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Chapter 3

Mathematics

CTEVT diploma and PCL entrance past exam questions

Past questions and answers

73 questions set from this chapter. Most repeated first.

  • 2079 · 1 mark

What is the total surface area of a cone when the base radius of the cone is 7 cm and the sum of the radius and slant height is 28 cm? (a) 231 cm² (b) 400 cm² (c) 462 cm² (d) 616 cm²

Similar questions: If the total surface area of (2075)

Answer

(d) 616 cm²

Given r=7r = 7 cm and r+l=28r + l = 28 cm.

TSA=πr(r+l)=227×7×28=616 cm2\text{TSA} = \pi r(r + l) = \frac{22}{7} \times 7 \times 28 = 616 \text{ cm}^2

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  • 2075 · 1 mark

If the total surface area of the cone is 704 cm2704\ cm^2 and its base radius is 7 cm, the slant height of the cone is (a) 24 cm (b) 22 cm (c) 25 cm (d) 44 cm

Similar questions: What is the total surface area (2079)

Answer

(c) 25 cm Total surface area =πr(r+l)=\pi r(r+l): 227×7 (7+l)=704⇒22(7+l)=704⇒7+l=32⇒l=25 cm\frac{22}{7}\times 7\,(7+l)=704 \Rightarrow 22(7+l)=704 \Rightarrow 7+l=32 \Rightarrow l=25\text{ cm}

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  • 2079 · 1 mark

The area of triangle XYZ is 45 cm². If XY=12XY = 12 cm and YZ=15YZ = 15 cm, what will be the value of ∠XYZ\angle XYZ? (a) 30° (b) 45° (c) 60° (d) 90°

Answer

(a) 30°

Area=12 XY⋅YZsin⁡∠XYZ\text{Area} = \tfrac{1}{2}\, XY \cdot YZ \sin\angle XYZ

45=12×12×15×sin⁡∠XYZ=90sin⁡∠XYZ45 = \tfrac{1}{2} \times 12 \times 15 \times \sin\angle XYZ = 90\sin\angle XYZ

So sin⁡∠XYZ=12\sin\angle XYZ = \frac{1}{2} and ∠XYZ=30°\angle XYZ = 30°.

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  • 2079 · 1 mark

If x=10ax = 10^a, y=10by = 10^b and xbya=100x^b y^a = 100, what is the value of abab? (a) 1 (b) 2\sqrt{2} (c) 2 (d) 10

Answer

(a) 1

xbya=(10a)b(10b)a=10ab⋅10ab=102abx^b y^a = (10^a)^b (10^b)^a = 10^{ab} \cdot 10^{ab} = 10^{2ab}

Since this equals 100=102100 = 10^2, 2ab=22ab = 2, so ab=1ab = 1.

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  • 2079 · 1 mark

If 3m+1−2=0\sqrt{3m+1} - 2 = 0, what is the value of mm? (a) 0 (b) -1 (c) 1 (d) 4

Answer

(c) 1

3m+1=2⇒3m+1=4⇒m=1\sqrt{3m+1} = 2 \Rightarrow 3m + 1 = 4 \Rightarrow m = 1

Check: 4−2=0\sqrt{4} - 2 = 0.

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  • 2079 · 1 mark

If m+n=7m + n = 7 and 2m−3n=42m - 3n = 4, what will be the value of m−nm - n? (a) 2.5 (b) 5 (c) 3 (d) 5.2

Answer

(c) 3

From m+n=7m + n = 7, m=7−nm = 7 - n. Then 2(7−n)−3n=42(7 - n) - 3n = 4 gives 14−5n=414 - 5n = 4, so n=2n = 2 and m=5m = 5.

m−n=5−2=3m - n = 5 - 2 = 3

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  • 2079 · 1 mark

6 black and 7 red marbles of the same size and shape are kept in a bag. Two marbles are drawn one at a time without replacement. What is the probability of getting a red marble at first and then a black marble? (a) 726\frac{7}{26} (b) 2726\frac{27}{26} (c) 7884\frac{78}{84} (d) 926\frac{9}{26}

Answer

(a) 726\frac{7}{26}

Total marbles = 13. Red first: 713\frac{7}{13}; then black from the remaining 12: 612\frac{6}{12}.

P=713×612=42156=726P = \frac{7}{13} \times \frac{6}{12} = \frac{42}{156} = \frac{7}{26}

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  • 2079 · 1 mark

If 10 is added to the number that is seven times the square of a natural number, the result is 353. What will be the number? (a) ±7\pm 7 (b) -7 (c) 7 (d) 9

Answer

(c) 7

Let the natural number be xx. Then 7x2+10=3537x^2 + 10 = 353, so 7x2=3437x^2 = 343 and x2=49x^2 = 49, giving x=±7x = \pm 7. Since the number is natural, x=7x = 7.

