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CAT 2023 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.

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Quantitative Ability

CAT 2023 Slot 2 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q45MCQIndices & Surds

Let aa, bb, mm and nn be natural numbers such that a>1a > 1 and b>1b > 1. If ambn=144145a^m b^n = 144^{145}, then the largest possible value of n−mn - m is
  1. 580
  2. 290
  3. 289
  4. 579
Answer and solution

Answer: (D) 579

Since 144=24×32144 = 2^4 \times 3^2, we have 144145=2580×3290144^{145} = 2^{580} \times 3^{290}. To make n−mn - m as large as possible, make nn large and mm small. How large can nn be? If bb is even, 2n2^n divides bnb^n, which divides 2580×32902^{580} \times 3^{290}, so n≤580n \le 580. If bb is odd, it is a power of 3, so 3n3^n divides bnb^n, which divides 2580×32902^{580} \times 3^{290}, so n≤290n \le 290. So n≤580n \le 580, with n=580n = 580 only when b=2b = 2. The smallest natural number mm is 1. Both extremes can happen together: take b=2b = 2, n=580n = 580, a=3290a = 3^{290}, m=1m = 1. Then ambn=3290×2580=144145a^m b^n = 3^{290} \times 2^{580} = 144^{145}, with a>1a > 1 and b>1b > 1. So the largest value of n−mn - m is 580−1=579580 - 1 = 579. Option A (580) would need m=0m = 0, but mm is a natural number, so m≥1m \ge 1. Hence, option D (579).

Q46MCQInequalities & Modulus

Any non-zero real numbers xx, yy such that y≠3y \neq 3 and xy<x+3y−3\dfrac{x}{y} < \dfrac{x+3}{y-3}, will satisfy the condition:
  1. xy<yx\dfrac{x}{y} < \dfrac{y}{x}
  2. If y<0y < 0, then −x<y-x < y
  3. If y>10y > 10, then −x>y-x > y
  4. If x<0x < 0, then −x<y-x < y
Answer and solution

Answer: (B) If y<0y < 0, then −x<y-x < y

Bring everything to one side: xy−x+3y−3<0⇒x(y−3)−y(x+3)y(y−3)<0⇒−3(x+y)y(y−3)<0\frac{x}{y} - \frac{x+3}{y-3} < 0 \Rightarrow \frac{x(y-3) - y(x+3)}{y(y-3)} < 0 \Rightarrow \frac{-3(x+y)}{y(y-3)} < 0. Dividing by −3-3 reverses the sign: x+yy(y−3)>0\frac{x+y}{y(y-3)} > 0. If y<0y < 0, then yy and y−3y - 3 are both negative, so y(y−3)>0y(y-3) > 0. The inequality then needs x+y>0x + y > 0, that is, −x<y-x < y. So option B always holds. Option D fails: take x=−5x = -5, y=1y = 1. Then xy=−5<1=x+3y−3\frac{x}{y} = -5 < 1 = \frac{x+3}{y-3}, so the condition holds, but −x=5>1=y-x = 5 > 1 = y. Option C fails: if y>10y > 10, then y(y−3)>0y(y-3) > 0, so again x+y>0x + y > 0, which gives −x<y-x < y, the opposite of C. Option A fails: x=5x = 5, y=4y = 4 satisfies the condition (54<8\frac{5}{4} < 8), but 54<45\frac{5}{4} < \frac{4}{5} is false. Hence, option B (If y<0y < 0, then −x<y-x < y).

Q47MCQProperties of Numbers

For any natural numbers mm, nn, and kk, such that kk divides both m+2nm + 2n and 3m+4n3m + 4n, kk must be a common divisor of
  1. mm and nn
  2. 2m2m and 3n3n
  3. mm and 2n2n
  4. 2m2m and nn
Answer and solution

Answer: (C) mm and 2n2n

If kk divides two numbers, it divides any integer combination of them. kk divides m+2nm + 2n, so it divides 3(m+2n)=3m+6n3(m + 2n) = 3m + 6n. It also divides 3m+4n3m + 4n, so it divides the difference (3m+6n)−(3m+4n)=2n(3m + 6n) - (3m + 4n) = 2n. Likewise, kk divides 2(m+2n)=2m+4n2(m + 2n) = 2m + 4n, so it divides (3m+4n)−(2m+4n)=m(3m + 4n) - (2m + 4n) = m. So kk is always a common divisor of mm and 2n2n. The other options need not hold. Take m=2m = 2 and n=1n = 1: then m+2n=4m + 2n = 4 and 3m+4n=103m + 4n = 10, so k=2k = 2 divides both. But k=2k = 2 does not divide n=1n = 1 (options A and D) or 3n=33n = 3 (option B). Hence, option C (mm and 2n2n).

