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CAT 2025 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.

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Quantitative Ability

CAT 2025 Slot 3 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q47MCQQuadratic & Polynomial Equations

If (x2+1x2)=25\left(x^2 + \dfrac{1}{x^2}\right) = 25 and x>0x > 0, then the value of (x7+1x7)\left(x^7 + \dfrac{1}{x^7}\right) is
  1. 44850344850\sqrt{3}
  2. 44853344853\sqrt{3}
  3. 44859344859\sqrt{3}
  4. 44856344856\sqrt{3}
Answer and solution

Answer: (B) 44853344853\sqrt{3}

Given x2+1x2=25x^2 + \dfrac{1}{x^2} = 25 and x>0x > 0. Then (x+1x)2=25+2=27\left(x + \dfrac{1}{x}\right)^2 = 25 + 2 = 27, so x+1x=33x + \dfrac{1}{x} = 3\sqrt{3} (positive, since x>0x > 0). Cubing: x3+1x3=(33)3−3(33)=813−93=723x^3 + \dfrac{1}{x^3} = (3\sqrt{3})^3 - 3(3\sqrt{3}) = 81\sqrt{3} - 9\sqrt{3} = 72\sqrt{3}. Squaring the given value: x4+1x4=252−2=623x^4 + \dfrac{1}{x^4} = 25^2 - 2 = 623. Now (x3+1x3)(x4+1x4)=x7+1x7+x+1x\left(x^3 + \dfrac{1}{x^3}\right)\left(x^4 + \dfrac{1}{x^4}\right) = x^7 + \dfrac{1}{x^7} + x + \dfrac{1}{x}. So x7+1x7=723×623−33=448563−33=448533x^7 + \dfrac{1}{x^7} = 72\sqrt{3} \times 623 - 3\sqrt{3} = 44856\sqrt{3} - 3\sqrt{3} = 44853\sqrt{3}. Option D (44856344856\sqrt{3}) is the product alone; it forgets to subtract the cross term x+1x=33x + \dfrac{1}{x} = 3\sqrt{3}. Hence, option B (44853344853\sqrt{3}).

Q48MCQFunctions & Graphs

For real values of xx, the range of the function f(x)=2x−32x2+4x−6f(x) = \dfrac{2x - 3}{2x^2 + 4x - 6} is
  1. (−∞,18]∪[1,∞)\left(-\infty, \dfrac{1}{8}\right] \cup [1, \infty)
  2. (−∞,14]∪[1,∞)\left(-\infty, \dfrac{1}{4}\right] \cup [1, \infty)
  3. (−∞,18]∪[12,∞)\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty\right)
  4. (−∞,14]∪[12,∞)\left(-\infty, \dfrac{1}{4}\right] \cup \left[\dfrac{1}{2}, \infty\right)
Answer and solution

Answer: (C) (−∞,18]∪[12,∞)\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty\right)

Let y=f(x)y = f(x). Then y(2x2+4x−6)=2x−3y(2x^2 + 4x - 6) = 2x - 3, i.e. 2yx2+(4y−2)x+(3−6y)=02yx^2 + (4y - 2)x + (3 - 6y) = 0. For y≠0y \ne 0 this is a quadratic in xx, and a real xx exists only when its discriminant is non-negative: (4y−2)2−8y(3−6y)=64y2−40y+4=4(8y−1)(2y−1)≥0(4y - 2)^2 - 8y(3 - 6y) = 64y^2 - 40y + 4 = 4(8y - 1)(2y - 1) \ge 0. So y≤18y \le \dfrac{1}{8} or y≥12y \ge \dfrac{1}{2}. The value y=0y = 0 is also reached, at x=32x = \dfrac{3}{2}. The denominator is 00 at x=1x = 1 and x=−3x = -3, but these never solve the quadratic: substituting them gives 1=01 = 0 and 9=09 = 0. The endpoints are reached: f(3)=324=18f(3) = \dfrac{3}{24} = \dfrac{1}{8} and f(0)=−3−6=12f(0) = \dfrac{-3}{-6} = \dfrac{1}{2}. So the range is (−∞,18]∪[12,∞)\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty\right). Option A has the right lower part but starts the upper part at 11, leaving out values such as f(0)=12f(0) = \dfrac{1}{2}. Hence, option C ((−∞,18]∪[12,∞)\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty\right)).

