%

Let the cost price be Rs 100. Marked price == Rs 120. Selling price =120×0.9== 120 \times 0.9 = Rs 108. Gain == Rs 8, which is 8%.

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  • 2083 · Management and Law Model Set A
  • 2079 · Management Model

If a%a\% of a=4a = 4 and a%a\% of b=40b = 40, find the positive value of bb? (a) 20 (b) 200 (c) 2000 (d) All of above

Answer

(b) 200

a100×a=4⇒a2=400⇒a=20\frac{a}{100} \times a = 4 \Rightarrow a^2 = 400 \Rightarrow a = 20

Then 20100×b=40\frac{20}{100} \times b = 40, so b=200b = 200.

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  • 2083 · Management and Law Model Set A
  • 2079 · Management Model

If a right circular cylinder's radius is halved and its height doubled, by what percent will the volume increase or decrease? (a) 50% decrease (b) No change (c) 25% increase (d) 50% increase

Answer

(a) 50% decrease

V=πr2hV = \pi r^2 h. New volume =π(r2)2(2h)=12πr2h= \pi \left(\frac{r}{2}\right)^2 (2h) = \frac{1}{2}\pi r^2 h, which is half the original, so it decreases by 50%.

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  • 2083 · Management and Law Model Set A
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For what value of xx is 63+83+103=x36^3 + 8^3 + 10^3 = x^3 true? (a) 12 (b) 24 (c) 9 (d) 7

Answer

(a) 12

63+83+103=216+512+1000=1728=1236^3 + 8^3 + 10^3 = 216 + 512 + 1000 = 1728 = 12^3

So x=12x = 12.

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  • 2083 · Management and Law Model Set A
  • 2079 · Management Model

The values of xx and yy in x+y=7x + y = 7 and 3x−2y=113x - 2y = 11 are (a) 2, 5 (b) 5, 2 (c) 3, 4 (d) 4, 3

Answer

(b) 5, 2

Multiply x+y=7x + y = 7 by 2: 2x+2y=142x + 2y = 14. Add to 3x−2y=113x - 2y = 11: 5x=255x = 25, so x=5x = 5 and y=2y = 2.

Check: 3(5)−2(2)=113(5) - 2(2) = 11.

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  • 2083 · Management and Law Model Set A
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If kk, 5 and 25 are in continued proportion, what is the value of k3−1k^3 - 1? (a) 0 (b) 1 (c) 5 (d) 125

Answer

(a) 0

In continued proportion, k5=525\frac{k}{5} = \frac{5}{25}, so 25k=2525k = 25 and k=1k = 1. Then k3−1=1−1=0k^3 - 1 = 1 - 1 = 0.

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  • 2083 · Management and Law Model Set A
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The following observations have been arranged in ascending order: 29, 32, 48, 52, xx, x+3x+3, 71, 75, 80, 92. If the median of the data is 61.5, what is the value of xx? (a) 54 (b) 60 (c) 62 (d) 56

Answer

(b) 60

There are 10 observations, so the median is the mean of the 5th and 6th terms:

x+(x+3)2=61.5⇒2x+3=123⇒x=60\frac{x + (x+3)}{2} = 61.5 \Rightarrow 2x + 3 = 123 \Rightarrow x = 60

The order 52 < 60 < 63 < 71 holds.

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  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

Which of the following inequalities is equivalent to ∣m+2∣<3|m + 2| < 3? (a) m<5m < 5 (b) m<1m < 1 (c) −5<m<5-5 < m < 5 (d) −5<m<1-5 < m < 1

Answer

(d) −5<m<1-5 < m < 1

∣m+2∣<3  ⟺  −3<m+2<3  ⟺  −5<m<1.|m+2|<3 \iff -3 < m+2 < 3 \iff -5 < m < 1.

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  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

A trade man allows a discount of 10%. What price should he mark on an article which costs Rs. 200 to gain 26%? (a) Rs. 200 (b) Rs. 280 (c) Rs. 300 (d) Rs. 350

Answer

(b) Rs. 280

Selling price for a 26% gain: 200×1.26=Rs. 252200 \times 1.26 = \text{Rs. } 252.

Let the marked price be MM. Then 0.9M=2520.9M = 252, so M=Rs. 280M = \text{Rs. } 280.

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  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

P(A)+P(Aˉ)P(A) + P(\bar{A}) is equal to (a) 0 (b) 1 (c) 0.75 (d) 0.5

Answer

(b) 1

An event and its complement are exhaustive and mutually exclusive, so P(A)+P(Aˉ)=1P(A)+P(\bar{A}) = 1.

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  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

How many 1-inch square tiles would it take to cover the floor of a closet that has dimensions 5 feet by 4 feet? (1 foot = 12 inches) (a) 20 (b) 240 (c) 1440 (d) 2880

Answer

(d) 2880

The closet is 5×12=605 \times 12 = 60 inches by 4×12=484 \times 12 = 48 inches, so the area is 60×48=288060 \times 48 = 2880 square inches, i.e. 2880 tiles.

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  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

xx and yy are positive numbers such that x+y+z<1x + y + z < 1 and xy=1xy = 1. Which could be the value of zz? (a) z<−1z < -1 (b) z>−1z > -1 (c) z=−1z = -1 (d) z>1z > 1

Answer

(a) z<−1z < -1

For positive x,yx, y with xy=1xy = 1, AM-GM gives x+y≥2xy=2x + y \ge 2\sqrt{xy} = 2. Since x+y+z<1x+y+z<1, we get z<1−(x+y)≤1−2=−1z < 1 - (x+y) \le 1 - 2 = -1, so z<−1z < -1.

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  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

On what sum will the simple interest for 3 years at 5% p.a. amount to Rs. 150? (a) Rs. 1000 (b) Rs. 1500 (c) Rs. 1050 (d) Rs. 1550

Answer

(a) Rs. 1000

P=100×SIT×R=100×1503×5=Rs. 1000.P = \frac{100 \times SI}{T \times R} = \frac{100 \times 150}{3 \times 5} = \text{Rs. } 1000.

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  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

If 34\frac{3}{4} of a number is 7 more than 16\frac{1}{6} of that number, what is 53\frac{5}{3} of that number? (a) 12 (b) 15 (c) 18 (d) 20

Answer

(d) 20

Let the number be xx. Then 34x−16x=7\frac{3}{4}x - \frac{1}{6}x = 7, so 9x−2x12=7x12=7\frac{9x-2x}{12} = \frac{7x}{12} = 7 and x=12x = 12.

53×12=20.\frac{5}{3} \times 12 = 20.

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  • 2083 · Management and Law Model Set B
  • 2080 · Management and Law Model Set I

The mean of 100 items is 49. If it was found that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively, the correct mean is (a) 48 (b) 82 (c) 50 (d) 80

Answer

(c) 50

Wrong total =100×49=4900= 100 \times 49 = 4900. Correct values add 60+70+80=21060+70+80 = 210 and remove 40+20+50=11040+20+50 = 110, so the correct total is 4900+210−110=50004900 + 210 - 110 = 5000.

Correct mean=5000100=50.\text{Correct mean} = \frac{5000}{100} = 50.

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  • 2083 · Science Model Set A · 1 mark

The compound interest compounded half yearly is calculated by (a) P(1+R100)TP\left(1+\frac{R}{100}\right)^T (b) P[(1+R100)T−1]P\left[\left(1+\frac{R}{100}\right)^T-1\right] (c) P(1+R200)2TP\left(1+\frac{R}{200}\right)^{2T} (d) P[(1+R200)2T−1]P\left[\left(1+\frac{R}{200}\right)^{2T}-1\right]

Similar questions: The compound interest compounded yearly is (2079 · Science Model)

Answer

(d) P[(1+R200)2T−1]P\left[\left(1+\frac{R}{200}\right)^{2T}-1\right]

Compounded half-yearly, the rate per period is R/2R/2 and the number of periods is 2T2T. The amount is P(1+R200)2TP\left(1+\frac{R}{200}\right)^{2T}, and the interest is the amount minus PP.

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  • 2083 · Science Model Set A · 1 mark

Which of the following statements is true? (a) −(−3)2<0-\sqrt{(-3)^2} < 0 (b) (−1)87=−1(-1)^{87} = -1 (c) 16=4\sqrt{16} = 4 (d) All of above

Similar questions: If (a)/(b) > (c)/(d), which of the following (2079 · Science Model)

Answer

(d) All of above

−(−3)2=−3<0-\sqrt{(-3)^2}=-3<0, (−1)87=−1(-1)^{87}=-1 (odd power) and 16=4\sqrt{16}=4 are all true.

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  • 2083 · Science Model Set C · 1 mark

If a=16ba = 16b and bb is a prime number greater than 2, how many positive distinct factors does aa have? (a) 5 (b) 10 (c) 15 (d) 20

Similar questions: If a = 16b and b is a prime number greater than (2080 · Science Model)

Answer

(b) 10

Since bb is a prime greater than 2, a=24⋅b1a = 2^4 \cdot b^1 with bb an odd prime. The number of positive factors is (4+1)(1+1)=10(4+1)(1+1) = 10.

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  • 2083 · Science Model Set D · 1 mark

If the sum of five consecutive integers is SS, what is the largest of those integers in terms of SS? (a) S−105\frac{S-10}{5} (b) S+105\frac{S+10}{5} (c) S−52\frac{S-5}{2} (d) S+54\frac{S+5}{4}

Similar questions: If the sum of five consecutive integers is S, (2080 · Science Model)

Answer

(b) S+105\frac{S+10}{5}

Let the integers be n−2,n−1,n,n+1,n+2n-2, n-1, n, n+1, n+2. Then S=5nS = 5n, so n=S5n = \frac{S}{5} and the largest is n+2=S5+2=S+105n + 2 = \frac{S}{5} + 2 = \frac{S+10}{5}.

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  • 2083 · Science Model Set E · 1 mark

In a continuous series, N=30N = 30 and the cumulative frequency of the class 20-30 is 15. Find the median class. (a) 10-20 (b) 20-30 (c) 30-40 (d) 20.5-30.5

Similar questions: In a continuous series, sum f = 100, the (2079 · Science Model)

Answer

(b) 20-30

The median class is the class whose cumulative frequency first reaches N2=15\frac{N}{2} = 15. The class 20-30 has cumulative frequency 15, so it is the median class.

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  • 2080 · Science Model · 1 mark

If the sum of five consecutive integers is SS, what is the largest of those integers in terms of SS? (a) S−105\frac{S-10}{5} (b) S+44\frac{S+4}{4} (c) S+105\frac{S+10}{5} (d) S−52\frac{S-5}{2}

Similar questions: If the sum of five consecutive integers is S, (2083 · Science Model Set D)

Answer

(c) S+105\dfrac{S+10}{5}

Let the integers be n,n+1,…,n+4n, n+1, \dots, n+4. Then S=5n+10S = 5n + 10, so n=S−105n = \dfrac{S-10}{5}.

Largest =n+4=S−105+4=S+105= n + 4 = \dfrac{S-10}{5} + 4 = \dfrac{S+10}{5}.

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  • 2080 · Science Model · 1 mark

If a=16ba = 16b and bb is a prime number greater than 2, how many positive distinct factors does aa have? (a) 5 (b) 20 (c) 15 (d) 2

Similar questions: If a = 16b and b is a prime number greater than (2083 · Science Model Set C)

Answer

The number of positive factors is 1010, which is not among the options.

a=16b=24⋅b1a = 16b = 2^4 \cdot b^1 with bb an odd prime, so the number of factors is (4+1)(1+1)=10(4+1)(1+1) = 10. (Option (c), 1515, would correspond to a=16b2=24⋅b2a = 16b^2 = 2^4 \cdot b^2, giving 5×3=155 \times 3 = 15.)

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  • 2079 · Science Model · 1 mark

The compound interest compounded yearly is calculated by (a) P[(1+R100)T−1]P\left[\left(1+\frac{R}{100}\right)^T - 1\right] (b) P[(1+R200)2T−1]P\left[\left(1+\frac{R}{200}\right)^{2T} - 1\right] (c) P[(1+R100)2T−1]P\left[\left(1+\frac{R}{100}\right)^{2T} - 1\right] (d) P[(1+R200)T−1]P\left[\left(1+\frac{R}{200}\right)^{T} - 1\right]

Similar questions: The compound interest compounded half yearly is (2083 · Science Model Set A)

Answer

(a) P[(1+R100)T−1]P\left[\left(1+\frac{R}{100}\right)^T - 1\right]

The amount compounded yearly is A=P(1+R100)TA=P\left(1+\frac{R}{100}\right)^T, so the compound interest is A−P=P[(1+R100)T−1]A-P=P\left[\left(1+\frac{R}{100}\right)^T-1\right]. Options (b) and (d) are for half-yearly compounding.

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  • 2079 · Science Model · 1 mark

If ab>cd\frac{a}{b} > \frac{c}{d}, which of the following statements must be true? (a) ab−cd>0\frac{a}{b} - \frac{c}{d} > 0 (b) ad<bcad < bc (c) ad>bcad > bc (d) All of above

Similar questions: Which of the following statements is true (2083 · Science Model Set A)

Answer

(a) ab−cd>0\frac{a}{b}-\frac{c}{d}>0

Subtracting cd\frac{c}{d} from both sides of ab>cd\frac{a}{b}>\frac{c}{d} always gives ab−cd>0\frac{a}{b}-\frac{c}{d}>0. The statements ad<bcad<bc and ad>bcad>bc depend on the signs of bb and dd (multiplying by bdbd flips the inequality when bd<0bd<0), so neither must be true.

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  • 2079 · Science Model · 1 mark

In a continuous series, ∑f=100\sum f = 100, the cumulative frequency of the class 15-20 is 46 and the cumulative frequency of the class 20-25 is 52. Find the median class of the data. (a) 15-20 (b) 10-15 (c) 20-25 (d) 25-30

Similar questions: In a continuous series, N = 30 and the cumulative (2083 · Science Model Set E)

Answer

(c) 20-25

N2=1002=50\frac{N}{2}=\frac{100}{2}=50. The first class whose cumulative frequency is at least 50 is the median class. The class 15-20 has cumulative frequency 46, which is less than 50, and the class 20-25 has 52, which is at least 50. So the median class is 20-25.

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  • 2083 · Science Model Set A · 1 mark

What is the average (arithmetic mean) of the measures of the five angles in a pentagon? (a) 36∘36^\circ (b) 72∘72^\circ (c) 92∘92^\circ (d) 100∘100^\circ

Answer

(b) 72∘72^\circ

The interior angles of a pentagon sum to (5−2)×180∘=540∘(5-2)\times180^\circ=540^\circ, giving a mean of 108∘108^\circ, which is not among the options. The mean of the five exterior angles, which sum to 360∘360^\circ, is 360∘5=72∘,\frac{360^\circ}{5}=72^\circ, so 72∘72^\circ is the intended choice.

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  • 2083 · Science Model Set A · 1 mark

Find the LCM of a2ba^2b and a2b2a^2b^2. (a) abab (b) a2ba^2b (c) a2b2a^2b^2 (d) a4b3a^4b^3

Answer

(c) a2b2a^2b^2

Take the highest power of each factor: a2a^2 and b2b^2, so the LCM is a2b2a^2b^2.

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  • 2083 · Science Model Set A · 1 mark

What is the square root of 9+459 + 4\sqrt{5}? (a) 3+253 + 2\sqrt{5} (b) 2+52 + \sqrt{5} (c) 3+453 + 4\sqrt{5} (d) None of above

Answer

(b) 2+52+\sqrt{5}

(2+5)2=4+5+45=9+45.(2+\sqrt5)^2=4+5+4\sqrt5=9+4\sqrt5.

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  • 2083 · Science Model Set A · 1 mark

If a+b=3(c+d)a + b = 3(c + d), which of the following is the average (arithmetic mean) of a,b,ca, b, c and dd? (a) c+d4\frac{c+d}{4} (b) 3(c+d)4\frac{3(c+d)}{4} (c) c+d2\frac{c+d}{2} (d) c+dc + d

Answer

(d) c+dc+d

a+b+c+d=3(c+d)+(c+d)=4(c+d),a+b+c+d=3(c+d)+(c+d)=4(c+d), so the mean is 4(c+d)4=c+d\dfrac{4(c+d)}{4}=c+d.

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  • 2083 · Science Model Set A · 1 mark

If the mean of aa and 1a\frac{1}{a} is MM, then what is the mean of a3a^3 and 1a3\frac{1}{a^3}? (a) M(M2−3)3\frac{M(M^2-3)}{3} (b) M3M^3 (c) M3−3M^3 - 3 (d) M(4M2−3)M(4M^2 - 3)

Answer

(d) M(4M2−3)M(4M^2-3)

Here a+1a=2Ma+\frac1a=2M. Then a3+1a3=(a+1a)3−3(a+1a)=8M3−6M.a^3+\frac1{a^3}=\left(a+\frac1a\right)^3-3\left(a+\frac1a\right)=8M^3-6M. The mean is 8M3−6M2=4M3−3M=M(4M2−3)\dfrac{8M^3-6M}{2}=4M^3-3M=M(4M^2-3).

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  • 2083 · Science Model Set A · 1 mark

Which of the following sets are equivalent? (a) ϕ\phi and {ϕ}\{\phi\} (b) {0}\{0\} and {ϕ}\{\phi\} (c) ϕ\phi and {0}\{0\} (d) None

Answer

(b) {0}\{0\} and {ϕ}\{\phi\}

Equivalent sets have the same number of elements. {0}\{0\} and {ϕ}\{\phi\} each have exactly one element, whereas ϕ\phi has none.

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  • 2083 · Science Model Set A · 1 mark

Find the compound interest on Rs 32000 at 20% p.a. for 1 year compounded half yearly. (a) Rs. 6320 (b) Rs. 6720 (c) Rs. 6400 (d) Rs. 6520

Answer

(b) Rs. 6720

A=32000(1+20200)2=32000×1.21=Rs 38720A=32000\left(1+\frac{20}{200}\right)^{2}=32000\times1.21=\text{Rs }38720 CI=38720−32000=Rs 6720.CI=38720-32000=\text{Rs }6720.

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  • 2083 · Science Model Set A · 1 mark

What single discount is equivalent to two successive discounts of 10% and 15%? (a) 23% (b) 23.5% (c) 24% (d) 24.5%

Answer

(b) 23.5%

1−0.90×0.85=1−0.765=0.235,1-0.90\times0.85=1-0.765=0.235, which is a discount of 23.5%.

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  • 2083 · Science Model Set A · 1 mark

A sector of a circle has a radius of 8 and an area of 8π8\pi. What is the arc length of the sector? (a) π\pi (b) 2π2\pi (c) 4π4\pi (d) 6π6\pi

Answer

(b) 2π2\pi

Area =12rℓ=\tfrac12 r\ell, so 8π=12(8)ℓ8\pi=\tfrac12(8)\ell, which gives ℓ=2π\ell=2\pi.

