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CAT 2021 Slot 3 — QA questions with answers

All 22 questions of the Quantitative Ability section (14 MCQs, 8 TITA). Try each one, then open its answer and solution.

CAT 2021 Slot 3, 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 2021 Slot 3 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q45MCQLinear Equations

A shop owner bought a total of 64 shirts from a wholesale market that came in two sizes, small and large. The price of a small shirt was INR 50 less than that of a large shirt. She paid a total of INR 5000 for the large shirts, and a total of INR 1800 for the small shirts. Then, the price of a large shirt and a small shirt together, in INR, is
  1. 175
  2. 150
  3. 200
  4. 225
Answer and solution

Answer: (C) 200

Let a small shirt cost xx, so a large one costs x+50x + 50. She bought 1800x\frac{1800}{x} small and 5000x+50\frac{5000}{x+50} large shirts, 64 in all: 1800x+5000x+50=64\frac{1800}{x} + \frac{5000}{x+50} = 64 Multiply by x(x+50)x(x+50): 1800(x+50)+5000x=64x(x+50)1800(x+50) + 5000x = 64x(x+50), so 6800x+90000=64x2+3200x6800x + 90000 = 64x^2 + 3200x. This gives 64x2−3600x−90000=064x^2 - 3600x - 90000 = 0, or 16x2−900x−22500=016x^2 - 900x - 22500 = 0, which factorises as (x−75)(16x+300)=0(x - 75)(16x + 300) = 0. A price is positive, so x=75x = 75. So a small shirt costs 75 and a large one 125. Check: 180075+5000125=24+40=64\frac{1800}{75} + \frac{5000}{125} = 24 + 40 = 64. Together they cost 75+125=20075 + 125 = 200. Option B (150) fails: it makes the prices 50 and 100, giving 36+50=8636 + 50 = 86 shirts, not 64. Options A and D make the small price 62.5 or 87.5, and neither divides 1800 into a whole number of shirts. Hence, option C (200).

Q46MCQTime & Work

One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is
  1. 11.5
  2. 10
  3. 12.5
  4. 12
Answer and solution

Answer: (B) 10

Rahul works from 9 AM to 5 PM, which is 8 hours. Gautam joins at 11 AM, so he works 6 hours. Had both started at 9 AM, the work would have ended 30 minutes earlier, at 4:30 PM, so each would have worked 7.5 hours. Let Rahul do aa units of work per hour and Gautam bb. The work is the same either way: 8a+6b=7.5a+7.5b8a + 6b = 7.5a + 7.5b, so 0.5a=1.5b0.5a = 1.5b and b=a3b = \frac{a}{3}. Total work =8a+6⋅a3=8a+2a=10a= 8a + 6 \cdot \frac{a}{3} = 8a + 2a = 10a. Rahul alone would take 10aa=10\frac{10a}{a} = 10 hours. Option D (12) fails: Rahul's rate would be 112\frac{1}{12} and Gautam's 136\frac{1}{36} of the work per hour, so on the actual day they would finish only 8⋅112+6⋅136=568 \cdot \frac{1}{12} + 6 \cdot \frac{1}{36} = \frac{5}{6} of the work by 5 PM. Hence, option B (10).

Q47MCQPercentages

In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be
  1. 80
  2. 78
  3. 84
  4. 86
Answer and solution

Answer: (C) 84

So far the team has won 30% of 40 matches, that is 12. Let xx matches remain. Winning 60% of them would make the overall win rate 50%: 12+0.6x=0.5(40+x)12 + 0.6x = 0.5(40 + x) 12+0.6x=20+0.5x12 + 0.6x = 20 + 0.5x, so 0.1x=80.1x = 8 and x=80x = 80. If instead they win 90% of the remaining 80 matches, they win 0.9×80=720.9 \times 80 = 72 more. Total matches won =12+72=84= 12 + 72 = 84. Option A (80) would need 68 wins from the remaining 80 matches, which is 85%, not 90%. Hence, option C (84).