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  • 2079 · 1 mark

The area of an equilateral triangle is 939\sqrt{3} cm². What is the perimeter of the triangle? (a) 6 cm (b) 9 cm (c) 18 cm (d) 27327\sqrt{3} cm

Answer

(c) 18 cm

34a2=93⇒a2=36⇒a=6 cm\frac{\sqrt{3}}{4}a^2 = 9\sqrt{3} \Rightarrow a^2 = 36 \Rightarrow a = 6 \text{ cm}

Perimeter=3a=18 cm\text{Perimeter} = 3a = 18 \text{ cm}

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  • 2079 · 1 mark

What is the result when x2−xy+y2x−y−x2+xy+y2x+y\frac{x^2 - xy + y^2}{x - y} - \frac{x^2 + xy + y^2}{x + y} is simplified? (a) 2(x3+x3)2(x^3 + x^3) (b) 2x3x2−y2\frac{2x^3}{x^2 - y^2} (c) 2y3x2−y2\frac{2y^3}{x^2 - y^2} (d) (2y)3x2−y2\frac{(2y)^3}{x^2 - y^2}

Answer

(c) 2y3x2−y2\dfrac{2y^3}{x^2 - y^2}

Take the common denominator (x−y)(x+y)=x2−y2(x-y)(x+y) = x^2 - y^2:

(x2−xy+y2)(x+y)−(x2+xy+y2)(x−y)x2−y2=(x3+y3)−(x3−y3)x2−y2=2y3x2−y2\frac{(x^2 - xy + y^2)(x+y) - (x^2 + xy + y^2)(x-y)}{x^2 - y^2} = \frac{(x^3 + y^3) - (x^3 - y^3)}{x^2 - y^2} = \frac{2y^3}{x^2 - y^2}

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  • 2079 · 1 mark

What is the highest common factor (H.C.F.) of two algebraic expressions (x+2)(x+2) and (x−2)(x-2)? (a) x+2x + 2 (b) x−2x - 2 (c) x2−4x^2 - 4 (d) 1

Answer

(d) 1

(x+2)(x+2) and (x−2)(x-2) are different linear factors with no common factor other than 1, so their H.C.F. is 1.

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  • 2079 · 1 mark

What is the order of the surd 3743\sqrt[4]{7}? (a) 1 (b) 2 (c) 3 (d) 4

Answer

(d) 4

The order of a surd an\sqrt[n]{a} is nn. In 3743\sqrt[4]{7} the root index is 4, so the order is 4.

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  • 2079 · 1 mark

What is the H.C.F. of x2−1x^2 - 1 and x2−3x+2x^2 - 3x + 2? (a) x−1x - 1 (b) x+1x + 1 (c) (x−1)(x−2)(x-1)(x-2) (d) (x+1)(x−1)(x+1)(x-1)

Answer

(a) x−1x - 1

x2−1=(x−1)(x+1),x2−3x+2=(x−1)(x−2)x^2-1=(x-1)(x+1),\qquad x^2-3x+2=(x-1)(x-2)

The only common factor is x−1x-1, so it is the H.C.F.

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  • 2079 · 1 mark

If A={2,3,5}A = \{2, 3, 5\} and B={1,4}B = \{1, 4\}, what is the value of n(A∩B)n(A \cap B)? (a) θ\theta (b) { }\{\ \} (c) 0 (d) 5

Answer

(c) 0

A={2,3,5}A=\{2,3,5\} and B={1,4}B=\{1,4\} have no common element, so A∩B=∅A\cap B=\varnothing and n(A∩B)=0n(A\cap B)=0.

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  • 2079 · 1 mark

How many different values of the variable of unit place digit in a two digit number? (a) 1 (b) 9 (c) 10 (d) yy

Answer

(c) 10

The unit place digit of a two-digit number can be any of 0,1,2,…,90,1,2,\dots,9, which gives 10 different values.

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  • 2079 · 1 mark

If 2x+1−4=02^{x+1} - 4 = 0, what is the value of xx? (a) 0 (b) -1 (c) 1 (d) 2

Answer

(c) 1

2x+1=4=22⇒x+1=2⇒x=12^{x+1}=4=2^2 \Rightarrow x+1=2 \Rightarrow x=1

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  • 2079 · 1 mark

Which one is the formula to calculate the curved surface area of a cylinder? (a) πrh\pi r h (b) 2πrh2\pi r h (c) 2πr2h2\pi r^2 h (d) 2π(r+h)2\pi(r + h)

Answer

(b) 2πrh2\pi r h

Unrolling the curved surface of a cylinder gives a rectangle of length 2πr2\pi r (the circumference) and width hh, so the curved surface area is 2πrh2\pi r h.

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  • 2079 · 1 mark

What is the relation between the area of a square and a triangle standing on the same base and between the same parallel lines? (a) Area of square = area of triangle (b) Area of square = 12\frac{1}{2} area of triangle (c) Area of square = 2 area of triangle (d) Area of triangle = 2 area of square

Answer

(c) Area of square = 2 area of triangle

Let the common base be bb and the height between the parallels be hh. The area of the triangle is 12bh\frac{1}{2}bh and the area of the parallelogram (square or rectangle) is bhbh, which is twice the triangle. So area of square = 2 area of triangle.

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  • 2079 · 1 mark

The marked price of an article is Rs. xx. If the price of the article after discount is Rs. yy, what percent discount is provided? (a) yx×100%\frac{y}{x} \times 100\% (b) xy×100%\frac{x}{y} \times 100\% (c) x−yx%\frac{x-y}{x}\% (d) x−yx×100%\frac{x-y}{x} \times 100\%

Answer

(d) x−yx×100%\frac{x-y}{x} \times 100\%

Discount =x−y= x-y, and discount percent is taken on the marked price, so it is x−yx×100%\frac{x-y}{x}\times 100\%.

