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CAT 2020 Slot 1 — QA questions with answers

All 26 questions of the Quantitative Ability section (18 MCQs, 8 TITA). Try each one, then open its answer and solution.

CAT 2020 Slot 1, timed like the real exam (120 minutes, 40 a section) and scored the CAT way: +3 right, −1 for a wrong MCQ, 0 for a wrong TITA, with the full post-mock analysis. Older official papers are in CATin Pro, with 80+ mocks and the full planner. The last three years of papers are free.

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Quantitative Ability

CAT 2020 Slot 1 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q51MCQLogarithms

If YY is a negative number such that 2Y2log⁡35=5log⁡232^{Y^2 \log_3 5} = 5^{\log_2 3}, then YY equals to:
  1. log⁡2(1/5)\log_2(1/5)
  2. log⁡2(1/3)\log_2(1/3)
  3. −log⁡2(1/5)-\log_2(1/5)
  4. −log⁡2(1/3)-\log_2(1/3)
Answer and solution

Answer: (B) log⁡2(1/3)\log_2(1/3)

Use alog⁡bc=clog⁡baa^{\log_b c} = c^{\log_b a} on the left side: 2Y2log⁡35=5Y2log⁡322^{Y^2 \log_3 5} = 5^{Y^2 \log_3 2}. So 5Y2log⁡32=5log⁡235^{Y^2 \log_3 2} = 5^{\log_2 3}. Equating the powers of 5 gives Y2log⁡32=log⁡23Y^2 \log_3 2 = \log_2 3. Since log⁡32=1log⁡23\log_3 2 = \frac{1}{\log_2 3}, this becomes Y2=(log⁡23)2Y^2 = (\log_2 3)^2, so Y=±log⁡23Y = \pm \log_2 3. YY is negative, so Y=−log⁡23=log⁡2(1/3)Y = -\log_2 3 = \log_2(1/3). Option D, −log⁡2(1/3)-\log_2(1/3), equals +log⁡23+\log_2 3, which is positive and so is ruled out. Options A and C involve log⁡25\log_2 5, which does not satisfy the equation. Hence, option B (log⁡2(1/3)\log_2(1/3)).

Q52TITALinear Equations

A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 62

Work backwards. If he has nn toffees before a child, the child gets n2+1\frac{n}{2} + 1 and n2−1\frac{n}{2} - 1 remain. So if mm remain after a child, there were 2(m+1)2(m + 1) before that child. After the 5th child, 0 remain. Before the 5th child: 2(0+1)=22(0 + 1) = 2. Before the 4th: 2(2+1)=62(2 + 1) = 6. Before the 3rd: 2(6+1)=142(6 + 1) = 14. Before the 2nd: 2(14+1)=302(14 + 1) = 30. Before the 1st: 2(30+1)=622(30 + 1) = 62. Check forwards: from 62 the first child gets 32, leaving 30; then 16, leaving 14; then 8, leaving 6; then 4, leaving 2; then 2, leaving 0. The answer is 62.

Q53TITADigits & Base Systems

How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 21

The product of the digits must be 3, 4, 5 or 6. No digit can be 0, or the product would be 0. Product 3: digits 1, 1, 3, which can be arranged in 3!2!=3\frac{3!}{2!} = 3 ways. Product 5: digits 1, 1, 5, in 3 ways. Product 4: digits 1, 1, 4 (3 ways) or 1, 2, 2 (3 ways), giving 6 numbers. Product 6: digits 1, 1, 6 (3 ways) or 1, 2, 3 (3!=63! = 6 ways), giving 9 numbers. Total =3+3+6+9=21= 3 + 3 + 6 + 9 = 21. The answer is 21.

