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CAT 2023 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 2023 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. Free — sign in and you come straight back to this paper.

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Quantitative Ability

CAT 2023 Slot 3 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q45MCQQuadratic & Polynomial Equations

If xx is a positive real number such that x8+(1x)8=47x^8 + \left(\dfrac{1}{x}\right)^8 = 47, then the value of x9+(1x)9x^9 + \left(\dfrac{1}{x}\right)^9 is
  1. 40540\sqrt{5}
  2. 30530\sqrt{5}
  3. 36536\sqrt{5}
  4. 34534\sqrt{5}
Answer and solution

Answer: (D) 34534\sqrt{5}

Let tn=xn+1xnt_n = x^n + \dfrac{1}{x^n}, so t2n=tn2−2t_{2n} = t_n^2 - 2. From t8=47t_8 = 47: t42=49t_4^2 = 49, so t4=7t_4 = 7; t22=9t_2^2 = 9, so t2=3t_2 = 3; t12=5t_1^2 = 5, so t1=5t_1 = \sqrt5 (all positive, since x>0x > 0). Use tn+1=t1tn−tn−1t_{n+1} = t_1 t_n - t_{n-1}: t3=35−5=25t_3 = 3\sqrt5 - \sqrt5 = 2\sqrt5, t5=75−25=55t_5 = 7\sqrt5 - 2\sqrt5 = 5\sqrt5, t6=25−7=18t_6 = 25 - 7 = 18, t7=185−55=135t_7 = 18\sqrt5 - 5\sqrt5 = 13\sqrt5, and t8=65−18=47t_8 = 65 - 18 = 47 checks. t9=t1t8−t7=475−135=345t_9 = t_1 t_8 - t_7 = 47\sqrt5 - 13\sqrt5 = 34\sqrt5. As a check, t9=t4t5−t1=355−5=345t_9 = t_4 t_5 - t_1 = 35\sqrt5 - \sqrt5 = 34\sqrt5. Adding t1t_1 instead of subtracting it gives the trap value 36536\sqrt5, option C. Hence, option D (34534\sqrt{5}).

Q46MCQIndices & Surds

Let nn and mm be two positive integers such that there are exactly 41 integers greater than 8m8^m and less than 8n8^n, which can be expressed as powers of 2. Then, the smallest possible value of n+mn + m is
  1. 42
  2. 44
  3. 14
  4. 16
Answer and solution

Answer: (D) 16

Write both bounds as powers of 2: 8m=23m8^m = 2^{3m} and 8n=23n8^n = 2^{3n}. The powers of 2 strictly between them are 23m+1,23m+2,…,23n−12^{3m+1}, 2^{3m+2}, \dots, 2^{3n-1}, which is (3n−1)−3m=3(n−m)−1(3n - 1) - 3m = 3(n - m) - 1 numbers. So 3(n−m)−1=413(n - m) - 1 = 41, which gives n−m=14n - m = 14. With m≥1m \ge 1, the sum n+m=2m+14n + m = 2m + 14 is smallest at m=1m = 1, n=15n = 15, giving n+m=16n + m = 16. C (14) is the difference n−mn - m, not the sum; it would need m=0m = 0, which is not a positive integer. Hence, option D (16).

Q47MCQLinear Equations

For some real numbers aa and bb, the system of equations x+y=4x + y = 4 and (a+5)x+(b2−15)y=8b(a+5)x+(b^2-15)y=8b has infinitely many solutions for xx and yy. Then, the maximum possible value of abab is
  1. 33
  2. 25
  3. 15
  4. 55
Answer and solution

Answer: (A) 33

Infinitely many solutions means the second equation is a multiple of the first, x+y=4x + y = 4. So a+51=b2−151=8b4=2b\dfrac{a+5}{1} = \dfrac{b^2-15}{1} = \dfrac{8b}{4} = 2b. From b2−15=2bb^2 - 15 = 2b: b2−2b−15=0b^2 - 2b - 15 = 0, so (b−5)(b+3)=0(b - 5)(b + 3) = 0 and b=5b = 5 or b=−3b = -3. Also a+5=2ba + 5 = 2b, so a=2b−5a = 2b - 5. b=5b = 5: a=5a = 5, ab=25ab = 25. b=−3b = -3: a=−11a = -11, ab=33ab = 33. The maximum is 33. B (25) is the value from b=5b = 5, which is smaller. Hence, option A (33).