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  • 2079 Paying · 1 mark

If X={1,4,6}X = \{1, 4, 6\} and Y={2,5}Y = \{2, 5\}, what is the value of n(X∪Y)n(X \cup Y)? (a) 0 (b) 2 (c) 3 (d) 5

Answer

(d) 5 X∪Y={1,2,4,5,6}X \cup Y = \{1, 2, 4, 5, 6\}, and the sets have no common element, so n(X∪Y)=3+2=5n(X \cup Y) = 3 + 2 = 5.

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  • 2079 Paying · 1 mark

The present population of a town is XX. If the population of the town increases every year by R%R\%, what will be the population of the town after YY years? (a) X(1+R100)TX\left(1 + \frac{R}{100}\right)^T (b) X(1−R100)TX\left(1 - \frac{R}{100}\right)^T (c) X(1+R100)YX\left(1 + \frac{R}{100}\right)^Y (d) X(1+R100)RX\left(1 + \frac{R}{100}\right)^R

Answer

(c) X(1+R100)YX\left(1 + \frac{R}{100}\right)^Y For growth at R%R\% per year, population after YY years is P=X(1+R100)YP = X\left(1 + \frac{R}{100}\right)^Y. Option (a) uses TT and (d) uses RR, neither of which is the number of years.

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  • 2079 Paying · 1 mark

How much more is Rs. 6,000 including 13% VAT on it than Rs. 6,500? (a) Rs. 200 (b) Rs. 280 (c) Rs. 300 (d) Rs. 380

Answer

(b) Rs. 280

Price with VAT: 6000×1.13=67806000 \times 1.13 = 6780.

Difference: 6780−6500=Rs. 2806780 - 6500 = \text{Rs. } 280.

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  • 2079 Paying · 1 mark

The present price of a television is Rs. 35,000. After how many years will it be Rs. 25,515 if its price reduces by 10% per annum? (a) 2 years (b) 2122\frac{1}{2} years (c) 3 years (d) 4 years

Answer

(c) 3 years

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

25515=35000(0.9)n⇒(0.9)n=0.729=(0.9)325515 = 35000(0.9)^n \Rightarrow (0.9)^n = 0.729 = (0.9)^3

So n=3n = 3 years.

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  • 2079 Paying · 1 mark

What is a triangle whose three sides are 4 cm, 6 cm and 12 cm respectively? (a) Acute angled triangle (b) Equilateral triangle (c) Obtuse angle triangle (d) No triangle

Answer

(d) No triangle

A triangle exists only if the sum of any two sides exceeds the third. Here 4+6=10<124 + 6 = 10 < 12, so these lengths cannot form a triangle.

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  • 2079 Paying · 1 mark

What is the vertical height of a square based pyramid having its base area 64 cm264\ cm^2 and volume 640 cm3640\ cm^3? (a) 5 cm (b) 8 cm (c) 10 cm (d) 30 cm

Answer

(d) 30 cm

V=13×base area×h⇒640=13×64×hV = \frac{1}{3} \times \text{base area} \times h \Rightarrow 640 = \frac{1}{3} \times 64 \times h

h=192064=30 cmh = \frac{1920}{64} = 30 \text{ cm}

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  • 2079 Paying · 1 mark

The total surface area of a hemisphere is 243π cm2243\pi\ cm^2. What will be its diameter? (a) 9 cm (b) 18 cm (c) 27 cm (d) 30 cm

Answer

(b) 18 cm

Total surface area of a hemisphere =3πr2= 3\pi r^2, so 3πr2=243π⇒r2=81⇒r=93\pi r^2 = 243\pi \Rightarrow r^2 = 81 \Rightarrow r = 9 cm.

Diameter =2r=18= 2r = 18 cm.

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  • 2079 Paying · 1 mark

What is the L.C.M. of two algebraic expressions (x+1)(x+1) and (x2−1)(x^2 - 1)? (a) x+1x + 1 (b) x2−1x^2 - 1 (c) (x2−1)(x+1)(x^2 - 1)(x + 1) (d) x3−1x^3 - 1

Answer

(b) x2−1x^2 - 1

Since x2−1=(x+1)(x−1)x^2 - 1 = (x+1)(x-1), it already contains the factor (x+1)(x+1). So the L.C.M. of (x+1)(x+1) and (x+1)(x−1)(x+1)(x-1) is (x+1)(x−1)=x2−1(x+1)(x-1) = x^2 - 1.

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  • 2079 Paying · 1 mark

What is the H.C.F. of x2−1x^2 - 1 and x2+3x+2x^2 + 3x + 2? (a) x+1x + 1 (b) x−1x - 1 (c) x2−1x^2 - 1 (d) x−2x - 2

Answer

(a) x+1x + 1

x2−1=(x+1)(x−1)x^2 - 1 = (x+1)(x-1) and x2+3x+2=(x+1)(x+2)x^2 + 3x + 2 = (x+1)(x+2).

The only common factor is x+1x + 1.