Q48MCQIndices & Surds

The sum of all possible values of xx satisfying the equation 24x2−22x2+x+16+22x+30=02^{4x^2} - 2^{2x^2 + x + 16} + 2^{2x + 30} = 0, is
  1. 3
  2. 32\dfrac{3}{2}
  3. 52\dfrac{5}{2}
  4. 12\dfrac{1}{2}
Answer and solution

Answer: (D) 12\dfrac{1}{2}

Let a=22x2a = 2^{2x^2} and b=2x+15b = 2^{x+15}. Then a2=24x2a^2 = 2^{4x^2}, b2=22x+30b^2 = 2^{2x+30} and 2ab=21⋅22x2⋅2x+15=22x2+x+162ab = 2^1 \cdot 2^{2x^2} \cdot 2^{x+15} = 2^{2x^2+x+16}. So the equation is a2−2ab+b2=0a^2 - 2ab + b^2 = 0, that is, (a−b)2=0(a - b)^2 = 0, so a=ba = b: 22x2=2x+15⇒2x2=x+15⇒2x2−x−15=02^{2x^2} = 2^{x+15} \Rightarrow 2x^2 = x + 15 \Rightarrow 2x^2 - x - 15 = 0. Factorising, 2x2−6x+5x−15=2x(x−3)+5(x−3)=(2x+5)(x−3)=02x^2 - 6x + 5x - 15 = 2x(x - 3) + 5(x - 3) = (2x + 5)(x - 3) = 0, so x=3x = 3 or x=−52x = -\frac{5}{2}. Both satisfy the original equation, since every step can be reversed. The sum of the values is 3−52=123 - \frac{5}{2} = \frac{1}{2}. Option A (3) counts only the root x=3x = 3 and misses x=−52x = -\frac{5}{2}. Hence, option D (12\dfrac{1}{2}).

Q49TITAProperties of Numbers

The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is

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

Answer and solution

Answer: 15

A number with exactly two factors other than 1 and itself has exactly 4 factors in all. A number has exactly 4 factors only in two forms: N=p3N = p^3 for a prime pp (factors 1,p,p2,p31, p, p^2, p^3), or N=pqN = pq for distinct primes pp and qq (factors 1,p,q,pq1, p, q, pq). Cubes of primes below 50: 23=82^3 = 8 and 33=273^3 = 27 (53=1255^3 = 125 is too big). That is 2 numbers. Products of two distinct primes below 50: With 2: 2×3,2×5,2×7,2×11,2×13,2×17,2×19,2×232 \times 3, 2 \times 5, 2 \times 7, 2 \times 11, 2 \times 13, 2 \times 17, 2 \times 19, 2 \times 23, which is 8 numbers. With 3 and a larger prime: 3×5,3×7,3×11,3×133 \times 5, 3 \times 7, 3 \times 11, 3 \times 13, which is 4 numbers (3×17=513 \times 17 = 51 is too big). With 5 and a larger prime: only 5×7=355 \times 7 = 35, which is 1 number. That gives 8+4+1=138 + 4 + 1 = 13 products. Total: 2+13=152 + 13 = 15. The answer is 15.

