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CAT 2021 Slot 2 — 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 2, 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 2 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q45TITAIndices & Surds

For all possible integers nn satisfying 2.25≤2+2n+2≤2022.25 \le 2 + 2^{n+2} \le 202, then the number of integer values of 3+3n+13 + 3^{n+1} is

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

Answer and solution

Answer: 7

Subtract 2 throughout: 0.25≤2n+2≤2000.25\le 2^{n+2}\le 200. Since 0.25=2−20.25=2^{-2} and 27=128≤200<256=282^7=128\le 200<256=2^8, the integer n+2n+2 runs from −2-2 to 77. So nn runs from −4-4 to 55, which is 10 integers. Now 3+3n+13+3^{n+1} is an integer only when n+1≥0n+1\ge 0, that is, n≥−1n\ge -1. For n≤−2n\le -2, 3n+13^{n+1} is a fraction such as 13\frac{1}{3} or 19\frac{1}{9}. So nn can be −1,0,1,2,3,4,5-1, 0, 1, 2, 3, 4, 5. These give values from 3+30=43+3^0=4 up to 3+36=7323+3^6=732, seven different integers. The answer is 7.

Q46MCQSequences & Series

Three positive integers x, y and z are in arithmetic progression. If y−x>2 and xyz=5(x+y+z), then z-x equals
  1. 8
  2. 12
  3. 14
  4. 10
Answer and solution

Answer: (C) 14

Let x=ax=a, y=a+dy=a+d and z=a+2dz=a+2d, with d=y−x>2d=y-x>2. Then x+y+z=3(a+d)x+y+z=3(a+d), so xyz=5(x+y+z)xyz=5(x+y+z) becomes a(a+d)(a+2d)=15(a+d)a(a+d)(a+2d)=15(a+d). Since a+d>0a+d>0, divide by it: a(a+2d)=15a(a+2d)=15. With positive integers and a<a+2da<a+2d, the factor pairs are 1×151\times 15 and 3×53\times 5. a=1a=1, a+2d=15a+2d=15 gives d=7d=7, which satisfies d>2d>2. a=3a=3, a+2d=5a+2d=5 gives d=1d=1, which fails d>2d>2. So x=1x=1, y=8y=8, z=15z=15. Check: 1×8×15=1201\times 8\times 15=120 and 5(1+8+15)=1205(1+8+15)=120. So z−x=14z-x=14. No other valid progression exists; for example, z−x=12z-x=12 would need d=6d=6 and a(a+12)=15a(a+12)=15, which has no positive integer solution. Hence, option C (14).

Q47TITADigits & Base Systems

For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

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Answer and solution

Answer: 4195

Let the digits be aa (thousands), bb (hundreds), cc (tens) and dd (units). The conditions are a+b+c=14a+b+c=14 (1), b+c+d=15b+c+d=15 (2) and c=d+4c=d+4 (3). Subtracting (1) from (2) gives d−a=1d-a=1, so d=a+1d=a+1. Then c=d+4=a+5c=d+4=a+5. Since cc is a digit, a+5≤9a+5\le 9, so a≤4a\le 4. To make the number as large as possible, take the largest thousands digit, a=4a=4. Then d=5d=5, c=9c=9, and from (1), b=14−4−9=1b=14-4-9=1. Check: 4+1+9=144+1+9=14, 1+9+5=151+9+5=15 and 9=5+49=5+4. The answer is 4195.