Q54MCQInequalities & Modulus

The number of real-valued solutions of the equation 2x+2−x=2−(x−2)22^x + 2^{-x} = 2-(x-2)^2 is:
  1. 1
  2. 2
  3. infinite
  4. 0
Answer and solution

Answer: (D) 0

By AM–GM, since 2x2^x and 2−x2^{-x} are positive, 2x+2−x≥22x⋅2−x=22^x + 2^{-x} \ge 2\sqrt{2^x \cdot 2^{-x}} = 2, with equality only at x=0x = 0. Since (x−2)2≥0(x - 2)^2 \ge 0, the right side satisfies 2−(x−2)2≤22 - (x - 2)^2 \le 2, with equality only at x=2x = 2. So the left side is always at least 2 and the right side at most 2. They can be equal only if both equal 2 at the same xx. But the left side needs x=0x = 0 and the right side needs x=2x = 2, so no real xx works. The graph shows this: the upper curve y=2x+2−xy = 2^x + 2^{-x} has its lowest point at (0,2)(0, 2), the downward parabola y=2−(x−2)2y = 2 - (x - 2)^2 has its highest point at (2,2)(2, 2), and the two never meet. Solution figure for question 54, CAT 2020 Slot 1 Option A (1) would need a common point, but at x=0x = 0 the right side is 2−4=−22 - 4 = -2, not 2. Hence, option D (0).

Q55MCQIndices & Surds

How many distinct positive integer-valued solutions exist to the equation (x2−7x+11)(x2−13x+42)=1(x^2-7x+11)^{(x^2-13x+42)} = 1 ?
  1. 8
  2. 4
  3. 2
  4. 6
Answer and solution

Answer: (D) 6

ab=1a^b = 1 in three cases. Exponent 0 with a non-zero base: x2−13x+42=(x−6)(x−7)=0x^2 - 13x + 42 = (x - 6)(x - 7) = 0 gives x=6,7x = 6, 7. The base there is 5 and 11, both non-zero. Base 1: x2−7x+11=1x^2 - 7x + 11 = 1 gives (x−2)(x−5)=0(x - 2)(x - 5) = 0, so x=2,5x = 2, 5. Base −1-1 with an even exponent: x2−7x+12=0x^2 - 7x + 12 = 0 gives x=3,4x = 3, 4. The exponent is 9−39+42=129 - 39 + 42 = 12 at x=3x = 3 and 16−52+42=616 - 52 + 42 = 6 at x=4x = 4. Both are even, so both work. The solutions are 2, 3, 4, 5, 6 and 7, which is six values. Option B (4) is what you get if you miss the base −1-1 case. Hence, option D (6).

Q56MCQMensuration

A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is
  1. 243
  2. 232
  3. 256
  4. 264
Answer and solution

Answer: (A) 243

The top piece is a small cone of height 27−18=927 - 18 = 9 cm, which is 13\frac{1}{3} of the full height. It is similar to the original cone, so its radius is also 13\frac{1}{3} of the base radius: rr against 3r3r. Solution figure for question 56, CAT 2020 Slot 1 Volumes of similar cones are in the ratio of the cubes of their heights, so the small cone is (13)3=127\left(\frac{1}{3}\right)^3 = \frac{1}{27} of the original volume VV. Small cone =V27= \frac{V}{27} and the lower piece (the frustum) =26V27= \frac{26V}{27}. Difference =26V27−V27=25V27=225= \frac{26V}{27} - \frac{V}{27} = \frac{25V}{27} = 225, so V=225×2725=243V = 225 \times \frac{27}{25} = 243 cc. Option C (256) would give a difference of 2527×256≈237\frac{25}{27} \times 256 \approx 237, not 225. Hence, option A (243).

Q57TITACoordinate Geometry

The area of the region satisfying the inequalities ∣x∣−y≤1,y≥0|x| - y \le 1, y \ge 0 and y≤1y \le 1 is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 3

The inequality ∣x∣−y≤1|x| - y \le 1 means ∣x∣≤y+1|x| \le y + 1, that is, −(y+1)≤x≤y+1-(y + 1) \le x \le y + 1. Together with 0≤y≤10 \le y \le 1, the region lies in the band between y=0y = 0 and y=1y = 1, between the lines y=−x−1y = -x - 1 and y=x−1y = x - 1. At y=0y = 0: ∣x∣≤1|x| \le 1, so −1≤x≤1-1 \le x \le 1, a width of 2. At y=1y = 1: ∣x∣≤2|x| \le 2, so −2≤x≤2-2 \le x \le 2, a width of 4. The width grows steadily with yy, so the region is a trapezium with vertices (−1,0)(-1, 0), (1,0)(1, 0), (2,1)(2, 1) and (−2,1)(-2, 1). It is the white region in the figure. Solution figure for question 57, CAT 2020 Slot 1 Its parallel sides are 2 and 4, and its height is 1. Area =12×(2+4)×1=3= \frac{1}{2} \times (2 + 4) \times 1 = 3. The answer is 3.