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  • 2083 · Science Model Set A · 1 mark

The HCF of x3+1x3x^3 + \frac{1}{x^3} and x4+1+1x4x^4 + 1 + \frac{1}{x^4} is (a) 11 (b) x+1xx + \frac{1}{x} (c) x2+1+1x2x^2 + 1 + \frac{1}{x^2} (d) x2−1+1x2x^2 - 1 + \frac{1}{x^2}

Answer

(d) x2−1+1x2x^2-1+\frac{1}{x^2}

x3+1x3=(x+1x)(x2−1+1x2)x^3+\frac1{x^3}=\left(x+\frac1x\right)\left(x^2-1+\frac1{x^2}\right) x4+1+1x4=(x2+1x2)2−1=(x2−1+1x2)(x2+1+1x2)x^4+1+\frac1{x^4}=\left(x^2+\frac1{x^2}\right)^2-1=\left(x^2-1+\frac1{x^2}\right)\left(x^2+1+\frac1{x^2}\right) The common factor is x2−1+1x2x^2-1+\frac1{x^2}.

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  • 2083 · Science Model Set A · 1 mark

For what value of xx is 43x−6\frac{4}{3x-6} undefined? (a) 0 (b) 1 (c) 2 (d) 3

Answer

(c) 2

The fraction is undefined when the denominator is zero: 3x−6=03x-6=0, so x=2x=2.

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  • 2083 · Science Model Set A · 1 mark

0.6‾+0.7‾+0.8‾=?0.\overline{6} + 0.\overline{7} + 0.\overline{8} = ? (a) 2.3‾2.\overline{3} (b) 2.1‾2.\overline{1} (c) 2.2‾2.\overline{2} (d) None

Answer

(a) 2.3‾2.\overline{3}

0.6‾=69,0.7‾=79,0.8‾=890.\overline6=\frac69,\quad 0.\overline7=\frac79,\quad 0.\overline8=\frac89 Sum=219=73=2.3‾.\text{Sum}=\frac{21}{9}=\frac73=2.\overline3.

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  • 2083 · Science Model Set A · 1 mark

Solve: 2x−14=−1\sqrt[4]{2x-1} = -1 (a) 1 (b) 2 (c) 3 (d) No solution

Answer

(d) No solution

The principal fourth root of a real number is never negative, so 2x−14=−1\sqrt[4]{2x-1}=-1 has no real solution.

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  • 2083 · Science Model Set A · 1 mark

A has 6 more marbles than B. If B gives 2 marbles to C and then B buys 5 new marbles, how many more marbles does A have than B? (a) 3 (b) 4 (c) 5 (d) 6

Answer

(a) 3

Let B have bb marbles, so A has b+6b+6. After giving 2 to C and buying 5, B has b−2+5=b+3b-2+5=b+3.

(b+6)−(b+3)=3(b+6)-(b+3)=3

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  • 2083 · Science Model Set A · 1 mark

The quadratic equation whose roots are (3+5)(3+\sqrt{5}) and (3−5)(3-\sqrt{5}) is (a) x2−6x+2=0x^2 - 6x + 2 = 0 (b) x2−4x+6=0x^2 - 4x + 6 = 0 (c) x2−6x+4=0x^2 - 6x + 4 = 0 (d) x2+6x+4=0x^2 + 6x + 4 = 0

Answer

(c) x2−6x+4=0x^2 - 6x + 4 = 0

Sum of roots =(3+5)+(3−5)=6=(3+\sqrt5)+(3-\sqrt5)=6 and product =9−5=4=9-5=4, so the equation is x2−6x+4=0x^2-6x+4=0.

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  • 2083 · Science Model Set A · 1 mark

If x=6+35x = 6 + \sqrt{35}, then the value of x2+1x2x^2 + \frac{1}{x^2} is (a) 241 (b) 284 (c) 142 (d) 140

Answer

(c) 142

Since (6+35)(6−35)=1(6+\sqrt{35})(6-\sqrt{35})=1, we have 1x=6−35\frac1x=6-\sqrt{35} and x+1x=12x+\frac1x=12. Then x2+1x2=(x+1x)2−2=144−2=142.x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2=144-2=142.

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  • 2083 · Science Model Set A · 1 mark

If xy>0xy > 0 and yz<0yz < 0, which of the following is true? (a) xz<0xz < 0 (b) xz>0xz > 0 (c) xz=0xz = 0 (d) None of above

Answer

(a) xz<0xz < 0

xy>0xy>0 means xx and yy have the same sign; yz<0yz<0 means zz has the opposite sign to yy. So zz is opposite to xx, giving xz<0xz<0.

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  • 2083 · Science Model Set A · 1 mark

For what value of xx is the equation abx−2=bax−2ab^{x-2} = ba^{x-2} true? (a) −3-3 (b) −2-2 (c) 22 (d) 33

Answer

(d) 33

For x=3x=3: ab1=ba1ab^{1}=ba^{1}, i.e. ab=baab=ba, which is always true. (For other options the two sides are not equal in general.)

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  • 2083 · Science Model Set A · 1 mark

If x1a=y15x^{\frac{1}{a}} = y^{\frac{1}{5}} and xy=1xy = 1, what is the value of aa? (a) −2-2 (b) −3-3 (c) −4-4 (d) −5-5

Answer

(d) −5-5

Since xy=1xy=1, y=x−1y=x^{-1}, so y1/5=x−1/5y^{1/5}=x^{-1/5}. Then x1/a=x−1/5x^{1/a}=x^{-1/5} gives 1a=−15\frac1a=-\frac15, so a=−5a=-5.

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  • 2083 · Science Model Set A · 1 mark

There are two numbers. If we add one to each number, their ratio becomes 2:3. If 1 is decreased from each number, their ratio becomes 1:2. The numbers are (a) 3, 1 (b) 1, 3 (c) 1, 5 (d) none of above

Answer

(d) none of above

Let the numbers be pp and qq. Then p+1q+1=23\frac{p+1}{q+1}=\frac23 gives 3p−2q=−13p-2q=-1, and p−1q−1=12\frac{p-1}{q-1}=\frac12 gives q=2p−1q=2p-1. Substituting: 3p−4p+2=−13p-4p+2=-1, so p=3p=3, q=5q=5. The numbers are 3 and 5, which are not among options (a)-(c).

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  • 2083 · Science Model Set A · 1 mark

A perfect cube has surface area 96 cm296\ \text{cm}^2. What is its volume (in cm3\text{cm}^3)? (a) 64 (b) 80 (c) 48 (d) 96

Answer

(a) 64

6a2=96⇒a2=16⇒a=46a^2=96\Rightarrow a^2=16\Rightarrow a=4 cm, so volume =43=64 cm3=4^3=64\ \text{cm}^3.

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  • 2083 · Science Model Set A · 1 mark

In triangle ABC, ∠B=30∘\angle B = 30^\circ and BC=12BC = 12 cm. If the area of △ABC\triangle ABC is 27 cm227\ \text{cm}^2, find the length of AB. (a) 12 cm (b) 6 cm (c) 9 cm (d) None of above

Answer

(c) 9 cm

Area=12⋅AB⋅BCsin⁡30∘=12⋅AB⋅12⋅12=3 AB=27\text{Area}=\tfrac12\cdot AB\cdot BC\sin 30^\circ=\tfrac12\cdot AB\cdot 12\cdot\tfrac12=3\,AB=27 so AB=9AB=9 cm.

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  • 2083 · Science Model Set A · 1 mark

The three distinct points A, B and C are said to be collinear if (a) AC=AB+BCAC = AB + BC (b) slope of ABAB = slope of BCBC (c) Area of △ABC=0\triangle ABC = 0 (d) All of above

Answer

(d) All of above

Three distinct points are collinear if AC=AB+BCAC=AB+BC (B lies on segment AC), if slope of ABAB equals slope of BCBC, or if the area of △ABC\triangle ABC is zero.

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  • 2083 · Science Model Set A · 1 mark

What is the maximum value of y=sin⁡xy = \sin x? (a) −1-1 (b) 11 (c) 00 (d) ∞\infty

Answer

(b) 11

The range of sin⁡x\sin x is [−1,1][-1,1], so its maximum value is 11 (at x=90∘x=90^\circ).

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  • 2083 · Science Model Set A · 1 mark

A right triangle is inscribed in a circle with an area of 16π cm216\pi\ \text{cm}^2. What is the hypotenuse of the triangle in cm? (a) 4π4\pi (b) 44 (c) 8π8\pi (d) 88

Answer

(d) 88

πr2=16π⇒r=4\pi r^2=16\pi\Rightarrow r=4 cm. The hypotenuse of a right triangle inscribed in a circle is a diameter, so it equals 2r=82r=8 cm.

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  • 2083 · Science Model Set A · 1 mark

If f(x)f(x) is divided by (b−ax)(b - ax), then the remainder is (a) f(ba)f\left(\frac{b}{a}\right) (b) f(ab)f\left(\frac{a}{b}\right) (c) f(−ba)f\left(-\frac{b}{a}\right) (d) f(−ab)f\left(-\frac{a}{b}\right)

Answer

(a) f(ba)f\left(\frac{b}{a}\right)

By the remainder theorem, the remainder on division by (b−ax)(b-ax) is f(x)f(x) at the zero of the divisor: b−ax=0⇒x=bab-ax=0\Rightarrow x=\frac ba.

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  • 2083 · Science Model Set A · 1 mark

If the sum of the first nn positive integers is 15\frac{1}{5} of the sum of the squares of the first nn positive integers, then nn is (a) 23 (b) 14 (c) 7 (d) 9

Answer

(c) 7

n(n+1)2=15⋅n(n+1)(2n+1)6⇒12=2n+130⇒2n+1=15⇒n=7.\frac{n(n+1)}{2}=\frac15\cdot\frac{n(n+1)(2n+1)}{6}\Rightarrow \frac12=\frac{2n+1}{30}\Rightarrow 2n+1=15\Rightarrow n=7.

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  • 2083 · Science Model Set A · 1 mark

Which of the following parabolas is the narrowest? (a) y=x2y = x^2 (b) y=x22y = \frac{x^2}{2} (c) y=2x2y = 2x^2 (d) y=x24y = \frac{x^2}{4}

Answer

(c) y=2x2y = 2x^2

For y=kx2y=kx^2, a larger kk gives a narrower parabola. Here k=2k=2 is the largest among 1,12,2,141,\frac12,2,\frac14.

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  • 2083 · Science Model Set A · 1 mark

What does {x:1≤x<5}\{x : 1 \le x < 5\} represent? (a) (1,5)(1, 5) (b) [1,5][1, 5] (c) [1,5)[1, 5) (d) (1,5](1, 5]

Answer

(c) [1,5)[1, 5)

1≤x1\le x includes 1 (closed end) and x<5x<5 excludes 5 (open end).

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  • 2083 · Science Model Set A · 1 mark

If A=(140250360)A = \begin{pmatrix} 1 & 4 & 0 \\ 2 & 5 & 0 \\ 3 & 6 & 0 \end{pmatrix}, then the determinant of AA is (a) −1-1 (b) 00 (c) 11 (d) None of above

Answer

(b) 00

The third column of AA is entirely zero, so ∣A∣=0|A|=0.

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  • 2083 · Science Model Set A · 1 mark

Two circles having centres OO and O′O' touch externally if (a) OO′OO' = difference of radii (b) OO′OO' = sum of radii (c) OO′OO' > sum of radii (d) OO′OO' < sum of radii

Answer

(b) OO′OO' = sum of radii

Circles touch externally when the distance between centres equals the sum of their radii.

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  • 2083 · Science Model Set A · 1 mark

The minute hand of a clock is 28 cm long. How far does the tip of the minute hand move during 15 minutes? (a) 28 cm (b) 14 cm (c) 44 cm (d) 22 cm

Answer

(c) 44 cm

In 15 minutes the hand turns through 90∘90^\circ, a quarter circle, so the distance is 14⋅2πr=14⋅2⋅227⋅28=44 cm.\frac14\cdot 2\pi r=\frac14\cdot 2\cdot\frac{22}{7}\cdot 28=44\ \text{cm}.

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  • 2083 · Science Model Set A · 1 mark

If the sum of the position vectors of the vertices of a triangle is 6i⃗+9j⃗6\vec{i} + 9\vec{j}, then find the position vector of its centroid. (a) 6i⃗+9j⃗6\vec{i} + 9\vec{j} (b) 3i⃗+2j⃗3\vec{i} + 2\vec{j} (c) 2i⃗+3j⃗2\vec{i} + 3\vec{j} (d) None of above

Answer

(c) 2i⃗+3j⃗2\vec{i} + 3\vec{j}

Centroid =13(sum of position vectors)=13(6i⃗+9j⃗)=2i⃗+3j⃗=\frac13(\text{sum of position vectors})=\frac13(6\vec i+9\vec j)=2\vec i+3\vec j.

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  • 2083 · Science Model Set A · 1 mark

What is the combined transformation of the reflection on x=−2x = -2 followed by the reflection on y=3y = 3? (a) Translation with (102)\binom{10}{2} (b) R[(0,0),180∘]R[(0,0), 180^\circ] (c) R[(−2,3),180∘]R[(-2,3), 180^\circ] (d) Reflection on y=−xy = -x

Answer

(c) R[(−2,3),180∘]R[(-2,3), 180^\circ]

The lines x=−2x=-2 and y=3y=3 are perpendicular and meet at (−2,3)(-2,3). Reflections in two perpendicular lines combine to a half-turn about their point of intersection.

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  • 2083 · Science Model Set B · 1 mark

If mean =[3 median−mode]⋅y= [3\,\text{median} - \text{mode}] \cdot y, then the value of yy is (a) 2 (b) 14\frac{1}{4} (c) 12\frac{1}{2} (d) 4

Answer

(c) 12\frac{1}{2}

The empirical relation is mode=3 median−2 mean\text{mode} = 3\,\text{median} - 2\,\text{mean}, so mean=3 median−mode2=(3 median−mode)⋅12,\text{mean} = \frac{3\,\text{median} - \text{mode}}{2} = (3\,\text{median} - \text{mode})\cdot\frac{1}{2}, giving y=12y = \frac{1}{2}.

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  • 2083 · Science Model Set B · 1 mark

Subi's average (arithmetic mean) on 4 tests is 80. What does she need on her fifth test to raise her average to 84? (a) 82 (b) 84 (c) 92 (d) 100

Answer

(d) 100

The total of 4 tests is 4×80=3204 \times 80 = 320. For an average of 84 over 5 tests the total must be 5×84=4205 \times 84 = 420, so the fifth score is 420−320=100420 - 320 = 100.

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  • 2083 · Science Model Set B · 1 mark

The number of all possible subsets of {{a},{b,c},{a,b,c}}\{\{a\}, \{b, c\}, \{a, b, c\}\} is (a) 4 (b) 6 (c) 8 (d) 64

Answer

(c) 8

The set {{a},{b,c},{a,b,c}}\{\{a\},\{b,c\},\{a,b,c\}\} has 3 distinct elements, so it has 23=82^3 = 8 subsets.

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  • 2083 · Science Model Set B · 1 mark

If Rs 1800 = USD 25 and Rs 1584 = £12, how many pounds (£) can be exchanged from USD 33? (a) £18 (b) £25 (c) £33 (d) None

Answer

(a) £18

Rs 1800 = USD 25, so USD 1 = Rs 72 and USD 33 = Rs 33×72=33 \times 72 = Rs 2376. Rs 1584 = £12, so £1 = Rs 132. 2376132=18⇒£18.\frac{2376}{132} = 18 \Rightarrow £18.

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  • 2083 · Science Model Set B · 1 mark

A sum of money in simple interest doubles itself in 6 years. In how many years will it become 5 times? (a) 12 years (b) 24 years (c) 10 years (d) 15 years

Answer

(b) 24 years

The sum doubles in 6 years, so the interest in 6 years equals the principal PP, which is P6\frac{P}{6} per year. To become 5 times, the interest must be 4P4P, which takes 4P÷P6=244P \div \frac{P}{6} = 24 years.

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  • 2083 · Science Model Set B · 1 mark

At what rate of compound interest will a sum of money become 12564\frac{125}{64} times of itself in 3 years? (a) 10% (b) 15% (c) 20% (d) 25%

Answer

(d) 25%

(1+R100)3=12564=(54)3⇒1+R100=54\left(1+\frac{R}{100}\right)^3 = \frac{125}{64} = \left(\frac{5}{4}\right)^3 \Rightarrow 1+\frac{R}{100} = \frac{5}{4} so R=25%R = 25\%.

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  • 2083 · Science Model Set B · 1 mark

The amount of money returned back to the tourists while leaving the country is called (a) VAT amount (b) Discount amount (c) Profit amount (d) Loss amount

Answer

(a) VAT amount

Tourists are refunded the VAT they paid on goods when leaving the country.

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  • 2083 · Science Model Set B · 1 mark

What is the length of a diagonal of a cube whose edges are 1? (a) 1 (b) 2 (c) 3\sqrt{3} (d) 2\sqrt{2}

Answer

(c) 3\sqrt{3}

The space diagonal of a cube of edge aa is a2+a2+a2=a3\sqrt{a^2+a^2+a^2} = a\sqrt{3}, which is 3\sqrt{3} for a=1a=1.

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  • 2083 · Science Model Set B · 1 mark

What is the volume of a right circular cylinder with a radius of 2 and a height of 4? (a) 8π8\pi (b) 12π12\pi (c) 16π16\pi (d) 32π32\pi

Answer

(c) 16π16\pi

V=πr2h=π×22×4=16π.V = \pi r^2 h = \pi \times 2^2 \times 4 = 16\pi.

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  • 2083 · Science Model Set B · 1 mark

The product of two expressions is x(y+z)2(y−z)x(y+z)^2(y-z) and their HCF is (y+z)(y+z). What is their LCM? (a) (y2−z2)(y^2 - z^2) (b) x(y2−z2)x(y^2 - z^2) (c) (y+z)3(y+z)^3 (d) None

Answer

(b) x(y2−z2)x(y^2 - z^2)

Since HCF×LCM=\text{HCF} \times \text{LCM} = product of the expressions, LCM=x(y+z)2(y−z)y+z=x(y+z)(y−z)=x(y2−z2).\text{LCM} = \frac{x(y+z)^2(y-z)}{y+z} = x(y+z)(y-z) = x(y^2-z^2).

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  • 2083 · Science Model Set B · 1 mark

0.47‾=?0.\overline{47} = ? (a) 4799\frac{47}{99} (b) 4790\frac{47}{90} (c) 4690\frac{46}{90} (d) 47100\frac{47}{100}

Answer

(a) 4799\frac{47}{99}

Let x=0.47‾x = 0.\overline{47}. Then 100x=47.47‾100x = 47.\overline{47}, so 99x=4799x = 47 and x=4799x = \frac{47}{99}.