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  • 2079 Paying · 1 mark

Which one is a quadratic equation in which the values of xx are 2 and −1-1? (a) x2+x−2=0x^2 + x - 2 = 0 (b) x2−x+2=0x^2 - x + 2 = 0 (c) x2−x−2=0x^2 - x - 2 = 0 (d) 2x2−x−1=02x^2 - x - 1 = 0

Answer

(c) x2−x−2=0x^2 - x - 2 = 0

Sum of roots =2+(−1)=1= 2 + (-1) = 1, product =2×(−1)=−2= 2 \times (-1) = -2.

Equation: x2−(sum)x+product=x2−x−2=0x^2 - (\text{sum})x + \text{product} = x^2 - x - 2 = 0. Check: (x−2)(x+1)=x2−x−2(x-2)(x+1) = x^2 - x - 2.

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  • 2079 Paying · 1 mark

If the sum of a positive number and its square is 30, what is the number? (a) 3 (b) 5 (c) ±5\pm 5 (d) 6

Answer

(b) 5

Let the number be xx: x+x2=30⇒x2+x−30=0⇒(x+6)(x−5)=0x + x^2 = 30 \Rightarrow x^2 + x - 30 = 0 \Rightarrow (x+6)(x-5) = 0.

So x=5x = 5 or −6-6; the number is positive, so x=5x = 5.

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  • 2079 Paying · 1 mark

What is the relation between the areas of the triangles standing on the same base and between the same parallel lines? (a) double (b) unequal (c) equal (d) half one

Answer

(c) equal

Triangles on the same base and between the same parallel lines have the same base and the same height, so their areas are equal.

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  • 2079 Paying · 1 mark

In △PQR\triangle PQR, PQ=7PQ = 7 cm, QR=16QR = 16 cm and the area of △PQR\triangle PQR is 283 cm228\sqrt{3}\ cm^2. What is the measure of ∠PQR\angle PQR? (a) 30° (b) 45° (c) 60° (d) 75°

Answer

(c) 60°

Area=12⋅PQ⋅QRsin⁡Q⇒283=12(7)(16)sin⁡Q=56sin⁡Q\text{Area} = \frac{1}{2} \cdot PQ \cdot QR \sin Q \Rightarrow 28\sqrt{3} = \frac{1}{2}(7)(16)\sin Q = 56 \sin Q

sin⁡Q=32⇒∠PQR=60°\sin Q = \frac{\sqrt{3}}{2} \Rightarrow \angle PQR = 60°

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  • 2079 Paying · 1 mark

What is the angle of elevation when a man observes the top of a tower of height 80380\sqrt{3} m from 240 m from the foot of the tower? (a) 30° (b) 45° (c) 60° (d) 75°

Answer

(a) 30°

tan⁡θ=heightdistance=803240=33=13\tan\theta = \frac{\text{height}}{\text{distance}} = \frac{80\sqrt{3}}{240} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}

So θ=30°\theta = 30°.

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  • 2079 Paying · 1 mark

The third quartile is calculated by the formula ........ (a) L+N3−c.ff×iL + \frac{\frac{N}{3} - c.f}{f} \times i (b) L+N4−c.ff×iL + \frac{\frac{N}{4} - c.f}{f} \times i (c) L+3N4−c.ff×iL + \frac{\frac{3N}{4} - c.f}{f} \times i (d) L+3N4+c.ff×iL + \frac{\frac{3N}{4} + c.f}{f} \times i

Answer

(c) L+3N4−c.ff×iL + \frac{\frac{3N}{4} - c.f}{f} \times i

The third quartile Q3Q_3 is the value at the 3N4\frac{3N}{4}th item; LL is the lower limit of the Q3Q_3 class, c.fc.f the cumulative frequency before it, ff its frequency and ii its class width.

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  • 2079 Paying · 1 mark

Two players, A and B play a tennis match. If the probability of winning the match by A is 13\frac{1}{3}, what is the probability of winning the match by B? (a) 23\frac{2}{3} (b) 13\frac{1}{3} (c) 56\frac{5}{6} (d) 16\frac{1}{6}

Answer

(a) 23\frac{2}{3}

In a match one player wins and the other loses (no draw), so P(B)=1−P(A)=1−13=23P(B) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}.

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  • 2076 · 1 mark

If 2sec⁡2α=tan⁡β+cot⁡β2\sec 2\alpha = \tan\beta + \cot\beta, then the value of (α+β)(\alpha + \beta) = (a) π2\frac{\pi}{2} (b) π3\frac{\pi}{3} (c) π4\frac{\pi}{4} (d) π6\frac{\pi}{6}

Answer

(c) π4\frac{\pi}{4}

tan⁡β+cot⁡β=1sin⁡βcos⁡β=2sin⁡2β\tan\beta+\cot\beta=\dfrac{1}{\sin\beta\cos\beta}=\dfrac{2}{\sin 2\beta}, so sec⁡2α=1sin⁡2β\sec 2\alpha=\dfrac{1}{\sin 2\beta}, i.e. cos⁡2α=sin⁡2β\cos 2\alpha=\sin 2\beta.

Hence 2α+2β=π22\alpha+2\beta=\dfrac{\pi}{2}, so α+β=π4\alpha+\beta=\dfrac{\pi}{4}.