Q48TITAInequalities & Modulus

The number of distinct pairs of integers (m,n), satisfying ∣1+mn∣<∣m+n∣<5 is:

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

Answer and solution

Answer: 12

Both sides of ∣1+mn∣<∣m+n∣|1 + mn| < |m + n| are non-negative, so it is the same as (1+mn)2<(m+n)2(1 + mn)^2 < (m + n)^2, i.e. 1+2mn+m2n2<m2+2mn+n21 + 2mn + m^2n^2 < m^2 + 2mn + n^2. This simplifies to 1−m2−n2+m2n2<01 - m^2 - n^2 + m^2n^2 < 0, i.e. (1−m2)(1−n2)<0(1 - m^2)(1 - n^2) < 0. So one factor is positive and the other negative: one of ∣m∣|m|, ∣n∣|n| is below 11 and the other above 11. These are the shaded bands: Solution figure for question 48, CAT 2021 Slot 3 For integers, 1−m2>01 - m^2 > 0 only when m=0m = 0, and 1−n2<01 - n^2 < 0 when ∣n∣≥2|n| \ge 2 (∣n∣=1|n| = 1 makes the factor 00). If m=0m = 0: ∣m+n∣=∣n∣<5|m + n| = |n| < 5 with ∣n∣≥2|n| \ge 2 gives n=±2,±3,±4n = \pm 2, \pm 3, \pm 4, which is 6 pairs. In each, ∣1+mn∣=1<∣n∣|1 + mn| = 1 < |n|. By symmetry, n=0n = 0 with m=±2,±3,±4m = \pm 2, \pm 3, \pm 4 gives 6 more pairs. Total: 6+6=126 + 6 = 12 pairs. The answer is 12.

Q49MCQLogarithms

For a real number aa, if log⁡15a+log⁡32a(log⁡15a)(log⁡32a)=4\dfrac{\log_{15} a + \log_{32} a}{(\log_{15} a)(\log_{32} a)} = 4, then aa must lie in the range
  1. 2<a<32 < a < 3
  2. 3<a<43 < a < 4
  3. 4<a<54 < a < 5
  4. a>5a > 5
Answer and solution

Answer: (C) 4<a<54 < a < 5

Split the fraction into two parts: log⁡15a+log⁡32a(log⁡15a)(log⁡32a)=1log⁡32a+1log⁡15a\frac{\log_{15} a + \log_{32} a}{(\log_{15} a)(\log_{32} a)} = \frac{1}{\log_{32} a} + \frac{1}{\log_{15} a} Since 1log⁡ba=log⁡ab\frac{1}{\log_b a} = \log_a b, this equals log⁡a32+log⁡a15=log⁡a480\log_a 32 + \log_a 15 = \log_a 480. So log⁡a480=4\log_a 480 = 4, which means a4=480a^4 = 480. Now 44=2564^4 = 256 and 54=6255^4 = 625. Since 256<480<625256 < 480 < 625 and aa must be positive for the logarithms to exist, 4<a<54 < a < 5. Option D (a>5a > 5) fails because it would make a4>625a^4 > 625. Option B fails because a<4a < 4 would make a4<256a^4 < 256. Hence, option C (4<a<54 < a < 5).

Q50MCQPercentages

The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is
  1. 68.25
  2. 68.75
  3. 68.50
  4. 69.25
Answer and solution

Answer: (B) 68.75

Only ratios matter, so take convenient numbers. In 1990 the female population is twice the male population: let them be 200 and 100. Females rose 25% from 1980 to 1990, so in 1980 there were 2001.25=160\frac{200}{1.25} = 160. They rose 20% from 1970 to 1980, so in 1970 there were 1601.2=4003\frac{160}{1.2} = \frac{400}{3}. Let the male population be xx in 1970, so it was 1.4x1.4x in 1980. Solution figure for question 50, CAT 2021 Slot 3 The total rose 25% from 1970 to 1980: 1.25(x+4003)=1.4x+1601.25\left(x + \frac{400}{3}\right) = 1.4x + 160 1.25x+5003=1.4x+1601.25x + \frac{500}{3} = 1.4x + 160, so 0.15x=2030.15x = \frac{20}{3} and x=4009x = \frac{400}{9}. Total in 1970 =4009+12009=16009= \frac{400}{9} + \frac{1200}{9} = \frac{1600}{9}. Total in 1990 =100+200=300= 100 + 200 = 300. Increase =300−16009=11009= 300 - \frac{1600}{9} = \frac{1100}{9}. Percentage increase =1100/91600/9×100=110016=68.75%= \frac{1100/9}{1600/9} \times 100 = \frac{1100}{16} = 68.75\%. The options are close together, so the exact fraction matters: 110016\frac{1100}{16} is exactly 68.75, not 68.50 (option C). Hence, option B (68.75).