Q50TITALogarithms

For some positive real number xx, if log⁡3(x)+log⁡x(25)log⁡x(0.008)=163\log_{\sqrt{3}}(x) + \dfrac{\log_x(25)}{\log_x(0.008)} = \dfrac{16}{3}, then the value of log⁡3(3x2)\log_3(3x^2) is

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

Answer and solution

Answer: 7

log⁡3x=2log⁡3x\log_{\sqrt3} x = 2\log_3 x. log⁡x25log⁡x0.008=log⁡25log⁡0.008=log⁡52log⁡5−3=−23\dfrac{\log_x 25}{\log_x 0.008} = \dfrac{\log 25}{\log 0.008} = \dfrac{\log 5^2}{\log 5^{-3}} = -\dfrac{2}{3} (as 0.008=11250.008 = \tfrac{1}{125}). So 2log⁡3x−23=1632\log_3 x - \dfrac23 = \dfrac{16}{3}, which gives log⁡3x=3\log_3 x = 3. log⁡3(3x2)=log⁡33+2log⁡3x=1+6=7\log_3(3x^2) = \log_3 3 + 2\log_3 x = 1 + 6 = 7.

Q51TITAQuadratic & Polynomial Equations

Let kk be the largest integer such that the equation (x−1)2+2kx+11=0(x-1)^2 + 2kx + 11 = 0 has no real roots. If yy is a positive real number, then the least possible value of k4y+9y\dfrac{k}{4y} + 9y is

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

Answer and solution

Answer: 6

Expand: (x−1)2+2kx+11=x2+(2k−2)x+12(x - 1)^2 + 2kx + 11 = x^2 + (2k - 2)x + 12. No real roots means the discriminant is negative: (2k−2)2−4⋅12<0⇒4(k−1)2<48⇒(k−1)2<12(2k - 2)^2 - 4 \cdot 12 < 0 \Rightarrow 4(k - 1)^2 < 48 \Rightarrow (k - 1)^2 < 12. Since k−1k - 1 is an integer, (k−1)2≤9(k - 1)^2 \le 9, so k−1≤3k - 1 \le 3. The largest kk is 4 (k=5k = 5 gives (k−1)2=16>12(k - 1)^2 = 16 > 12). Then k4y+9y=44y+9y=1y+9y\frac{k}{4y} + 9y = \frac{4}{4y} + 9y = \frac{1}{y} + 9y. For y>0y > 0, by AM-GM: 1y+9y≥21y⋅9y=29=6\frac{1}{y} + 9y \ge 2\sqrt{\frac{1}{y} \cdot 9y} = 2\sqrt{9} = 6. Equality holds when 1y=9y\frac{1}{y} = 9y, that is, y=13y = \frac{1}{3}, which is a positive real number. Check: 3+9×13=3+3=63 + 9 \times \frac{1}{3} = 3 + 3 = 6. So the least value is 6. The answer is 6.

Q52MCQTime & Work

Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is
  1. 90
  2. 120
  3. 75
  4. 60
Answer and solution

Answer: (A) 90

Let Pipe A fill the tank in xx hours, so Pipe B empties it in x−1x - 1 hours, and let Pipe C fill it in yy hours. All three together fill it in 2 hours: 1x−1x−1+1y=12\frac{1}{x} - \frac{1}{x-1} + \frac{1}{y} = \frac{1}{2} (1) B runs for 1 hour and C for 1+114=941 + 1\frac{1}{4} = \frac{9}{4} hours: 94y−1x−1=1\frac{9}{4y} - \frac{1}{x-1} = 1 (2) Subtracting (2) from (1): 1x−54y=−12\frac{1}{x} - \frac{5}{4y} = -\frac{1}{2}, so 1y=45(1x+12)\frac{1}{y} = \frac{4}{5}\left(\frac{1}{x} + \frac{1}{2}\right). Putting this into (2): 95x+910−1x−1=1\frac{9}{5x} + \frac{9}{10} - \frac{1}{x-1} = 1. Multiplying by 10x(x−1)10x(x-1): 18(x−1)−10x=x(x−1)18(x - 1) - 10x = x(x - 1), so x2−9x+18=0x^2 - 9x + 18 = 0 and x=3x = 3 or 66. A takes less than 5 hours, so x=3x = 3. Then (2) gives 94y=1+12\frac{9}{4y} = 1 + \frac{1}{2}, so y=32y = \frac{3}{2} hours =90= 90 minutes. Check (1): 13−12+23=12\frac{1}{3} - \frac{1}{2} + \frac{2}{3} = \frac{1}{2}. Option B (120 minutes, y=2y = 2) would make (2) give x=9x = 9, breaking both (1) and the 5-hour limit. Hence, option A (90).