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  • 2076 · 1 mark

If (A+B)=π4(A + B) = \frac{\pi}{4}, then (cot⁡A−1)(cot⁡B−1)(\cot A - 1)(\cot B - 1) = (a) 1 (b) 12 (c) 2 (d) −1-1

Answer

(c) 2

Since cot⁡(A+B)=1\cot(A+B)=1, we have cot⁡Acot⁡B−1cot⁡A+cot⁡B=1\dfrac{\cot A\cot B-1}{\cot A+\cot B}=1, so cot⁡Acot⁡B−1=cot⁡A+cot⁡B\cot A\cot B-1=\cot A+\cot B.

Then (cot⁡A−1)(cot⁡B−1)=cot⁡Acot⁡B−(cot⁡A+cot⁡B)+1=1+1=2(\cot A-1)(\cot B-1)=\cot A\cot B-(\cot A+\cot B)+1=1+1=2.

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  • 2076 · 1 mark

In △ABC\triangle ABC, if 2cos⁡A=sin⁡Bsin⁡C2\cos A = \frac{\sin B}{\sin C}, then the △\triangle is: (a) Right angled triangle (b) Equilateral triangle (c) Isosceles triangle (d) Scalene triangle

Answer

(c) Isosceles triangle

2cos⁡Asin⁡C=sin⁡B=sin⁡(A+C)=sin⁡Acos⁡C+cos⁡Asin⁡C2\cos A\sin C=\sin B=\sin(A+C)=\sin A\cos C+\cos A\sin C.

So cos⁡Asin⁡C−sin⁡Acos⁡C=0\cos A\sin C-\sin A\cos C=0, i.e. sin⁡(C−A)=0\sin(C-A)=0, giving C=AC=A. Two angles are equal, so the triangle is isosceles.

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  • 2076 · 1 mark

The number of non-empty proper subsets of set A={1,2,3,4}A = \{1, 2, 3, 4\} is: (a) 16 (b) 14 (c) 15 (d) 13

Answer

(b) 14

A set with 4 elements has 24=162^4=16 subsets. Removing the empty set and the set itself gives 16−2=1416-2=14 non-empty proper subsets.

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  • 2076 · 1 mark

If f(x+1)=x+2f(x + 1) = x + 2, then f(x−1)f(x - 1) = (a) 2 (b) xx (c) x+1x + 1 (d) x−1x - 1

Answer

(b) xx

Put t=x+1t=x+1: f(t)=t−1+2=t+1f(t)=t-1+2=t+1. So f(x−1)=(x−1)+1=xf(x-1)=(x-1)+1=x.

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  • 2076 · 1 mark

For any two square matrices A and B, which of the following is not true? (a) (A+B)T=AT+BT(A + B)^T = A^T + B^T (b) (AB)T=BTAT(AB)^T = B^T A^T (c) A(B+C)=AB+ACA(B + C) = AB + AC (d) (A−B)T=BT−AT(A - B)^T = B^T - A^T

Answer

(d) (A−B)T=BT−AT(A-B)^T = B^T - A^T

The correct rule is (A−B)T=AT−BT(A-B)^T=A^T-B^T, so the statement in (d) is not true. The other three are standard properties.

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  • 2076 · 1 mark

If AB=CAB = C, then the order of matrices A, B and C are: (a) A2×3A_{2\times3}, B2×3B_{2\times3}, C2×3C_{2\times3} (b) A3×2A_{3\times2}, B2×3B_{2\times3}, C3×2C_{3\times2} (c) A3×3A_{3\times3}, B3×2B_{3\times2}, C3×2C_{3\times2} (d) A3×2A_{3\times2}, B2×3B_{2\times3}, C3×3C_{3\times3}

Answer

(d) A3×2A_{3\times2}, B2×3B_{2\times3}, C3×3C_{3\times3}

For ABAB to exist, the columns of AA must equal the rows of BB, and the product has the rows of AA and the columns of BB. Here (3×2)(2×3)=3×3(3\times2)(2\times3)=3\times3.

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  • 2076 · 1 mark

If 1q+r,1r+p,1p+q\frac{1}{q+r}, \frac{1}{r+p}, \frac{1}{p+q} are in AP, then: (a) p2,q2,r2p^2, q^2, r^2 are in AP (b) p,q,rp, q, r are in GP (c) p2,q2,r2p^2, q^2, r^2 are in GP (d) p2,q2,r2p^2, q^2, r^2 are in HP

Answer

(a) p2,q2,r2p^2, q^2, r^2 are in AP

Since the terms are in AP, 1r+p−1q+r=1p+q−1r+p\dfrac{1}{r+p}-\dfrac{1}{q+r}=\dfrac{1}{p+q}-\dfrac{1}{r+p}, so q−p(r+p)(q+r)=r−q(p+q)(r+p)\dfrac{q-p}{(r+p)(q+r)}=\dfrac{r-q}{(p+q)(r+p)}.

Cancelling (r+p)(r+p) gives (q−p)(p+q)=(r−q)(q+r)(q-p)(p+q)=(r-q)(q+r), i.e. q2−p2=r2−q2q^2-p^2=r^2-q^2. Hence p2,q2,r2p^2,q^2,r^2 are in AP.