Q48MCQLogarithms

For a real number xx, if 12,log⁡3(2x−9)log⁡34\dfrac{1}{2}, \dfrac{\log_3(2^x - 9)}{\log_3 4}, and log⁡5(2x+172)log⁡54\dfrac{\log_5\left(2^x + \dfrac{17}{2}\right)}{\log_5 4} are in an arithmetic progression, then the common difference is
  1. log⁡4(32)\log_4\left(\dfrac{3}{2}\right)
  2. log⁡47\log_4 7
  3. log⁡4(232)\log_4\left(\dfrac{23}{2}\right)
  4. log⁡4(72)\log_4\left(\dfrac{7}{2}\right)
Answer and solution

Answer: (D) log⁡4(72)\log_4\left(\dfrac{7}{2}\right)

By change of base, log⁡3(2x−9)log⁡34=log⁡4(2x−9)\dfrac{\log_3(2^x - 9)}{\log_3 4} = \log_4(2^x - 9) and log⁡5(2x+172)log⁡54=log⁡4(2x+172)\dfrac{\log_5(2^x + \frac{17}{2})}{\log_5 4} = \log_4(2^x + \frac{17}{2}). In an AP, twice the middle term equals the sum of the other two. Since 12=log⁡42\frac12 = \log_4 2: 2log⁡4(2x−9)=log⁡42+log⁡4(2x+172)=log⁡4(2x+1+17)2\log_4(2^x - 9) = \log_4 2 + \log_4(2^x + \frac{17}{2}) = \log_4(2^{x+1} + 17). Let t=2xt = 2^x: (t−9)2=2t+17(t - 9)^2 = 2t + 17, so t2−20t+64=0t^2 - 20t + 64 = 0 and t=16t = 16 or 44. The log needs t>9t > 9, so t=16t = 16. The terms are 12\frac12, log⁡47\log_4 7 and log⁡4492\log_4 \frac{49}{2}. The common difference is log⁡47−log⁡42=log⁡472\log_4 7 - \log_4 2 = \log_4 \frac{7}{2}. B (log⁡47\log_4 7) is the middle term itself, not the difference. Hence, option D (log⁡4(72)\log_4\left(\dfrac{7}{2}\right)).

Q49TITAIndices & Surds

Let nn be any natural number such that 5n−1<3n+15^{n-1} < 3^{n + 1}. Then, the least integer value of mm that satisfies 3n+1<2n+m3^{n+1} < 2^{n+m} for each such nn, is

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

Answer and solution

Answer: 5

5n−1<3n+1  ⟺  (53)n−1<95^{n-1} < 3^{n+1} \iff \left(\tfrac{5}{3}\right)^{n-1} < 9. Since (5/3)4≈7.7<9(5/3)^4 \approx 7.7 < 9 and (5/3)5≈12.9>9(5/3)^5 \approx 12.9 > 9, this holds exactly for n=1,2,3,4,5n = 1, 2, 3, 4, 5. We need 2n+m>3n+12^{n+m} > 3^{n+1} for each of these. The tightest case is n=5n = 5: 36=7293^6 = 729, so 25+m>7292^{5+m} > 729 needs 5+m≥105 + m \ge 10, i.e. m≥5m \ge 5 (29=5122^9 = 512 is too small). Check m=5m = 5 for the rest: n=1n=1: 26=64>92^6 = 64 > 9; n=2n=2: 128>27128 > 27; n=3n=3: 256>81256 > 81; n=4n=4: 512>243512 > 243. The least mm is 55.

Q50TITAProperties of Numbers

The sum of the first two natural numbers, each having 15 factors (including 1 and the number itself), is

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Answer and solution

Answer: 468

A number with 15 factors has the form p14p^{14} or p4q2p^4 q^2 (since 15=15×1=5×315 = 15 \times 1 = 5 \times 3). The smallest such numbers: 24⋅32=1442^4 \cdot 3^2 = 144, 34⋅22=3243^4 \cdot 2^2 = 324, 24⋅52=4002^4 \cdot 5^2 = 400, ..., and 214=163842^{14} = 16384 is far larger. The first two are 144144 and 324324; their sum is 468468.