Q58MCQMensuration

On a rectangular metal sheet of area 135 sq in, a circle is painted such that the circle touches two opposite sides. If the area of the sheet left unpainted is two-thirds of the painted area then the perimeter of the rectangle in inches is
  1. 3π(5+12/π)3\sqrt{\pi}(5+12/\pi)
  2. 4π(3+9/π)4\sqrt{\pi}(3+9/\pi)
  3. 3π(52+6π)3\sqrt{\pi}\left(\dfrac{5}{2}+\dfrac{6}{\pi}\right)
  4. 5π(3+9/π)5\sqrt{\pi}(3+9/\pi)
Answer and solution

Answer: (A) 3π(5+12/π)3\sqrt{\pi}(5+12/\pi)

The circle touches two opposite sides, so its diameter equals the distance between them. That must be the shorter side, since the circle could not fit across the longer one. Let the radius be rr, so the shorter side is 2r2r. Solution figure for question 58, CAT 2020 Slot 1 Unpainted =23= \frac{2}{3} of painted, so the sheet is 53\frac{5}{3} of the circle: 53πr2=135\frac{5}{3}\pi r^2 = 135. Then πr2=81\pi r^2 = 81 and r=9πr = \frac{9}{\sqrt{\pi}}. Shorter side =2r=18π= 2r = \frac{18}{\sqrt{\pi}}. Longer side =1352r=135π18=15π2= \frac{135}{2r} = \frac{135\sqrt{\pi}}{18} = \frac{15\sqrt{\pi}}{2}. Perimeter =2(15π2+18π)=15π+36π=3π(5+12π)= 2\left(\frac{15\sqrt{\pi}}{2} + \frac{18}{\sqrt{\pi}}\right) = 15\sqrt{\pi} + \frac{36}{\sqrt{\pi}} = 3\sqrt{\pi}\left(5 + \frac{12}{\pi}\right). Option C, 3π(52+6π)3\sqrt{\pi}\left(\frac{5}{2} + \frac{6}{\pi}\right), is exactly half of this. It is the semi-perimeter, not the perimeter. Hence, option A (3π(5+12/π)3\sqrt{\pi}(5+12/\pi)).

Q59MCQPolygons & Circles

A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is
  1. 6π25\frac{6\pi}{25}
  2. 5π18\frac{5\pi}{18}
  3. 3π25\frac{3\pi}{25}
  4. 2π15\frac{2\pi}{15}
Answer and solution

Answer: (A) 6π25\frac{6\pi}{25}

The diagonals bisect each other at right angles at O, giving four right triangles with legs 6 and 8. So each side of the rhombus is 62+82=10\sqrt{6^2 + 8^2} = 10 cm. The inscribed circle is centred at O and touches side AD at T, so its radius rr is OT, the perpendicular from O to AD. Solution figure for question 59, CAT 2020 Slot 1 In right triangle AOD, twice the area is both OA×ODOA \times OD and AD×OTAD \times OT, so 6×8=10×r6 \times 8 = 10 \times r, giving r=4.8r = 4.8 cm. Area of the rhombus =12×12×16=96= \frac{1}{2} \times 12 \times 16 = 96 sq cm. Circle area =π(4.8)2=23.04π= \pi (4.8)^2 = 23.04\pi. Ratio =23.04π96=6π25= \frac{23.04\pi}{96} = \frac{6\pi}{25}. Option C, 3π25\frac{3\pi}{25}, is half the correct ratio. It comes from dividing by the full product of the diagonals, 192, instead of half of it. Hence, option A (6π25\frac{6\pi}{25}).