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  • 2076 · 1 mark

If mthm^{th} term of an AP is nn and nthn^{th} term is mm, then rthr^{th} term is: (a) m−n−rm - n - r (b) m+n−rm + n - r (c) m−n+rm - n + r (d) n−r−mn - r - m

Answer

(b) m+n−rm+n-r

a+(m−1)d=na+(m-1)d=n and a+(n−1)d=ma+(n-1)d=m. Subtracting: (m−n)d=n−m(m-n)d=n-m, so d=−1d=-1 and a=m+n−1a=m+n-1.

Tr=a+(r−1)d=m+n−1−(r−1)=m+n−rT_r=a+(r-1)d=m+n-1-(r-1)=m+n-r.

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  • 2076 · 1 mark

If α\alpha and β\beta are roots of the equation x2−cp(x+1)−c=0x^2 - cp(x + 1) - c = 0 then (1+α)(1+β)(1 + \alpha)(1 + \beta) = (a) 1+p1 + p (b) 1−p1 - p (c) 1+c1 + c (d) 1−c1 - c

Answer

(d) 1−c1-c

The equation is x2−cpx−(cp+c)=0x^2-cpx-(cp+c)=0, so α+β=cp\alpha+\beta=cp and αβ=−(cp+c)\alpha\beta=-(cp+c).

(1+α)(1+β)=1+(α+β)+αβ=1+cp−cp−c=1−c(1+\alpha)(1+\beta)=1+(\alpha+\beta)+\alpha\beta=1+cp-cp-c=1-c.

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  • 2076 · 1 mark

If 7 is added to the square of a natural number then the result is 43. What is the number? (a) 50\sqrt{50} (b) ±7\pm 7 (c) ±6\pm 6 (d) 6

Answer

(d) 6

Let the number be nn: n2+7=43n^2+7=43, so n2=36n^2=36 and n=±6n=\pm6. Since the number is natural, n=6n=6.

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  • 2076 · 1 mark

The length of three sides of a triangle are xx cm, yy cm, and zz cm respectively. What is the area of the triangle? (a) xyz cm2xyz\ cm^2 (b) 12xyz cm2\frac{1}{2}xyz\ cm^2 (c) s(s−x)(s−y)(s−z) cm2\sqrt{s(s-x)(s-y)(s-z)}\ cm^2 (d) 12xy cm2\frac{1}{2}xy\ cm^2

Answer

(c) s(s−x)(s−y)(s−z) cm2\sqrt{s(s-x)(s-y)(s-z)}\ cm^2

This is Heron's formula, where s=x+y+z2s=\dfrac{x+y+z}{2} is the semi-perimeter.

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  • 2076 · 1 mark

If set A={a,b,e,f}A = \{a, b, e, f\} and set B={b,e,g,a}B = \{b, e, g, a\}, what is n(B−A)n(B - A)? (a) 1 (b) 2 (c) 4 (d) 6

Answer

(a) 1

B−AB-A contains the elements of BB not in AA: B−A={g}B-A=\{g\}, so n(B−A)=1n(B-A)=1.

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  • 2076 · 1 mark

If x+y+z=0x + y + z = 0, then x3+y3+z3x^3 + y^3 + z^3 is equal to (a) xyzxyz (b) 2xyz2xyz (c) 3xyz3xyz (d) 4xyz4xyz

Answer

(c) 3xyz3xyz

Using x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx), the right side is 0 when x+y+z=0x+y+z=0. So x3+y3+z3=3xyzx^3+y^3+z^3=3xyz.

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  • 2076 · 1 mark

In a two digits' number, the sum of the digits is 7 and difference is 3. What will be the digit of ten's place? (a) 2 (b) 4 (c) 5 (d) 7

Answer

(c) 5

Let the tens digit be xx and the units digit yy: x+y=7x+y=7, x−y=3x-y=3. Adding: 2x=102x=10, so x=5x=5 and y=2y=2. The number is 52.

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  • 2076 · 1 mark

The values of the variable xx in a quadratic equation are 2 and 3. What will be the quadratic equation? (a) x2−x−6x^2 - x - 6 (b) x2−2x+6x^2 - 2x + 6 (c) x2−5x+6x^2 - 5x + 6 (d) x2−5x−6x^2 - 5x - 6

Answer

(c) x2−5x+6x^2-5x+6

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

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  • 2076 · 1 mark

The mean weight of 150 students is 60 kg. The mean weight of boys is 70 kg and that of girls is 55 kg. The number of girls is: (a) 50 (b) 150 (c) 25 (d) 100

Answer

(d) 100

Let girls =g=g, boys =150−g=150-g. Total weight =150×60=9000=150\times60=9000.

70(150−g)+55g=9000⇒10500−15g=9000⇒g=10070(150-g)+55g=9000 \Rightarrow 10500-15g=9000 \Rightarrow g=100.

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  • 2076 · 1 mark

What is the probability of getting a number divisible by 2 or a cube number when a fair die is thrown? (a) 1/3 (b) 1/2 (c) 2/3 (d) 1/6

Answer

(c) 2/3

Numbers divisible by 2: {2,4,6}\{2,4,6\}. Cube numbers on a die: {1}\{1\} (since 8>68>6). The union is {1,2,4,6}\{1,2,4,6\}, which has 4 outcomes.