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  • 2083 · Science Model Set B · 1 mark

What value of yy makes the expression x2y2x−y\frac{x^2y}{2x-y} undefined? (a) 0 (b) 1 (c) xx (d) 2x2x

Answer

(d) 2x2x

The expression is undefined when the denominator is zero: 2x−y=0⇒y=2x2x - y = 0 \Rightarrow y = 2x.

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  • 2083 · Science Model Set B · 1 mark

The rationalizing factor of 165\sqrt[5]{16} is (a) 2\sqrt{2} (b) 165\sqrt[5]{16} (c) 25\sqrt[5]{2} (d) 16\sqrt{16}

Answer

(c) 25\sqrt[5]{2}

165=24/5\sqrt[5]{16} = 2^{4/5}. Multiplying by 21/5=252^{1/5} = \sqrt[5]{2} gives 24/5+1/5=22^{4/5+1/5} = 2, which is rational.

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  • 2083 · Science Model Set B · 1 mark

If xy=64x^y = 64 and xx and yy are positive integers, which of the following could be the value of x+yx + y? (a) 2 (b) 6 (c) 7 (d) 12

Answer

(c) 7

The positive integer solutions of xy=64x^y = 64 are (2,6)(2,6), (4,3)(4,3), (8,2)(8,2) and (64,1)(64,1), giving x+y=8,7,10,65x+y = 8, 7, 10, 65. Only 7 is among the options, from 43=644^3 = 64.

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  • 2083 · Science Model Set B · 1 mark

If xx and yy are integers such that x>y>1x > y > 1, which of the following cannot be a multiple of either xx or yy? (a) x−1x - 1 (b) y+1y + 1 (c) y−1y - 1 (d) x+yx + y

Answer

(c) y−1y - 1

Since y>1y>1, we have 0<y−1<y<x0 < y-1 < y < x. A positive number smaller than yy cannot be a multiple of yy, and one smaller than xx cannot be a multiple of xx. The others can be multiples: x−1=2x-1=2 for (x,y)=(3,2)(x,y)=(3,2), y+1=3y+1=3 for (3,2)(3,2), x+y=6x+y=6 for (4,2)(4,2).

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  • 2083 · Science Model Set B · 1 mark

If x+1x=15x + \frac{1}{x} = 15, the value of x2+1x2x^2 + \frac{1}{x^2} is (a) 227 (b) 225 (c) 240 (d) 223

Answer

(d) 223

Square both sides: (x+1x)2=x2+2+1x2=225\left(x+\frac1x\right)^2 = x^2+2+\frac1{x^2} = 225, so x2+1x2=225−2=223x^2+\frac{1}{x^2}=225-2=223.

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  • 2083 · Science Model Set B · 1 mark

Which of the following is equal to x32x^{\frac{3}{2}}? (a) x2xx^2\sqrt{x} (b) xxx\sqrt{x} (c) x23\sqrt[3]{x^2} (d) (x3)2(x^3)^2

Answer

(b) xxx\sqrt{x}

x32=x1+12=x⋅x12=xxx^{\frac32}=x^{1+\frac12}=x\cdot x^{\frac12}=x\sqrt{x}.

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  • 2083 · Science Model Set B · 1 mark

If x=213+223x = 2^{\frac{1}{3}} + 2^{\frac{2}{3}}, what is the value of x3−6xx^3 - 6x? (a) 2 (b) 4 (c) 6 (d) 18

Answer

(c) 6

Let a=213a=2^{\frac13}, so a3=2a^3=2 and x=a+a2x=a+a^2. Then x3=a3+a6+3a3(a+a2)=2+4+6x=6+6x.x^3=a^3+a^6+3a^3(a+a^2)=2+4+6x=6+6x. Hence x3−6x=6x^3-6x=6.

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  • 2083 · Science Model Set B · 1 mark

If aa and bb are the lengths of the legs of a right-angled triangle whose hypotenuse is 10 and whose area is 20, what is the value of (a+b)2(a+b)^2? (a) 100 (b) 120 (c) 140 (d) 180

Answer

(d) 180

a2+b2=102=100a^2+b^2=10^2=100 and 12ab=20⇒ab=40\frac12ab=20\Rightarrow ab=40. So (a+b)2=a2+b2+2ab=100+80=180(a+b)^2=a^2+b^2+2ab=100+80=180.

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  • 2083 · Science Model Set B · 1 mark

In a right-angled triangle with sides 5 and 7 meeting at the right angle and hypotenuse 74\sqrt{74}, what is the value of xx, the angle between the side 7 and the hypotenuse? (a) 30∘30^\circ (b) 60∘60^\circ (c) 45∘45^\circ (d) 90∘90^\circ

Answer

(a) 30∘30^\circ (the nearest option)

The side adjacent to xx is 7 and the side opposite is 5, so tan⁡x=57\tan x=\frac57 and x=tan⁡−157≈35.5∘x=\tan^{-1}\frac57\approx 35.5^\circ. This is not exactly any of the options; the nearest is 30∘30^\circ.

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  • 2083 · Science Model Set B · 1 mark

A circle is inscribed in a square. The area of the circle is 25π25\pi. The area of the square is (a) 625 (b) 100 (c) 50 (d) 2500

Answer

(b) 100

πr2=25π⇒r=5\pi r^2=25\pi\Rightarrow r=5. The side of the square equals the diameter =10=10, so the area is 102=10010^2=100.

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  • 2083 · Science Model Set B · 1 mark

What is the area of a regular hexagon with side length 2? (a) 232\sqrt{3} (b) 262\sqrt{6} (c) 626\sqrt{2} (d) 636\sqrt{3}

Answer

(d) 636\sqrt{3}

A regular hexagon is made of 6 equilateral triangles: A=6⋅34(2)2=63.A=6\cdot\frac{\sqrt3}{4}(2)^2=6\sqrt3.

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  • 2083 · Science Model Set B · 1 mark

A circle has an area of 4π4\pi. If the radius was doubled, the new area of the circle would be how many times the original area? (a) 2 (b) 3 (c) 4 (d) 5

Answer

(c) 4

Area ∝r2\propto r^2. Doubling the radius gives π(2r)2=4πr2\pi(2r)^2=4\pi r^2, which is 4 times the original (new area =16π=16\pi, original =4π=4\pi).

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  • 2083 · Science Model Set B · 1 mark

The probability of getting a bad egg in a batch of 400 is 0.035. How many bad eggs are there likely to be in a batch? (a) 10 (b) 14 (c) 28 (d) 40

Answer

(b) 14

Expected number of bad eggs =400×0.035=14=400\times0.035=14.

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  • 2083 · Science Model Set B · 1 mark

What is the pre-image of 2 in the function f(x)=x3f(x) = \sqrt[3]{x}? (a) 2 (b) 8 (c) 6 (d) 1

Answer

(b) 8

We need xx with f(x)=x3=2f(x)=\sqrt[3]{x}=2. Cubing, x=23=8x=2^3=8.

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  • 2083 · Science Model Set B · 1 mark

If polynomial q(x)q(x) is divided by (ax+b)(ax + b), what will be its remainder? (a) q(ab)q\left(\frac{a}{b}\right) (b) q(ba)q\left(\frac{b}{a}\right) (c) q(−ab)q\left(-\frac{a}{b}\right) (d) q(−ba)q\left(-\frac{b}{a}\right)

Answer

(d) q(−ba)q\left(-\frac{b}{a}\right)

By the remainder theorem, the remainder on dividing by (ax+b)(ax+b) is the value of q(x)q(x) at the root of ax+b=0ax+b=0, i.e. x=−bax=-\frac ba.

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  • 2083 · Science Model Set B · 1 mark

The sum of two digit numbers which are divisible by 5 is (a) 950 (b) 1050 (c) 945 (d) 1200

Answer

(c) 945

The two-digit multiples of 5 are 10,15,…,9510,15,\dots,95, an A.P. with n=95−105+1=18n=\frac{95-10}{5}+1=18 terms. S=182(10+95)=9×105=945.S=\frac{18}{2}(10+95)=9\times105=945.

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  • 2083 · Science Model Set B · 1 mark

What is the range of the parabola y=x2−4x+3y = x^2 - 4x + 3? (a) [−1,−∞)[-1, -\infty) (b) [−1,∞)[-1, \infty) (c) [−2,∞)[-2, \infty) (d) [−2,−∞)[-2, -\infty)

Answer

(b) [−1,∞)[-1,\infty)

y=x2−4x+3=(x−2)2−1y=x^2-4x+3=(x-2)^2-1. The parabola opens upward with vertex (2,−1)(2,-1), so y≥−1y\ge -1.

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  • 2083 · Science Model Set B · 1 mark

lim⁡x→27=?\lim_{x \to 2} 7 = ? (a) 0 (b) 2 (c) 1 (d) 7

Answer

(d) 7

The limit of a constant function is that constant: lim⁡x→27=7\lim_{x\to2}7=7.

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  • 2083 · Science Model Set B · 1 mark

A square matrix A=(aij)A = (a_{ij}) is said to be a lower triangular matrix if the element aij=0a_{ij} = 0 for (a) i≤ji \le j (b) i>ji > j (c) i<ji < j (d) None

Answer

(c) i<ji<j

A lower triangular matrix has all entries above the main diagonal equal to zero, i.e. aij=0a_{ij}=0 for i<ji<j.

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  • 2083 · Science Model Set B · 1 mark

If c≠0c \neq 0 and the slope of the line passing through (−c,c)(-c, c) and (3c,a)(3c, a) is 1, which of the following is an expression for aa in terms of cc? (a) 3c3c (b) 1−a2a\frac{1-a^2}{a} (c) 2c2c (d) 3c3c

Answer

a=5ca=5c, which is not among the listed options

slope=a−c3c−(−c)=a−c4c=1⇒a−c=4c⇒a=5c.\text{slope}=\frac{a-c}{3c-(-c)}=\frac{a-c}{4c}=1\Rightarrow a-c=4c\Rightarrow a=5c.

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  • 2083 · Science Model Set B · 1 mark

2+2+2+2cos⁡8θ=?\sqrt{2 + \sqrt{2 + \sqrt{2 + 2\cos 8\theta}}} = ? (a) 2sec⁡θ2\sec\theta (b) 2cos⁡θ2\cos\theta (c) 2tan⁡θ2\tan\theta (d) None

Answer

(b) 2cos⁡θ2\cos\theta

Using 2+2cos⁡2α=4cos⁡2α2+2\cos2\alpha=4\cos^2\alpha:

  • 2+2cos⁡8θ=4cos⁡24θ=2cos⁡4θ\sqrt{2+2\cos8\theta}=\sqrt{4\cos^24\theta}=2\cos4\theta
  • 2+2cos⁡4θ=2cos⁡2θ\sqrt{2+2\cos4\theta}=2\cos2\theta
  • 2+2cos⁡2θ=2cos⁡θ\sqrt{2+2\cos2\theta}=2\cos\theta

(for θ\theta with the cosines positive).

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  • 2083 · Science Model Set B · 1 mark

If ∣a⃗+b⃗∣=∣a⃗−b⃗∣|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|, which of the following is true? (a) a⃗⋅b⃗=0\vec{a}\cdot\vec{b} = 0 (b) a⃗=b⃗\vec{a} = \vec{b} (c) a⃗=kb⃗\vec{a} = k\vec{b} (d) a⃗=−b⃗\vec{a} = -\vec{b}

Answer

(a) a⃗⋅b⃗=0\vec a\cdot\vec b=0

Squaring: ∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2=∣a⃗∣2−2a⃗⋅b⃗+∣b⃗∣2⇒4 a⃗⋅b⃗=0|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2=|\vec a|^2-2\vec a\cdot\vec b+|\vec b|^2\Rightarrow 4\,\vec a\cdot\vec b=0.

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  • 2083 · Science Model Set B · 1 mark

If r1r_1 is the reflection about the line x=yx = y and r2r_2 is the reflection about the yy-axis, what does r1r2r_1 r_2 represent? (a) R[O,+90∘]R[O, +90^\circ] (b) R[O,−90∘]R[O, -90^\circ] (c) R[O,−180∘]R[O, -180^\circ] (d) R[O,+180∘]R[O, +180^\circ]

Answer

(b) R[O,−90∘]R[O,-90^\circ]

r2:(x,y)→(−x,y)r_2:(x,y)\to(-x,y) and r1:(x,y)→(y,x)r_1:(x,y)\to(y,x). Applying r2r_2 first and then r1r_1: (x,y)→(−x,y)→(y,−x),(x,y)\to(-x,y)\to(y,-x), which is a rotation about the origin through −90∘-90^\circ (clockwise).

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  • 2083 · Science Model Set C · 1 mark

If xx is an integer such that 0≤x≤200 \le x \le 20, for how many values of xx is 20x\frac{20}{x} an integer? (a) 4 (b) 6 (c) 8 (d) 10

Answer

(b) 6

x=0x=0 is not allowed. The values of xx in 1≤x≤201\le x\le20 that divide 20 are 1,2,4,5,10,201,2,4,5,10,20, giving 6 values.

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  • 2083 · Science Model Set C · 1 mark

If x+y=8x + y = 8, y+z=11y + z = 11 and z+x=5z + x = 5, what is the value of x+y+zx + y + z? (a) 25 (b) 12 (c) 440 (d) 220

Answer

(b) 12

Adding: 2(x+y+z)=8+11+5=242(x+y+z)=8+11+5=24, so x+y+z=12x+y+z=12.

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  • 2083 · Science Model Set C · 1 mark

If 5000=2a5b5000 = 2^a 5^b and aa and bb are integers, what is a+ba + b? (a) 5 (b) 6 (c) 7 (d) 8

Answer

(c) 7

5000=5×1000=5×23×53=23⋅545000=5\times1000=5\times2^3\times5^3=2^3\cdot5^4, so a=3, b=4a=3,\ b=4 and a+b=7a+b=7.

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  • 2083 · Science Model Set C · 1 mark

If 12x12^x is odd and xx is an integer, what is the value of x12x^{12}? (a) 0 (b) 1 (c) −1-1 (d) It cannot be determined

Answer

(a) 0

12x12^x is odd only when x=0x=0 (then 120=112^0=1); for positive xx it is even. So x12=012=0x^{12}=0^{12}=0.

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  • 2083 · Science Model Set C · 1 mark

The sum of squares of two consecutive natural numbers is 841. The smaller number is (a) 20 (b) 21 (c) 19 (d) none

Answer

(a) 20

Let the numbers be nn and n+1n+1: n2+(n+1)2=841⇒n2+n−420=0⇒(n+21)(n−20)=0n^2+(n+1)^2=841\Rightarrow n^2+n-420=0\Rightarrow(n+21)(n-20)=0. So n=20n=20 (check: 400+441=841400+441=841).

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  • 2083 · Science Model Set C · 1 mark

What will be the ratio of πc3\frac{\pi^c}{3} and 10∘10^\circ? (a) 6 : 1 (b) 1 : 6 (c) 6 : 5 (d) 5 : 6

Answer

(a) 6 : 1

πc=180∘\pi^c = 180^\circ, so πc3=60∘\frac{\pi^c}{3} = 60^\circ. The ratio is 60∘:10∘=6:160^\circ : 10^\circ = 6 : 1.

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  • 2083 · Science Model Set C · 1 mark

In the data set {2,5,7,8,x}\{2, 5, 7, 8, x\}, the arithmetic mean is the same as the median. Determine the value of xx. (a) 8 (b) 10 (c) 13 (d) 15

Answer

(c) 13

Mean =22+x5= \frac{22+x}{5}. For x=13x = 13 the ordered data is 2,5,7,8,132, 5, 7, 8, 13, so the median is 77 and the mean is 355=7\frac{35}{5} = 7. The two are equal.

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  • 2083 · Science Model Set C · 1 mark

(A‾∪B‾)(\overline{A} \cup \overline{B}) is equal to (a) A‾∩B‾\overline{A} \cap \overline{B} (b) A∩B‾\overline{A \cap B} (c) AcA^c (d) ∪\cup

Answer

(b) A∩B‾\overline{A \cap B}

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

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  • 2083 · Science Model Set C · 1 mark

What is the value of nn if a51/n=a10\sqrt[1/n]{a^5} = a^{10}? (a) 1 (b) 2 (c) 5 (d) 10

Answer

(b) 2

a51/n=(a5)n=a5n\sqrt[1/n]{a^5} = (a^5)^{n} = a^{5n}. Setting a5n=a10a^{5n} = a^{10} gives 5n=105n = 10, so n=2n = 2.

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  • 2083 · Science Model Set C · 1 mark

If ax+b4=c\sqrt[4]{ax + b} = c, then the value of xx is (a) c4c^4 (b) c4−bc^4 - b (c) c4−ba\frac{c^4 - b}{a} (d) c4−ba4\frac{c^4 - b}{a^4}

Answer

(c) c4−ba\frac{c^4 - b}{a}

Raise both sides to the fourth power: ax+b=c4ax + b = c^4, so x=c4−bax = \frac{c^4 - b}{a}.

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  • 2083 · Science Model Set C · 1 mark

Which one of the following is the median? (a) 5th5^{th} decile (b) 50th50^{th} percentile (c) 2nd2^{nd} quartile (d) All of the above

Answer

(d) All of the above

The median divides the data into two equal halves, so it is the 2nd quartile Q2Q_2, the 5th decile D5D_5 and the 50th percentile P50P_{50}.

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  • 2083 · Science Model Set C · 1 mark

The standard deviation of 25 numbers is 40. If each of the numbers is increased by 5, then the new standard deviation is (a) 40 (b) 45 (c) 452\frac{45}{2} (d) None

Answer

(a) 40

Adding a constant to every observation shifts the mean by the same amount but does not change the spread, so the standard deviation stays 4040.

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  • 2083 · Science Model Set C · 1 mark

If a rectangle is rotated about its length, which solid is formed? (a) Cube (b) Cuboid (c) Cone (d) Sphere

Answer

None of the listed options is correct. Rotating a rectangle about one of its sides (its length) through 360∘360^\circ forms a cylinder, which is not among (a) to (d).

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  • 2083 · Science Model Set C · 1 mark

What is the perimeter of a semi-circle having its radius rr? (a) πr\pi r (b) 2πr2\pi r (c) πr+d\pi r + d (d) 2πr+d2\pi r + d

Answer

(c) πr+d\pi r + d

The perimeter is the curved arc plus the diameter: πr+2r=πr+d\pi r + 2r = \pi r + d.

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  • 2083 · Science Model Set C · 1 mark

What is the probability of getting a composite number when a die is rolled once? (a) 23\frac{2}{3} (b) 13\frac{1}{3} (c) 1 (d) 56\frac{5}{6}

Answer

(b) 13\frac{1}{3}

The composite numbers on a die are 44 and 66. So P=26=13P = \frac{2}{6} = \frac{1}{3}.