Q60MCQAverages, Mixtures & Alligations

Among 100 students, x1x_1 have birthdays in January, x2x_2 have birthdays in February, and so on. If x0=max⁡(x1,x2,…,x12)x_0 = \max(x_1, x_2, \ldots, x_{12}), then the smallest possible value of x0x_0 is
  1. 8
  2. 9
  3. 10
  4. 12
Answer and solution

Answer: (B) 9

The 12 monthly counts add to 100. If every month had at most 8 students, the total would be at most 12×8=96<10012 \times 8 = 96 < 100. So some month has at least 9, and x0≥9x_0 \ge 9. 9 can be achieved: put 9 students in four months and 8 in the other eight. This gives 4×9+8×8=36+64=1004 \times 9 + 8 \times 8 = 36 + 64 = 100, with a maximum of 9. Option A (8) fails because 12 months of at most 8 students hold only 96 people. Hence, option B (9).

Q61MCQTime, Speed & Distance

A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is
  1. 100
  2. 90
  3. 80
  4. 70
Answer and solution

Answer: (B) 90

Let the cars meet tt minutes after starting, with speeds v1v_1 (Car 1) and v2v_2 (Car 2). After meeting, Car 1 covers the stretch Car 2 had covered, in 45 minutes: v2t=45v1v_2 t = 45 v_1. After meeting, Car 2 covers the stretch Car 1 had covered, in 20 minutes: v1t=20v2v_1 t = 20 v_2. Multiplying the two equations: v1v2t2=900 v1v2v_1 v_2 t^2 = 900\, v_1 v_2, so t2=900t^2 = 900 and t=30t = 30 minutes. Then v1v2=20t=23\dfrac{v_1}{v_2} = \dfrac{20}{t} = \dfrac{2}{3}, and with v1=60v_1 = 60 km/hr, v2=60×32=90v_2 = 60 \times \dfrac{3}{2} = 90 km/hr. Check: in 30 minutes Car 1 covers 3030 km, which Car 2 covers in 20 minutes at 9090 km/hr. With option A (100 km/hr) the two equations disagree: v1t=20v2v_1 t = 20 v_2 gives t=3313t = 33\tfrac{1}{3} minutes, but v2t=45v1v_2 t = 45 v_1 gives t=27t = 27 minutes. Hence, option B (90).

Q62TITAProfit, Loss & Discount

A person spent Rs 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at 20% profit and the laptop at 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 20000

Let the purchase price of the desktop computer be xx and the laptop computer be yy.
Total cost price (CP): x+y=50000x + y = 50000.

He sold the desktop at a 20% profit, so its selling price is 1.2x1.2x.
He sold the laptop at a 10% loss, so its selling price is 0.9y0.9y.
Overall, he made a 2% profit on the total investment of 50000, so the total selling price (SP) is 50000×1.02=5100050000 \times 1.02 = 51000.

We have the equation:
1.2x+0.9y=510001.2x + 0.9y = 51000

Multiply the first equation by 0.9:
0.9x+0.9y=450000.9x + 0.9y = 45000

Subtract this from the second equation:
(1.2x+0.9y)−(0.9x+0.9y)=51000−45000(1.2x + 0.9y) - (0.9x + 0.9y) = 51000 - 45000
0.3x=60000.3x = 6000
x=60000.3=20000x = \frac{6000}{0.3} = 20000.

The purchase price of the desktop is Rs. 20000.

Q63TITAAverages, Mixtures & Alligations

A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 8

Initially, the 40-litre solution has dye and water in the ratio 2:32:3.
Amount of Dye = 25×40=16\frac{2}{5} \times 40 = 16 litres.
Amount of Water = 35×40=24\frac{3}{5} \times 40 = 24 litres.

Water is added to change the proportion to 2:52:5.
Since the amount of dye remains the same (16 litres), let the new amount of water be WW.
16W=25  ⟹  2W=80  ⟹  W=40\frac{16}{W} = \frac{2}{5} \implies 2W = 80 \implies W = 40 litres.
(This means 16 litres of water was added, making the total volume 16+40=5616 + 40 = 56 litres).