Q48MCQPercentages

Raj invested ₹ 10000 in a fund. At the end of first year, he incurred a loss but his balance was more than ₹ 5000. This balance, when invested for another year, grew and the percentage of growth in the second year was five times the percentage of loss in the first year. If the gain of Raj from the initial investment over the two year period is 35%, then the percentage of loss in the first year is
  1. 5
  2. 15
  3. 17
  4. 10
Answer and solution

Answer: (D) 10

Let the loss in the first year be x%x\%. After one year the balance is 10000(1−x100)10000\left(1-\frac{x}{100}\right), and this is more than 5000, so x<50x<50. In the second year it grows by 5x%5x\%. The overall gain is 35%, so the final amount is 13500: 10000(1−x100)(1+5x100)=1350010000\left(1-\frac{x}{100}\right)\left(1+\frac{5x}{100}\right)=13500 (100−x)(100+5x)=13500(100-x)(100+5x)=13500 10000+400x−5x2=1350010000+400x-5x^2=13500 x2−80x+700=0  ⟹  (x−10)(x−70)=0x^2-80x+700=0 \implies (x-10)(x-70)=0. So x=10x=10 or x=70x=70, and x<50x<50 gives x=10x=10. Check: 10000×0.9×1.5=1350010000\times 0.9\times 1.5=13500. Option B (15) fails: 0.85×1.75=1.48750.85\times 1.75=1.4875, a gain of about 49%, not 35%. Hence, option D (10).

Q49TITAPermutations & Combinations

The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is

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Answer and solution

Answer: 1000

The balloons, pencils and erasers are shared out independently, so multiply the number of ways for each. The number of ways to share nn identical items among 3 children, with no minimum, is (n+22)\binom{n+2}{2}. Balloons: give each child 4 first, using 12. The remaining 3 balloons can be shared in (52)=10\binom{5}{2}=10 ways. Pencils: give each child 1 first, using 3. The remaining 3 pencils can be shared in (52)=10\binom{5}{2}=10 ways. Erasers: there is no minimum, so the 3 erasers can be shared in (52)=10\binom{5}{2}=10 ways. Total =10×10×10=1000=10\times 10\times 10=1000. The answer is 1000.

Q50MCQTime, Speed & Distance

Two trains A and B were moving in opposite directions, their speeds being in the ratio 5 : 3. The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other. It took another 69 seconds for the rear ends of the trains to cross each other. The ratio of length of train A to that of train B is
  1. 3:2
  2. 5:3
  3. 2:3
  4. 2:1
Answer and solution

Answer: (A) 3:2

Let the speeds be 5k5k and 3k3k. The trains move in opposite directions, so their relative speed is 8k8k. Treat train B as standing still and train A as moving at 8k8k. When the front ends meet, A's front is at B's front. A's front reaches B's rear after covering B's length, which takes 46 s, so LB=8k×46L_B=8k\times 46. The rear ends cross when A's rear passes B's rear. By then A has covered LB+LAL_B+L_A in relative terms, which takes 46+69=11546+69=115 s. So LA+LB=8k×115L_A+L_B=8k\times 115, which gives LA=8k×69L_A=8k\times 69. Therefore LALB=6946=32\frac{L_A}{L_B}=\frac{69}{46}=\frac{3}{2}. The speed ratio 5 : 3 does not affect the answer, because only the relative speed matters. Option C (2:3) is the same ratio inverted, which comes from swapping the two time intervals. Hence, option A (3:2).

Q51MCQQuadratic & Polynomial Equations

Suppose one of the roots of the equation ax2−bx+c=0 is 2+ 3, Where a,b and c are rational numbers and a=0. If b=c3 then ∣a∣ equals.
  1. 1
  2. 2
  3. 3
  4. 4
Answer and solution

Answer: (B) 2

The coefficients aa, bb and cc are rational, so an irrational root comes with its conjugate. The roots are 2+32+\sqrt{3} and 2−32-\sqrt{3}. For ax2−bx+c=0ax^2-bx+c=0, the sum of the roots is ba\frac{b}{a} and the product is ca\frac{c}{a}. Sum: (2+3)+(2−3)=4(2+\sqrt{3})+(2-\sqrt{3})=4, so b=4ab=4a. Product: (2+3)(2−3)=4−3=1(2+\sqrt{3})(2-\sqrt{3})=4-3=1, so c=ac=a. Now b=c3b=c^3 gives 4a=a34a=a^3. Since a≠0a\ne 0, a2=4a^2=4, so a=2a=2 or a=−2a=-2, and ∣a∣=2|a|=2. Option D (4) is the value of a2a^2, not of ∣a∣|a|. Hence, option B (2).