Q53MCQSimple & Compound Interest

Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. He repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year. Then, the total interest, in rupees, paid over the three years is nearest to
  1. 45311
  2. 51311
  3. 33130
  4. 40991
Answer and solution

Answer: (B) 51311

8% a year compounded half-yearly is 4% per half-year, a factor of 1.041.04 each half-year. After the first year (two half-years): 200000×1.042=216320200000 \times 1.04^2 = 216320. Interest so far: Rs 16320. He repays Rs 10320, so 216320−10320=206000216320 - 10320 = 206000 is outstanding. This grows for two more years (four half-years): 206000×1.044=206000×1.16985856≈240990.86206000 \times 1.04^4 = 206000 \times 1.16985856 \approx 240990.86. Interest in these two years: 240990.86−206000=34990.86240990.86 - 206000 = 34990.86. Total interest: 16320+34990.86=51310.86≈5131116320 + 34990.86 = 51310.86 \approx 51311. Check: he pays 10320+240990.86=251310.8610320 + 240990.86 = 251310.86 in all on a loan of 200000, which is the same Rs 51310.86 of interest. Option D (40991) is 240991−200000240991 - 200000: it leaves out the Rs 10320 paid after the first year. Hence, option B (51311).

Q54MCQTime, Speed & Distance

Ravi is driving at a speed of 40 km/h on a road. Vijay is 54 meters behind Ravi and driving in the same direction as Ravi. Ashok is driving along the same road from the opposite direction at a speed of 50 km/h and is 225 meters away from Ravi. The speed, in km/h, at which Vijay should drive so that all the three cross each other at the same time, is
  1. 58.8
  2. 67.2
  3. 61.6
  4. 64.4
Answer and solution

Answer: (C) 61.6

Convert to m/s by multiplying by 518\frac{5}{18}: Ravi goes at 40×518=100940 \times \frac{5}{18} = \frac{100}{9} m/s and Ashok at 50×518=125950 \times \frac{5}{18} = \frac{125}{9} m/s. Ravi and Ashok approach each other, so their relative speed is 1009+1259=25\frac{100}{9} + \frac{125}{9} = 25 m/s. They meet after 22525=9\frac{225}{25} = 9 seconds. In those 9 seconds Ravi covers 1009×9=100\frac{100}{9} \times 9 = 100 m. Vijay starts 54 m behind Ravi, so to reach the same point at the same moment he must cover 100+54=154100 + 54 = 154 m in 9 seconds. Vijay's speed is 1549\frac{154}{9} m/s =1549×185=3085=61.6= \frac{154}{9} \times \frac{18}{5} = \frac{308}{5} = 61.6 km/h. Check: Vijay must gain 54 m on Ravi in 9 s, which is 6 m/s or 21.6 km/h, and 40+21.6=61.640 + 21.6 = 61.6. Option A (58.8 km/h) is too slow: it covers only 58.8×518×9=14758.8 \times \frac{5}{18} \times 9 = 147 m in 9 seconds, 7 m short. Hence, option C (61.6).

Q55MCQProfit, Loss & Discount

Minu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit. Then, Kanu sells it back to Minu at 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs.500, then the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu is
  1. 35.42%
  2. 52%
  3. 31.25%
  4. 26%
Answer and solution

Answer: (C) 31.25%

Minu buys the sunglasses for Rs 1000 and sells them to Kanu at 20% profit: 1000×1.2=12001000 \times 1.2 = 1200. Her profit on this sale is Rs 200. Kanu sells them back to Minu at a 20% loss on his cost of Rs 1200: 1200×0.8=9601200 \times 0.8 = 960. So Minu's new cost is Rs 960. Her total profit is Rs 500, so the sale to Tanu must bring 500−200=300500 - 200 = 300. She sells to Tanu for 960+300=1260960 + 300 = 1260. Profit percentage on this sale: 300960×100%=31.25%\frac{300}{960} \times 100\% = 31.25\%. Check: she paid 1000+960=19601000 + 960 = 1960 and received 1200+1260=24601200 + 1260 = 2460, a profit of Rs 500. Option B (52%) credits the whole Rs 500 to the last sale (500960≈52%\frac{500}{960} \approx 52\%), but Rs 200 of it was earned from Kanu. Hence, option C (31.25%).