P=46=23P=\dfrac{4}{6}=\dfrac{2}{3}.

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  • 2076 · 1 mark

What is the H.C.F. of relative prime numbers a, b and c? (a) bc (b) abc (c) ac (d) 1

Answer

(d) 1

Relatively prime (co-prime) numbers have no common factor other than 1, so their H.C.F. is 1.

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  • 2076 · 1 mark

What is the order of the surd 3x\sqrt{3x}? (a) 1 (b) 2 (c) 3 (d) 0

Answer

(b) 2

3x=(3x)1/2\sqrt{3x}=(3x)^{1/2} is a square root, so the order (index) of the surd is 2.

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  • 2075 · 1 mark

For what value of xx, ax\sqrt[x]{a} equals a3a^3? (a) 1 (b) 1/31/3 (c) 2/32/3 (d) 3

Answer

(b) 1/31/3 ax=a1/x\sqrt[x]{a}=a^{1/x}, so a1/x=a3a^{1/x}=a^3 gives 1/x=31/x=3, hence x=13x=\dfrac{1}{3}.

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  • 2075 · 1 mark

If the length of the rectangle is twice the breadth and its perimeter is 60 cm. Find its length. (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm

Answer

(b) 20 cm Let breadth =b=b, length =2b=2b. Perimeter =2(2b+b)=6b=60=2(2b+b)=6b=60, so b=10b=10 cm and length =2b=20=2b=20 cm.

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  • 2075 · 1 mark

If in a certain figure, AB is a tangent, OB is the radius and angle OAB=30°OAB = 30°, what is the value of angle AOB? (a) 90° (b) 60° (c) 30° (d) 15°

Answer

(b) 60° A tangent is perpendicular to the radius at the point of contact, so ∠OBA=90∘\angle OBA=90^\circ. In △OAB\triangle OAB: ∠AOB=180∘−90∘−30∘=60∘\angle AOB=180^\circ-90^\circ-30^\circ=60^\circ

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  • 2075 · 1 mark

What is the area of a triangle whose sides are 16 cm, 30 cm and 34 cm? (a) 37.9 cm237.9\ cm^2 (b) 75.8 cm275.8\ cm^2 (c) 120 cm2120\ cm^2 (d) 240 cm2240\ cm^2

Answer

(d) 240 cm2240\ cm^2 Since 162+302=256+900=1156=34216^2+30^2=256+900=1156=34^2, it is a right triangle with legs 16 and 30: A=12×16×30=240 cm2A=\tfrac12\times16\times30=240\ cm^2

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  • 2075 · 1 mark

What is the probability of getting 6 on a dice and head on a coin when a dice is rolled and a coin is tossed simultaneously? (a) 1/81/8 (b) 2/32/3 (c) 1/121/12 (d) 3/23/2

Answer

(c) 1/121/12 The events are independent: P(6)=16P(6)=\tfrac16 and P(head)=12P(\text{head})=\tfrac12, so P=16×12=112P=\frac16\times\frac12=\frac1{12}

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  • 2075 · 1 mark

If U.S. $1 = Rs. 105.52, how much dollar can be exchanged with Rs. 90312? (a) $855.87 (b) $90206.48 (c) $90417.52 (d) $9529722.2

Answer

(a) $855.87 Dollars=90312105.52≈855.9\text{Dollars}=\frac{90312}{105.52}\approx 855.9 which matches option (a) $855.87.

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  • 2075 · 1 mark

If A and B can do a piece of work in 10 days and 20 days respectively, how many part of the work do they work together in 3 days? (a) 1/101/10 (b) 1/201/20 (c) 3/203/20 (d) 9/209/20

Answer

(d) 9/209/20 Work done together in one day =110+120=320=\frac1{10}+\frac1{20}=\frac3{20}, so in 3 days it is 3×320=9203\times\frac3{20}=\frac9{20}.

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  • 2075 · 1 mark

What percent of discount should be given on a doll costing Rs. 1800 so that a customer can purchase it at Rs. 1500? (a) 13% (b) 16.66% (c) 20% (d) 83.33%

Answer

(b) 16.66% Discount =1800−1500==1800-1500= Rs 300. Discount %=3001800×100=16.66%\text{Discount \%}=\frac{300}{1800}\times100=16.66\%

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  • 2075 · 1 mark

What does the formula P(1+R100)n−1P\left(1 + \frac{R}{100}\right)^n - 1 represent? (a) compound amount (b) compound interest (c) simple interest (d) simple amount

Answer

(b) compound interest The compound amount is A=P(1+R100)nA=P\left(1+\frac{R}{100}\right)^n, and compound interest is CI=A−P=P[(1+R100)n−1]CI=A-P=P\left[\left(1+\frac{R}{100}\right)^n-1\right], which is what the formula represents.