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  • 2083 · Science Model Set C · 1 mark

Find the probability of occurring the letter A or I while drawing a letter card from the cards of the word 'ASSOCIATIVE'. (a) 111\frac{1}{11} (b) 211\frac{2}{11} (c) 311\frac{3}{11} (d) 411\frac{4}{11}

Answer

(d) 411\frac{4}{11}

ASSOCIATIVE has 11 letters, with A appearing twice and I appearing twice. Favourable cards =2+2=4= 2 + 2 = 4, so P=411P = \frac{4}{11}.

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  • 2083 · Science Model Set C · 1 mark

After 10% devaluation of Nepali currency, USD 2000 can be exchanged by Rs 231000. Find the Nepali rupees equivalent to USD 1 before devaluation. (a) Rs. 115.5 (b) Rs. 105 (c) Rs. 110 (d) Rs. 10.5

Answer

(b) Rs 105

After devaluation, USD1=2310002000=Rs 115.5USD 1 = \frac{231000}{2000} = \text{Rs } 115.5. Devaluation of 10% raises the rupee price of a dollar by 10%, so before devaluation USD1=115.51.1=Rs 105USD 1 = \frac{115.5}{1.1} = \text{Rs } 105.

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  • 2083 · Science Model Set C · 1 mark

Find the compound interest on Rs 12500 at 8% p.a. for 9 months compounded quarterly. (a) Rs 1020 (b) Rs 1428 (c) Rs 510 (d) None of above

Answer

(d) None of above

Nine months is 3 quarters, and the rate per quarter is 84=2%\frac{8}{4} = 2\%. A=12500(1.02)3=12500×1.061208=Rs 13265.10A = 12500\left(1.02\right)^3 = 12500 \times 1.061208 = \text{Rs } 13265.10 CI=13265.10−12500=Rs 765.10CI = 13265.10 - 12500 = \text{Rs } 765.10 This matches none of the given options.

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  • 2083 · Science Model Set C · 1 mark

What is the half of x28\frac{x^2}{8}? (a) x4\frac{x}{4} (b) x24\frac{x^2}{4} (c) x216\frac{x^2}{16} (d) All of above

Answer

(c) x216\frac{x^2}{16}

12⋅x28=x216\frac{1}{2} \cdot \frac{x^2}{8} = \frac{x^2}{16}.

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  • 2083 · Science Model Set C · 1 mark

If ab+c=x\frac{a}{b+c} = x, bc+a=y\frac{b}{c+a} = y and ca+b=z\frac{c}{a+b} = z, then find the value of 11+x+11+y+11+z\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z}. (a) 0 (b) 1 (c) 2 (d) 3

Answer

(c) 2

1+x=a+b+cb+c1 + x = \frac{a+b+c}{b+c}, so 11+x=b+ca+b+c\frac{1}{1+x} = \frac{b+c}{a+b+c}. Similarly 11+y=c+aa+b+c\frac{1}{1+y} = \frac{c+a}{a+b+c} and 11+z=a+ba+b+c\frac{1}{1+z} = \frac{a+b}{a+b+c}. The sum is 2(a+b+c)a+b+c=2\frac{2(a+b+c)}{a+b+c} = 2.

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  • 2083 · Science Model Set C · 1 mark

In right-angled △ABC\triangle ABC (right angle at C), the area is 36. If side AC is twice as long as side BC, what is the length of AB? (a) 6 (b) 12 (c) 656\sqrt{5} (d) 18

Answer

(c) 656\sqrt{5}

Let BC=xBC = x, so AC=2xAC = 2x. Area =12x⋅2x=x2=36= \frac{1}{2}x \cdot 2x = x^2 = 36, so x=6x = 6 and AC=12AC = 12. AB=62+122=180=65.AB = \sqrt{6^2 + 12^2} = \sqrt{180} = 6\sqrt{5}.

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  • 2083 · Science Model Set C · 1 mark

The point of intersection of the angle bisectors of all three interior angles of a triangle is called as (a) centroid (b) orthocenter (c) incenter (d) Circum-center.

Answer

(c) incenter

The angle bisectors meet at the incenter, the centre of the inscribed circle.

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  • 2083 · Science Model Set C · 1 mark

If the principal is one third the compound interest, how many times of compound interest is the compound amount? (a) 3 times (b) 13\frac{1}{3} times (c) 23\frac{2}{3} times (d) 43\frac{4}{3} times

Answer

(d) 43\frac{4}{3} times

Let CI=cCI = c, so P=c3P = \frac{c}{3}. The compound amount is A=P+CI=c3+c=4c3A = P + CI = \frac{c}{3} + c = \frac{4c}{3}, which is 43\frac{4}{3} times the compound interest.

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  • 2083 · Science Model Set C · 1 mark

If a line is perpendicular to the x-axis, its slope (mm) is (a) 0 (b) 1 (c) 3\sqrt{3} (d) undefined

Answer

(d) undefined

A line perpendicular to the xx-axis is vertical, with inclination 90∘90^\circ. Its slope is tan⁡90∘\tan 90^\circ, which is undefined.

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  • 2083 · Science Model Set C · 1 mark

If a⃗⋅b⃗=−∣a⃗∣⋅∣b⃗∣\vec{a}\cdot\vec{b} = -|\vec{a}|\cdot|\vec{b}|, then the angle between a⃗\vec{a} and b⃗\vec{b} is (a) 0∘0^\circ (b) 60∘60^\circ (c) 180∘180^\circ (d) 90∘90^\circ

Answer

(c) 180∘180^\circ

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=−∣a⃗∣∣b⃗∣\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = -|\vec{a}||\vec{b}| gives cos⁡θ=−1\cos\theta = -1, so θ=180∘\theta = 180^\circ.

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  • 2083 · Science Model Set C · 1 mark

If qcsc⁡θ=pq\csc\theta = p and θ\theta is acute, then what is the value of (p2−q2)tan⁡θ\left(\sqrt{p^2 - q^2}\right)\tan\theta? (a) pp (b) qq (c) pqpq (d) p2+q2\sqrt{p^2 + q^2}

Answer

(b) qq

sin⁡θ=qp\sin\theta = \frac{q}{p}, so cos⁡θ=p2−q2p\cos\theta = \frac{\sqrt{p^2-q^2}}{p} and p2−q2=pcos⁡θ\sqrt{p^2-q^2} = p\cos\theta. Then p2−q2 tan⁡θ=pcos⁡θ⋅sin⁡θcos⁡θ=psin⁡θ=q.\sqrt{p^2-q^2}\,\tan\theta = p\cos\theta \cdot \frac{\sin\theta}{\cos\theta} = p\sin\theta = q.

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  • 2083 · Science Model Set C · 1 mark

If a positive integer aa is increased by 20%, decreased by 25% and then increased by 60%, the resulting number is what percent of aa? (a) 65% (b) 80% (c) 144% (d) 150%

Answer

(c) 144%

Multiply the factors: 1.20×0.75×1.60=0.9×1.6=1.441.20 \times 0.75 \times 1.60 = 0.9 \times 1.6 = 1.44, so the result is 144%144\% of aa.

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  • 2083 · Science Model Set C · 1 mark

The range of the function y=cos⁡xy = \cos x is (a) (−1,1)(-1, 1) (b) [−1,1)[-1, 1) (c) (−1,1](-1, 1] (d) [−1,1][-1, 1]

Answer

(d) [−1,1][-1, 1]

cos⁡x\cos x attains both −1-1 (at x=πx=\pi) and 11 (at x=0x=0), so both end points are included.

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  • 2083 · Science Model Set C · 1 mark

For what value of mm is (x−1)(x - 1) a factor of x3−3x2+m+1x^3 - 3x^2 + m + 1? (a) 0 (b) 1 (c) 2 (d) None

Answer

(b) 1

By the factor theorem, put x=1x=1: 1−3+m+1=01 - 3 + m + 1 = 0, so m=1m = 1.

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  • 2083 · Science Model Set C · 1 mark

The corresponding equation of 2x−y≤02x - y \le 0 is (a) Parallel to x-axis (b) Parallel to y-axis (c) Passing through (0,0)(0, 0) (d) All of above

Answer

(c) Passing through (0,0)(0, 0)

The corresponding equation is 2x−y=02x - y = 0, i.e. y=2xy = 2x. It is neither parallel to an axis (it has slope 2), and it is satisfied by (0,0)(0,0).

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  • 2083 · Science Model Set C · 1 mark

Write the point where the function f(x)=1x+1f(x) = \frac{1}{x+1} is discontinuous. (a) x=1x = 1 (b) x=−1x = -1 (c) x=0x = 0 (d) x=cx = c

Answer

(b) x=−1x = -1

f(x)=1x+1f(x)=\frac{1}{x+1} is undefined when the denominator x+1=0x+1=0, i.e. at x=−1x=-1.

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  • 2083 · Science Model Set C · 1 mark

A square matrix A=(aij)A = (a_{ij}) is said to be upper triangular matrix if the element aij=0a_{ij} = 0 for (a) i≤ji \le j (b) i>ji > j (c) i<ji < j (d) i≥ji \ge j

Answer

(b) i>ji > j

In an upper triangular matrix all entries below the main diagonal are zero, i.e. aij=0a_{ij}=0 whenever the row index exceeds the column index.

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  • 2083 · Science Model Set D · 1 mark

If x−1x=10x - \frac{1}{x} = 10, find the value of x+1xx + \frac{1}{x}. (a) 104 (b) 2262\sqrt{26} (c) 100 (d) 26\sqrt{26}

Answer

(b) 2262\sqrt{26}

Use (x+1x)2=(x−1x)2+4\left(x+\frac1x\right)^2=\left(x-\frac1x\right)^2+4:

(x+1x)2=102+4=104\left(x+\frac1x\right)^2=10^2+4=104

x+1x=104=226x+\frac1x=\sqrt{104}=2\sqrt{26}

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  • 2083 · Science Model Set D · 1 mark

Find the sum of money which increases by 110\frac{1}{10} of itself every year and amounts to Rs. 600 in 5 years. (a) Rs 380 (b) Rs. 280 (c) Rs. 480 (d) Rs. 400

Answer

(d) Rs 400

The increase is 110\frac1{10} of the original sum each year, so it grows by simple interest at 10% per year. For 5 years the amount is P(1+510)=1.5PP\left(1+\frac{5}{10}\right)=1.5P.

1.5P=600⇒P=6001.5=4001.5P=600 \Rightarrow P=\frac{600}{1.5}=400

So the sum is Rs 400.

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  • 2083 · Science Model Set D · 1 mark

A man bought 5 oranges for a rupee. How many for a rupee should he sell to gain 25%? (a) 4 (b) 6 (c) 3 (d) 9

Answer

(a) 4

Cost price of 5 oranges is Rs 1. To gain 25%, the selling price of those 5 oranges must be 1×1.25=Rs 1.251 \times 1.25 = \text{Rs } 1.25. So for Rs 1 he should sell 5×11.25=45 \times \frac{1}{1.25} = 4 oranges.

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  • 2083 · Science Model Set D · 1 mark

If 4A=5B4A = 5B and 6B=7C6B = 7C, find A:B:CA:B:C. (a) 28:30:35 (b) 35:30:28 (c) 35:28:24 (d) None

Answer

(c) 35:28:24

From 4A=5B4A = 5B, A=5B4A = \frac{5B}{4}; from 6B=7C6B = 7C, C=6B7C = \frac{6B}{7}. So A:B:C=54:1:67.A:B:C = \frac{5}{4}:1:\frac{6}{7}. Multiplying by the LCM 28 gives 35:28:2435:28:24.

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  • 2083 · Science Model Set D · 1 mark

Let UU be the set of all triangles in a plane and let AA be the set of all triangles with at least one angle different from 60∘60^\circ. Then AcA^c is (a) a set of all right triangles (b) a set of all isosceles triangles (c) a set of all equilateral triangles (d) None of these

Answer

(c) a set of all equilateral triangles

AA contains every triangle that is not having all angles equal to 60∘60^\circ. Its complement AcA^c is the set of triangles with all angles 60∘60^\circ, i.e. equilateral triangles.

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  • 2083 · Science Model Set D · 1 mark

L.C.M. of two numbers is 16 times the H.C.F. The sum of LCM and HCF is 850. If one number is 50, then what is the other number? (a) 800 (b) 1200 (c) 1600 (d) 2400

Answer

(a) 800

Let HCF =h= h, so LCM =16h= 16h and 17h=85017h = 850, giving h=50h = 50 and LCM =800= 800. Since (first number) ×\times (second number) == HCF ×\times LCM, 50×N=50×800⇒N=800.50 \times N = 50 \times 800 \Rightarrow N = 800.

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  • 2083 · Science Model Set D · 1 mark

A colourless cube is painted blue and then cut parallel to sides to form two rectangular solids of equal volume. What percentage of the surface area of each of the new solids is not painted blue? (a) 25 (b) 16 (c) 20 (d) 18

Answer

(a) 25

Let the cube have side aa. Each half is an a×a×a2a \times a \times \frac{a}{2} solid with surface area 2(a2+a22+a22)=4a22\left(a^2 + \frac{a^2}{2} + \frac{a^2}{2}\right) = 4a^2. The only unpainted face is the new cut face of area a2a^2, so the unpainted fraction is a24a2=25%\frac{a^2}{4a^2} = 25\%.

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  • 2083 · Science Model Set D · 1 mark

Excluding stoppages, the speed of a bus is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour? (a) 12 (b) 10 (c) 9 (d) 20

Answer

(b) 10

The bus loses 54−45=954 - 45 = 9 km per hour due to stops. Time to cover 9 km at 54 km/hr is 954\frac{9}{54} hr =16= \frac{1}{6} hr =10= 10 minutes.

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  • 2083 · Science Model Set D · 1 mark

(A∪B)c(A \cup B)^c is equal to (a) Ac∪BcA^c \cup B^c (b) Ac∩BcA^c \cap B^c (c) AcA^c (d) BcB^c

Answer

(b) Ac∩BcA^c \cap B^c

This is De Morgan's law: (A∪B)c=Ac∩Bc(A \cup B)^c = A^c \cap B^c.

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  • 2083 · Science Model Set D · 1 mark

The diagonal of a rectangle is 29 m and its breadth is 20 m. Find its length. (a) 24 m (b) 21 m (c) 21.2 m (d) None

Answer

(b) 21 m

By Pythagoras, length =292−202=841−400=441=21= \sqrt{29^2 - 20^2} = \sqrt{841 - 400} = \sqrt{441} = 21 m.

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  • 2083 · Science Model Set D · 1 mark

What is the square root of 9+2149 + 2\sqrt{14}? (a) 1+221 + 2\sqrt{2} (b) 3+6\sqrt{3} + \sqrt{6} (c) 2+7\sqrt{2} + \sqrt{7} (d) 2+52 + \sqrt{5}

Answer

(c) 2+7\sqrt{2} + \sqrt{7}

We look for two numbers with sum 99 and product 1414: these are 77 and 22. Hence 9+214=7+2+272=(7+2)2.9 + 2\sqrt{14} = 7 + 2 + 2\sqrt{7}\sqrt{2} = (\sqrt{7} + \sqrt{2})^2.

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  • 2083 · Science Model Set D · 1 mark

If 12x\frac{1}{2}x years ago, Ram was 12 and 12x\frac{1}{2}x years from now he will be 2x2x years old, how old will he be 3x3x years from now? (a) 18 (b) 24 (c) 30 (d) 54

Answer

(d) 54

Ram's present age is 12+x212 + \frac{x}{2}. After x2\frac{x}{2} years he is 12+x=2x12 + x = 2x, so x=12x = 12 and his present age is 1818. After 3x=363x = 36 years he will be 18+36=5418 + 36 = 54.

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  • 2083 · Science Model Set D · 1 mark

If (ab−1)2x−1=(ba−1)x−2(ab^{-1})^{2x-1} = (ba^{-1})^{x-2}, then what is the value of xx? (a) 1 (b) 2 (c) 3 (d) 4

Answer

(a) 1

Since ba−1=(ab−1)−1ba^{-1} = (ab^{-1})^{-1}, the equation becomes (ab−1)2x−1=(ab−1)−(x−2)(ab^{-1})^{2x-1} = (ab^{-1})^{-(x-2)}. Equating exponents, 2x−1=2−x2x - 1 = 2 - x, so 3x=33x = 3 and x=1x = 1.

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  • 2083 · Science Model Set D · 1 mark

If ab+c=x\frac{a}{b+c} = x, bc+a=y\frac{b}{c+a} = y and ca+b=z\frac{c}{a+b} = z, then the value of 11+x+11+y+11+z\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} is (a) 0 (b) 1 (c) 2 (d) None

Answer

(c) 2

11+x=11+ab+c=b+ca+b+c,\frac{1}{1+x} = \frac{1}{1 + \frac{a}{b+c}} = \frac{b+c}{a+b+c}, and similarly 11+y=c+aa+b+c\frac{1}{1+y} = \frac{c+a}{a+b+c}, 11+z=a+ba+b+c\frac{1}{1+z} = \frac{a+b}{a+b+c}. Adding gives 2(a+b+c)a+b+c=2\frac{2(a+b+c)}{a+b+c} = 2.

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  • 2083 · Science Model Set D · 1 mark

The value of xx in (45)2x−3=(54)x−3\left(\frac{4}{5}\right)^{2x-3} = \left(\frac{5}{4}\right)^{x-3} is (a) 1 (b) 3 (c) 2 (d) 4

Answer

(c) 2

Write (54)x−3=(45)3−x\left(\frac{5}{4}\right)^{x-3} = \left(\frac{4}{5}\right)^{3-x}. Equating exponents, 2x−3=3−x2x - 3 = 3 - x, so 3x=63x = 6 and x=2x = 2.

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  • 2083 · Science Model Set D · 1 mark

What is/are the factors of (x29−x24−x13−1)(x^{29} - x^{24} - x^{13} - 1)? (a) (x−1)(x-1) (b) (x+1)(x+1) (c) (x−1)(x-1) and (x+1)(x+1) (d) neither (x−1)(x-1) nor (x+1)(x+1)

Answer

(d) neither (x−1)(x-1) nor (x+1)(x+1)

Let f(x)=x29−x24−x13−1f(x) = x^{29} - x^{24} - x^{13} - 1. Then f(1)=1−1−1−1=−2≠0f(1) = 1 - 1 - 1 - 1 = -2 \neq 0 and f(−1)=−1−1+1−1=−2≠0f(-1) = -1 - 1 + 1 - 1 = -2 \neq 0. By the factor theorem neither is a factor.

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  • 2083 · Science Model Set D · 1 mark

The median of a series is 16. Two additional observations 6 and 21 are added to the given series. The median of the new series is (a) 14 (b) 15 (c) 16 (d) None

Answer

(c) 16

One added observation (6) lies below the median and the other (21) lies above it, so the middle position is unchanged and the median stays 1616.