One-fourth (14\frac{1}{4}) of this diluted solution is taken out.
Dye removed = 14×16=4\frac{1}{4} \times 16 = 4 litres. Dye remaining = 16−4=1216 - 4 = 12 litres.
Water removed = 14×40=10\frac{1}{4} \times 40 = 10 litres. Water remaining = 40−10=3040 - 10 = 30 litres.

We need to bring the proportion back to 2:32:3 by adding dye.
Let the amount of dye to be added be dd.
12+d30=23  ⟹  3(12+d)=60  ⟹  36+3d=60  ⟹  3d=24  ⟹  d=8\frac{12 + d}{30} = \frac{2}{3} \implies 3(12 + d) = 60 \implies 36 + 3d = 60 \implies 3d = 24 \implies d = 8.

Hence, 8 litres of dye must be added.

Q64MCQIndices & Surds

If x=(4096)7+43x = (4096)^{7 + 4\sqrt{3}}, then which of the following equals to 64?
  1. x7x23\frac{x^7}{x^{2\sqrt{3}}}
  2. x72x43\frac{x^\frac{7}{2}}{x^{4\sqrt{3}}}
  3. x72x23\frac{x^\frac{7}{2}}{x^{2\sqrt{3}}}
  4. x7x43\frac{x^7}{x^{4\sqrt{3}}}
Answer and solution

Answer: (C) x72x23\frac{x^\frac{7}{2}}{x^{2\sqrt{3}}}

Given x=(4096)7+43x = (4096)^{7 + 4\sqrt{3}}.
Note that 4096=2124096 = 2^{12}. So, x=(212)7+43=212(7+43)x = \left(2^{12}\right)^{7 + 4\sqrt{3}} = 2^{12(7 + 4\sqrt{3})}.

Let's check the options to see which equals 64=2664 = 2^6.
We can evaluate the expression x7x43\frac{x^7}{x^{4\sqrt{3}}} (Option D):
x7x43=x7−43\frac{x^7}{x^{4\sqrt{3}}} = x^{7 - 4\sqrt{3}}
Substitute x=212(7+43)x = 2^{12(7 + 4\sqrt{3})}:
x7−43=(212(7+43))7−43x^{7 - 4\sqrt{3}} = \left(2^{12(7 + 4\sqrt{3})}\right)^{7 - 4\sqrt{3}}
=212(7+43)(7−43)= 2^{12(7 + 4\sqrt{3})(7 - 4\sqrt{3})}

Using the identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2:
(7+43)(7−43)=72−(43)2=49−48=1(7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1.

Thus, x7−43=212(1)=212=4096x^{7 - 4\sqrt{3}} = 2^{12(1)} = 2^{12} = 4096. This is not 64.

Now check Option C: x7/2x23\frac{x^{7/2}}{x^{2\sqrt{3}}}
x7/2x23=x72−23=x12(7−43)=(x7−43)12\frac{x^{7/2}}{x^{2\sqrt{3}}} = x^{\frac{7}{2} - 2\sqrt{3}} = x^{\frac{1}{2}(7 - 4\sqrt{3})} = \left(x^{7 - 4\sqrt{3}}\right)^{\frac{1}{2}}
From above, we know x7−43=4096=212x^{7 - 4\sqrt{3}} = 4096 = 2^{12}.
So, (212)1/2=26=64\left(2^{12}\right)^{1/2} = 2^6 = 64.

Hence, Option C is the correct answer.

Q65MCQRatios, Proportions & Partnership

An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A. B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg. of the metal C is
  1. 48
  2. 84
  3. 70
  4. 96
Answer and solution

Answer: (B) 84

The weight of each metal is its volume times its weight per unit volume. Take volumes 3v,4v,7v3v, 4v, 7v and unit weights 5w,2w,6w5w, 2w, 6w. The weights are then A: 15vw15vw, B: 8vw8vw and C: 42vw42vw, a total of 65vw65vw. So C makes up 4265\frac{42}{65} of the alloy by weight: 4265×130=84\frac{42}{65} \times 130 = 84 kg. Using the volume ratio alone would give 714×130=65\frac{7}{14} \times 130 = 65 kg, which ignores the different unit weights. Option C (70) would make C only 713\frac{7}{13} of the weight, less than its true share of 4265\frac{42}{65}. Hence, option B (84).