Q51TITAQuadratic & Polynomial Equations

A quadratic equation x2+bx+c=0x^2 + bx + c = 0 has two real roots. If the difference between the reciprocals of the roots is 13\dfrac{1}{3}, and the sum of the reciprocals of the squares of the roots is 59\dfrac{5}{9}, then the largest possible value of (b+c)(b + c) is

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Answer and solution

Answer: 9

Let u,vu, v be the reciprocals of the roots. Then (u−v)2=19(u - v)^2 = \dfrac19 and u2+v2=59u^2 + v^2 = \dfrac59. So 2uv=59−19=49⇒uv=292uv = \dfrac59 - \dfrac19 = \dfrac49 \Rightarrow uv = \dfrac29, and (u+v)2=59+49=1⇒u+v=±1(u + v)^2 = \dfrac59 + \dfrac49 = 1 \Rightarrow u + v = \pm 1. For x2+bx+c=0x^2 + bx + c = 0: uv=1c⇒c=92uv = \dfrac{1}{c} \Rightarrow c = \dfrac92, and u+v=−bc=±1⇒b=∓92u + v = \dfrac{-b}{c} = \pm1 \Rightarrow b = \mp\dfrac92. The largest b+cb + c is 92+92=9\dfrac92 + \dfrac92 = 9 (the roots are real, since 1−4⋅29>01 - 4 \cdot \tfrac29 > 0).

Q52MCQProfit, Loss & Discount

A merchant purchases a cloth at a rate of Rs.100 per meter and receives 5 cm length of cloth free for every 100 cm length of cloth purchased by him. He sells the same cloth at a rate of Rs.110 per meter but cheats his customers by giving 95 cm length of cloth for every 100 cm length of cloth purchased by the customers. If the merchant provides a 5% discount, the resulting profit earned by him is
  1. 4.2%
  2. 9.7%
  3. 15.5%
  4. 16%
Answer and solution

Answer: (C) 15.5%

Buying: for every Rs.100 he pays, he receives 105 cm, so 1 m actually costs him 1001.05\dfrac{100}{1.05} rupees. Selling: the price is Rs.110 per metre less 5%, so the customer pays Rs.104.50 per metre charged. But he hands over only 95 cm, so he earns 104.50.95=110\dfrac{104.5}{0.95} = 110 rupees per metre actually given. Profit =110100/1.05−1=110×1.05100−1=1.155−1=15.5%= \dfrac{110}{100/1.05} - 1 = \dfrac{110 \times 1.05}{100} - 1 = 1.155 - 1 = 15.5\%. B (9.7%) ignores the short measure: 104.5×1.05100−1≈9.7%\dfrac{104.5 \times 1.05}{100} - 1 \approx 9.7\%. Hence, option C (15.5%).

Q53MCQTime & Work

Rahul, Rakshita and Gurmeet, working together, would have taken more than 7 days to finish a job. On the other hand, Rahul and Gurmeet, working together would have taken less than 15 days to finish the job. However, they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. If Rakshita had worked alone on the job then the number of days she would have taken to finish the job, cannot be
  1. 20
  2. 17
  3. 16
  4. 21
Answer and solution

Answer: (D) 21

Let Rakshita's daily rate be KK and the combined daily rate of Rahul and Gurmeet be XX (as fractions of the job). All three worked 6 days, then Rakshita worked 3 more days: 6(X+K)+3K=16(X + K) + 3K = 1, so X=1−9K6X = \dfrac{1 - 9K}{6}. Rahul and Gurmeet together take less than 15 days, so X>115X > \dfrac{1}{15}: 1−9K>251 - 9K > \dfrac{2}{5}, which gives K<115K < \dfrac{1}{15}. All three together take more than 7 days, so X+K<17X + K < \dfrac17: 1−3K6<17\dfrac{1 - 3K}{6} < \dfrac17, so 1−3K<671 - 3K < \dfrac67, which gives K>121K > \dfrac{1}{21}. So Rakshita alone would take 1K\dfrac{1}{K} days, strictly between 15 and 21. The values 20, 17 and 16 all lie in this range, so each is possible. But 21 would need K=121K = \dfrac{1}{21}, and then all three together take exactly 7 days, not more than 7. Hence, option D (21).