Q49MCQDigits & Base Systems

For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is
  1. 811
  2. 4078
  3. 735
  4. 3289
Answer and solution

Answer: (A) 811

Let the number be abcd‾\overline{abcd} with a≥1a \geq 1. The conditions are a+b+c=15a + b + c = 15, b+c+d=16b + c + d = 16 and c=d+6c = d + 6. Subtracting the first from the second gives d−a=1d - a = 1, so d=a+1d = a + 1 and c=d+6=a+7c = d + 6 = a + 7. Since c≤9c \leq 9, a≤2a \leq 2; and a≥1a \geq 1. So a=1a = 1 or a=2a = 2. If a=1a = 1: d=2d = 2, c=8c = 8, b=15−1−8=6b = 15 - 1 - 8 = 6, giving 1682. If a=2a = 2: d=3d = 3, c=9c = 9, b=15−2−9=4b = 15 - 2 - 9 = 4, giving 2493. Check 2493: 2+4+9=152 + 4 + 9 = 15, 4+9+3=164 + 9 + 3 = 16 and 9=3+69 = 3 + 6. No other number fits. Difference =2493−1682=811= 2493 - 1682 = 811. Because only these two numbers are possible, the difference is fixed. A value such as 735 (option C) would need a different pair of numbers, and no other pair satisfies all three conditions. Hence, option A (811).

Q50MCQTime & Work

Teams A, B, and C consist of five, eight, and ten members, respectively, such that every member within a team is equally productive. Working separately, teams A, B, and C can complete a certain job in 40 hours, 50 hours, and 4 hours, respectively. Two members from team A, three members from team B, and one member from team C together start the job, and the member from team C leaves after 23 hours. The number of additional member(s) from team B, that would be required to replace the member from team C, to finish the job in the next one hour, is
  1. 2
  2. 4
  3. 1
  4. 3
Answer and solution

Answer: (A) 2

Take the job as 200 units (LCM of 40, 50 and 4). Team A does 200÷40=5200 \div 40 = 5 units/hour, so each of its 5 members does 1 unit/hour. Team B does 200÷50=4200 \div 50 = 4 units/hour, so each of its 8 members does 0.5 unit/hour. Team C does 200÷4=50200 \div 4 = 50 units/hour, so each of its 10 members does 5 units/hour. 2 members of A, 3 of B and 1 of C together do 2(1)+3(0.5)+1(5)=8.52(1) + 3(0.5) + 1(5) = 8.5 units/hour. In 23 hours they do 8.5×23=195.58.5 \times 23 = 195.5 units, leaving 200−195.5=4.5200 - 195.5 = 4.5 units. The C member leaves. Let xx extra B members join so that the rest is done in 1 hour: 2(1)+(3+x)(0.5)=4.5⇒(3+x)(0.5)=2.5⇒3+x=5⇒x=22(1) + (3 + x)(0.5) = 4.5 \Rightarrow (3 + x)(0.5) = 2.5 \Rightarrow 3 + x = 5 \Rightarrow x = 2. With only 1 extra member (option C), the rate is 2+4(0.5)=42 + 4(0.5) = 4 units/hour, short of the 4.5 units needed in one hour. Hence, option A (2).

Q51MCQAverages, Mixtures & Alligations

The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is
  1. 50000
  2. 45000
  3. 40000
  4. 54000
Answer and solution

Answer: (D) 54000

Let each manager's average salary be MM and each engineer's be EE (in rupees). Total salary: 5M+25E=30×60000=18000005M + 25E = 30 \times 60000 = 1800000. After the managers' 20% raise, the average rises by 5% to 6300063000, so the new total is 30×63000=189000030 \times 63000 = 1890000: 5(1.2M)+25E=6M+25E=18900005(1.2M) + 25E = 6M + 25E = 1890000. Subtracting the first equation gives M=90000M = 90000. Then 25E=1800000−5(90000)=135000025E = 1800000 - 5(90000) = 1350000, so E=54000E = 54000. Check option A (50000): then 5M=1800000−1250000=5500005M = 1800000 - 1250000 = 550000, and a 20% raise for the managers would add 110000110000, not the required 1890000−1800000=900001890000 - 1800000 = 90000. Hence, option D (54000).