Q66TITAQuadratic & Polynomial Equations

The number of distinct real roots of the equation (x+1x)2−3(x+1x)+2=0(x + \frac{1}{x})^2 - 3(x + \frac{1}{x}) + 2 = 0 equals

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 1

Let a=x+1xa = x + \frac{1}{x}. Then a2−3a+2=0a^2 - 3a + 2 = 0, so (a−1)(a−2)=0(a - 1)(a - 2) = 0, and a=1a = 1 or a=2a = 2. If x+1x=1x + \frac{1}{x} = 1, then x2−x+1=0x^2 - x + 1 = 0. Its discriminant is 1−4=−3<01 - 4 = -3 < 0, so there is no real root. If x+1x=2x + \frac{1}{x} = 2, then x2−2x+1=0x^2 - 2x + 1 = 0, that is (x−1)2=0(x - 1)^2 = 0, so x=1x = 1. This is a repeated root, counted once. So the equation has exactly one distinct real root, x=1x = 1. The answer is 1.

Q67TITALogarithms

If log⁡45=(log⁡4y)(log⁡65)\log_4 5 = (\log_4 y)(\log_6 \sqrt{5}), then yy equals

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 36

Given: log⁡45=(log⁡4y)(log⁡65)\log_4 5 = (\log_4 y)(\log_6 \sqrt{5})

We can rewrite the equation by converting all logarithms using the change of base formula to natural logs:
log⁡5log⁡4=(log⁡ylog⁡4)(log⁡5log⁡6)\frac{\log 5}{\log 4} = \left( \frac{\log y}{\log 4} \right) \left( \frac{\log \sqrt{5}}{\log 6} \right)
log⁡5log⁡4=log⁡ylog⁡4×12log⁡5log⁡6\frac{\log 5}{\log 4} = \frac{\log y}{\log 4} \times \frac{\frac{1}{2}\log 5}{\log 6}

Cancelling log⁡5log⁡4\frac{\log 5}{\log 4} from both sides (since it is non-zero):
1=log⁡y×1/2log⁡6=log⁡y2log⁡61 = \log y \times \frac{1/2}{\log 6} = \frac{\log y}{2\log 6}

2log⁡6=log⁡y2\log 6 = \log y
log⁡62=log⁡y\log 6^2 = \log y
log⁡36=log⁡y  ⟹  y=36\log 36 = \log y \implies y = 36.

Q68MCQTime, Speed & Distance

Leaving home at the same time, Amal reaches the office at 10:15 am if he travels at 8 km/hr, and at 9:40 am if he travels at 15 km/hr. Leaving home at 9.10 am, at what speed, in km/hr, must he travel so as to reach office exactly at 10 am?
  1. 13
  2. 12
  3. 14
  4. 11
Answer and solution

Answer: (B) 12

Let the distance be dd km. At 8 km/h he arrives 35 minutes later than at 15 km/h (9:40 to 10:15), so d8−d15=3560=712\frac{d}{8} - \frac{d}{15} = \frac{35}{60} = \frac{7}{12}. Then 7d120=712\frac{7d}{120} = \frac{7}{12}, so d=10d = 10 km. Leaving at 9:10 and arriving at 10:00 gives 50 minutes, which is 56\frac{5}{6} hour. Speed =105/6=12= \frac{10}{5/6} = 12 km/h. Option C (14) takes 1014\frac{10}{14} hour, about 43 minutes, so he would arrive about 9:53, too early. Option D (11) takes about 55 minutes, so he would arrive late. Hence, option B (12).