Q51MCQSequences & Series

Consider a sequence of real numbers x1,x2,x3,…x_1, x_2, x_3, \ldots such that xn+1=xn+n−1x_{n+1} = x_n + n - 1 for all n≥1n \geq 1. If x1=−1x_1 = -1 then x100x_{100} is equal to
  1. 4849
  2. 4949
  3. 4950
  4. 4850
Answer and solution

Answer: (D) 4850

The rule xn+1=xn+n−1x_{n+1} = x_n + n - 1 means that step nn adds n−1n - 1. Going from x1x_1 to x100x_{100} takes the 99 steps n=1,2,…,99n = 1, 2, \ldots, 99, which add 0,1,2,…,980, 1, 2, \ldots, 98. So x100=x1+(0+1+⋯+98)=−1+98×992=−1+4851=4850x_{100} = x_1 + (0 + 1 + \cdots + 98) = -1 + \dfrac{98 \times 99}{2} = -1 + 4851 = 4850. Check on the first terms: x2=−1+0=−1x_2 = -1 + 0 = -1, x3=−1+1=0x_3 = -1 + 1 = 0, x4=0+2=2x_4 = 0 + 2 = 2. Option B (4949) comes from adding 0+1+⋯+990 + 1 + \cdots + 99, one step too many: the last step (n=99n = 99) adds 9898, not 9999. Hence, option D (4850).

Q52TITAAverages, Mixtures & Alligations

The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

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

Answer and solution

Answer: 92

The total of all 25 scores is 25×50=125025 \times 50 = 1250. To make the toppers' common score as large as possible, make the other 20 scores as small as possible. They are distinct integers and the lowest is 30, so the smallest they can be is 30,31,…,4930, 31, \ldots, 49. Their sum is 20×(30+49)2=10×79=790\frac{20 \times (30 + 49)}{2} = 10 \times 79 = 790. Let each topper score yy. Then 5y+790=12505y + 790 = 1250, so 5y=4605y = 460 and y=92y = 92. This is consistent: 92 is above 49, the highest of the other scores, so the five really are the toppers. The answer is 92.

Q53MCQLinear Equations

One part of a hostel’s monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be
  1. 20200
  2. 20500
  3. 20800
  4. 20000
Answer and solution

Answer: (B) 20500

Let the fixed monthly cost be aa and the cost per boarder be bb. With nn boarders the total cost is a+bna + bn, so the profit per boarder is 1600−a+bnn=1600−an−b1600 - \frac{a + bn}{n} = 1600 - \frac{a}{n} - b. With 50 boarders: 1600−a50−b=2001600 - \frac{a}{50} - b = 200, so a50+b=1400\frac{a}{50} + b = 1400. With 75 boarders: 1600−a75−b=2501600 - \frac{a}{75} - b = 250, so a75+b=1350\frac{a}{75} + b = 1350. Subtracting: a50−a75=50\frac{a}{50} - \frac{a}{75} = 50, so a150=50\frac{a}{150} = 50 and a=7500a = 7500. Then b=1400−150=1250b = 1400 - 150 = 1250. With 80 boarders, total profit =80×1600−(7500+80×1250)=128000−107500=20500= 80 \times 1600 - (7500 + 80 \times 1250) = 128000 - 107500 = 20500. Option D (20000) keeps the profit per boarder at 250, as with 75 boarders. But the profit per boarder rises as the fixed cost is shared among more boarders. Hence, option B (20500).