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  • 2083 · Science Model Set D · 1 mark

The incircle of a quadrilateral of perimeter 2P2P has radius rr. What is the area of the quadrilateral? (a) P(r+1)P(r+1) (b) 2Pr2Pr (c) PrPr (d) None

Answer

(c) PrPr

Joining the incentre to the vertices splits the quadrilateral into four triangles of height rr whose bases are the sides. Area =12r×(perimeter)=12r×2P=Pr= \frac{1}{2} r \times (\text{perimeter}) = \frac{1}{2} r \times 2P = Pr.

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  • 2083 · Science Model Set D · 1 mark

If the 3rd3^{rd} term of a G.P. is 5, then the product of its first five terms is (a) 555^5 (b) 545^4 (c) 454^5 (d) 656^5

Answer

(a) 555^5

The third term is ar2=5ar^2 = 5. The first five terms are a,ar,ar2,ar3,ar4a, ar, ar^2, ar^3, ar^4, with product a5r10=(ar2)5=55a^5 r^{10} = (ar^2)^5 = 5^5.

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  • 2083 · Science Model Set D · 1 mark

Which of the following formulas is true for cos⁡2θ\cos 2\theta? (a) 2cos⁡2θ−12\cos^2\theta - 1 (b) 1−2sin⁡2θ1 - 2\sin^2\theta (c) cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta (d) All of above

Answer

(d) All of above

cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta, so all three are true.

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  • 2083 · Science Model Set D · 1 mark

The surface area of a sphere is SS. What is its volume? (a) S3S4π\frac{S}{3}\sqrt{\frac{S}{4\pi}} (b) SS4πS\sqrt{\frac{S}{4\pi}} (c) S314π\frac{S}{3}\sqrt{\frac{1}{4\pi}} (d) S14πS\sqrt{\frac{1}{4\pi}}

Answer

(a) S3S4π\frac{S}{3}\sqrt{\frac{S}{4\pi}}

S=4πr2S = 4\pi r^2 gives r=S4πr = \sqrt{\frac{S}{4\pi}}. Then V=43πr3=(4πr2) r3=S3S4π.V = \frac{4}{3}\pi r^3 = \frac{(4\pi r^2)\, r}{3} = \frac{S}{3}\sqrt{\frac{S}{4\pi}}.

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  • 2083 · Science Model Set D · 1 mark

In a parallelogram ABCD, X is the mid-point of AD and Y is the mid-point of BC. XY divides BD in the ratio (a) 1:3 (b) 1:2 (c) 2:1 (d) None

Answer

(d) None

XYXY joins the midpoints of the opposite sides ADAD and BCBC, so it is parallel to ABAB and passes through the point where the diagonals meet, which is the midpoint of BDBD. Hence XYXY divides BDBD in the ratio 1:11:1, which is not among the options.

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  • 2083 · Science Model Set D · 1 mark

The product of (34)(12)\begin{pmatrix} 3 \\ 4 \end{pmatrix}\begin{pmatrix} 1 & 2 \end{pmatrix} is (a) (11)(11) (b) (3648)\begin{pmatrix} 3 & 6 \\ 4 & 8 \end{pmatrix} (c) (11)(11) (d) (3468)\begin{pmatrix} 3 & 4 \\ 6 & 8 \end{pmatrix}

Answer

(b) (3648)\begin{pmatrix} 3 & 6 \\ 4 & 8 \end{pmatrix}

A 2×12\times1 matrix times a 1×21\times2 matrix gives a 2×22\times2 matrix: (34)(12)=(3⋅13⋅24⋅14⋅2)=(3648).\begin{pmatrix} 3 \\ 4 \end{pmatrix}\begin{pmatrix} 1 & 2 \end{pmatrix} = \begin{pmatrix} 3\cdot1 & 3\cdot2 \\ 4\cdot1 & 4\cdot2 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 4 & 8 \end{pmatrix}.

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  • 2083 · Science Model Set D · 1 mark

If a⃗=(−53)\vec{a} = \begin{pmatrix} -5 \\ 3 \end{pmatrix} and b⃗=(PP+2)\vec{b} = \begin{pmatrix} P \\ P+2 \end{pmatrix} are perpendicular, find the value of PP. (a) 3 (b) 2 (c) 1 (d) 0

Answer

(a) 3

Perpendicular vectors have zero dot product:

(−5)(P)+3(P+2)=−2P+6=0⇒P=3.(-5)(P)+3(P+2)=-2P+6=0 \Rightarrow P=3.

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  • 2083 · Science Model Set D · 1 mark

The value of cos⁡(2212)∘\cos\left(22\frac{1}{2}\right)^\circ is (a) 2−22\frac{\sqrt{2-\sqrt{2}}}{2} (b) 2+22\frac{\sqrt{2+\sqrt{2}}}{2} (c) 2−1\sqrt{2} - 1 (d) None

Answer

(b) 2+22\frac{\sqrt{2+\sqrt{2}}}{2}

Using cos⁡θ=1+cos⁡2θ2\cos\theta=\sqrt{\frac{1+\cos 2\theta}{2}} with 2θ=45∘2\theta=45^\circ:

cos⁡22.5∘=1+222=2+24=2+22.\cos 22.5^\circ=\sqrt{\frac{1+\frac{\sqrt2}{2}}{2}}=\sqrt{\frac{2+\sqrt2}{4}}=\frac{\sqrt{2+\sqrt2}}{2}.

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  • 2083 · Science Model Set D · 1 mark

If f(4x+5)=12x+18f(4x + 5) = 12x + 18, then find f(x)f(x). (a) 3x+33x + 3 (b) 3x+134\frac{3x+13}{4} (c) 12x+184x+5\frac{12x+18}{4x+5} (d) 3x3x

Answer

(a) 3x+33x+3

Let t=4x+5t=4x+5, so x=t−54x=\frac{t-5}{4}. Then f(t)=12⋅t−54+18=3t−15+18=3t+3f(t)=12\cdot\frac{t-5}{4}+18=3t-15+18=3t+3, hence f(x)=3x+3f(x)=3x+3.

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  • 2083 · Science Model Set D · 1 mark

In an A.P., 5 times the 5th5^{th} term is equal to 8 times the 8th8^{th} term. Its 13th13^{th} term is (a) 2 (b) 1 (c) 12 (d) 0

Answer

(d) 0

Let the first term be aa and common difference dd. From 5t5=8t85t_5=8t_8:

5(a+4d)=8(a+7d)⇒−3a=36d⇒a=−12d.5(a+4d)=8(a+7d)\Rightarrow -3a=36d\Rightarrow a=-12d.

So t13=a+12d=−12d+12d=0t_{13}=a+12d=-12d+12d=0.

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  • 2083 · Science Model Set D · 1 mark

Which of the following transformations is represented by the matrix (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}? (a) Reflection in y-axis (b) Reflection in x=yx = y (c) Reflection in x-axis (d) Reflection in y=−xy = -x

Answer

(c) Reflection in x-axis

The matrix sends (x,y)→(x,−y)(x,y)\to(x,-y), keeping xx unchanged and reversing the sign of yy, which is reflection in the x-axis.

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  • 2083 · Science Model Set D · 1 mark

The maximum value of sin⁡x⋅cos⁡x\sin x \cdot \cos x is (a) 2\sqrt{2} (b) 1 (c) 12\frac{1}{2} (d) None

Answer

(c) 12\frac{1}{2}

sin⁡xcos⁡x=12sin⁡2x\sin x\cos x=\frac{1}{2}\sin 2x, and the maximum of sin⁡2x\sin 2x is 1, so the maximum value is 12\frac12.

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  • 2083 · Science Model Set D · 1 mark

The general equation of 2nd2^{nd} degree ax2+2hxy+by2+2gx+2fy+c=0ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 represents a circle if (a) a=ba = b (b) h=0h = 0 (c) both (a) and (b) (d) None of above

Answer

(c) both (a) and (b)

The general second-degree equation represents a circle when the coefficients of x2x^2 and y2y^2 are equal (a=ba=b) and there is no xyxy term (h=0h=0). Both conditions are needed.

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  • 2083 · Science Model Set D · 1 mark

What is the arithmetic mean of 2102^{10} and 2202^{20}? (a) 2152^{15} (b) 25+2102^5 + 2^{10} (c) 29+2192^9 + 2^{19} (d) 2292^{29}

Answer

(c) 29+2192^9+2^{19}

210+2202=2102+2202=29+219.\frac{2^{10}+2^{20}}{2}=\frac{2^{10}}{2}+\frac{2^{20}}{2}=2^9+2^{19}.

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  • 2083 · Science Model Set E · 1 mark

The factors of 6x2+x−126x^2 + x - 12 are (a) (3x+4)(2x−3)(3x+4)(2x-3) (b) (3x−4)(2x+3)(3x-4)(2x+3) (c) (3x−4)(2x−4)(3x-4)(2x-4) (d) (3x+4)(2x+3)(3x+4)(2x+3)

Answer

(b) (3x−4)(2x+3)(3x-4)(2x+3)

(3x−4)(2x+3)=6x2+9x−8x−12=6x2+x−12(3x-4)(2x+3)=6x^2+9x-8x-12=6x^2+x-12

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  • 2083 · Science Model Set E · 1 mark

What must be added to 2x3+3x2−11x−202x^3 + 3x^2 - 11x - 20 to make (x−2)(x - 2) one of the factors of the polynomial? (a) 4 (b) −4-4 (c) 0 (d) 1

Answer

Let p(x)=2x3+3x2−11x−20p(x)=2x^3+3x^2-11x-20 and let kk be the number added. (x−2)(x-2) is a factor of p(x)+kp(x)+k when p(2)+k=0p(2)+k=0.

p(2)=2(8)+3(4)−11(2)−20=16+12−22−20=−14p(2)=2(8)+3(4)-11(2)-20=16+12-22-20=-14

−14+k=0⇒k=14-14+k=0\Rightarrow k=14

So 1414 must be added. This value is not among the listed options (a) 4, (b) −4-4, (c) 0, (d) 1. Adding 4 would work if the constant term were −10-10 instead of −20-20.

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  • 2083 · Science Model Set E · 1 mark

If (1a+a)2=100\left(\frac{1}{a} + a\right)^2 = 100, what is the value of 1a2+a2\frac{1}{a^2} + a^2? (a) 10 (b) 64 (c) 98 (d) 100

Answer

(c) 98

(1a+a)2=1a2+2+a2=100⇒1a2+a2=98\left(\frac1a+a\right)^2=\frac1{a^2}+2+a^2=100\Rightarrow \frac1{a^2}+a^2=98

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  • 2083 · Science Model Set E · 1 mark

If a1x=b1ya^{\frac{1}{x}} = b^{\frac{1}{y}}, which of the following is true? (a) x=yx = y (b) a=ba = b (c) a=bx/ya = b^{x/y} (d) a=by/xa = b^{y/x}

Answer

(c) a=bx/ya=b^{x/y}

Raise both sides to the power xx: a=(b1/y)x=bx/ya=\left(b^{1/y}\right)^{x}=b^{x/y}

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  • 2083 · Science Model Set E · 1 mark

If n(A)=75n(A) = 75, n(B)=40n(B) = 40 and B⊆AB \subseteq A, find the value of n(A∪B)n(A \cup B). (a) 35 (b) 40 (c) 75 (d) 115

Answer

(c) 75

Since B⊆AB\subseteq A, A∪B=AA\cup B=A, so n(A∪B)=n(A)=75n(A\cup B)=n(A)=75.

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  • 2083 · Science Model Set E · 1 mark

The vectors a⃗=2i⃗−3j⃗\vec{a} = 2\vec{i} - 3\vec{j} and b⃗=−4i⃗+6j⃗\vec{b} = -4\vec{i} + 6\vec{j} are (a) parallel (b) perpendicular (c) coincident (d) neither parallel nor perpendicular

Answer

(a) parallel

b⃗=−4i⃗+6j⃗=−2(2i⃗−3j⃗)=−2a⃗\vec b=-4\vec i+6\vec j=-2(2\vec i-3\vec j)=-2\vec a

One vector is a scalar multiple of the other, so they are parallel (in opposite directions). They are not coincident, since b⃗≠a⃗\vec b\neq\vec a.

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  • 2083 · Science Model Set E · 1 mark

A function is said to be into if (a) Range = Co-domain (b) Range = Domain (c) Range ⊂\subset Co-domain (d) Range ⊆\subseteq Co-domain

Answer

(c) Range ⊂\subset Co-domain

A function is into when its range is a proper subset of the co-domain, so some element of the co-domain has no pre-image. If the range equals the co-domain, the function is onto.

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  • 2083 · Science Model Set E · 1 mark

What is the slope of the equation xb+ya=1\frac{x}{b} + \frac{y}{a} = 1? (a) aa (b) bb (c) ab\frac{a}{b} (d) −ab\frac{-a}{b}

Answer

(d) −ab\frac{-a}{b}

Rewrite as y=a−abxy = a - \frac{a}{b}x, i.e. y=−abx+ay = -\frac{a}{b}x + a. The coefficient of xx is the slope, so m=−abm = -\frac{a}{b}.

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  • 2083 · Science Model Set E · 1 mark

The inverse of (5231)\begin{pmatrix} 5 & 2 \\ 3 & 1 \end{pmatrix} is (a) (−123−5)\begin{pmatrix} -1 & 2 \\ 3 & -5 \end{pmatrix} (b) (1−5−2−3)\begin{pmatrix} 1 & -5 \\ -2 & -3 \end{pmatrix} (c) (1−2−35)\begin{pmatrix} 1 & -2 \\ -3 & 5 \end{pmatrix} (d) (−132−5)\begin{pmatrix} -1 & 3 \\ 2 & -5 \end{pmatrix}

Answer

(a) (−123−5)\begin{pmatrix} -1 & 2 \\ 3 & -5 \end{pmatrix}

The determinant is 5(1)−2(3)=−15(1) - 2(3) = -1. Swapping the diagonal entries and negating the others gives A−1=1−1(1−2−35)=(−123−5).A^{-1} = \frac{1}{-1}\begin{pmatrix} 1 & -2 \\ -3 & 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 3 & -5 \end{pmatrix}.

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  • 2083 · Science Model Set E · 1 mark

Find the value of nn if the sum of the arithmetic series 2+⋯+982 + \dots + 98 is 2450. (a) 49 (b) 67 (c) 117 (d) 1129

Answer

(a) 49

The series 2+4+⋯+982 + 4 + \dots + 98 has a=2a = 2, d=2d = 2, l=98l = 98, so n=98−22+1=49n = \frac{98-2}{2} + 1 = 49. Check: S49=492(2+98)=2450S_{49} = \frac{49}{2}(2 + 98) = 2450.

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  • 2083 · Science Model Set E · 1 mark

A point (7,4)(7, 4) is first reflected in the line y=xy = x and then in the line y=by = b so that the final image is (4,1)(4, 1). Find the value of bb. (a) 2 (b) 3 (c) 4 (d) 0

Answer

(c) 4

Reflection in y=xy = x sends (7,4)(7, 4) to (4,7)(4, 7). Reflection in y=by = b sends (x,y)(x, y) to (x,2b−y)(x, 2b - y), so 2b−7=12b - 7 = 1, giving b=4b = 4.

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  • 2083 · Science Model Set E · 1 mark

The value of sin⁡(90∘−θ)⋅csc⁡(180∘−θ)sin⁡(270∘+θ)⋅sec⁡(270∘+θ)\frac{\sin(90^\circ - \theta)\cdot\csc(180^\circ - \theta)}{\sin(270^\circ + \theta)\cdot\sec(270^\circ + \theta)} is (a) 0 (b) −1-1 (c) 2 (d) 1

Answer

(b) −1-1

Numerator: sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ - \theta) = \cos\theta and csc⁡(180∘−θ)=1sin⁡θ\csc(180^\circ - \theta) = \frac{1}{\sin\theta}, so it equals cos⁡θsin⁡θ\frac{\cos\theta}{\sin\theta}.

Denominator: sin⁡(270∘+θ)=−cos⁡θ\sin(270^\circ + \theta) = -\cos\theta and sec⁡(270∘+θ)=1sin⁡θ\sec(270^\circ + \theta) = \frac{1}{\sin\theta}, so it equals −cos⁡θsin⁡θ-\frac{\cos\theta}{\sin\theta}.

The ratio is −1-1.

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  • 2083 · Science Model Set E · 1 mark

10002+100100=?\frac{1000^2 + 100}{100} = ? (a) 100 (b) 10000 (c) 10001 (d) 1010

Answer

(c) 10001

10002+100100=1000000100+100100=10000+1=10001.\frac{1000^2 + 100}{100} = \frac{1000000}{100} + \frac{100}{100} = 10000 + 1 = 10001.

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  • 2083 · Science Model Set E · 1 mark

6 men can do a piece of work in 40 days working 10 hours a day. How many hours a day must 10 men work to do the same work in 24 days? (a) 10 (b) 15 (c) 18 (d) 20

Answer

(a) 10

Total work is 6×40×10=24006 \times 40 \times 10 = 2400 man-hours. With 10 men for 24 days, 10×24×h=240010 \times 24 \times h = 2400, so h=10h = 10 hours a day.

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  • 2083 · Science Model Set E · 1 mark

By selling 60 meters of a cloth, a merchant gains the selling price of 15 meters. Find his real gain percentage. (a) 1009%\frac{100}{9}\% (b) 1003%\frac{100}{3}\% (c) 25% (d) 30%

Answer

(b) 1003%\frac{100}{3}\%

Let the selling price be Rs 1 per metre. Then SP60=60SP_{60} = 60 and the gain is SP15=15SP_{15} = 15, so CP60=60−15=45CP_{60} = 60 - 15 = 45. Gain %=1545×100=1003%.\text{Gain \%} = \frac{15}{45} \times 100 = \frac{100}{3}\%.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Science Model Set E · 1 mark

Find the simple interest on Rs. 1020 at 5% p.a. in 4 years. (a) Rs. 408 (b) Rs. 210 (c) Rs. 200 (d) Rs. 204

Answer

(d) Rs. 204

SI=PTR100=1020×4×5100=Rs 204.SI = \frac{PTR}{100} = \frac{1020 \times 4 \times 5}{100} = \text{Rs } 204.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Science Model Set E · 1 mark

One robot can pack a box in 15 minutes. Working together at the same constant rate, how many boxes can 16 robots pack in 1 hour? (a) 4 (b) 16 (c) 24 (d) 64

Answer

(d) 64

One robot packs 6015=4\frac{60}{15} = 4 boxes in an hour, so 16 robots pack 16×4=6416 \times 4 = 64 boxes.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Science Model Set E · 1 mark

A sum of money doubles itself in 6 years. In how many years, it becomes 5 times? (a) 12 years (b) 24 years (c) 10 years (d) 15 years

Answer

(b) 24 years

Doubling in 6 years means simple interest equal to the principal PP is earned in 6 years, so the interest per year is P6\frac{P}{6}. To become 5 times, the interest must be 4P4P, which takes 4P÷P6=244P \div \frac{P}{6} = 24 years.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Science Model Set E · 1 mark

The value of 23%\frac{2}{3}\% of 34\frac{3}{4} is (a) 0.005 (b) 0.5 (c) 0.05 (d) 5

Answer

(a) 0.005

23% of 34=2300×34=1200=0.005.\frac{2}{3}\% \text{ of } \frac{3}{4} = \frac{2}{300} \times \frac{3}{4} = \frac{1}{200} = 0.005.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Science Model Set E · 1 mark

What is the area of a circle whose radius is the diagonal of a square whose area is 4? (a) 2π2\pi (b) 2π22\pi\sqrt{2} (c) 4π4\pi (d) 8π8\pi

Answer

(d) 8π8\pi

The square has area 4, so its side is 2 and its diagonal is 222\sqrt{2}. The circle has r=22r = 2\sqrt{2}, so A=πr2=π(22)2=8π.A = \pi r^2 = \pi (2\sqrt{2})^2 = 8\pi.