Q56MCQRatios, Proportions & Partnership

The price of a precious stone is directly proportional to the square of its weight. Sita has a precious stone weighing 18 units. If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000. Then, the price of the original precious stone is
  1. 1944000
  2. 972000
  3. 1620000
  4. 1296000
Answer and solution

Answer: (D) 1296000

Let price =kw2= kw^2. The original stone costs k×182=324kk \times 18^2 = 324k. The four pieces have distinct positive integer weights summing to 18, and their total price is kk times the sum of their squares. Lowest total: the sum of squares is smallest when the weights are as equal as possible. Four distinct integers closest to 18÷4=4.518 \div 4 = 4.5 are 3, 4, 5, 6, giving 9+16+25+36=869 + 16 + 25 + 36 = 86, so the price is 86k86k. Highest total: make one piece as large as possible by keeping the other three as small as possible: 1, 2, 3 and 12. This gives 1+4+9+144=1581 + 4 + 9 + 144 = 158, so the price is 158k158k. Difference: 158k−86k=72k=288000158k - 86k = 72k = 288000, so k=4000k = 4000. Original price: 324×4000=1296000324 \times 4000 = 1296000. Option A (1944000) would need k=6000k = 6000, making the difference 72×6000=43200072 \times 6000 = 432000, not 288000. Hence, option D (1296000).

Q57MCQAverages, Mixtures & Alligations

In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the nonmanufacturing employees is
  1. 6:5
  2. 4:5
  3. 5:4
  4. 5:6
Answer and solution

Answer: (B) 4:5

Let the company have 100x100x employees and a total salary of 6y6y. Manufacturing: 20x20x employees share 16×6y=y\frac{1}{6} \times 6y = y, so their average salary is y20x\frac{y}{20x}. Non-manufacturing: 80x80x employees share 6y−y=5y6y - y = 5y, so their average salary is 5y80x=y16x\frac{5y}{80x} = \frac{y}{16x}. Ratio: y20x:y16x=120:116=16:20=4:5\frac{y}{20x} : \frac{y}{16x} = \frac{1}{20} : \frac{1}{16} = 16 : 20 = 4 : 5. Check: manufacturing staff are 20% of the employees but receive only about 16.7% of the salary, so their average must be lower, as 4:54 : 5 shows. Option C (5:4) is the same ratio reversed; it would make manufacturing employees earn more on average. Hence, option B (4:5).

Q58TITALinear Equations

If a certain amount of money is divided equally among nn persons, each one receives Rs 352. However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330. Then, the maximum possible value of nn is

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

Answer and solution

Answer: 16

The total amount is 352n352n. If two persons get Rs 506 each, the remaining 352n−1012352n - 1012 is shared equally by the other n−2n - 2 persons, and each gets at most Rs 330: 352n−1012n−2≤330\frac{352n - 1012}{n - 2} \le 330 352n−1012≤330n−660352n - 1012 \le 330n - 660 22n≤35222n \le 352, so n≤16n \le 16. Check n=16n = 16: the total is 352×16=5632352 \times 16 = 5632. After Rs 1012 for the two persons, Rs 4620 is left for 14 persons, exactly Rs 330 each. That is allowed, since each may receive less than or equal to Rs 330. So the maximum possible value of nn is 16. The answer is 16.

Q59TITAProfit, Loss & Discount

Jayant bought a certain number of white shirts at the rate of Rs 1000 per piece and a certain number of blue shirts at the rate of Rs 1125 per piece. For each shirt, he then set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10% and made a total profit of Rs.51000. If he bought both colors of shirts, then the maximum possible total number of shirts that he could have bought is