Q52MCQProfit, Loss & Discount

The monthly sales of a product from January to April were 120, 135, 150 and 165 units, respectively. The cost price of the product was Rs. 240 per unit, and a fixed marked price was used for the product in all the four months. Discounts of 20%, 10% and 5% were given on the marked price per unit in January, February and March, respectively, while no discounts were given in April. If the total profit from January to April was Rs. 138825, then the marked price per unit, in rupees, was
  1. 525
  2. 515
  3. 520
  4. 510
Answer and solution

Answer: (A) 525

Let the marked price be MM rupees; the cost price is Rs 240 per unit. Selling prices per unit: January 0.8M0.8M, February 0.9M0.9M, March 0.95M0.95M, April MM. Total revenue =120(0.8M)+135(0.9M)+150(0.95M)+165M=96M+121.5M+142.5M+165M=525M= 120(0.8M) + 135(0.9M) + 150(0.95M) + 165M = 96M + 121.5M + 142.5M + 165M = 525M. Total units =120+135+150+165=570= 120 + 135 + 150 + 165 = 570, so total cost =570×240=136800= 570 \times 240 = 136800. Profit: 525M−136800=138825⇒525M=275625⇒M=525525M - 136800 = 138825 \Rightarrow 525M = 275625 \Rightarrow M = 525. Check option C (520): the profit would be 525×520−136800=273000−136800=136200525 \times 520 - 136800 = 273000 - 136800 = 136200, short of 138825. Hence, option A (525).

Q53MCQSet Theory

In a class of 150 students, 75 students chose physics, 111students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemistry is equal to the number of students who chose both chemistry and mathematics, and this is half the number of students who chose both physics and mathematics. The maximum possible number of students who chose physics but not mathematics, is
  1. 35
  2. 30
  3. 40
  4. 55
Answer and solution

Answer: (A) 35

Let ∣P∩C∣=∣M∩C∣=u|P \cap C| = |M \cap C| = u, ∣P∩M∣=2u|P \cap M| = 2u and ∣P∩M∩C∣=x|P \cap M \cap C| = x, with x≥1x \geq 1. Inclusion–exclusion: 150=75+111+40−(2u+u+u)+x150 = 75 + 111 + 40 - (2u + u + u) + x, so 4u−x=764u - x = 76, i.e. x=4u−76x = 4u - 76. Since x≥1x \geq 1, u≥19.25u \geq 19.25, so u≥20u \geq 20. Since x≤ux \leq u (the triple overlap cannot exceed a pair overlap), 3u≤763u \leq 76, so u≤25u \leq 25. Physics but not mathematics =∣P∣−∣P∩M∣=75−2u= |P| - |P \cap M| = 75 - 2u. This is largest when uu is smallest, u=20u = 20, which gives x=4x = 4. Check the regions for u=20u = 20, x=4x = 4: physics only 19, mathematics only 55, chemistry only 4, P and M only 36, P and C only 16, M and C only 16, all three 4. They total 150 and none is negative. So the maximum is 75−40=3575 - 40 = 35. Option C (40) would need u=17.5u = 17.5, which is not a whole number, and any u<20u < 20 makes x<1x < 1. Hence, option A (35).

Q54MCQQuadratic & Polynomial Equations

If f(x)=(x2+3x)(x2+3x+2)f(x) = (x^2 + 3x)(x^2 + 3x + 2), then the sum of all real roots of the equation (f(x)+1)=9701\sqrt{(f(x) + 1)} = 9701 is
  1. −3-3
  2. 33
  3. 66
  4. −6-6
Answer and solution

Answer: (A) −3-3

Let t=x2+3xt = x^2 + 3x. Then f(x)=t(t+2)f(x) = t(t + 2), so f(x)+1=t2+2t+1=(t+1)2f(x) + 1 = t^2 + 2t + 1 = (t + 1)^2. The equation becomes (t+1)2=∣t+1∣=9701\sqrt{(t + 1)^2} = |t + 1| = 9701. Case 1: t+1=9701t + 1 = 9701, so x2+3x−9700=0x^2 + 3x - 9700 = 0. Its discriminant is 9+38800=38809>09 + 38800 = 38809 > 0, so it has two real roots, and their sum is −3-3. Case 2: t+1=−9701t + 1 = -9701, so x2+3x+9702=0x^2 + 3x + 9702 = 0. Its discriminant is 9−38808<09 - 38808 < 0, so it has no real roots. The sum of all real roots is therefore −3-3. Option D (−6-6) adds the root sums of both quadratics, but the second quadratic has no real roots. Hence, option A (−3-3).