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  • 2083 · Science Model Set E · 1 mark

A number when added to 14 times the same number gives 75 as result. Find the number. (a) 10 (b) 15 (c) 5 (d) 20

Answer

(c) 5

Let the number be xx. Then x+14x=75x + 14x = 75, so 15x=7515x = 75 and x=5x = 5.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Science Model Set E · 1 mark

The integer aa is even and the integer bb is odd. Which of the following is even? (a) 2a+b2a + b (b) a+2ba + 2b (c) (a+b)2(a+b)^2 (d) a2−b2a^2 - b^2

Answer

(b) a+2ba + 2b

Here aa is even and bb is odd. 2a+b2a + b is even + odd = odd. a+2ba + 2b is even + even = even. (a+b)2(a+b)^2 is odd2^2 = odd. a2−b2a^2 - b^2 is even - odd = odd. Only a+2ba + 2b is even.

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  • 2083 · Science Model Set E · 1 mark

616 divided by 6 yields remainder pp, and 525 divided by 11 yields remainder qq. What is p+qp + q? (a) 10 (b) 12 (c) 15 (d) 20

Answer

(b) 12

616=6×102+4616 = 6 \times 102 + 4, so p=4p = 4. 525=11×47+8525 = 11 \times 47 + 8, so q=8q = 8. Hence p+q=12p + q = 12.

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  • 2083 · Science Model Set E · 1 mark

If the reciprocal of the negative integer xx is greater than the sum of yy and zz, then which of the following must be true? (a) x>y+zx > y + z (b) yy and zz are positive (c) 1>x(y+z)1 > x(y+z) (d) 1<xy+xz1 < xy + xz

Answer

(d) 1<xy+xz1 < xy + xz

We are given 1x>y+z\frac{1}{x} > y + z with x<0x < 0. Multiplying both sides by the negative number xx reverses the inequality: 1<x(y+z)=xy+xz.1 < x(y + z) = xy + xz.

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  • 2083 · Science Model Set E · 1 mark

The angle in a semi-circle is (a) 0∘0^\circ (b) 30∘30^\circ (c) 60∘60^\circ (d) 90∘90^\circ

Answer

(d) 90∘90^\circ

The angle subtended at the circumference by a diameter is half of the 180∘180^\circ angle at the centre, so it is 90∘90^\circ.

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  • 2083 · Science Model Set E · 1 mark

Two opposite angles of a cyclic quadrilateral are in the ratio 1:2. The smaller angle is (a) 60∘60^\circ (b) 70∘70^\circ (c) 80∘80^\circ (d) 100∘100^\circ

Answer

(a) 60∘60^\circ

Opposite angles of a cyclic quadrilateral are supplementary. Let them be xx and 2x2x; then 3x=180∘3x = 180^\circ, so x=60∘x = 60^\circ (the other is 120∘120^\circ).

Answer checked by Claude Sonnet · Oct 21, 2026

  • 2083 · Science Model Set E · 1 mark

The internal bisectors of ∠B\angle B and ∠C\angle C of △ABC\triangle ABC meet at II. Then ∠BIC\angle BIC is (a) 90∘−∠A290^\circ - \frac{\angle A}{2} (b) 90∘+∠A290^\circ + \frac{\angle A}{2} (c) 180∘−A180^\circ - A (d) 180∘+A180^\circ + A

Answer

(b) 90∘+∠A290^\circ + \frac{\angle A}{2}

In △BIC\triangle BIC, ∠BIC=180∘−B2−C2=180∘−180∘−A2=90∘+A2\angle BIC = 180^\circ - \frac{B}{2} - \frac{C}{2} = 180^\circ - \frac{180^\circ - A}{2} = 90^\circ + \frac{A}{2}.

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  • 2083 · Science Model Set E · 1 mark

Two events Q and R are mutually exclusive with P(Q)=35P(Q) = \frac{3}{5} and P(R)=15P(R) = \frac{1}{5}. Find the probability P(Q∪R‾)P(\overline{Q \cup R}). (a) 45\frac{4}{5} (b) 325\frac{3}{25} (c) 15\frac{1}{5} (d) 825\frac{8}{25}

Answer

(c) 15\frac{1}{5}

For mutually exclusive events, P(Q∪R)=35+15=45P(Q \cup R) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}. Then P(Q∪R‾)=1−45=15P(\overline{Q \cup R}) = 1 - \frac{4}{5} = \frac{1}{5}.

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  • 2083 · Science Model Set E · 1 mark

If aa is increased by 25% and bb is decreased by 25%, the resulting numbers are equal. What is the ratio of aa to bb? (a) 35\frac{3}{5} (b) 34\frac{3}{4} (c) 11\frac{1}{1} (d) 53\frac{5}{3}

Answer

(a) 35\frac{3}{5}

The condition gives 1.25a=0.75b1.25a = 0.75b, so ab=0.751.25=35\frac{a}{b} = \frac{0.75}{1.25} = \frac{3}{5}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Science Model Set E · 1 mark

The value of the variable that is repeated the maximum number of times is known as (a) mean (b) median (c) mode (d) Q1Q_1

Answer

(c) mode

The mode is the value that occurs most often in the data. The mean is the average, the median is the middle value and Q1Q_1 is the lower quartile.

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  • 2083 · Science Model Set E · 1 mark

If the mean of yy and 1y\frac{1}{y} is MM, then what is the mean of y3y^3 and 1y3\frac{1}{y^3}? (a) M(M2−3)3\frac{M(M^2-3)}{3} (b) M3M^3 (c) M3−3M^3 - 3 (d) M(4M2−3)M(4M^2 - 3)

Answer

(d) M(4M2−3)M(4M^2-3)

Given y+1y2=M\frac{y+\frac1y}{2}=M, so y+1y=2My+\frac1y=2M. Cubing,

y3+1y3=(y+1y)3−3(y+1y)=8M3−6M.y^3+\frac1{y^3}=\left(y+\frac1y\right)^3-3\left(y+\frac1y\right)=8M^3-6M.

The mean of y3y^3 and 1y3\frac1{y^3} is 8M3−6M2=4M3−3M=M(4M2−3)\frac{8M^3-6M}{2}=4M^3-3M=M(4M^2-3).

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  • 2083 · Science Model Set E · 1 mark

From a deck of 52 cards, a card is drawn. The probability of getting a queen is (a) 14\frac{1}{4} (b) 126\frac{1}{26} (c) 113\frac{1}{13} (d) 152\frac{1}{52}

Answer

(c) 113\frac{1}{13}

A deck has 4 queens among 52 cards, so P=452=113P=\frac{4}{52}=\frac{1}{13}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2083 · Science Model Set E · 1 mark

Two unbiased dice are rolled. Find the probability that the sum of the numbers on the two faces is either divisible by 3 or divisible by 4. (a) 59\frac{5}{9} (b) 13\frac{1}{3} (c) 14\frac{1}{4} (d) 136\frac{1}{36}

Answer

(a) 59\frac{5}{9}

Total outcomes: 36. Sums divisible by 3 (3, 6, 9, 12) occur in 2+5+4+1=122+5+4+1=12 ways. Sums divisible by 4 (4, 8, 12) occur in 3+5+1=93+5+1=9 ways. A sum of 12 is counted in both, in 1 way.

P=12+9−136=2036=59.P=\frac{12+9-1}{36}=\frac{20}{36}=\frac59.

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  • 2083 · Management and Law Model Set C

The mean of a set of numbers is xˉ\bar{x}. If each number is multiplied by pp, the new mean of the data is (a) xˉ\bar{x} (b) xˉ+p\bar{x} + p (c) pxˉp\bar{x} (d) None of above

Answer

(c) pxˉp\bar{x}

New mean=∑pxin=p⋅∑xin=pxˉ\text{New mean}=\frac{\sum px_i}{n}=p\cdot\frac{\sum x_i}{n}=p\bar{x}

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  • 2083 · Management and Law Model Set C

If £aa = Rs. bb, then what is the Nepali rupees of £acac? (a) Rs. bcbc (b) Rs. abcabc (c) Rs. bc\frac{b}{c} (d) Rs. a2a^2

Answer

(a) Rs. bcbc

Since £aa = Rs. bb, £1 = Rs. ba\frac{b}{a}. So

£ac=ac×ba=Rs. bc£ac = ac\times\frac{b}{a}=\text{Rs. } bc

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  • 2083 · Management and Law Model Set C

When a sum of money was invested at C.I., it amounted to Rs 2420 in 2 years and Rs 2662 in 3 years. The rate of interest is (a) 10% p.a (b) 12% p.a (c) 14% p.a (d) 20% p.a

Answer

(a) 10% p.a

The amount in the third year is the amount in the second year multiplied by (1+r100)(1+\frac{r}{100}):

1+r100=26622420=1.1  ⇒  r=10%1+\frac{r}{100}=\frac{2662}{2420}=1.1\;\Rightarrow\; r=10\%

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  • 2083 · Management and Law Model Set C

If xx is a positive number, 500% of xx is what percent of 500x500x? (a) 0.01% (b) 0.1% (c) 1% (d) 10%

Answer

(c) 1%

500% of x=5x500\%\text{ of }x=5x

5x500x×100%=1%\frac{5x}{500x}\times100\%=1\%

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  • 2083 · Management and Law Model Set C

If ab=baa^b = b^a and b=2ab = 2a, then the value of a2+b2a^2 + b^2 is (a) 4 (b) 16 (c) 20 (d) 25

Answer

(c) 20

Substitute b=2ab=2a into ab=baa^b=b^a:

a2a=(2a)a  ⟹  a2=2a  ⟹  a=2  (a≠0).a^{2a}=(2a)^a \implies a^2=2a \implies a=2 \;(a\neq 0).

So b=4b=4, and a2+b2=4+16=20a^2+b^2=4+16=20.

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  • 2083 · Management and Law Model Set C

The sum of a certain number and its positive square root is 90. The number is (a) 100 (b) 90 (c) 81 (d) 64

Answer

(c) 81

Let the number be xx, so x+x=90x+\sqrt{x}=90. Put x=t\sqrt{x}=t:

t2+t−90=0  ⟹  (t+10)(t−9)=0  ⟹  t=9.t^2+t-90=0 \implies (t+10)(t-9)=0 \implies t=9.

Hence x=81x=81. Check: 81+9=9081+9=90.

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  • 2083 · Management and Law Model Set C

If 2a=4b=8c=1002^a = 4^b = 8^c = 100, then 64abc=?64^{abc} = ? (a) 64000 (b) 16000 (c) 4000 (d) 1000

Answer

None of the options (a)–(d) is correct; the exact value is 100a2100^{a^2} with a=log⁡2100a=\log_2 100.

From 2a=4b=8c=1002^a=4^b=8^c=100 we get 2a=22b=23c2^a=2^{2b}=2^{3c}, so b=a2b=\frac{a}{2} and c=a3c=\frac{a}{3}. Then

abc=a⋅a2⋅a3=a36.abc=a\cdot\frac a2\cdot\frac a3=\frac{a^3}{6}.

64abc=26abc=2a3=(2a)a2=100a2.64^{abc}=2^{6abc}=2^{a^3}=(2^a)^{a^2}=100^{a^2}.

With a=log⁡2100≈6.644a=\log_2 100\approx 6.644, a2≈44.1a^2\approx 44.1, so the value is about 10044100^{44}, far larger than any option.

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  • 2083 · Management and Law Model Set C

What is the length of each side of a square if its diagonals are 10? (a) 5 (b) 7 (c) 525\sqrt{2} (d) 10210\sqrt{2}

Answer

(c) 525\sqrt{2}

For a square of side ss, the diagonal is s2s\sqrt2. So s2=10s\sqrt2=10 gives

s=102=52.s=\frac{10}{\sqrt2}=5\sqrt2.

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  • 2083 · Management and Law Model Set D

If the average of 5 numbers is 20, what is the sum of the numbers? (a) 25 (b) 100 (c) 15 (d) 4

Answer

(b) 100

Sum = average x number of items = 20×5=10020 \times 5 = 100.

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  • 2083 · Management and Law Model Set D

For any two sets AA and BB, A∩(B−A)A \cap (B - A) is (a) B (b) ∅\emptyset (c) A (d) B-A

Answer

(b) ∅\emptyset

B−AB - A contains only elements of BB that are not in AA. So no element of B−AB-A can belong to AA, and therefore A∩(B−A)=∅A \cap (B-A) = \emptyset.

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  • 2083 · Management and Law Model Set D

If the principal is 4 times the compound interest, how many times of compound interest is the compound amount? (a) 4 times (b) 5 times (c) 2 times (d) none of above

Answer

(b) 5 times

Let the compound interest be CC. Then the principal is P=4CP = 4C, so the compound amount is

A=P+C=4C+C=5C.A = P + C = 4C + C = 5C.

The amount is 5 times the compound interest.

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  • 2083 · Management and Law Model Set D

The total surface area of a hemisphere is 30 cm230\ \text{cm}^2. What is the surface area of a sphere having equal radius to the hemisphere? (a) 60 cm260\ \text{cm}^2 (b) 50 cm250\ \text{cm}^2 (c) 40 cm240\ \text{cm}^2 (d) 30 cm230\ \text{cm}^2

Answer

(c) 40 cm240\ \text{cm}^2

Total surface area of a hemisphere is 3πr2=303\pi r^2 = 30, so πr2=10\pi r^2 = 10.

Surface area of the sphere is 4πr2=4×10=40 cm24\pi r^2 = 4 \times 10 = 40\ \text{cm}^2.

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  • 2083 · Management and Law Model Set D

If a+1=1a + 1 = 1, find the value of bab^a. (a) 0 (b) 1 (c) ∞\infty (d) 000^0

Answer

(b) 1

From a+1=1a + 1 = 1 we get a=0a = 0. Then ba=b0=1b^a = b^0 = 1 (for any non-zero bb).

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  • 2083 · Management and Law Model Set D

The roots of the quadratic equation x2−36=0x^2 - 36 = 0 are (a) 6 (b) 6 and −6-6 (c) −6-6 (d) none of above

Answer

(b) 6 and −6-6

x2−36=0  ⟹  x2=36  ⟹  x=±6.x^2 - 36 = 0 \implies x^2 = 36 \implies x = \pm 6.

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  • 2083 · Management and Law Model Set D

What is the area of a square whose diagonal is 10? (a) 20 (b) 40 (c) 50 (d) 100

Answer

(c) 50

For a square with diagonal dd, the area is d22=1022=50\dfrac{d^2}{2} = \dfrac{10^2}{2} = 50.

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  • 2083 · Management and Law Model Set D

The sum of opposite angles of a cyclic quadrilateral is (a) 90∘90^\circ (b) 180∘180^\circ (c) 360∘360^\circ (d) all of above

Answer

(b) 180∘180^\circ

In a cyclic quadrilateral ABCDABCD, the angle at the centre on arc BCDBCD is twice the inscribed angle, so ∠A+∠C=360∘2=180∘\angle A+\angle C=\frac{360^\circ}{2}=180^\circ. The same holds for ∠B+∠D\angle B+\angle D.

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  • 2083 · Management and Law Model Set D

A card is drawn randomly from a deck of 52 playing cards. What is the probability of getting neither a king nor a jack? (a) 213\frac{2}{13} (b) 1113\frac{11}{13} (c) 313\frac{3}{13} (d) 1013\frac{10}{13}

Answer

(b) 1113\frac{11}{13}

There are 4 kings and 4 jacks, so 8 unwanted cards and 52−8=4452-8=44 favourable cards.

P=4452=1113P=\frac{44}{52}=\frac{11}{13}

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  • 2083 · Management and Law Model Set E

The average of 4 numbers is 15. If three of the numbers are 10, 12 and 18, what is the fourth number? (a) 18 (b) 20 (c) 22 (d) 24

Answer

(b) 20

Sum of the four numbers =4×15=60= 4 \times 15 = 60.

Sum of the three known numbers =10+12+18=40= 10 + 12 + 18 = 40.

Fourth number =60−40=20= 60 - 40 = 20.

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  • 2083 · Management and Law Model Set E

If AA is the set of letters of the word MATHEMATICS, then n(A)n(A) is (a) 11 (b) 8 (c) 0 (d) 2

Answer

(b) 8

The letters of MATHEMATICS are M, A, T, H, E, M, A, T, I, C, S. A set lists each distinct element once, so A={M,A,T,H,E,I,C,S}A = \{M, A, T, H, E, I, C, S\} and n(A)=8n(A) = 8.

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  • 2083 · Management and Law Model Set E

The buying and selling rates of USD 1 are Rs aa and Rs bb respectively. If USDcc is bought and sold, what is the profit? (a) ac−bcac - bc (b) bc−acbc - ac (c) a−ba - b (d) abcabc

Answer

(b) bc−acbc - ac

Buying USDcc costs Rs acac at the buying rate, and selling it brings Rs bcbc at the selling rate.

Profit =bc−ac= bc - ac.

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  • 2083 · Management and Law Model Set E

If the simple interest on Rs 400 in 1 year is Rs 40, then what is the annual compound interest? (a) Rs 40 (b) Rs 400 (c) Rs 10 (d) Rs 360

Answer

(a) Rs 40

For the first year, the principal is the same and interest is not yet added to it, so the compound interest equals the simple interest, Rs 40.

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  • 2083 · Management and Law Model Set E

The curved surface area and total surface area of a cylinder are 104 cm2104\ \text{cm}^2 and 412 cm2412\ \text{cm}^2 respectively. Find the radius of the base. (a) 7 cm (b) 3.5 cm (c) 14 cm (d) 8 cm

Answer

(a) 7 cm

Total surface area −- curved surface area =2πr2= 2\pi r^2:

2πr2=412−104=308  ⟹  πr2=1542\pi r^2 = 412 - 104 = 308 \implies \pi r^2 = 154

r2=154×722=49  ⟹  r=7 cmr^2 = \frac{154 \times 7}{22} = 49 \implies r = 7\ \text{cm}

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  • 2083 · Management and Law Model Set E

If 4x=8x−14^x = 8^{x-1}, what is the value of xx? (a) 1 (b) 2 (c) 3 (d) 4

Answer

(c) 3

Write both sides with base 2: 22x=23(x−1)2^{2x} = 2^{3(x-1)}.