Q52TITAAverages, Mixtures & Alligations

From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

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Answer and solution

Answer: 45

Let the capacity be VV litres. Each time 9 litres are drawn and replaced with water, the milk left is multiplied by (1−9V)\left(1 - \frac{9}{V}\right). After two such operations, milk =V(1−9V)2= V\left(1 - \frac{9}{V}\right)^2. Milk : water =16:9= 16 : 9, so milk is 1625\frac{16}{25} of the container: (1−9V)2=1625\left(1 - \frac{9}{V}\right)^2 = \frac{16}{25}, so 1−9V=451 - \frac{9}{V} = \frac{4}{5}. The negative root, −45-\frac{4}{5}, would give V=5V = 5, which is impossible because 9 litres are drawn each time. So 9V=15\frac{9}{V} = \frac{1}{5} and V=45V = 45. The answer is 45.

Q53MCQPolygons & Circles

If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is
  1. 37+13\sqrt{37} + \sqrt{13}
  2. 13+12\sqrt{13} + \sqrt{12}
  3. 37+132\frac{\sqrt{37} + \sqrt{13}}{2}
  4. 13+122\frac{\sqrt{13} + \sqrt{12}}{2}
Answer and solution

Answer: (A) 37+13\sqrt{37} + \sqrt{13}

The diagonals of a rhombus bisect each other at right angles. In the figure they meet at E, with AE=EC=aAE = EC = a and BE=ED=bBE = ED = b. Solution figure for question 53, CAT 2021 Slot 2 Each side is the hypotenuse of a right triangle such as AEB, so a2+b2=52=25a^2 + b^2 = 5^2 = 25. The rhombus is made of four such right triangles, each of area 12ab\frac{1}{2}ab, so its area is 2ab=122ab = 12 and ab=6ab = 6. Then (a+b)2=25+12=37(a + b)^2 = 25 + 12 = 37, so a+b=37a + b = \sqrt{37}, and (a−b)2=25−12=13(a - b)^2 = 25 - 12 = 13, so the two half-diagonals differ by 13\sqrt{13}. The longer half-diagonal is therefore 37+132\frac{\sqrt{37} + \sqrt{13}}{2}, and the longer diagonal is twice that, 37+13\sqrt{37} + \sqrt{13}. Check: the shorter diagonal is 37−13\sqrt{37} - \sqrt{13}, and 12(37+13)(37−13)=12(37−13)=12\frac{1}{2}(\sqrt{37} + \sqrt{13})(\sqrt{37} - \sqrt{13}) = \frac{1}{2}(37 - 13) = 12, the given area. Option C is only the half-diagonal, a leg of the right triangle, not the full diagonal. Hence, option A (37+13\sqrt{37} + \sqrt{13}).

Q54TITALogarithms

If log⁡2[3+log⁡3{4+log⁡4(x−1)}]−2=0\log_2[3+\log_3\{4+\log_4(x-1)\}]-2=0 then 4x4x equals

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Answer and solution

Answer: 5

Undo the logarithms one layer at a time, from the outside in. Since log⁡2[3+log⁡3{4+log⁡4(x−1)}]=2\log_2[3+\log_3\{4+\log_4(x-1)\}] = 2, we get 3+log⁡3{4+log⁡4(x−1)}=22=43+\log_3\{4+\log_4(x-1)\} = 2^2 = 4. So log⁡3{4+log⁡4(x−1)}=1\log_3\{4+\log_4(x-1)\} = 1, which gives 4+log⁡4(x−1)=31=34+\log_4(x-1) = 3^1 = 3. Then log⁡4(x−1)=−1\log_4(x-1) = -1, so x−1=4−1=14x-1 = 4^{-1} = \frac{1}{4} and x=54x = \frac{5}{4}. Since x−1>0x - 1 > 0, every logarithm is defined. Therefore 4x=4×54=54x = 4 \times \frac{5}{4} = 5. The answer is 5.