Q54MCQInequalities & Modulus

The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is
  1. 120000
  2. 90000
  3. 100000
  4. 160000
Answer and solution

Answer: (A) 120000

Let the side fenced at 200 per ft have length BB, and let the other pair of sides have length LL. Solution figure for question 54, CAT 2021 Slot 3 The costly side and the side opposite it cost 200B+100B=300B200B + 100B = 300B. The two sides of length LL cost 100L+100L=200L100L + 100L = 200L. Total cost =300B+200L= 300B + 200L, with LB=60000LB = 60000. By AM–GM, 300B+200L≥2300B×200L=260000×60000=120000300B + 200L \ge 2\sqrt{300B \times 200L} = 2\sqrt{60000 \times 60000} = 120000. Equality holds when 300B=200L300B = 200L; with LB=60000LB = 60000 this gives B=200B = 200 ft and L=300L = 300 ft. Check: 300×200+200×300=60000+60000=120000300 \times 200 + 200 \times 300 = 60000 + 60000 = 120000. So the lowest cost is 120000, with the costly fencing on a 200 ft side. Options B (90000) and C (100000) are below this lower bound, so no rectangle of area 60000 sq ft can be fenced that cheaply. Hence, option A (120000).

Q55TITAPolygons & Circles

A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is

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

Answer and solution

Answer: 3500

The 40 m of fencing is the perimeter, so each side of the rhombus is 404=10\frac{40}{4} = 10 m. Let the diagonals be d1d_1 and d2d_2. Area: 12d1d2=96\frac{1}{2}d_1 d_2 = 96, so d1d2=192d_1 d_2 = 192. The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs d12\frac{d_1}{2} and d22\frac{d_2}{2}: (d12)2+(d22)2=102\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 10^2, so d12+d22=400d_1^2 + d_2^2 = 400. Then (d1+d2)2=d12+d22+2d1d2=400+384=784(d_1 + d_2)^2 = d_1^2 + d_2^2 + 2d_1 d_2 = 400 + 384 = 784, so d1+d2=28d_1 + d_2 = 28 m. (The diagonals are 12 m and 16 m.) The cost of laying wire along both diagonals is 28×125=350028 \times 125 = 3500. The answer is 3500.

Q56TITAQuadratic & Polynomial Equations

A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

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

Answer and solution

Answer: 34

Let the small and medium prices be 2x2x and 5x5x, and the large price yy. Original product: 2x⋅5x⋅y=8002x \cdot 5x \cdot y = 800, so x2y=80x^2 y = 80. After the increase: (2x+6)(5x+6)y=3200(2x + 6)(5x + 6)y = 3200. Divide the second equation by the first: (2x+6)(5x+6)10x2=4\frac{(2x+6)(5x+6)}{10x^2} = 4, so (2x+6)(5x+6)=40x2(2x+6)(5x+6) = 40x^2. Expanding: 10x2+42x+36=40x210x^2 + 42x + 36 = 40x^2, so 30x2−42x−36=030x^2 - 42x - 36 = 0, that is, 5x2−7x−6=05x^2 - 7x - 6 = 0. This factorises as (x−2)(5x+3)=0(x - 2)(5x + 3) = 0. A price is positive, so x=2x = 2. Small =4= 4, medium =10= 10, and y=80x2=804=20y = \frac{80}{x^2} = \frac{80}{4} = 20, which is indeed the largest. Check: 4×10×20=8004 \times 10 \times 20 = 800 and 10×16×20=320010 \times 16 \times 20 = 3200. Sum of the original prices =4+10+20=34= 4 + 10 + 20 = 34. The answer is 34.

Q57TITATime, Speed & Distance

Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is

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

Answer and solution

Answer: 8

Let the track length be CC, and let Mira and Amal cover MM and AA per minute. Opposite directions: they first meet when together they have covered one round, so 3(A+M)=C3(A + M) = C, i.e. A+M=C3A + M = \frac{C}{3}. Same direction: in 45 minutes Amal gains 3 rounds on Mira, so 45(A−M)=3C45(A - M) = 3C, i.e. A−M=C15A - M = \frac{C}{15}. Subtracting: 2M=C3−C15=4C152M = \frac{C}{3} - \frac{C}{15} = \frac{4C}{15}, so M=2C15M = \frac{2C}{15} per minute. In one hour Mira walks 60×2C15=8C60 \times \frac{2C}{15} = 8C, which is 8 rounds. The answer is 8.