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

Answer and solution

Answer: 407

Let him buy mm white and nn blue shirts. Total cost =1000m+1125n= 1000m + 1125n. The marked price is 1.251.25 times the average cost, and a 10% discount leaves 1.25×0.9=1.1251.25 \times 0.9 = 1.125 times the average cost. So every shirt sells for 18\frac{1}{8} more than the average cost, and the total profit is 18\frac{1}{8} of the total cost: 1000m+1125n8=51000⇒1000m+1125n=408000\frac{1000m + 1125n}{8} = 51000 \Rightarrow 1000m + 1125n = 408000. Dividing by 125: 8m+9n=32648m + 9n = 3264. Then m+n=3264−9n8+n=408−n8m + n = \frac{3264 - 9n}{8} + n = 408 - \frac{n}{8}, which is largest when nn is smallest. For mm to be a whole number, 3264−9n3264 - 9n must be divisible by 8. Since 3264=8×4083264 = 8 \times 408, 9n9n must be divisible by 8, so nn is a multiple of 8. He bought both colours, so n≥1n \ge 1, and the least value is n=8n = 8. Then m=3264−728=399m = \frac{3264 - 72}{8} = 399, and m+n=399+8=407m + n = 399 + 8 = 407. The answer is 407.

Q60TITAAverages, Mixtures & Alligations

A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process of replacing 4 liters of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is

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

Answer and solution

Answer: 7

Each time, 4 litres of the 40-litre mixture, one tenth of it, is removed, so one tenth of the milk goes and 910\frac{9}{10} of it remains. After nn replacements the milk is 40×(910)n40 \times \left(\frac{9}{10}\right)^n litres. The total stays 40 litres, so milk is less than water when milk is below 20 litres: 40×(0.9)n<20⇒(0.9)n<0.540 \times (0.9)^n < 20 \Rightarrow (0.9)^n < 0.5. Powers of 0.9: 0.92=0.810.9^2 = 0.81, 0.93=0.7290.9^3 = 0.729, 0.94=0.65610.9^4 = 0.6561, 0.95≈0.59050.9^5 \approx 0.5905, 0.96≈0.53140.9^6 \approx 0.5314, 0.97≈0.47830.9^7 \approx 0.4783. After 6 replacements the milk is about 40×0.5314≈21.340 \times 0.5314 \approx 21.3 litres, still more than the water. After 7 it is about 40×0.4783≈19.140 \times 0.4783 \approx 19.1 litres, less than the water. The answer is 7.

Q61MCQPolygons & Circles

A triangle is drawn with its vertices on the circle CC such that one of its sides is a diameter of CC and the other two sides have their lengths in the ratio a:ba : b. If the radius of the circle is rr, then the area of the triangle is
  1. abr22(a2+b2)\dfrac{abr^2}{2(a^2+b^2)}
  2. 2abr2a2+b2\dfrac{2abr^2}{a^2+b^2}
  3. 4abr2a2+b2\dfrac{4abr^2}{a^2+b^2}
  4. abr2a2+b2\dfrac{abr^2}{a^2+b^2}
Answer and solution

Answer: (B) 2abr2a2+b2\dfrac{2abr^2}{a^2+b^2}

A triangle with a diameter as one side has a right angle opposite it (the angle in a semicircle). Let BC be the diameter, so ∠BAC=90∘\angle BAC = 90^\circ and BC=2rBC = 2r. Solution figure for question 61, CAT 2023 Slot 2 The other two sides are in the ratio a:ba : b, so let AB=kaAB = ka and AC=kbAC = kb for some k>0k > 0. By Pythagoras, AB2+AC2=BC2AB^2 + AC^2 = BC^2: k2(a2+b2)=4r2k^2(a^2 + b^2) = 4r^2, so k2=4r2a2+b2k^2 = \frac{4r^2}{a^2 + b^2}. The area of a right triangle is half the product of its legs: Area =12×ka×kb=12k2ab=12×4r2a2+b2×ab=2abr2a2+b2= \frac{1}{2} \times ka \times kb = \frac{1}{2}k^2 ab = \frac{1}{2} \times \frac{4r^2}{a^2 + b^2} \times ab = \frac{2abr^2}{a^2 + b^2}. Check with a=b=1a = b = 1: the triangle is right-angled and isosceles on a hypotenuse 2r2r with height rr, so its area is 12×2r×r=r2\frac{1}{2} \times 2r \times r = r^2, and the formula gives 2r22=r2\frac{2r^2}{2} = r^2. Option C would give 2r22r^2, twice too much. Hence, option B (2abr2a2+b2\dfrac{2abr^2}{a^2+b^2}).