Q69MCQTime, Speed & Distance

A train travelled at one-third of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to
  1. 50
  2. 58
  3. 67
  4. 61
Answer and solution

Answer: (C) 67

At one-third speed the trip takes 3 times as long. It arrives 30 minutes late, so 3T−T=303T - T = 30, giving the usual time T=15T = 15 minutes. On the return trip, the train runs 5 minutes at its usual speed SS, covering 515=13\frac{5}{15} = \frac{1}{3} of the route. It then stops for 4 minutes, so 9 of the 15 minutes are used and 6 minutes remain for the other 23\frac{2}{3} of the route. At usual speed, 23\frac{2}{3} of the route takes 10 minutes. To do it in 6 minutes, the speed must be 106=53\frac{10}{6} = \frac{5}{3} times SS. The increase is 23\frac{2}{3} of SS, about 66.7%66.7\%, which is nearest to 67. Option D (61) is too small: at 1.61S1.61S the remaining part takes about 10/1.61≈6.210 / 1.61 \approx 6.2 minutes, so the train would be late. Hence, option C (67).

Q70MCQDigits & Base Systems

The mean of all 4-digit even natural numbers of the form 'aabb', where a>0a>0, is
  1. 4466
  2. 5050
  3. 4864
  4. 5544
Answer and solution

Answer: (D) 5544

The number aabb=1100a+11baabb = 1100a + 11b, where aa runs over 1 to 9 and bb over the even digits 0, 2, 4, 6, 8. Every pair gives one number, so there are 9×5=459 \times 5 = 45 numbers. Every value of aa is paired with every value of bb, so the mean of 1100a+11b1100a + 11b is 1100×(mean of a)+11×(mean of b)1100 \times (\text{mean of } a) + 11 \times (\text{mean of } b). The mean of aa is 5, and the mean of bb is 0+2+4+6+85=4\frac{0 + 2 + 4 + 6 + 8}{5} = 4. Mean =1100×5+11×4=5500+44=5544= 1100 \times 5 + 11 \times 4 = 5500 + 44 = 5544. Option A (4466) equals 1100×4+11×61100 \times 4 + 11 \times 6, which would need a mean of 4 for aa and 6 for bb, but the means are 5 and 4. Hence, option D (5544).

Q71MCQTime, Speed & Distance

Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to
  1. 87
  2. 82
  3. 78
  4. 75
Answer and solution

Answer: (B) 82

Let the train's speed be SS km/h and its length LL m. When passing a walker, the train covers its own length at its speed minus the walker's speed. L=(S−2)×518×90=25(S−2)L = (S - 2) \times \frac{5}{18} \times 90 = 25(S - 2) L=(S−4)×518×100=2509(S−4)L = (S - 4) \times \frac{5}{18} \times 100 = \frac{250}{9}(S - 4) Equating: 9(S−2)=10(S−4)9(S - 2) = 10(S - 4), so S=22S = 22 km/h and L=25×20=500L = 25 \times 20 = 500 m. To pass a post, the train covers 500 m at 22×518=55922 \times \frac{5}{18} = \frac{55}{9} m/s. This takes 500×955=90011≈81.8\frac{500 \times 9}{55} = \frac{900}{11} \approx 81.8 s. 81.8 s is much nearer 82 than option C (78) or option A (87). Hence, option B (82).

Q72MCQProperties of Numbers

If a,ba, b and cc are positive integers such that ab=432ab = 432, bc=96bc = 96 and c<9c < 9, then the smallest possible value of a+b+ca + b + c is
  1. 49
  2. 56
  3. 59
  4. 46
Answer and solution

Answer: (D) 46

cc must divide 96 and be less than 9, so cc is 1, 2, 3, 4, 6 or 8. Then b=96cb = \frac{96}{c}, and a=432ba = \frac{432}{b} must be a whole number. c=8c = 8: b=12b = 12, a=36a = 36, sum 56. c=6c = 6: b=16b = 16, a=27a = 27, sum 49. c=4c = 4: b=24b = 24, a=18a = 18, sum 46. c=3c = 3: b=32b = 32, a=13.5a = 13.5, not allowed. c=2c = 2: b=48b = 48, a=9a = 9, sum 59. c=1c = 1: b=96b = 96, a=4.5a = 4.5, not allowed. The smallest sum is 46, with a=18a = 18, b=24b = 24, c=4c = 4. Option A (49), from c=6c = 6, is the next smallest but is larger. Hence, option D (46).