Q55MCQPolygons & Circles

The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is
  1. 30
  2. 40
  3. 25
  4. 20
Answer and solution

Answer: (A) 30

Let AB=xAB = x. Since ABPD is a parallelogram, DP=AB=xDP = AB = x. P is the midpoint of CD, so PC=DP=xPC = DP = x and CD=2xCD = 2x. Solution figure for question 55, CAT 2021 Slot 2 Let hh be the height of the trapezium, the distance between AB and CD. The parallelogram ABPD has base DP=xDP = x and height hh, so its area is xhxh. Triangle BPC has base PC=xPC = x on CD and its vertex B on AB, so its height is also hh and its area is 12xh\frac{1}{2}xh. Given: xh−12xh=10xh - \frac{1}{2}xh = 10, so 12xh=10\frac{1}{2}xh = 10 and xh=20xh = 20. The trapezium is the parallelogram plus the triangle, so its area is 20+10=3020 + 10 = 30. As a check, 12(AB+CD)h=12(x+2x)h=32×20=30\frac{1}{2}(AB + CD)h = \frac{1}{2}(x + 2x)h = \frac{3}{2} \times 20 = 30. Option D (20) is only the area of the parallelogram ABPD, not of the whole trapezium. Hence, option A (30).

Q56MCQFunctions & Graphs

For all real values of xx, the range of the function f(x)=x2+2x+42x2+4x+9f(x) = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} is:
  1. [49,89]\left[\frac{4}{9}, \frac{8}{9}\right]
  2. [37,89)\left[\frac{3}{7}, \frac{8}{9}\right)
  3. (37,12)\left(\frac{3}{7}, \frac{1}{2}\right)
  4. [37,12)\left[\frac{3}{7}, \frac{1}{2}\right)
Answer and solution

Answer: (D) [37,12)\left[\frac{3}{7}, \frac{1}{2}\right)

Write q=2x2+4x+9q = 2x^2 + 4x + 9. The numerator is x2+2x+4=q−12x^2 + 2x + 4 = \frac{q - 1}{2}, so f(x)=q−12q=12−12qf(x) = \frac{q-1}{2q} = \frac{1}{2} - \frac{1}{2q}. Now q=2(x+1)2+7q = 2(x+1)^2 + 7, so q≥7q \ge 7, with q=7q = 7 exactly when x=−1x = -1. Lower end: ff is smallest when qq is smallest. At x=−1x = -1, f=12−114=37f = \frac{1}{2} - \frac{1}{14} = \frac{3}{7}. This value is reached, so 37\frac{3}{7} is included. Upper end: as xx grows large, q→∞q \to \infty and 12q→0\frac{1}{2q} \to 0, so ff approaches 12\frac{1}{2}. But 12q\frac{1}{2q} is always positive, so ff never equals 12\frac{1}{2}, and that end is open. So the range is [37,12)\left[\frac{3}{7}, \frac{1}{2}\right). Option C has the right end points but leaves out 37\frac{3}{7}, which is reached at x=−1x = -1. Option B's upper end 89\frac{8}{9} is too large, since f<12f < \frac{1}{2} always. Hence, option D ([37,12)\left[\frac{3}{7}, \frac{1}{2}\right)).