Q62MCQPolygons & Circles

In a rectangle ABCDABCD, AB=9AB = 9 cm and BC=6BC = 6 cm. PP and QQ are two points on BCBC such that the areas of the figures ABPABP, APQAPQ, and AQCDAQCD are in geometric progression. If the area of the figure AQCDAQCD is four times the area of triangle ABPABP, then BP:PQ:QCBP : PQ : QC is
  1. 1:2:4
  2. 1:2:1
  3. 2:4:1
  4. 1:1:2
Answer and solution

Answer: (C) 2:4:1

The areas of ABP, APQ and AQCD are in GP, and the third is 4 times the first, so the common ratio is 4=2\sqrt{4} = 2. Let the areas be kk, 2k2k and 4k4k. Join AC. Solution figure for question 62, CAT 2023 Slot 2 Triangles ABP, APQ and AQC have their bases on BC and the same height AB, so their areas are in the ratio BP:PQ:QCBP : PQ : QC. The diagonal AC halves the rectangle, so triangles ABC and ACD have equal areas. Let the area of triangle AQC be xx. Then area of ABC =k+2k+x=3k+x= k + 2k + x = 3k + x, and area of ACD == area of AQCD −x=4k−x- x = 4k - x. Setting them equal: 3k+x=4k−x3k + x = 4k - x, so x=k2x = \frac{k}{2}. So BP:PQ:QC=k:2k:k2=2:4:1BP : PQ : QC = k : 2k : \frac{k}{2} = 2 : 4 : 1. Check: the rectangle's area is 9×6=54=7k9 \times 6 = 54 = 7k, so BP=2k9=127BP = \frac{2k}{9} = \frac{12}{7}, PQ=247PQ = \frac{24}{7} and QC=67QC = \frac{6}{7}, which add up to 6=BC6 = BC. Option A (1:2:4) copies the ratio of the three areas, but AQCD is a quadrilateral, not a triangle on QC. Hence, option C (2:4:1).

Q63TITACoordinate Geometry

The area of the quadrilateral bounded by the Y-axis, the line x=5x = 5, and the lines ∣x−y∣−∣x−5∣=2|x - y| - |x - 5| = 2, is

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

Answer and solution

Answer: 45

The region lies between the Y-axis (x=0x = 0) and x=5x = 5, so take 0≤x≤50 \le x \le 5. There ∣x−5∣=5−x|x - 5| = 5 - x, and the equation becomes ∣x−y∣=2+(5−x)=7−x|x - y| = 2 + (5 - x) = 7 - x. So x−y=7−xx - y = 7 - x or x−y=−(7−x)x - y = -(7 - x), which gives two lines: y=2x−7y = 2x - 7 and y=7y = 7. At x=0x = 0 they give y=−7y = -7 and y=7y = 7; at x=5x = 5 they give y=3y = 3 and y=7y = 7. So the quadrilateral has vertices A (0,7)(0, 7), B (5,7)(5, 7), D (5,3)(5, 3) and E (0,−7)(0, -7). Solution figure for question 63, CAT 2023 Slot 2 Split it along y=3y = 3, with C at (0,3)(0, 3): Rectangle ABDC: 5×(7−3)=205 \times (7 - 3) = 20. Triangle CDE: base CE =3−(−7)=10= 3 - (-7) = 10 along the Y-axis and height 5, so its area is 12×10×5=25\frac{1}{2} \times 10 \times 5 = 25. Total area =20+25=45= 20 + 25 = 45. Equivalently, it is a trapezium with parallel sides 14 and 4, a distance 5 apart: 12(14+4)×5=45\frac{1}{2}(14 + 4) \times 5 = 45. The answer is 45.