Q54MCQPercentages

The population of a town in 2020 was 100000. The population decreased by y%y\% from the year 2020 to 2021, and increased by x%x\% from the year 2021 to 2022, where xx and yy are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between xx and yy is 10, then the lowest possible population of the town in 2021 was
  1. 72000
  2. 74000
  3. 73000
  4. 75000
Answer and solution

Answer: (C) 73000

The 2022 population is 100000(1−y100)(1+x100)100000\left(1 - \dfrac{y}{100}\right)\left(1 + \dfrac{x}{100}\right). For it to exceed 100000, the increase must be bigger than the decrease, so x>yx > y and x=y+10x = y + 10. The condition becomes (100−y)(110+y)>10000(100 - y)(110 + y) > 10000, i.e. 11000−10y−y2>1000011000 - 10y - y^2 > 10000, i.e. y2+10y−1000<0y^2 + 10y - 1000 < 0. So y<−5+1025≈27.02y < -5 + \sqrt{1025} \approx 27.02. As yy is a natural number, y≤27y \le 27. The 2021 population is 100000(1−y100)100000\left(1 - \dfrac{y}{100}\right), which is lowest when yy is largest. At y=27y = 27 it is 7300073000. Check: x=37x = 37, and 73000×1.37=100010>10000073000 \times 1.37 = 100010 > 100000. Option A (72000) would need y=28y = 28 and x=38x = 38, but then the 2022 population is 72000×1.38=9936072000 \times 1.38 = 99360, less than 100000. Options B and D are larger than 73000, so they are not the lowest. Hence, option C (73000).

Q55MCQAverages, Mixtures & Alligations

Anil mixes cocoa with sugar in the ratio 3 : 2 to prepare mixture A, and coffee with sugar in the ratio 7 : 3 to prepare mixture B. He combines mixtures A and B in the ratio 2 : 3 to make a new mixture C. If he mixes C with an equal amount of milk to make a drink, then the percentage of sugar in this drink will be
  1. 17
  2. 16
  3. 21
  4. 24
Answer and solution

Answer: (A) 17

Mixture A is cocoa : sugar =3:2= 3 : 2, so it is 25=40%\dfrac25 = 40\% sugar. Mixture B is coffee : sugar =7:3= 7 : 3, so it is 310=30%\dfrac{3}{10} = 30\% sugar. Mixture C takes A and B in the ratio 2:32 : 3. In 5 parts of C, the sugar is 2×0.4+3×0.3=0.8+0.9=1.72 \times 0.4 + 3 \times 0.3 = 0.8 + 0.9 = 1.7 parts, so C is 1.75=34%\dfrac{1.7}{5} = 34\% sugar. Adding an equal amount of milk doubles the volume but adds no sugar, so the drink is 342=17%\dfrac{34}{2} = 17\% sugar. The other options fail: 16%, 21% and 24% would need C to be 32%, 42% or 48% sugar, but C is 34% sugar. Hence, option A (17).

Q56MCQAverages, Mixtures & Alligations

There are three persons A, B and C in a room. If a person D joins the room, the average weight of the persons in the room reduces by xx kg. Instead of D, if person E joins the room, the average weight of the persons in the room increases by 2x2x kg. If the weight of E is 12 kg more than that of D, then the value of xx is
  1. 2
  2. 0.5
  3. 1
  4. 1.5
Answer and solution

Answer: (C) 1

Let SS be the total weight of A, B and C, so their average is S3\dfrac{S}{3}. When D joins, the average of the four falls by xx: S+D4=S3−x\dfrac{S + D}{4} = \dfrac{S}{3} - x. When E joins instead, the average rises by 2x2x: S+E4=S3+2x\dfrac{S + E}{4} = \dfrac{S}{3} + 2x. Subtracting the first equation from the second: E−D4=2x−(−x)=3x\dfrac{E - D}{4} = 2x - (-x) = 3x. With E−D=12E - D = 12: 3=3x3 = 3x, so x=1x = 1. Option D (1.5) comes from setting E−D4\dfrac{E - D}{4} equal to 2x2x only. But one new average is xx below the old one and the other is 2x2x above it, so the gap between them is 3x3x. Hence, option C (1).