Q57MCQSequences & Series

For a sequence of real numbers x1,x2,…,xnx_1, x_2, \ldots, x_n, if x1−x2+x3−⋯+(−1)n+1xn=n2+2nx_1 - x_2 + x_3 - \cdots + (-1)^{n+1}x_n = n^2 + 2n for all natural numbers nn, then the sum x49+x50x_{49} + x_{50} equals
  1. 200
  2. 2
  3. -200
  4. -2
Answer and solution

Answer: (D) -2

Let Sn=x1−x2+x3−⋯+(−1)n+1xn=n2+2nS_n = x_1 - x_2 + x_3 - \cdots + (-1)^{n+1}x_n = n^2 + 2n. Subtracting consecutive sums leaves only the last term: (−1)n+1xn=Sn−Sn−1=(n2+2n)−((n−1)2+2(n−1))=2n+1(-1)^{n+1}x_n = S_n - S_{n-1} = (n^2 + 2n) - \big((n-1)^2 + 2(n-1)\big) = 2n + 1. For n=49n = 49 the sign is (−1)50=+1(-1)^{50} = +1, so x49=99x_{49} = 99. For n=50n = 50 the sign is (−1)51=−1(-1)^{51} = -1, so −x50=101-x_{50} = 101 and x50=−101x_{50} = -101. So x49+x50=99−101=−2x_{49} + x_{50} = 99 - 101 = -2. Option A (200) comes from ignoring the alternating sign and adding 9999 and 101101. But x50x_{50} enters the sum with a minus sign, so x50x_{50} itself is −101-101. Hence, option D (-2).

Q58MCQInequalities & Modulus

For a real number x the condition ∣3x−20∣+∣3x−40∣=20 necessarily holds if
  1. 10<x<15
  2. 9<x<14
  3. 7<x<12
  4. 6<x<11
Answer and solution

Answer: (C) 7<x<12

The expressions inside the moduli change sign at x=203x = \frac{20}{3} and x=403x = \frac{40}{3}. If 203≤x≤403\frac{20}{3} \le x \le \frac{40}{3}: ∣3x−20∣+∣3x−40∣=(3x−20)+(40−3x)=20|3x-20| + |3x-40| = (3x - 20) + (40 - 3x) = 20. The equation holds for every such xx. If x>403x > \frac{40}{3}: the sum is 6x−606x - 60, which equals 20 only at x=403x = \frac{40}{3}, so it fails for every larger xx. If x<203x < \frac{20}{3}: the sum is 60−6x60 - 6x, which equals 20 only at x=203x = \frac{20}{3}, so it fails for every smaller xx. So the condition holds exactly when 203≤x≤403\frac{20}{3} \le x \le \frac{40}{3}, that is, about 6.67≤x≤13.336.67 \le x \le 13.33. 'Necessarily holds' means every xx in the option's interval must lie in this range. 7<x<127 < x < 12 lies inside it. Option D fails because it includes values such as x=6.5x = 6.5, below 203\frac{20}{3}; options A and B include values above 13.33, such as x=13.5x = 13.5. Hence, option C (7<x<12).

Q59MCQTime & Work

Anil can paint a house in 60 days while Bimal can paint it in 84 days. Anil starts painting and after 10 days, Bimal and Charu join him. Together, they complete the painting in 14 more days. If they are paid a total of ₹ 21000 for the job, then the share of Charu, in INR, proportionate to the work done by him, is
  1. 9000
  2. 9200
  3. 9100
  4. 9150
Answer and solution

Answer: (C) 9100

Anil works for all 10+14=2410 + 14 = 24 days; Bimal and Charu work for 14 days each. Anil does 160\frac{1}{60} of the job per day, so in 24 days he does 2460=25\frac{24}{60} = \frac{2}{5} of it. Bimal does 184\frac{1}{84} per day, so in 14 days he does 1484=16\frac{14}{84} = \frac{1}{6}. Charu does the rest: 1−25−16=30−12−530=13301 - \frac{2}{5} - \frac{1}{6} = \frac{30 - 12 - 5}{30} = \frac{13}{30}. Payment is in proportion to work done, so Charu gets 1330×21000=9100\frac{13}{30} \times 21000 = 9100. Option A (9000) would mean Charu did 37\frac{3}{7} of the work, but his share is exactly 1330\frac{13}{30}. Hence, option C (9100).