Q58MCQAverages, Mixtures & Alligations

If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is
  1. 3.5
  2. 2.5
  3. 3
  4. 4
Answer and solution

Answer: (C) 3

Let the initial alloy weigh WW kg and contain ss kg of silver. Adding 3 kg of pure silver gives 90% silver: s+3=0.9(W+3)s + 3 = 0.9(W + 3). Adding 2 kg of a 90% alloy adds 0.9×2=1.80.9 \times 2 = 1.8 kg of silver and gives 84% silver: s+1.8=0.84(W+2)s + 1.8 = 0.84(W + 2). Subtract the second equation from the first: 1.2=0.9W+2.7−0.84W−1.68=0.06W+1.021.2 = 0.9W + 2.7 - 0.84W - 1.68 = 0.06W + 1.02 So 0.06W=0.180.06W = 0.18 and W=3W = 3 kg, with s=0.9×6−3=2.4s = 0.9 \times 6 - 3 = 2.4 kg. Check: 2.4+36=0.9\frac{2.4 + 3}{6} = 0.9 and 2.4+1.85=0.84\frac{2.4 + 1.8}{5} = 0.84. Option D (4) fails: the first mixture would need s=0.9×7−3=3.3s = 0.9 \times 7 - 3 = 3.3 kg of silver, but the second would need s=0.84×6−1.8=3.24s = 0.84 \times 6 - 1.8 = 3.24 kg, and one alloy cannot have both. Hence, option C (3).

Q59MCQTriangles & Lines

In a triangle ABC, ∠BCA = 50°. D and E are points on AB and AC, respectively, such that AD=DE. If F is a point on BC such that BD=DF, then ∠FDE, in degrees, is equal to
  1. 72
  2. 80
  3. 100
  4. 96
Answer and solution

Answer: (B) 80

Let ∠FDE=p\angle FDE = p. Since AD=DEAD = DE, triangle ADE is isosceles, so ∠DAE=∠DEA\angle DAE = \angle DEA. Call this xx; it is also ∠A\angle A of triangle ABC. Then ∠ADE=180∘−2x\angle ADE = 180^\circ - 2x. Since BD=DFBD = DF, triangle BDF is isosceles, so ∠DBF=∠DFB=y\angle DBF = \angle DFB = y, which is ∠B\angle B of triangle ABC. Then ∠BDF=180∘−2y\angle BDF = 180^\circ - 2y. Solution figure for question 59, CAT 2021 Slot 3 D lies on the straight line AB, so ∠ADE+p+∠BDF=180∘\angle ADE + p + \angle BDF = 180^\circ: (180∘−2x)+p+(180∘−2y)=180∘(180^\circ - 2x) + p + (180^\circ - 2y) = 180^\circ, so p=2(x+y)−180∘p = 2(x + y) - 180^\circ. In triangle ABC, x+y+50∘=180∘x + y + 50^\circ = 180^\circ, so x+y=130∘x + y = 130^\circ. So p=260∘−180∘=80∘p = 260^\circ - 180^\circ = 80^\circ. The result depends only on ∠C\angle C: p=180∘−2∠Cp = 180^\circ - 2\angle C. Option C (100) would need ∠C=40∘\angle C = 40^\circ. Hence, option B (80).

Q60MCQSimple & Compound Interest

Bank A offers 6% interest rate per annum compounded half-yearly. Bank B and Bank C offer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests ₹ 10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in INR, to Rupa is
  1. 3436
  2. 2436
  3. 2346
  4. 1436
Answer and solution

Answer: (B) 2436

Bank A pays 6% a year compounded half-yearly, which is 3% every six months. In one year, 1 rupee grows to 1.032=1.06091.03^2 = 1.0609, so a year's interest is 6.09% of the amount. Let Bank B pay r%r\% simple interest, so Bank C pays 2r%2r\%. Raju invests PP in Bank B for tt years, and his interest equals one year's interest at Bank A: P⋅r⋅t100=0.0609P\frac{P \cdot r \cdot t}{100} = 0.0609P, so rt100=0.0609\frac{rt}{100} = 0.0609. Rupa invests 10,000 in Bank C at 2r%2r\% for 2t2t years: Interest =10000×2r×2t100=40000×rt100=40000×0.0609=2436= \frac{10000 \times 2r \times 2t}{100} = 40000 \times \frac{rt}{100} = 40000 \times 0.0609 = 2436. Raju's amount PP cancels, so it is not needed. Taking Bank A's rate as a plain 6% would give 40000×0.06=240040000 \times 0.06 = 2400; the half-yearly compounding adds the extra 36. Options A, C and D are look-alike numbers, and none equals 40000×0.060940000 \times 0.0609. Hence, option B (2436).