Q57MCQTime, Speed & Distance

A boat takes 2 hours to travel downstream a river from port A to port B, and 3 hours to return to port A. Another boat takes a total of 6 hours to travel from port B to port A and return to port B. If the speeds of the boats and the river are constant, then the time, in hours, taken by the slower boat to travel from port A to port B is
  1. 12(5−2)12(\sqrt{5} - 2)
  2. 3(3+5)3(3 + \sqrt{5})
  3. 3(5−1)3(\sqrt{5} - 1)
  4. 3(3−5)3(3 - \sqrt{5})
Answer and solution

Answer: (D) 3(3−5)3(3 - \sqrt{5})

Let the river flow from A to B at speed rr, and let AB=dAB = d. First boat (speed bb): downstream A to B takes 2 hours and upstream B to A takes 3 hours, so d=2(b+r)=3(b−r)d = 2(b + r) = 3(b - r). This gives b=5rb = 5r and d=12rd = 12r. Second boat (speed cc): B to A is upstream and the return is downstream, 6 hours in all, so 12rc−r+12rc+r=6⇒24rc=6(c2−r2)⇒c2−4rc−r2=0\dfrac{12r}{c - r} + \dfrac{12r}{c + r} = 6 \Rightarrow 24rc = 6(c^2 - r^2) \Rightarrow c^2 - 4rc - r^2 = 0. So c=(2+5)r≈4.24rc = (2 + \sqrt5)r \approx 4.24r, which is less than 5r5r. The second boat is the slower one. Its time from A to B, downstream, is 12rc+r=123+5=12(3−5)4=3(3−5)≈2.29\dfrac{12r}{c + r} = \dfrac{12}{3 + \sqrt5} = \dfrac{12(3 - \sqrt5)}{4} = 3(3 - \sqrt5) \approx 2.29 hours. Option C, 3(5−1)≈3.713(\sqrt5 - 1) \approx 3.71 hours, is this boat's upstream time from B to A, 12rc−r=121+5\dfrac{12r}{c - r} = \dfrac{12}{1 + \sqrt5}, not the downstream trip asked for. Hence, option D (3(3−5)3(3 - \sqrt{5})).

Q58TITARatios, Proportions & Partnership

The number of coins collected per week by two coin-collectors A and B are in the ratio 3 : 4. If the total number of coins collected by A in 5 weeks is a multiple of 7, and the total number of coins collected by B in 3 weeks is a multiple of 24, then the minimum possible number of coins collected by A in one week is

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

Answer and solution

Answer: 42

Let A collect 3k3k and B collect 4k4k coins per week. A in 5 weeks: 15k15k is a multiple of 7, so kk is a multiple of 7. B in 3 weeks: 12k12k is a multiple of 24, so kk is even. The least kk is 1414, so A collects at least 3×14=423 \times 14 = 42 coins a week.

Q59TITAPercentages

A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is

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Answer and solution

Answer: 340

Let the stock be NN, with mangoes 0.4N0.4N, bananas BB and apples AA (A+B=0.6NA + B = 0.6N). Sold: 0.2N+96+0.4A=0.5N⇒A=0.75N−2400.2N + 96 + 0.4A = 0.5N \Rightarrow A = 0.75N - 240, so B=240−0.15NB = 240 - 0.15N. Whole numbers of fruit: half the mangoes, 0.2N0.2N, is a whole number, so 5∣N5 \mid N; 0.4A0.4A is a whole number, so 5∣A5 \mid A; and A=0.75N−240A = 0.75N - 240 is a whole number, so 4∣N4 \mid N. Hence NN is a multiple of 20. We also need A≥5A \ge 5 and B≥96B \ge 96. A≥5⇒0.75N≥245⇒N≥326.7A \ge 5 \Rightarrow 0.75N \ge 245 \Rightarrow N \ge 326.7, so the smallest multiple of 20 is N=340N = 340: then A=15A = 15 and B=189≥96B = 189 \ge 96. Check: sold 68+96+6=170=50%68 + 96 + 6 = 170 = 50\% of 340.

Q60TITATime & Work

Gautam and Suhani, working together, can finish a job in 20 days. If Gautam does only 60% of his usual work on a day, Suhani must do 150% of her usual work on that day to exactly make up for it. Then, the number of days required by the faster worker to complete the job working alone is

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Answer and solution

Answer: 36

Let their daily rates be gg and ss: g+s=120g + s = \dfrac{1}{20}. Making up the shortfall: 0.6g+1.5s=g+s⇒0.5s=0.4g⇒s=0.8g0.6g + 1.5s = g + s \Rightarrow 0.5s = 0.4g \Rightarrow s = 0.8g. So 1.8g=120⇒g=1361.8g = \dfrac{1}{20} \Rightarrow g = \dfrac{1}{36}. Gautam is the faster worker and takes 3636 days alone.