Q55MCQTime, Speed & Distance

Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was
  1. 12
  2. 20
  3. 18
  4. 15
Answer and solution

Answer: (D) 15

The planned journey takes 6 hours (5 pm to 11 pm) at speed vv. Let the stop happen when TT planned hours remain, so the remaining distance is vTvT. Stop of 20 minutes: he covers vTvT at v+3v + 3 in T−13T - \dfrac{1}{3} hours. vT=(v+3)(T−13)⇒3T=v3+1⇒T=v9+13vT = (v + 3)\left(T - \dfrac{1}{3}\right) \Rightarrow 3T = \dfrac{v}{3} + 1 \Rightarrow T = \dfrac{v}{9} + \dfrac{1}{3}. Stop of 30 minutes: he covers vTvT at v+5v + 5 in T−12T - \dfrac{1}{2} hours. vT=(v+5)(T−12)⇒5T=v2+52⇒T=v10+12vT = (v + 5)\left(T - \dfrac{1}{2}\right) \Rightarrow 5T = \dfrac{v}{2} + \dfrac{5}{2} \Rightarrow T = \dfrac{v}{10} + \dfrac{1}{2}. Equating: v9+13=v10+12⇒v90=16⇒v=15\dfrac{v}{9} + \dfrac{1}{3} = \dfrac{v}{10} + \dfrac{1}{2} \Rightarrow \dfrac{v}{90} = \dfrac{1}{6} \Rightarrow v = 15. Check: T=2T = 2, so 30 km remain. At 18 km/h that takes 100 minutes, plus the 20-minute stop makes 2 hours; at 20 km/h it takes 90 minutes, plus 30 minutes makes 2 hours. Option C (18) gives T=73T = \dfrac{7}{3} from the first case but T=2.3T = 2.3 from the second, so it cannot satisfy both. Hence, option D (15).

Q56TITATime, Speed & Distance

Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. She takes exactly 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, she would take to walk from A to B and run from B to C, is

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

Answer and solution

Answer: 444

Let the walking speed be ww; the running speed is then 1.4w1.4w. 3 hours 30 minutes =3.5= 3.5 hours. Walking from B to C takes 3.5 hours, so BC=3.5wBC = 3.5w. Running from B to A takes 3.5 hours, so BA=3.5×1.4w=4.9wBA = 3.5 \times 1.4w = 4.9w. Walking from A to B takes 4.9ww=4.9\dfrac{4.9w}{w} = 4.9 hours =294= 294 minutes. Running from B to C takes 3.5w1.4w=2.5\dfrac{3.5w}{1.4w} = 2.5 hours =150= 150 minutes. Total time =294+150=444= 294 + 150 = 444 minutes. The answer is 444.

Q57MCQDigits & Base Systems

The sum of all the digits of the number (1050+1025−123)(10^{50} + 10^{25} - 123) is
  1. 324
  2. 255
  3. 212
  4. 221
Answer and solution

Answer: (D) 221

Write the number as 1050+(1025−123)10^{50} + (10^{25} - 123). 102510^{25} is 1 followed by 25 zeros, so 1025−12310^{25} - 123 is a 25-digit number: twenty-two 9s followed by 877. (Compare 105−123=9987710^{5} - 123 = 99877: two 9s, then 877.) 105010^{50} is 1 followed by 50 zeros, a 51-digit number. Adding the 25-digit number fills its last 25 places, so the result is a 1, then 25 zeros, then twenty-two 9s, then 877, which is 1+25+25=511 + 25 + 25 = 51 digits in all. Sum of digits =1+22×9+8+7+7=1+198+22=221= 1 + 22 \times 9 + 8 + 7 + 7 = 1 + 198 + 22 = 221. Option C (212) is 9 less, which is what you get by counting only twenty-one 9s. Hence, option D (221).