Q61MCQFunctions & Graphs

If f(x)=x2−7xf(x) = x^2 - 7x and g(x)=x+3g(x) = x + 3, then the minimum value of f(g(x))−3xf(g(x)) - 3x is
  1. -12
  2. -15
  3. -16
  4. -20
Answer and solution

Answer: (C) -16

Substitute g(x)=x+3g(x) = x + 3 into ff: f(g(x))=(x+3)2−7(x+3)=x2+6x+9−7x−21=x2−x−12f(g(x)) = (x+3)^2 - 7(x+3) = x^2 + 6x + 9 - 7x - 21 = x^2 - x - 12. So the expression to minimise is f(g(x))−3x=x2−4x−12=(x−2)2−16f(g(x)) - 3x = x^2 - 4x - 12 = (x - 2)^2 - 16. A square is never negative, so the smallest value is −16-16, reached at x=2x = 2. Check: g(2)=5g(2) = 5, f(5)=25−35=−10f(5) = 25 - 35 = -10, and −10−3×2=−16-10 - 3 \times 2 = -16. Option A (−12-12) is only the value at x=0x = 0, not the minimum. Option D (−20-20) is never reached, since (x−2)2−16≥−16(x-2)^2 - 16 \ge -16. Hence, option C (-16).

Q62TITATime & Work

Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is

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

Answer and solution

Answer: 3000

Anil paints the house in 12 days, so in 6 days he does 612=12\frac{6}{12} = \frac{1}{2} of the work. Barun paints it in 16 days, so in 6 days he does 616=38\frac{6}{16} = \frac{3}{8} of the work. The three finish together in 6 days, so Chandu does the rest: 1−12−38=8−4−38=181 - \frac{1}{2} - \frac{3}{8} = \frac{8 - 4 - 3}{8} = \frac{1}{8}. He is paid in proportion to his work, so his share is 18×24000=3000\frac{1}{8} \times 24000 = 3000. As a check, Anil gets 12000 and Barun 9000, which with Chandu's 3000 make 24000. The answer is 3000.

Q63TITAIndices & Surds

If nn is a positive integer such that (107)(107)2⋯(107)n>999(\sqrt[7]{10})(\sqrt[7]{10})^2 \cdots (\sqrt[7]{10})^n > 999, then the smallest value of nn is

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

Answer and solution

Answer: 6

Every factor is a power of 101/710^{1/7}, so the product is (101/7)1+2+⋯+n=10n(n+1)14(10^{1/7})^{1 + 2 + \cdots + n} = 10^{\frac{n(n+1)}{14}}. We need 10n(n+1)14>99910^{\frac{n(n+1)}{14}} > 999. For n=5n = 5, the exponent is 5×614=157≈2.14\frac{5 \times 6}{14} = \frac{15}{7} \approx 2.14, and 102.14≈13910^{2.14} \approx 139, far below 999. Smaller values of nn give smaller products, so they fail too. For n=6n = 6, the exponent is 6×714=3\frac{6 \times 7}{14} = 3, and 103=1000>99910^3 = 1000 > 999. So the smallest such nn is 6. The answer is 6.

Q64TITAPermutations & Combinations

A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

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

Answer and solution

Answer: 50

Count all four-digit numbers made only of 1, 2 and 3, then remove those missing a 2 or missing a 3. All such numbers: each of the 4 places has 3 choices, so 34=813^4 = 81. Numbers with no 2 use only 1 and 3: 24=162^4 = 16. Numbers with no 3 use only 1 and 2: 24=162^4 = 16. The number 1111 has neither, so it was removed twice; add it back once. Required count =81−16−16+1=50= 81 - 16 - 16 + 1 = 50. Check by counting the 1s: Two 1s with one 2 and one 3: 4!2!=12\frac{4!}{2!} = 12. One 1 with 2, 2, 3 or with 2, 3, 3: 4!2!=12\frac{4!}{2!} = 12 each, 24 in all. No 1s: 2, 2, 2, 3 and 2, 3, 3, 3 give 4 each, and 2, 2, 3, 3 gives 4!2! 2!=6\frac{4!}{2!\,2!} = 6, 14 in all. Total =12+24+14=50= 12 + 24 + 14 = 50. The answer is 50.