Q61MCQTriangles & Lines

Let △ABC\triangle ABC be an isosceles triangle such that ABAB and ACAC are of equal length. ADAD is the altitude from AA on BCBC and BEBE is the altitude from BB on ACAC. If ADAD and BEBE intersect at OO such that ∠AOB=105∘\angle AOB = 105^\circ, then ADBE\dfrac{AD}{BE} equals
  1. sin⁡15∘\sin 15^\circ
  2. cos⁡15∘\cos 15^\circ
  3. 2cos⁡15∘2 \cos 15^\circ
  4. 2sin⁡15∘2 \sin 15^\circ
Answer and solution

Answer: (C) 2cos⁡15∘2 \cos 15^\circ

OO is where two altitudes meet, so it is the orthocentre of the triangle. Solution figure for question 61, CAT 2023 Slot 3 In quadrilateral ODCEODCE, the angles at DD and EE are right angles, so ∠DOE=180∘−∠C\angle DOE = 180^\circ - \angle C. ∠AOB\angle AOB is vertically opposite ∠DOE\angle DOE, so 105∘=180∘−∠C105^\circ = 180^\circ - \angle C, giving ∠C=75∘\angle C = 75^\circ. As AB=ACAB = AC, ∠B=∠C=75∘\angle B = \angle C = 75^\circ, so ∠A=30∘\angle A = 30^\circ. In an isosceles triangle, the altitude ADAD also bisects ∠A\angle A. So in right triangle ABDABD, ∠BAD=15∘\angle BAD = 15^\circ and AD=ABcos⁡15∘AD = AB\cos 15^\circ. In right triangle ABEABE, BEBE is opposite the 30∘30^\circ angle at AA, so BE=ABsin⁡30∘=AB2BE = AB \sin 30^\circ = \dfrac{AB}{2}. So ADBE=ABcos⁡15∘AB/2=2cos⁡15∘\dfrac{AD}{BE} = \dfrac{AB\cos 15^\circ}{AB/2} = 2\cos 15^\circ. Option B, cos⁡15∘\cos 15^\circ, drops the factor 2: it would need BE=ABBE = AB, but BEBE is only half of ABAB. Hence, option C (2cos⁡15∘2 \cos 15^\circ).

Q62MCQPolygons & Circles

A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is
  1. 2:12 : 1
  2. 1:11 : 1
  3. 5:1\sqrt{5} : 1
  4. 2:1\sqrt{2} : 1
Answer and solution

Answer: (A) 2:12 : 1

Place the rectangle with its base on the diameter, centred at the centre OO of the semicircle. Let the half-base be xx and the height yy. A top corner lies on the arc, so the line from OO to it is a radius. Solution figure for question 62, CAT 2023 Slot 3 So x2+y2=22=4x^2 + y^2 = 2^2 = 4, and the area is 2x⋅y=2xy2x \cdot y = 2xy. Since (x−y)2≥0(x - y)^2 \ge 0, 2xy≤x2+y2=42xy \le x^2 + y^2 = 4, with equality only when x=yx = y. So the largest area is 4, when x=y=2x = y = \sqrt2. The sides of the rectangle are then the base 2x=222x = 2\sqrt2 and the height y=2y = \sqrt2, a ratio of 2:12 : 1. Option B (1:11 : 1) is the ratio of the half-base to the height. The base is the full width 2x2x, twice the half-base, so the rectangle is not a square. Hence, option A (2:12 : 1).

Q63TITAPolygons & Circles

In a regular polygon, any interior angle exceeds the exterior angle by 120 degrees. Then, the number of diagonals of this polygon is

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Answer and solution

Answer: 54

Interior −- exterior =120∘= 120^\circ and interior ++ exterior =180∘= 180^\circ, so the exterior angle is 30∘30^\circ and the polygon has 36030=12\dfrac{360}{30} = 12 sides. Number of diagonals =n(n−3)2=12×92=54= \dfrac{n(n - 3)}{2} = \dfrac{12 \times 9}{2} = 54.