Q60TITAPercentages

A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic, then the number of non-metallic balls in the box is

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Answer and solution

Answer: 250

Let the number of white balls be x and black balls be y So we get x+y =450       (1) Now metallic black balls = 0.5y Metallic white balls = 0.4x From condition 0.4x=0.5y we get 4x-5y=0     (2) Solving (1) and (2) we get x=250 and y =200 Now number of Non Metallic balls = 0.6x+0.5y = 150+100 = 250

Q61TITAAverages, Mixtures & Alligations

In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is

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Answer and solution

Answer: 10

Let nn be the total number of matches after the next 10 are played, so he has played n−10n - 10 so far. Let xx be the goals he has scored so far. If he scores 1 goal in the next 10 matches: x+1n=0.15\frac{x + 1}{n} = 0.15, so x+1=0.15nx + 1 = 0.15n. (1) If he scores 2 goals: x+2n=0.2\frac{x + 2}{n} = 0.2, so x+2=0.2nx + 2 = 0.2n. (2) Subtracting (1) from (2): 1=0.05n1 = 0.05n, so n=20n = 20. Matches already played =n−10=20−10=10= n - 10 = 20 - 10 = 10. Check: x=0.15×20−1=2x = 0.15 \times 20 - 1 = 2 goals so far, and 2+220=0.2\frac{2 + 2}{20} = 0.2. The answer is 10.

Q62MCQProfit, Loss & Discount

A person buys tea of three different qualities at ₹ 800, ₹ 500, and ₹ 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at ₹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is
  1. 653
  2. 688
  3. 692
  4. 675
Answer and solution

Answer: (B) 688

Take the quantities in the ratio 2 : 3 : 5 as 12 kg, 18 kg and 30 kg, a total of 60 kg, so that one-sixth is a whole number. Cost price =800×12+500×18+300×30=9600+9000+9000=27600= 800 \times 12 + 500 \times 18 + 300 \times 30 = 9600 + 9000 + 9000 = 27600. For a 50% overall profit, total revenue must be 1.5×27600=414001.5 \times 27600 = 41400. One-sixth of the mixture, 10 kg, is sold at ₹700 per kg, giving 10×700=700010 \times 700 = 7000. The remaining 50 kg must bring in 41400−7000=3440041400 - 7000 = 34400, so its price is 3440050=688\frac{34400}{50} = 688 per kg. Option C (692) is close, but it gives 7000+50×692=416007000 + 50 \times 692 = 41600, more than the 41400 needed, so the profit would be above 50%. Hence, option B (688).

Q63MCQQuadratic & Polynomial Equations

Consider the pair of equations: x2−xy−x=22x^2 - xy - x = 22 and y2−xy+y=34y^2 - xy + y = 34. If x>yx > y, then x−yx - y equals
  1. 6
  2. 4
  3. 7
  4. 8
Answer and solution

Answer: (D) 8

Add the two equations: (x2−xy−x)+(y2−xy+y)=22+34(x^2 - xy - x) + (y^2 - xy + y) = 22 + 34, so x2−2xy+y2−x+y=56x^2 - 2xy + y^2 - x + y = 56. This is (x−y)2−(x−y)=56(x - y)^2 - (x - y) = 56. Put t=x−yt = x - y: t2−t−56=0t^2 - t - 56 = 0, so (t−8)(t+7)=0(t - 8)(t + 7) = 0 and t=8t = 8 or t=−7t = -7. Since x>yx > y, t=x−yt = x - y must be positive, so t=−7t = -7 is rejected and x−y=8x - y = 8. Check: with x−y=8x - y = 8, the first equation x(x−y−1)=22x(x - y - 1) = 22 gives x=227x = \frac{22}{7}, and the second y(y−x+1)=34y(y - x + 1) = 34 gives y=−347y = -\frac{34}{7}; indeed x−y=8x - y = 8. Option C (7) is the size of the rejected root t=−7t = -7, which would need x<yx < y. Hence, option D (8).