Q58TITATriangles & Lines

A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the sum of the lengths, in cm, of all three altitudes of the triangle ABC is

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

Answer and solution

Answer: 126

Triangle ABC has AB=AC=50AB = AC = 50 cm and BC=80BC = 80 cm. Area by Heron's formula: s=50+50+802=90s = \dfrac{50 + 50 + 80}{2} = 90, so the area is 90×40×40×10=1440000=1200\sqrt{90 \times 40 \times 40 \times 10} = \sqrt{1440000} = 1200 cm². (Check: the altitude from A bisects BC, so it is 502−402=30\sqrt{50^2 - 40^2} = 30 cm, and 12×80×30=1200\dfrac{1}{2} \times 80 \times 30 = 1200.) Altitude from A to BC: 2×120080=30\dfrac{2 \times 1200}{80} = 30 cm. Altitude from C to AB: 2×120050=48\dfrac{2 \times 1200}{50} = 48 cm. The altitude from B to AC is also 48 cm, because AC=ABAC = AB. Sum of the three altitudes =30+48+48=126= 30 + 48 + 48 = 126 cm. The answer is 126.

Q59TITAAverages, Mixtures & Alligations

Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After stirring, the same volume is taken out from B and poured into A. If the resultant ratio of alcohol and water in A is 15 : 4, then the volume, in litres, initially taken out from A is

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

Answer and solution

Answer: 16

Vessel A starts with 60 L of alcohol and vessel B with 60 L of water. Let xx L be moved each time. After the first transfer, A has 60−x60 - x L of alcohol, and B has 60 L of water and xx L of alcohol, 60+x60 + x L in all. The xx L taken from B has the same make-up, so it carries 60x60+x\dfrac{60x}{60 + x} L of water and x260+x\dfrac{x^2}{60 + x} L of alcohol. A ends with 60 L again, and all its water comes from this pour. Since alcohol : water in A is 15 : 4, the water in A is 60×419=2401960 \times \dfrac{4}{19} = \dfrac{240}{19} L. So 60x60+x=24019\dfrac{60x}{60 + x} = \dfrac{240}{19} ⇒1140x=240(60+x)=14400+240x\Rightarrow 1140x = 240(60 + x) = 14400 + 240x ⇒900x=14400⇒x=16\Rightarrow 900x = 14400 \Rightarrow x = 16. Check: water in A =60×1676=24019= \dfrac{60 \times 16}{76} = \dfrac{240}{19} L and alcohol =60−24019=90019= 60 - \dfrac{240}{19} = \dfrac{900}{19} L, a ratio of 900:240=15:4900 : 240 = 15 : 4. The answer is 16.

Q60TITASequences & Series

In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is

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

Answer and solution

Answer: 65

Given: Sum of 4th, 7th and 10th terms = 99.
(a+3d)+(a+6d)+(a+9d)=99⇒3a+18d=99⇒a+6d=33…(1)(a + 3d) + (a + 6d) + (a + 9d) = 99 \Rightarrow 3a + 18d = 99 \Rightarrow a + 6d = 33 \dots (1)

Also, sum of first 14 terms = 497:
S14=142(2a+13d)=7(2a+13d)=497⇒2a+13d=71…(2)S_{14} = \dfrac{14}{2}(2a + 13d) = 7(2a + 13d) = 497 \Rightarrow 2a + 13d = 71 \dots (2)

From (1) × 2: 2a+12d=662a + 12d = 66. Subtract from (2):
d=5d = 5
Substituting back: a+6(5)=33⇒a=3a + 6(5) = 33 \Rightarrow a = 3.

Sum of first 5 terms:
S5=52(2a+4d)=52(6+20)=52×26=65S_5 = \dfrac{5}{2}(2a + 4d) = \dfrac{5}{2}(6 + 20) = \dfrac{5}{2} \times 26 = \mathbf{65}.