Q64MCQSequences & Series

Let both the series a1,a2,a3…a_1, a_2, a_3 \dots and b1,b2,b3…b_1, b_2, b_3 \dots be in arithmetic progression such that the common differences of both the series are prime numbers. If a5=b9a_5 = b_9, a19=b19a_{19} = b_{19} and b2=0b_2 = 0, then a11a_{11} equals
  1. 86
  2. 79
  3. 83
  4. 84
Answer and solution

Answer: (B) 79

Let the first terms be aa and bb, and the common differences dd and ee, both prime. b2=b+e=0b_2 = b + e = 0, so b=−eb = -e. a5=b9a_5 = b_9: a+4d=b+8e=7ea + 4d = b + 8e = 7e. a19=b19a_{19} = b_{19}: a+18d=b+18e=17ea + 18d = b + 18e = 17e. Subtracting the first from the second: 14d=10e14d = 10e, i.e. 7d=5e7d = 5e. So 77 divides 5e5e, hence 77 divides ee; as ee is prime, e=7e = 7 and then d=5d = 5. Then a=7e−4d=49−20=29a = 7e - 4d = 49 - 20 = 29, and a11=a+10d=29+50=79a_{11} = a + 10d = 29 + 50 = 79. Check: a19=29+90=119a_{19} = 29 + 90 = 119 and b19=−7+18×7=119b_{19} = -7 + 18 \times 7 = 119. Option D (84) is a+11da + 11d, which is a12a_{12}, one term too far. Hence, option B (79).

Q65MCQQuadratic & Polynomial Equations

If p2+q2−29=2pq−20=52−2pqp^2 + q^2 - 29 = 2pq - 20 = 52 - 2pq, then the difference between the maximum and minimum possible value of (p3−q3)(p^3 - q^3) is
  1. 243
  2. 486
  3. 378
  4. 189
Answer and solution

Answer: (C) 378

From 2pq−20=52−2pq2pq - 20 = 52 - 2pq: 4pq=724pq = 72, so pq=18pq = 18. From p2+q2−29=2pq−20p^2 + q^2 - 29 = 2pq - 20: p2+q2=2pq+9=36+9=45p^2 + q^2 = 2pq + 9 = 36 + 9 = 45. Then (p−q)2=p2+q2−2pq=45−36=9(p - q)^2 = p^2 + q^2 - 2pq = 45 - 36 = 9, so p−q=3p - q = 3 or −3-3. Now p3−q3=(p−q)(p2+pq+q2)=(p−q)(45+18)=63(p−q)p^3 - q^3 = (p - q)(p^2 + pq + q^2) = (p - q)(45 + 18) = 63(p - q). So p3−q3p^3 - q^3 is 63×3=18963 \times 3 = 189 or 63×(−3)=−18963 \times (-3) = -189. Both occur: p=6p = 6, q=3q = 3 gives 189, and p=3p = 3, q=6q = 6 gives −189-189. The difference between the maximum and the minimum is 189−(−189)=378189 - (-189) = 378. Option D (189) is the maximum value alone, not the difference. Hence, option C (378).

Q66TITASequences & Series

Let ana_n and bnb_n be two sequences such that an=13+6(n−1)a_n = 13 + 6(n-1) and bn=15+7(n−1)b_n = 15 + 7(n-1) for all natural numbers nn. Then, the largest three digit integer that is common to both these sequences, is

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

Answer and solution

Answer: 967

an=13+6(n−1)a_n = 13 + 6(n - 1) gives 13, 19, 25, 31, 37, 43, …, and bn=15+7(n−1)b_n = 15 + 7(n - 1) gives 15, 22, 29, 36, 43, … The first common term is 43. From there, a number stays in both sequences only if the step is a multiple of both 6 and 7, so the common terms rise in steps of lcm(6,7)=42\text{lcm}(6, 7) = 42: common terms =43+42j= 43 + 42j, for j=0,1,2,…j = 0, 1, 2, \dots For the largest three-digit one: 43+42j≤999⇒42j≤956⇒j≤22.7643 + 42j \le 999 \Rightarrow 42j \le 956 \Rightarrow j \le 22.76, so j=22j = 22. The term is 43+42×22=43+924=96743 + 42 \times 22 = 43 + 924 = 967. Check: 967=13+6×159967 = 13 + 6 \times 159 and 967=15+7×136967 = 15 + 7 \times 136, so it lies in both sequences. The next common term, 967+42=1009967 + 42 = 1009, has four digits. The answer is 967.