Q73MCQPercentages

In a group of people, 28% of the members are young while the rest are old. If 65% of the members are literates, and 25% of the literates are young, then the percentage of old people among the illiterates is nearest to
  1. 62
  2. 55
  3. 59
  4. 66
Answer and solution

Answer: (D) 66

Take 100 people: 28 young and 72 old; 65 literate and 35 illiterate. Young literates =25%= 25\% of 65=16.2565 = 16.25. Young illiterates =28−16.25=11.75= 28 - 16.25 = 11.75. Old illiterates =35−11.75=23.25= 35 - 11.75 = 23.25. Share of old people among the illiterates =23.2535×100≈66.4%= \frac{23.25}{35} \times 100 \approx 66.4\%, which is nearest to 66. The next closest option, A (62), is more than 4 points away from 66.4. Hence, option D (66).

Q74TITASimple & Compound Interest

Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru’s investment, will their balances, i.e., principal plus accumulated interest, be equal?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 12

Let TT be the number of years after Veeru's investment when their balances become equal.
Veeru invested Rs 10000 at 5% simple interest.
Amount for Veeru after TT years = 10000+(10000×5×T100)=10000+500T10000 + \left(\frac{10000 \times 5 \times T}{100}\right) = 10000 + 500T.

Joy invested Rs 8000 at 10% simple interest exactly 2 years after Veeru.
So, the time period for Joy's investment is (T−2)(T - 2) years.
Amount for Joy after (T−2)(T - 2) years = 8000+(8000×10×(T−2)100)=8000+800(T−2)=8000+800T−1600=6400+800T8000 + \left(\frac{8000 \times 10 \times (T - 2)}{100}\right) = 8000 + 800(T - 2) = 8000 + 800T - 1600 = 6400 + 800T.

Equating their balances:
10000+500T=6400+800T10000 + 500T = 6400 + 800T
10000−6400=800T−500T10000 - 6400 = 800T - 500T
3600=300T  ⟹  T=123600 = 300T \implies T = 12.

Hence, their balances will be equal 12 years after Veeru's investment.

Q75MCQFunctions & Graphs

If f(5+x)=f(5−x)f(5+x) = f(5-x) for every real xx, and f(x)=0f(x)=0 has four distinct real roots, then the sum of these roots is
  1. 0
  2. 40
  3. 10
  4. 20
Answer and solution

Answer: (D) 20

f(5+x)=f(5−x)f(5 + x) = f(5 - x) means the graph of ff is symmetric about x=5x = 5. If rr is a root, put x=r−5x = r - 5: then f(10−r)=f(r)=0f(10 - r) = f(r) = 0, so 10−r10 - r is also a root. So the roots pair up as rr and 10−r10 - r, and each pair sums to 10. Only x=5x = 5 would pair with itself. If 5 were a root, the other three roots would have to pair up among themselves, which is impossible. So the four roots form two pairs. Sum =10+10=20= 10 + 10 = 20. Option C (10) counts only one pair. Hence, option D (20).

Q76MCQLinear Equations

Let A,BA, B and CC be three positive integers such that the sum of AA and the mean of BB and CC is 5. In addition, the sum of BB and the mean of AA and CC is 7. Then the sum of AA and BB is
  1. 5
  2. 4
  3. 6
  4. 7
Answer and solution

Answer: (C) 6

A+B+C2=5A + \frac{B + C}{2} = 5 gives 2A+B+C=102A + B + C = 10. B+A+C2=7B + \frac{A + C}{2} = 7 gives A+2B+C=14A + 2B + C = 14. Subtracting: B−A=4B - A = 4. Putting B=A+4B = A + 4 into the first equation gives 3A+C=63A + C = 6, so C=6−3AC = 6 - 3A. CC must be a positive integer, so A=1A = 1, which gives B=5B = 5 and C=3C = 3. So A+B=6A + B = 6. Option D (7) cannot work: since B=A+4B = A + 4, A+B=2A+4A + B = 2A + 4 is always even. Hence, option C (6).