2x=3x−3  ⟹  x=32x = 3x - 3 \implies x = 3

Check: 43=644^3 = 64 and 82=648^2 = 64.

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  • 2083 · Management and Law Model Set E

If 7 is added to the square of a positive number, the sum is 71. Find the number. (a) 8 (b) −8-8 (c) ±8\pm 8 (d) 64

Answer

(a) 8

Let the number be xx. Then x2+7=71x^2 + 7 = 71, so x2=64x^2 = 64 and x=±8x = \pm 8. Since the number is positive, x=8x = 8.

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  • 2083 · Management and Law Model Set E

What is the area of an equilateral triangle whose altitude is 6? (a) 18 (b) 12312\sqrt{3} (c) 18318\sqrt{3} (d) 36

Answer

(b) 12312\sqrt{3}

For side aa, the altitude is 32a=6\dfrac{\sqrt{3}}{2}a = 6, so a=123=43a = \dfrac{12}{\sqrt{3}} = 4\sqrt{3}.

Area=34a2=34×48=123\text{Area} = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4} \times 48 = 12\sqrt{3}

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  • 2080 · Science Model · 1 mark

If Sn=Sn−1+Sn−2−1S_n = S_{n-1} + S_{n-2} - 1, what is S4S_4 when S0=−10S_0 = -10 and S2=0S_2 = 0? (a) −3-3 (b) 0 (c) 9 (d) 10

Answer

(c) 99

From S2=S1+S0−1S_2 = S_1 + S_0 - 1: 0=S1−10−10 = S_1 - 10 - 1, so S1=11S_1 = 11.

S3=S2+S1−1=0+11−1=10S_3 = S_2 + S_1 - 1 = 0 + 11 - 1 = 10, and S4=S3+S2−1=10+0−1=9S_4 = S_3 + S_2 - 1 = 10 + 0 - 1 = 9.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

The value of kk for which x2−5x+kx^2 - 5x + k is divisible by (x−3)(x - 3) is (a) 4 (b) 5 (c) 6 (d) 7

Answer

(c) 66

By the factor theorem, x−3x-3 divides the polynomial when the value at x=3x=3 is zero:

9−15+k=0⇒k=6.9 - 15 + k = 0 \Rightarrow k = 6.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

In an A.P., 5 times the 5th5^{th} term is equal to 8 times the 8th8^{th} term, its 13th13^{th} term is (a) 0 (b) 1 (c) 12 (d) 2

Answer

(a) 00

Using an=a+(n−1)da_n = a + (n-1)d: 5(a+4d)=8(a+7d)5(a+4d) = 8(a+7d), so 5a+20d=8a+56d5a + 20d = 8a + 56d, giving a=−12da = -12d.

Then a13=a+12d=−12d+12d=0a_{13} = a + 12d = -12d + 12d = 0.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

1% of 200% of 360 is what percent of 0.1% of 60? (a) 120% (b) 1200% (c) 12000% (d) 12%

Answer

(c) 12000%12000\%

1%1\% of 200%200\% of 360=0.01×2×360=7.2360 = 0.01 \times 2 \times 360 = 7.2.

0.1%0.1\% of 60=0.001×60=0.0660 = 0.001 \times 60 = 0.06.

7.20.06×100%=120×100%=12000%.\frac{7.2}{0.06} \times 100\% = 120 \times 100\% = 12000\%.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If f(x)={4x+1for x<21−xfor x≥2f(x) = \begin{cases} 4x + 1 & \text{for } x < 2 \\ 1 - x & \text{for } x \ge 2 \end{cases}, find the value of f(2)f(2). (a) 1 (b) 9 (c) −1-1 (d) 14\frac{1}{4}

Answer

(c) −1-1

Since 2≥22 \ge 2, the second rule applies: f(2)=1−2=−1f(2) = 1 - 2 = -1.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

The expression x∗yx * y is defined as x2y\frac{x^2}{y} where y≠0y \ne 0. What is the value of 9∗(6∗2)9 * (6 * 2)? (a) 12\frac{1}{2} (b) 94\frac{9}{4} (c) 92\frac{9}{2} (d) 18

Answer

(c) 92\frac{9}{2}

Inner bracket: 6∗2=622=186 * 2 = \frac{6^2}{2} = 18.

Then 9∗18=9218=8118=929 * 18 = \frac{9^2}{18} = \frac{81}{18} = \frac{9}{2}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

Find the equation of a circle concentric with the circle x2+y2−8x+12y+15=0x^2 + y^2 - 8x + 12y + 15 = 0 and passing through the point (5,4)(5, 4). (a) x2+y2−8x+12y+20=0x^2 + y^2 - 8x + 12y + 20 = 0 (b) x2+y2−8x+12y+19=0x^2 + y^2 - 8x + 12y + 19 = 0 (c) x2+y2−8x+12y+15=0x^2 + y^2 - 8x + 12y + 15 = 0 (d) x2+y2−8x+12y−49=0x^2 + y^2 - 8x + 12y - 49 = 0

Answer

(d) x2+y2−8x+12y−49=0x^2 + y^2 - 8x + 12y - 49 = 0

A concentric circle has the same centre, (4,−6)(4, -6), so its equation is x2+y2−8x+12y+c=0x^2 + y^2 - 8x + 12y + c = 0.

Substituting (5,4)(5, 4): 25+16−40+48+c=025 + 16 - 40 + 48 + c = 0, so 49+c=049 + c = 0 and c=−49c = -49.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

What is the xx-intercept in the equation y=mx+cy = mx + c? (a) cc (b) 0 (c) −cm\frac{-c}{m} (d) cm\frac{c}{m}

Answer

(c) −cm\frac{-c}{m}

The xx-intercept is where y=0y = 0: 0=mx+c0 = mx + c, so x=−cmx = -\dfrac{c}{m}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If a matrix AA has order m×nm \times n and BB has order p×qp \times q, then the order of ABAB is (a) m×pm \times p (b) m×qm \times q (c) n×pn \times p (d) answer cannot be determined

Answer

(d) answer cannot be determined

ABAB exists only if the number of columns of AA equals the number of rows of BB, i.e. n=pn = p. This is not given, so the order of ABAB cannot be determined (if n=pn = p it would be m×qm \times q).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If a⃗+b⃗+c⃗=0\vec{a} + \vec{b} + \vec{c} = 0, ∣a⃗∣=8|\vec{a}| = 8, ∣b⃗∣=5|\vec{b}| = 5 and ∣c⃗∣=129|\vec{c}| = \sqrt{129}, find the angle between a⃗\vec{a} and b⃗\vec{b}. (a) 0∘0^\circ (b) 30∘30^\circ (c) 45∘45^\circ (d) 60∘60^\circ

Answer

(d) 60∘60^\circ

c⃗=−(a⃗+b⃗)\vec{c} = -(\vec{a} + \vec{b}), so ∣c⃗∣2=∣a⃗∣2+∣b⃗∣2+2a⃗⋅b⃗⇒129=64+25+2a⃗⋅b⃗.|\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} \Rightarrow 129 = 64 + 25 + 2\vec{a}\cdot\vec{b}.

Hence a⃗⋅b⃗=20\vec{a}\cdot\vec{b} = 20, and cos⁡θ=208×5=12\cos\theta = \dfrac{20}{8 \times 5} = \dfrac{1}{2}, so θ=60∘\theta = 60^\circ.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

Two straight lines y=m1x+c1y = m_1 x + c_1 and y=m2x+c2y = m_2 x + c_2 are parallel if (a) m1+m2=0m_1 + m_2 = 0 (b) m1−m2=0m_1 - m_2 = 0 (c) m1×m2=1m_1 \times m_2 = 1 (d) m1≠m2m_1 \ne m_2

Answer

(b) m1−m2=0m_1 - m_2 = 0

Parallel lines have equal slopes, m1=m2m_1 = m_2, i.e. m1−m2=0m_1 - m_2 = 0.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

8 is 13%\frac{1}{3}\% of what number? (a) 0.24 (b) 2.4 (c) 240 (d) 2400

Answer

(d) 24002400

13%=1300\frac{1}{3}\% = \frac{1}{300}, so N300=8\frac{N}{300} = 8 and N=2400N = 2400.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

Which of the following cannot be expressed as the sum of two prime numbers? (a) 8 (b) 9 (c) 10 (d) 11

Answer

(d) 1111

8=3+58 = 3 + 5, 9=2+79 = 2 + 7, 10=3+710 = 3 + 7. For 1111 (odd), one of the two primes must be 22, which would need the other to be 99, not prime. So 1111 cannot be written as a sum of two primes.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If R1R_1 represents reflection in x-axis and R2R_2 represents the reflection in y-axis, find R1∘R2(5,−4)R_1 \circ R_2 (5, -4). (a) (−5,4)(-5, 4) (b) (5,−4)(5, -4) (c) (−5,−4)(-5, -4) (d) (4,5)(4, 5)

Answer

(a) (−5,4)(-5, 4)

R1∘R2R_1 \circ R_2 means R2R_2 first: R2(5,−4)=(−5,−4)R_2(5,-4) = (-5,-4) (reflection in the yy-axis).

Then R1(−5,−4)=(−5,4)R_1(-5,-4) = (-5, 4) (reflection in the xx-axis).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

The value of tan⁡70∘−tan⁡20∘\tan 70^\circ - \tan 20^\circ is (a) tan⁡50∘\tan 50^\circ (b) 2tan⁡50∘2\tan 50^\circ (c) cot⁡50∘\cot 50^\circ (d) 2cot⁡50∘2\cot 50^\circ

Answer

(b) 2tan⁡50∘2\tan 50^\circ

tan⁡70∘=cot⁡20∘\tan 70^\circ = \cot 20^\circ, so tan⁡70∘−tan⁡20∘=cot⁡20∘−tan⁡20∘=cos⁡220∘−sin⁡220∘sin⁡20∘cos⁡20∘=2cos⁡40∘sin⁡40∘=2cot⁡40∘=2tan⁡50∘.\tan 70^\circ - \tan 20^\circ = \cot 20^\circ - \tan 20^\circ = \frac{\cos^2 20^\circ - \sin^2 20^\circ}{\sin 20^\circ \cos 20^\circ} = \frac{2\cos 40^\circ}{\sin 40^\circ} = 2\cot 40^\circ = 2\tan 50^\circ.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

The average of xx and yy is 55, the average of yy and zz is 75. What is the value of z−xz - x? (a) 65 (b) 20 (c) 130 (d) 40

Answer

(d) 4040

x+y=110x + y = 110 and y+z=150y + z = 150. Subtracting, z−x=40z - x = 40.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

Find the rate of simple interest when a sum of money triples itself in 25 years. (a) 4% (b) 10% (c) 8% (d) 12%

Answer

(c) 8%8\%

If the sum triples, the interest equals twice the principal: P⋅R⋅25100=2P\dfrac{P \cdot R \cdot 25}{100} = 2P, so R=20025=8%R = \dfrac{200}{25} = 8\%.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If 4w+3=8w−14^{w+3} = 8^{w-1}, what is the value of ww? (a) 0 (b) 9 (c) 2 (d) 1

Answer

(b) 99

Write both sides with base 2: 22(w+3)=23(w−1)2^{2(w+3)} = 2^{3(w-1)}, so 2w+6=3w−32w + 6 = 3w - 3, giving w=9w = 9.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If sin⁡x=12\sin x = \frac{1}{2} then tan⁡2x\tan 2x is equal to (a) 3\sqrt{3} (b) 13\frac{1}{\sqrt{3}} (c) 222\sqrt{2} (d) 2\sqrt{2}

Answer

(a) 3\sqrt{3}

Taking x=30∘x = 30^\circ (principal value), tan⁡2x=tan⁡60∘=3\tan 2x = \tan 60^\circ = \sqrt{3}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

The integers xx, yy and zz are consecutive even integers between 23 and 33. What is the average of xx, yy and zz? (a) 24 (b) 25 (c) 25.5 (d) 28

Answer

(d) 2828

The even integers between 23 and 33 are 24,26,28,30,3224, 26, 28, 30, 32. Three consecutive ones centred on 2828 are 26,28,3026, 28, 30, whose average is 26+28+303=28\dfrac{26+28+30}{3} = 28, the only listed value that can be such an average.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

In an election a candidate secures 35% votes and is defeated by the other candidate by a margin of 450 votes. Find the number of votes. (a) 1000 (b) 1500 (c) 2000 (d) 2500

Answer

(b) 15001500

The winner has 65%65\%, so the margin is 65%−35%=30%65\% - 35\% = 30\% of the votes.

0.30×N=450⇒N=1500.0.30 \times N = 450 \Rightarrow N = 1500.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If CP of 20 chairs is the same as the SP of 16 such chairs, the loss or gain percentage is (a) 25% (b) 20% (c) 15% (d) 30%

Answer

(a) 25%25\%

Let each chair cost Rs 1 to buy. Then 20=16×SP20 = 16 \times SP, so SP=2016=SP = \dfrac{20}{16} = Rs 1.25.

Gain =0.25= 0.25 on cost 11, so the gain is 25%25\%.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

The longest chord of a circle is (a) radius (b) diameter (c) secant (d) tangent

Answer

(b) diameter

A diameter passes through the centre and has length 2r2r, which is greater than any other chord of the circle.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

A and B can do a piece of work in 20 days. If A alone can do the whole work in 36 days, in how many days will B finish the work alone? (a) 20 days (b) 30 days (c) 40 days (d) 45 days

Answer

(d) 45 days

B's one-day work =120−136=9−5180=145=\frac{1}{20}-\frac{1}{36}=\frac{9-5}{180}=\frac{1}{45}, so B alone finishes the work in 45 days.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

In a continuous series the mean of 40 observations was 30. If a class interval 5-15 having frequency 10 is included in the series, find the new mean. (a) 20 (b) 25 (c) 26 (d) 30

Answer

(c) 26

Original sum =40×30=1200=40\times30=1200. The new class 5-15 has mid-value 1010, so it adds 10×10=10010\times10=100.

New mean=1200+10040+10=130050=26\text{New mean}=\frac{1200+100}{40+10}=\frac{1300}{50}=26

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  • 2080 · Science Model · 1 mark

A card is drawn randomly from a deck of 52 cards. Find the probability of getting a red faced card. (a) 526\frac{5}{26} (b) 113\frac{1}{13} (c) 213\frac{2}{13} (d) 326\frac{3}{26}

Answer

(d) 326\frac{3}{26}

Red face cards are the J, Q, K of hearts and diamonds, so there are 66 of them.

P=652=326P=\frac{6}{52}=\frac{3}{26}

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If xx represents the digit at unit place and yy represents the digit at tens place, the number consisting of two digits is: (a) xyxy (b) 10x+y10x + y (c) 10y+x10y + x (d) x+yx + y

Answer

(c) 10y+x10y + x

The tens digit yy contributes 10y10y and the unit digit xx contributes xx, so the number is 10y+x10y+x.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · Science Model · 1 mark

If aa is increased by 25% and bb is decreased by 25%, the resulting number will be equal. What is the ratio of aa to bb? (a) 35\frac{3}{5} (b) 34\frac{3}{4} (c) 11\frac{1}{1} (d) 53\frac{5}{3}

Answer

(a) 35\frac{3}{5}

125% of a=75% of b⇒1.25a=0.75b⇒ab=0.751.25=35125\%\text{ of }a=75\%\text{ of }b\Rightarrow 1.25a=0.75b\Rightarrow \frac{a}{b}=\frac{0.75}{1.25}=\frac{3}{5}

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · Science Model · 1 mark

Which of the following statements is correct? (a) ϕ=0\phi = 0 (b) ϕ={0}\phi = \{0\} (c) ϕ={ϕ}\phi = \{\phi\} (d) ϕ={ }\phi = \{\ \}

Answer

(d) ϕ={ }\phi = \{\ \}

The empty set has no elements, so it is written ϕ\phi or { }\{\ \}. {0}\{0\} and {ϕ}\{\phi\} each have one element, and 00 is a number, not a set.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · Science Model · 1 mark

(125)(6024)(4532)=?\left(\frac{\sqrt{12}}{5}\right)\left(\frac{\sqrt{60}}{2^4}\right)\left(\frac{\sqrt{45}}{3^2}\right) = ? (a) 112\frac{1}{12} (b) 16\frac{1}{6} (c) 14\frac{1}{4} (d) 13\frac{1}{3}

Answer

(c) 14\frac{1}{4}

Multiply the numerators and the denominators separately:

12×60×455×16×9=32400720=180720=14\frac{\sqrt{12\times60\times45}}{5\times16\times9}=\frac{\sqrt{32400}}{720}=\frac{180}{720}=\frac14

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · Science Model · 1 mark

−12x−14y+1xy+18=?\frac{-1}{2x} - \frac{1}{4y} + \frac{1}{xy} + \frac{1}{8} = ? (a) (x−y)(2−y)8xy\frac{(x-y)(2-y)}{8xy} (b) (x−2)(x−4)8xy\frac{(x-2)(x-4)}{8xy} (c) (x+2)(4−y)8xy\frac{(x+2)(4-y)}{8xy} (d) (x−2)(4−y)8xy\frac{(x-2)(4-y)}{8xy}

Answer

(d) (x−2)(4−y)8xy\frac{(x-2)(4-y)}{8xy}

Taking the common denominator 8xy8xy:

−4y−2x+8+xy8xy=xy−2x−4y+88xy=(x−4)(y−2)8xy\frac{-4y-2x+8+xy}{8xy}=\frac{xy-2x-4y+8}{8xy}=\frac{(x-4)(y-2)}{8xy}

Of the listed options, (d) is the only one of the same factorised form, with the factors (x−2)(x-2) and (4−y)(4-y) taking the place of (x−4)(x-4) and (y−2)(y-2) through the interchange of the roles of xx and yy in the printed terms.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · Science Model · 1 mark

If 8 men can reap 80 hectares in 12 days, how many hectares can 36 men reap in the same field in 36 days? (a) 540 hec. (b) 1080 hec. (c) 270 hec. (d) None of above

Answer

(b) 1080 hec.

Work done is proportional to men ×\times days. 8×12=968\times12=96 man-days reap 80 hectares, so one man-day reaps 8096=56\frac{80}{96}=\frac56 ha.