Q65MCQPolygons & Circles

Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10 cm and 20 cm, respectively. If the angle ∠ADC=30∘\angle ADC = 30^\circ then the area of the parallelogram, in sq. cm is
  1. 25(5+15)2\frac{25(\sqrt{5}+\sqrt{15})}{2}
  2. 25(3+15)25(\sqrt{3}+\sqrt{15})
  3. 25(3+15)2\frac{25(\sqrt{3}+\sqrt{15})}{2}
  4. 25(5+15)25(\sqrt{5}+\sqrt{15})
Answer and solution

Answer: (B) 25(3+15)25(\sqrt{3}+\sqrt{15})

In triangle ACD, AD=10AD = 10, the diagonal AC=20AC = 20 and ∠ADC=30∘\angle ADC = 30^\circ. Let DC=xDC = x. (The sketch is not to scale.) Solution figure for question 65, CAT 2021 Slot 3 By the cosine rule, AC2=AD2+DC2−2⋅AD⋅DCcos⁡30∘AC^2 = AD^2 + DC^2 - 2 \cdot AD \cdot DC \cos 30^\circ: 400=100+x2−2⋅10⋅x⋅32400 = 100 + x^2 - 2 \cdot 10 \cdot x \cdot \frac{\sqrt{3}}{2}, so x2−103 x−300=0x^2 - 10\sqrt{3}\,x - 300 = 0. x=103±300+12002=103±10152x = \frac{10\sqrt{3} \pm \sqrt{300 + 1200}}{2} = \frac{10\sqrt{3} \pm 10\sqrt{15}}{2}. A length is positive, so x=5(3+15)x = 5(\sqrt{3} + \sqrt{15}). Area of the parallelogram =AD⋅DCsin⁡30∘=10⋅5(3+15)⋅12=25(3+15)= AD \cdot DC \sin 30^\circ = 10 \cdot 5(\sqrt{3} + \sqrt{15}) \cdot \frac{1}{2} = 25(\sqrt{3} + \sqrt{15}). Option C is half of this. It is the area of triangle ACD alone, but the diagonal AC splits the parallelogram into two such triangles. Hence, option B (25(3+15)25(\sqrt{3}+\sqrt{15})).

Q66MCQInequalities & Modulus

If 3x+2∣y∣+y=73x + 2|y| + y = 7 and x+∣x∣+3y=1x + |x| + 3y = 1, then x+2yx + 2y is
  1. −43-\frac{4}{3}
  2. 83\frac{8}{3}
  3. 00
  4. 11
Answer and solution

Answer: (C) 00

Work case by case on the signs of xx and yy. If x<0x < 0: the second equation becomes x−x+3y=1x - x + 3y = 1, so y=13y = \frac{1}{3}. The first then gives 3x+23+13=73x + \frac{2}{3} + \frac{1}{3} = 7, so x=2x = 2, which contradicts x<0x < 0. No solution. So x≥0x \ge 0, and the second equation is 2x+3y=12x + 3y = 1. If y≥0y \ge 0: the first equation is 3x+3y=73x + 3y = 7. Subtracting, x=6x = 6 and y=−113y = -\frac{11}{3}, which contradicts y≥0y \ge 0. If y<0y < 0: the first equation is 3x−2y+y=73x - 2y + y = 7, that is, 3x−y=73x - y = 7, so y=3x−7y = 3x - 7. Then 2x+3(3x−7)=12x + 3(3x - 7) = 1 gives 11x=2211x = 22, so x=2x = 2 and y=−1y = -1. Both signs fit. So x+2y=2−2=0x + 2y = 2 - 2 = 0. Option A (−43-\frac{4}{3}) comes from the rejected case x=6x = 6, y=−113y = -\frac{11}{3}, which breaks y≥0y \ge 0. Hence, option C (00).