Q61TITAProperties of Numbers

Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose value lies between 150 and 500. If p is not less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is

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

Answer and solution

Answer: 397

We have p+q+r=900p + q + r = 900, rr a perfect square between 150 and 500, and 0.3q≤p≤0.7q0.3q \leq p \leq 0.7q. The possible values of rr run from 132=16913^2 = 169 to 222=48422^2 = 484. Maximum pp: make p+q=900−rp + q = 900 - r as large as possible (r=169r = 169, so p+q=731p + q = 731) and give pp its largest share, p=0.7qp = 0.7q. 1.7q=731⇒q=4301.7q = 731 \Rightarrow q = 430 and p=0.7×430=301p = 0.7 \times 430 = 301. Both are natural numbers, so this is attainable. Minimum pp: make p+qp + q as small as possible (r=484r = 484, so p+q=416p + q = 416) and give pp its smallest share, p=0.3qp = 0.3q. 1.3q=416⇒q=3201.3q = 416 \Rightarrow q = 320 and p=0.3×320=96p = 0.3 \times 320 = 96. Sum of the maximum and minimum values of pp =301+96=397= 301 + 96 = 397. The answer is 397.

Q62TITARatios, Proportions & Partnership

The ratio of the number of coins in boxes A and B was 17:7. After 108 coins were shifted from box A to box B, this ratio became 37:20. The number of coins that needs to be shifted further from A to B, to make this ratio 1:1, is

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

Answer and solution

Answer: 272

Let the coins in boxes A and B be 17x17x and 7x7x. After 108 coins move from A to B: 17x−1087x+108=3720\dfrac{17x - 108}{7x + 108} = \dfrac{37}{20}. 20(17x−108)=37(7x+108)⇒340x−2160=259x+3996⇒81x=6156⇒x=7620(17x - 108) = 37(7x + 108) \Rightarrow 340x - 2160 = 259x + 3996 \Rightarrow 81x = 6156 \Rightarrow x = 76. Now A has 17(76)−108=118417(76) - 108 = 1184 coins and B has 7(76)+108=6407(76) + 108 = 640 coins. (Check: 1184:640=37:201184 : 640 = 37 : 20.) The total, 1824, does not change, so for a 1 : 1 ratio each box needs 912 coins. Coins still to move from A to B =1184−912=272= 1184 - 912 = 272. The answer is 272.

Q63MCQTime & Work

The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled up completely by pipe A in 15 hours. If all the three pipes are used simultaneously to fill up this empty tank, the time, in minutes, required to fill up the entire tank completely is nearest to
  1. 71
  2. 78
  3. 73
  4. 76
Answer and solution

Answer: (C) 73

The flow rates of A, B and C are in the ratio 4 : 9 : 36, so take them as 4, 9 and 36 units per hour. A fills the tank in 15 hours, so the tank holds 4×15=604 \times 15 = 60 units. All three together fill 4+9+36=494 + 9 + 36 = 49 units per hour. Time taken =6049= \dfrac{60}{49} hours =360049= \dfrac{3600}{49} minutes ≈73.47\approx 73.47 minutes. The nearest option is 73. Option A (71) and option D (76) are each more than 2 minutes away from 73.47, so neither is the nearest. Hence, option C (73).

Q64MCQPolygons & Circles

ABCD is a trapezium in which AB is parallel to DC, AD is perpendicular to AB, and AB = 3DC. If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm, then the area, in sq. cm, of the trapezium is
  1. 30√3
  2. 54
  3. 48
  4. 36√2
Answer and solution

Answer: (C) 48

AD is perpendicular to both parallel sides, and the circle touches AB and DC, so the height AD equals the circle's diameter: AD=2×3=6AD = 2 \times 3 = 6 cm. Solution figure for question 64, CAT 2025 Slot 3 Let DC=xDC = x, so AB=3xAB = 3x. A circle can be inscribed in a quadrilateral only if the sums of opposite sides are equal: AB+DC=AD+BCAB + DC = AD + BC. So 4x=6+BC4x = 6 + BC, giving BC=4x−6BC = 4x - 6. Drop a perpendicular CE from C to AB. Then CE=AD=6CE = AD = 6, AE=DC=xAE = DC = x and EB=3x−x=2xEB = 3x - x = 2x. In right triangle CEB: BC2=EB2+CE2BC^2 = EB^2 + CE^2, so (4x−6)2=(2x)2+36(4x - 6)^2 = (2x)^2 + 36. 16x2−48x+36=4x2+36⇒12x2=48x⇒x=416x^2 - 48x + 36 = 4x^2 + 36 \Rightarrow 12x^2 = 48x \Rightarrow x = 4. So DC=4DC = 4, AB=12AB = 12 and BC=10BC = 10 (check: 82+62=1028^2 + 6^2 = 10^2). Area =12(AB+DC)×AD=12(12+4)×6=48= \dfrac{1}{2}(AB + DC) \times AD = \dfrac{1}{2}(12 + 4) \times 6 = 48 cm². Option B (54) would need AB+DC=18AB + DC = 18, i.e. x=4.5x = 4.5; then BC=12BC = 12, but EB2+CE2=81+36=117≠144EB^2 + CE^2 = 81 + 36 = 117 \neq 144. Hence, option C (48).