Q64TITATriangles & Lines

Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is

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

Answer and solution

Answer: 30

Let AD=2xAD = 2x and BD=xBD = x, so AB=3xAB = 3x. Let AE=2yAE = 2y and CE=3yCE = 3y, so AC=5yAC = 5y. Solution figure for question 64, CAT 2021 Slot 2 Triangles ADE and ABC share angle A, so each area is 12×\frac{1}{2} \times (the two sides at A) ×sin⁡A\times \sin A. Area of ADE =12⋅2x⋅2y⋅sin⁡A=2xysin⁡A=8= \frac{1}{2} \cdot 2x \cdot 2y \cdot \sin A = 2xy\sin A = 8, so xysin⁡A=4xy\sin A = 4. Area of ABC =12⋅3x⋅5y⋅sin⁡A=152xysin⁡A=152×4=30= \frac{1}{2} \cdot 3x \cdot 5y \cdot \sin A = \frac{15}{2}xy\sin A = \frac{15}{2} \times 4 = 30. Equivalently, [ADE][ABC]=ADAB⋅AEAC=23⋅25=415\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{2}{3} \cdot \frac{2}{5} = \frac{4}{15}, so [ABC]=8×154=30[ABC] = 8 \times \frac{15}{4} = 30 sq cm. The answer is 30.

Q65MCQRatios, Proportions & Partnership

Anil, Bobby, and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil’s share of investment is 70%. His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%. Chintu’s share of profit increases by ₹ 80 if the overall profit goes up from 15% to 17%. The amount, in INR, invested by Bobby is
  1. 2000
  2. 2400
  3. 2200
  4. 1800
Answer and solution

Answer: (A) 2000

Profit is shared in proportion to investment, so when the overall profit rate changes, each person's share changes in the same proportion. Let the total investment be TT. Anil holds 70%. When the profit rate falls from 18% to 15%, the total profit falls by 3% of TT, and Anil's share falls by 0.7×0.03T=4200.7 \times 0.03T = 420. So 0.03T=4200.7=6000.03T = \frac{420}{0.7} = 600 and T=20000T = 20000. When the rate rises from 15% to 17%, the total profit rises by 2% of TT, which is 400. Chintu's share rises by 80, that is 80400=20%\frac{80}{400} = 20\% of the increase, so Chintu holds 20% of the investment. Bobby holds the remaining 100%−70%−20%=10%100\% - 70\% - 20\% = 10\%, which is 0.1×20000=20000.1 \times 20000 = 2000. Option B (2400) would need Bobby to hold 12%, but Anil and Chintu together hold 90%, leaving only 10%. Hence, option A (2000).

Q66MCQTime & Work

Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is
  1. 144
  2. 140
  3. 264
  4. 120
Answer and solution

Answer: (A) 144

Let pipe A fill aa and pipe B drain bb of water per hour, and let the tank hold VV. First case: A runs from 2 pm to 10 pm (8 hours) and B from 3 pm to 10 pm (7 hours), so 8a−7b=V8a - 7b = V. Second case: A runs from 2 pm to 6 pm (4 hours) and B from 4 pm to 6 pm (2 hours), so 4a−2b=V4a - 2b = V. Equating: 8a−7b=4a−2b8a - 7b = 4a - 2b, so 4a=5b4a = 5b and b=0.8ab = 0.8a. Then V=4a−1.6a=2.4aV = 4a - 1.6a = 2.4a. A alone takes Va=2.4\dfrac{V}{a} = 2.4 hours =144= 144 minutes. Option D (120 minutes, i.e. V=2aV = 2a) fails: the second case would then give b=ab = a, and in the first case the tank would get only 8a−7a=a8a - 7a = a, half of it. Hence, option A (144).