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  • 2075 · 1 mark

The area of an isosceles triangle is given by: (a) 14b4a2−b2\frac{1}{4}b\sqrt{4a^2 - b^2} (b) 14a4a2−a2\frac{1}{4}a\sqrt{4a^2 - a^2} (c) 14b4a2+b2\frac{1}{4}b\sqrt{4a^2 + b^2} (d) 14a4a2+a2\frac{1}{4}a\sqrt{4a^2 + a^2}

Answer

(a) 14b4a2−b2\frac{1}{4}b\sqrt{4a^2-b^2} For an isosceles triangle with equal sides aa and base bb, the height is a2−b2/4\sqrt{a^2-b^2/4}, so A=12ba2−b24=14b4a2−b2A=\frac12 b\sqrt{a^2-\frac{b^2}{4}}=\frac14 b\sqrt{4a^2-b^2}

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  • 2075 · 1 mark

The compound interest of Rs. 1500 at 5% p.a. for 2 years will be: (a) Rs. 150.75 (b) Rs. 151.375 (c) Rs. 152.75 (d) Rs. 153.75

Answer

(d) Rs. 153.75

Amount after 2 years: A=1500(1+5100)2=1500×1.1025=Rs 1653.75A = 1500\left(1+\frac{5}{100}\right)^2 = 1500 \times 1.1025 = \text{Rs } 1653.75.

Compound interest =1653.75−1500=Rs 153.75= 1653.75 - 1500 = \text{Rs } 153.75.

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  • 2075 · 1 mark

The factors of x4+4x^4 + 4 are: (a) (x+2)(x2+2x+4)(x+2)(x^2+2x+4) (b) (x+2)(x2−2x+2)(x+2)(x^2-2x+2) (c) (x−2x+2)(x+2x+2)(x-2x+2)(x+2x+2) (d) (x2−2x+2)(x2+2x+2)(x^2-2x+2)(x^2+2x+2)

Answer

(d) (x2−2x+2)(x2+2x+2)(x^2-2x+2)(x^2+2x+2)

Add and subtract 4x24x^2 to complete a square:

x4+4=(x2+2)2−4x2=(x2+2−2x)(x2+2+2x).x^4+4 = (x^2+2)^2 - 4x^2 = (x^2+2-2x)(x^2+2+2x).

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  • 2075 · 1 mark

If the ratio of two digit numbers is 5:6 and their sum is 33, then the numbers will be (a) 15 and 18 (b) 21 and 12 (c) 9 and 24 (d) 45 and 54

Answer

(a) 15 and 18

Let the numbers be 5k5k and 6k6k. Then 11k=3311k = 33, so k=3k = 3, giving 1515 and 1818. Check: 15:18=5:615:18 = 5:6 and 15+18=3315+18=33.

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  • 2075 · 1 mark

The values of xx in the equation x2−5x+6=0x^2 - 5x + 6 = 0 are: (a) 2 and 3 (b) -2 and 3 (c) 2 and -3 (d) -2 and -3

Answer

(a) 2 and 3

x2−5x+6=(x−2)(x−3)=0⇒x=2 or 3.x^2-5x+6 = (x-2)(x-3) = 0 \Rightarrow x = 2 \text{ or } 3.

Check: sum of roots =5=5, product =6=6.

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  • 2075 · 1 mark

The value of sin⁡50°−sin⁡70°+sin⁡10°\sin 50° - \sin 70° + \sin 10° = (a) 1 (b) -1 (c) 0 (d) 2

Answer

(c) 0

sin⁡50°+sin⁡10°=2sin⁡30°cos⁡20°=cos⁡20°=sin⁡70°.\sin 50° + \sin 10° = 2\sin 30°\cos 20° = \cos 20° = \sin 70°.

So sin⁡50°−sin⁡70°+sin⁡10°=sin⁡70°−sin⁡70°=0\sin 50° - \sin 70° + \sin 10° = \sin 70° - \sin 70° = 0.

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  • 2075 · 1 mark

The sum of two positive numbers is 60 and their difference is 20. Find the numbers. (a) 40 and 20 (b) 45 and 15 (c) 33 and 13 (d) 25 and 35

Answer

(a) 40 and 20

Let the numbers be xx and yy with x+y=60x+y=60 and x−y=20x-y=20. Adding gives 2x=802x=80, so x=40x=40 and y=20y=20.

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  • 2075 · 1 mark

What type of sets are A={1,2,3,4,0}A = \{1, 2, 3, 4, 0\} and B={10,5,6,7,8}B = \{10, 5, 6, 7, 8\}? (a) Equal (b) Disjoint (c) Connected (d) Infinite

Answer

(b) Disjoint

AA and BB have no element in common (A∩B=∅A\cap B=\varnothing), so they are disjoint sets. They are finite and not equal.

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  • 2075 · 1 mark

The item of a statistical data that divides it into two equal parts (halves) is called: (a) Mode (b) Quartile (c) Median (d) Mean

Answer

(c) Median

The median is the middle value that divides ordered data into two equal halves. Quartiles divide it into four parts.

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Questions from CTEVT 2079 Scholarship seats entrance paper (English, Science, Mathematics) (Scholarship seats entrance, 2079 BS), CTEVT 2079 Full-paying seats entrance paper (English, Science, Mathematics) (Full-paying seats entrance, 2079 BS), CTEVT 2076 Scholarship seats entrance paper (English, Science, Mathematics) (Scholarship seats entrance, 2076 BS) and CTEVT 2075 Scholarship seats entrance paper (English, Science, Mathematics) (Scholarship seats entrance, 2075 BS). Answers are written for this site; check them against your class notes.