Q64MCQSequences & Series

The value of 1+(1+13)14+(1+13+19)116+(1+13+19+127)164+⋯1 + \left(1 + \dfrac{1}{3}\right)\dfrac{1}{4} + \left(1 + \dfrac{1}{3} + \dfrac{1}{9}\right)\dfrac{1}{16} + \left(1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27}\right)\dfrac{1}{64} + \cdots is
  1. 1513\dfrac{15}{13}
  2. 2712\dfrac{27}{12}
  3. 158\dfrac{15}{8}
  4. 1611\dfrac{16}{11}
Answer and solution

Answer: (D) 1611\dfrac{16}{11}

The term with index nn (n=0,1,2,…n = 0, 1, 2, \dots) is (1+13+⋯+13n)14n\left(1 + \dfrac13 + \dots + \dfrac{1}{3^n}\right)\dfrac{1}{4^n}. The bracket is a geometric sum: 1+13+⋯+13n=32(1−13n+1)1 + \dfrac13 + \dots + \dfrac{1}{3^n} = \dfrac32\left(1 - \dfrac{1}{3^{n+1}}\right). So the term is 32(14n−13⋅112n)\dfrac32\left(\dfrac{1}{4^n} - \dfrac13 \cdot \dfrac{1}{12^n}\right). Summing the two infinite geometric series: ∑14n=11−1/4=43\sum \dfrac{1}{4^n} = \dfrac{1}{1 - 1/4} = \dfrac43 and ∑112n=11−1/12=1211\sum \dfrac{1}{12^n} = \dfrac{1}{1 - 1/12} = \dfrac{12}{11}. Sum =32(43−13⋅1211)=32(43−411)=32⋅3233=1611≈1.45= \dfrac32\left(\dfrac43 - \dfrac13 \cdot \dfrac{12}{11}\right) = \dfrac32\left(\dfrac43 - \dfrac{4}{11}\right) = \dfrac32 \cdot \dfrac{32}{33} = \dfrac{16}{11} \approx 1.45. A quick check on the options: the first two terms are 1+131 + \dfrac13, so the sum exceeds 43\dfrac43, which rules out 1513\dfrac{15}{13}. Every bracket is less than 32\dfrac32, so the sum is less than 1+32(14+116+… )=1.51 + \dfrac32\left(\dfrac14 + \dfrac{1}{16} + \dots\right) = 1.5, which rules out 158\dfrac{15}{8} and 2712\dfrac{27}{12}. Hence, option D (1611\dfrac{16}{11}).

Q65MCQSequences & Series

Let an=46+8na_n = 46 + 8n and bn=98+4nb_n = 98 + 4n be two sequences for natural numbers n≤100n \leq 100. Then, the sum of all terms common to both the sequences is
  1. 14900
  2. 14798
  3. 15000
  4. 14602
Answer and solution

Answer: (A) 14900

an=46+8na_n = 46 + 8n for n=1,…,100n = 1, \dots, 100 gives 54,62,…,84654, 62, \dots, 846: every number from 54 to 846 that leaves remainder 6 on division by 8. bn=98+4nb_n = 98 + 4n gives 102,106,…,498102, 106, \dots, 498: every number from 102 to 498 that leaves remainder 2 on division by 4. A number that leaves remainder 6 on division by 8 also leaves remainder 2 on division by 4. So the common terms are the numbers of the form 8k+68k + 6 from 102 to 498: 102,110,…,494102, 110, \dots, 494 (498 leaves remainder 2 on division by 8, so the last one is 494). There are 494−1028+1=50\dfrac{494 - 102}{8} + 1 = 50 of them, and their sum is 502(102+494)=25×596=14900\dfrac{50}{2}(102 + 494) = 25 \times 596 = 14900. Option B (14798) is this sum without the first common term, 102. But 102 is both b1b_1 and a7=46+56a_7 = 46 + 56, so it must be counted. Hence, option A (14900).

Q66TITAFunctions & Graphs

Suppose f(x,y)f(x, y) is a real-valued function such that f(3x+2y,2x−5y)=19xf(3x + 2y, 2x - 5y) = 19x, for all real numbers xx and yy. The value of xx for which f(x,2x)=27f(x, 2x) = 27, is

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

Answer and solution

Answer: 3

Let u=3x+2yu = 3x + 2y and v=2x−5yv = 2x - 5y. Then 5u+2v=15x+10y+4x−10y=19x5u + 2v = 15x + 10y + 4x - 10y = 19x. So f(u,v)=5u+2vf(u, v) = 5u + 2v for all real u,vu, v. f(x,2x)=5x+4x=9x=27⇒x=3f(x, 2x) = 5x + 4x = 9x = 27 \Rightarrow x = 3.