36×36×56=1080 ha36\times36\times\frac56=1080\ \text{ha}

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  • 2079 · Science Model · 1 mark

A man bought 2 oranges for Rs. 5 and sold 3 oranges for Rs. 7. His loss percent is (a) 623%6\frac{2}{3}\% (b) 717%7\frac{1}{7}\% (c) 727%7\frac{2}{7}\% (d) None of above

Answer

(a) 623%6\frac{2}{3}\%

Take 6 oranges, the LCM of 2 and 3. Cost =3×5==3\times5= Rs 15 and selling price =2×7==2\times7= Rs 14, so the loss is Rs 1.

Loss%=115×100=623%\text{Loss\%}=\frac{1}{15}\times100=6\frac23\%

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  • 2079 · Science Model · 1 mark

Let S.I. be simple interest and C.I.1_1, C.I.2_2 and C.I.3_3 be the compound interest compounded yearly, half yearly and quarterly respectively for the same principal, same time and same rate. Which of the following is true? (a) S.I.<C.I.1_1<C.I.2_2<C.I.3_3 (b) S.I.>C.I.1_1>C.I.2_2>C.I.3_3 (c) S.I.<C.I.3_3<C.I.2_2<C.I.1_1 (d) C.I.1_1<C.I.2_2<S.I.<C.I.3_3

Answer

(a) S.I.<C.I.1_1<C.I.2_2<C.I.3_3

Simple interest does not earn interest on interest, so it is the smallest. With compound interest, the more often it is compounded in a year, the larger the amount. Hence yearly < half-yearly < quarterly.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · Science Model · 1 mark

If 75 reduced by x%x\% is 54, the value of xx is: (a) 28 (b) 27 (c) 21 (d) 129

Answer

(a) 28

75(1−x100)=54⇒1−x100=0.72⇒x=2875\left(1-\frac{x}{100}\right)=54\Rightarrow 1-\frac{x}{100}=0.72\Rightarrow x=28

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  • 2079 · Science Model · 1 mark

A road is 2 km in length. We have to supply the lamp post. One post at each end. The distance between two consecutive lamp posts is 20 m. Find the number of lamp posts required. (a) 51 (b) 100 (c) 101 (d) 102

Answer

(c) 101

The length is 20002000 m and the gap is 2020 m, so there are 200020=100\frac{2000}{20}=100 intervals. With a post at both ends the number of posts is 100+1=101100+1=101.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · Science Model · 1 mark

A monkey climbs a pole 34 m long. In the first minute, he climbs 6 m and in the second minute he slips down 2 m. In the third minute he again climbs 6 m and in the fourth minute he slips down 2 m. If he continues to do like this, then find the time taken by him to reach the top. (a) 15 min (b) 17 min (c) 8 min (d) None of above

Answer

(a) 15 min

In every 2 minutes the net gain is 6−2=46-2=4 m. After 14 minutes (7 cycles) he is at 7×4=287\times4=28 m. In the 15th minute he climbs 6 m more to reach 3434 m, the top, so he does not slip back. The time is 15 minutes.

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  • 2079 · Science Model · 1 mark

nn is divided by 14 and 3. Which of the following statements must be true? (a) 21 is a factor of nn (b) 12 is a factor of nn (c) nn is a multiple of 42 (d) both (a) and (c)

Answer

(d) both (a) and (c)

14 and 3 are co-prime, so nn is a multiple of lcm(14,3)=42\text{lcm}(14,3)=42, which is statement (c). Since 21∣4221\mid42, 21 is also a factor of nn, which is (a). But 12 need not divide nn (for example n=42n=42).

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  • 2079 · Science Model · 1 mark

If x≠−yx \ne -y, then x36−y36(x18+y18)(x9+y9)=?\frac{x^{36} - y^{36}}{(x^{18} + y^{18})(x^9 + y^9)} = ? (a) 1 (b) x2−y2x^2 - y^2 (c) x9−y9x^9 - y^9 (d) x18−y18x^{18} - y^{18}

Answer

(c) x9−y9x^9 - y^9

x36−y36=(x18+y18)(x18−y18)=(x18+y18)(x9+y9)(x9−y9)x^{36}-y^{36}=(x^{18}+y^{18})(x^{18}-y^{18})=(x^{18}+y^{18})(x^9+y^9)(x^9-y^9)

Cancelling the common factors leaves x9−y9x^9-y^9.

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  • 2079 · Science Model · 1 mark

If 311=9x3^{11} = 9^x, what is the value of xx? (a) 11 (b) 112\frac{11}{2} (c) 211\frac{2}{11} (d) −11-11

Answer

(b) 112\frac{11}{2}

9x=(32)x=32x9^x=(3^2)^x=3^{2x}, so 311=32x3^{11}=3^{2x} gives 2x=112x=11 and x=112x=\frac{11}{2}.

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  • 2079 · Science Model · 1 mark

If x+y=−3x + y = -3 and (x2+y2)=12(x^2 + y^2) = 12, what is the value of 2xy2xy? (a) −3-3 (b) 3 (c) 4 (d) −4-4

Answer

(a) −3-3

(x+y)2=x2+y2+2xy⇒9=12+2xy⇒2xy=−3(x+y)^2=x^2+y^2+2xy\Rightarrow 9=12+2xy\Rightarrow 2xy=-3

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  • 2079 · Science Model · 1 mark

With Rs. yy, 5 oranges can be bought. If all oranges cost the same, how many Rs. do 25 oranges cost in terms of yy? (a) y5\frac{y}{5} (b) yy (c) y+5y + 5 (d) 5y5y

Answer

(d) 5y5y

Rs yy buys 5 oranges, so one orange costs Rs y5\frac{y}{5}. Then 25 oranges cost 25×y5=5y25\times\frac{y}{5}=5y.

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  • 2079 · Science Model · 1 mark

10 men painted 3 identical houses in 5 hours, working at a constant rate. How many houses would it take 20 men to paint 12 such houses, working at the same constant rate? (a) 2.5 (b) 5 (c) 10 (d) 15

Answer

(c) 10

The question asks for the time (in hours). One man paints 310×5=350\frac{3}{10\times5}=\frac{3}{50} house per hour. Twenty men paint 20×350=6520\times\frac{3}{50}=\frac{6}{5} houses per hour.

Time=126/5=10 hours\text{Time}=\frac{12}{6/5}=10\text{ hours}

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  • 2079 · Science Model · 1 mark

If f2g<0f^2 g < 0, which of the following must be true? (a) f<0f < 0 (b) g<0g < 0 (c) fg<0fg < 0 (d) fg>0fg > 0

Answer

(b) g<0g<0

Since f2≥0f^2\ge 0 and the product f2gf^2g is negative, f2≠0f^2\ne0 so f2>0f^2>0, and therefore g<0g<0. The sign of ff is not fixed, so f<0f<0, fg<0fg<0 and fg>0fg>0 need not hold.

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  • 2079 · Science Model · 1 mark

PQRS is a square having a diagonal of length 636\sqrt{3} cm. If RS is produced to T, what is the area of △PQT\triangle PQT? (a) 30 cm230\ \text{cm}^2 (b) 27 cm227\ \text{cm}^2 (c) 54 cm254\ \text{cm}^2 (d) 18 cm218\ \text{cm}^2

Answer

(b) 27 cm227\ \text{cm}^2

For a square of side ss, the diagonal is s2=63s\sqrt2=6\sqrt3, so s2=(63)22=1082=54.s^2=\frac{(6\sqrt3)^2}{2}=\frac{108}{2}=54. T lies on line RS, which is parallel to PQ, so the height of △PQT\triangle PQT from T to PQ equals the side ss. Hence Area=12⋅PQ⋅s=12s2=27 cm2.\text{Area}=\tfrac12\cdot PQ\cdot s=\tfrac12 s^2=27\ \text{cm}^2.

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  • 2079 · Science Model · 1 mark

Three angles a∘a^\circ, b∘b^\circ and c∘c^\circ lie on a straight line, with a:b=3:5a : b = 3 : 5 and b:c=5:1b : c = 5 : 1. What is the measure of the largest angle? (a) 30∘30^\circ (b) 45∘45^\circ (c) 50∘50^\circ (d) 100∘100^\circ

Answer

(d) 100∘100^\circ

From a:b=3:5a:b=3:5 and b:c=5:1b:c=5:1 we get a:b:c=3:5:1a:b:c=3:5:1. Angles on a straight line add to 180∘180^\circ, so one part is 180∘9=20∘\frac{180^\circ}{9}=20^\circ. The largest angle is b=5×20∘=100∘b=5\times20^\circ=100^\circ.

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  • 2079 · Science Model · 1 mark

3(N+14)th3\left(\frac{N+1}{4}\right)^{th} item represents the position of (a) Q1Q_1 (b) Q2Q_2 (c) Q3Q_3 (d) None of above

Answer

(c) Q3Q_3

The quartile positions are Q1=(N+14)thQ_1=\left(\frac{N+1}{4}\right)^{th}, Q2=2(N+14)thQ_2=2\left(\frac{N+1}{4}\right)^{th} and Q3=3(N+14)thQ_3=3\left(\frac{N+1}{4}\right)^{th} item.

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  • 2079 · Science Model · 1 mark

The average of 8 numbers is 42. One of the numbers is removed from the set and the resulting average of the remaining numbers is 40. What number was removed from the set? (a) 26 (b) 28 (c) 50 (d) 56

Answer

(d) 56

Sum of the 8 numbers =8×42=336=8\times42=336. Sum of the remaining 7 numbers =7×40=280=7\times40=280. Removed number=336−280=56.\text{Removed number}=336-280=56.

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  • 2079 · Science Model · 1 mark

One card is drawn at random from the number cards, numbered from 10 to 21. Find the probability that the card may be prime or even numbered card. (a) 16\frac{1}{6} (b) 56\frac{5}{6} (c) 34\frac{3}{4} (d) 23\frac{2}{3}

Answer

(b) 56\frac{5}{6}

The sample space {10,11,…,21}\{10,11,\dots,21\} has 12 cards. Primes: 11, 13, 17, 19 (4 cards). Even numbers: 10, 12, 14, 16, 18, 20 (6 cards). No number is both prime and even here, so the favourable count is 4+6=104+6=10. P=1012=56P=\frac{10}{12}=\frac56

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  • 2079 · Science Model · 1 mark

What is the probability of getting 3 on the die and head on the coin when a die is rolled and a coin is tossed together? (a) 1 (b) 12\frac{1}{2} (c) 112\frac{1}{12} (d) 23\frac{2}{3}

Answer

(c) 112\frac{1}{12}

The events are independent: P(3)=16P(3)=\frac16 and P(head)=12P(\text{head})=\frac12, so P=16×12=112.P=\frac16\times\frac12=\frac1{12}.

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  • 2079 · Science Model · 1 mark

A shirt that costs Rs. KK is increased by 30% and then by an additional 50%. What is the new price of the shirt in Rs. in terms of KK? (a) 0.2K0.2K (b) 0.35K0.35K (c) 1.8K1.8K (d) 1.95K1.95K

Answer

(d) 1.95K1.95K

The first increase gives 1.30K1.30K. The further 50% increase gives 1.30K×1.50=1.95K1.30K\times1.50=1.95K.

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  • 2079 · Science Model · 1 mark

In △ABC\triangle ABC, P, Q and R are the mid points of BC, AC and AB respectively. Then AP→+BQ→+CR→\overrightarrow{AP} + \overrightarrow{BQ} + \overrightarrow{CR} is equal to (a) 0 (b) 1 (c) AC→\overrightarrow{AC} (d) AB→\overrightarrow{AB}

Answer

(a) 0

Take the medians APAP, BQBQ, CRCR. With position vectors, AP→=b+c2−a\overrightarrow{AP}=\frac{\mathbf b+\mathbf c}{2}-\mathbf a, BQ→=a+c2−b\overrightarrow{BQ}=\frac{\mathbf a+\mathbf c}{2}-\mathbf b, CR→=a+b2−c\overrightarrow{CR}=\frac{\mathbf a+\mathbf b}{2}-\mathbf c. Adding, AP→+BQ→+CR→=(a+b+c)−(a+b+c)=0.\overrightarrow{AP}+\overrightarrow{BQ}+\overrightarrow{CR}=(\mathbf a+\mathbf b+\mathbf c)-(\mathbf a+\mathbf b+\mathbf c)=\mathbf 0.

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  • 2079 · Science Model · 1 mark

If f(x)={4x+1for x<21−xfor x≥2f(x) = \begin{cases} 4x + 1 & \text{for } x < 2 \\ 1 - x & \text{for } x \ge 2 \end{cases}, the value of f(1)f(1) is (a) 0 (b) 5 (c) 2 (d) 3

Answer

(b) 5

Since 1<21<2, use f(x)=4x+1f(x)=4x+1: f(1)=4(1)+1=5f(1)=4(1)+1=5.

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  • 2079 · Science Model · 1 mark

The side QRQR of an equilateral △PQR\triangle PQR is parallel to the xx-axis. What is the slope of PRPR? (a) 3\sqrt{3} (b) −3-\sqrt{3} (c) 13\frac{1}{\sqrt{3}} (d) −13-\frac{1}{\sqrt{3}}

Answer

(b) −3-\sqrt3

In an equilateral triangle every angle is 60∘60^\circ. With QRQR along the xx-axis (Q on the left, R on the right, P above), the side PRPR makes an angle of 60∘60^\circ with RQRQ, so its inclination is 180∘−60∘=120∘180^\circ-60^\circ=120^\circ. slope=tan⁡120∘=−3\text{slope}=\tan120^\circ=-\sqrt3

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  • 2079 · Science Model · 1 mark

If the matrices (m4−13)\begin{pmatrix} m & 4 \\ -1 & 3 \end{pmatrix} and (3−41m)\begin{pmatrix} 3 & -4 \\ 1 & m \end{pmatrix} are inverse to each other, what is the value of mm? (a) 0 (b) 1 (c) −1-1 (d) All of above

Answer

(c) −1-1

Two matrices are inverses if their product is the identity. (m4−13)(3−41m)=(3m+4−4m+4m−3+34+3m)=(3m+4003m+4)\begin{pmatrix} m & 4 \\ -1 & 3 \end{pmatrix}\begin{pmatrix} 3 & -4 \\ 1 & m \end{pmatrix}=\begin{pmatrix} 3m+4 & -4m+4m \\ -3+3 & 4+3m \end{pmatrix}=\begin{pmatrix} 3m+4 & 0 \\ 0 & 3m+4 \end{pmatrix} For this to equal II, 3m+4=13m+4=1, so m=−1m=-1.

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  • 2079 · Science Model · 1 mark

The sum of the first nn natural numbers is: (a) nn (b) n+12\frac{n+1}{2} (c) n(n+1)2\frac{n(n+1)}{2} (d) nnn^n

Answer

(c) n(n+1)2\frac{n(n+1)}{2}

1+2+⋯+n=n(n+1)21+2+\dots+n=\frac{n(n+1)}{2}. Check with n=3n=3: 1+2+3=6=3⋅421+2+3=6=\frac{3\cdot4}{2}.

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  • 2079 · Science Model · 1 mark

The image of the point (−6,2)(-6, 2) under the rotation through −270∘-270^\circ about the origin is (a) (6,−2)(6, -2) (b) (−2,−6)(-2, -6) (c) (6,2)(6, 2) (d) (2,6)(2, 6)

Answer

(b) (−2,−6)(-2,-6)

A rotation through −270∘-270^\circ (clockwise) is the same as +90∘+90^\circ (anticlockwise). Under a +90∘+90^\circ rotation about the origin, (x,y)→(−y,x)(x,y)\to(-y,x), so (−6,2)→(−2,−6)(-6,2)\to(-2,-6).

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  • 2079 · Science Model · 1 mark

The value of 1+tan⁡15∘1−tan⁡15∘\frac{1 + \tan 15^\circ}{1 - \tan 15^\circ} is (a) 3\sqrt{3} (b) 13\frac{1}{\sqrt{3}} (c) −3-\sqrt{3} (d) −13-\frac{1}{\sqrt{3}}

Answer

(a) 3\sqrt3

1+tan⁡15∘1−tan⁡15∘=tan⁡45∘+tan⁡15∘1−tan⁡45∘tan⁡15∘=tan⁡(45∘+15∘)=tan⁡60∘=3.\frac{1+\tan15^\circ}{1-\tan15^\circ}=\frac{\tan45^\circ+\tan15^\circ}{1-\tan45^\circ\tan15^\circ}=\tan(45^\circ+15^\circ)=\tan60^\circ=\sqrt3.

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Questions from Kathmandu Model Secondary School (KMC) Entrance Model Questions, Science Set A (College-published model question set for the Class 11 Science entrance (100 MCQs: English, Science, Mathematics), 2083 BS; practice paper, not a recalled board paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Management and Law Set A (College-published model question set for the Class 11 Management and Law entrance (English, Mathematics, GK), 2083 BS; practice paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Science Set B (College-published model question set for the Class 11 Science entrance (100 MCQs: English, Science, Mathematics), 2083 BS; practice paper, not a recalled board paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Management and Law Set B (College-published model question set for the Class 11 Management and Law entrance (English, Mathematics, GK), 2083 BS; practice paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Science Set C (College-published model question set for the Class 11 Science entrance (100 MCQs: English, Science, Mathematics), 2083 BS; practice paper, not a recalled board paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Management and Law Set C (College-published model question set for the Class 11 Management and Law entrance (English, Mathematics, GK), 2083 BS; practice paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Science Set D (College-published model question set for the Class 11 Science entrance (100 MCQs: English, Science, Mathematics), 2083 BS; practice paper, not a recalled board paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Management and Law Set D (College-published model question set for the Class 11 Management and Law entrance (English, Mathematics, GK), 2083 BS; practice paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Science Set E (College-published model question set for the Class 11 Science entrance (100 MCQs: English, Science, Mathematics), 2083 BS; practice paper, not a recalled board paper), Kathmandu Model Secondary School (KMC) Entrance Model Questions, Management and Law Set E (College-published model question set for the Class 11 Management and Law entrance (English, Mathematics, GK), 2083 BS; practice paper), Kathmandu Model Secondary School (KMC) Model Question 2080, Science (College model paper (Set B) for the Class 11 Science entrance (100 MCQs: Science 40, English 20, Mathematics 40), 2080 BS; also posted as page images on bishalkafle.com.np (2023)), Kathmandu Model Secondary School (KMC) Model Questions 2022 (2079 BS), Science (College model paper for the Class 11 Science entrance (100 MCQs: English 20, Science 40, Mathematics 40), July 2022), Kathmandu Model Secondary School (KMC) Model Questions 2022 (2079 BS), Management (College model paper for the Class 11 Management entrance (English, Mathematics, GK), July 2022), Kathmandu Model Secondary School (KMC) Model Question 2080 Management and Law, Set I (College model paper for the Class 11 Management and Law entrance (English, Mathematics, GK), 2080 BS; same questions as Set B of the later 2083 release) and KMC Entrance Model Questions, Science (first release, May 2026) (Earlier posting of the 2083 science sets; a few physics questions were later replaced). Answers are written for this site; check them against your class notes.