Q65MCQTriangles & Lines

In ∆ABC, AB = AC = 12 cm and D is a point on side BC such that AD = 8 cm. If AD is extended to a point E such that ∠ACB = ∠AEB, then the length, in cm, of AE is
  1. 16
  2. 18
  3. 20
  4. 14
Answer and solution

Answer: (B) 18

Since AB=ACAB = AC, ∠ABC=∠ACB\angle ABC = \angle ACB. We are given ∠AEB=∠ACB\angle AEB = \angle ACB, so ∠AEB=∠ABC=∠ABD\angle AEB = \angle ABC = \angle ABD. E lies on AD extended, so triangles ABD and AEB share the angle at A (∠BAD=∠BAE\angle BAD = \angle BAE). With two pairs of equal angles, triangle ABD is similar to triangle AEB (A matches A, B matches E, and D matches B). Hence ABAE=ADAB⇒AB2=AD×AE\dfrac{AB}{AE} = \dfrac{AD}{AB} \Rightarrow AB^2 = AD \times AE. So AE=1228=1448=18AE = \dfrac{12^2}{8} = \dfrac{144}{8} = 18 cm. This holds wherever D lies on BC; D need not be the foot of the perpendicular from A. Option C (20) would need AB2=8×20=160AB^2 = 8 \times 20 = 160, but AB2=144AB^2 = 144. Hence, option B (18).

Q66TITAIndices & Surds

If 1212x×424x+12×52y=84z×2012x×2433x−612^{12x} \times 4^{24x+12} \times 5^{2y} = 8^{4z} \times 20^{12x} \times 243^{3x-6}, where xx, yy and zz are natural numbers, then x+y+zx + y + z equals

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

Answer and solution

Answer: 112

Given equation: 1212x×424x+12×52y=84z×2012x×2433x−612^{12x} \times 4^{24x + 12} \times 5^{2y} = 8^{4z} \times 20^{12x} \times 243^{3x - 6}

Expressing all bases in terms of prime factors (2,3,5)(2, 3, 5):
(3×22)12x×(22)24x+12×52y=(23)4z×(22×5)12x×(35)3x−6(3 \times 2^2)^{12x} \times (2^2)^{24x + 12} \times 5^{2y} = (2^3)^{4z} \times (2^2 \times 5)^{12x} \times (3^5)^{3x - 6}

Expanding the exponents:
312x×224x×248x+24×52y=212z×224x×512x×315x−303^{12x} \times 2^{24x} \times 2^{48x + 24} \times 5^{2y} = 2^{12z} \times 2^{24x} \times 5^{12x} \times 3^{15x - 30}

Combining powers of the same base on the LHS:
272x+24×312x×52y=212z+24x×315x−30×512x2^{72x + 24} \times 3^{12x} \times 5^{2y} = 2^{12z + 24x} \times 3^{15x - 30} \times 5^{12x}

Equating the exponents of corresponding prime bases:
1) For base 3: 12x=15x−30⇒3x=30⇒x=1012x = 15x - 30 \Rightarrow 3x = 30 \Rightarrow x = 10
2) For base 5: 2y=12x⇒y=6x=6(10)=602y = 12x \Rightarrow y = 6x = 6(10) = 60
3) For base 2: 72x+24=12z+24x⇒48x+24=12z⇒4x+2=z72x + 24 = 12z + 24x \Rightarrow 48x + 24 = 12z \Rightarrow 4x + 2 = z
Since x=10x = 10, z=4(10)+2=42z = 4(10) + 2 = 42

Therefore, x+y+z=10+60+42=112x + y + z = 10 + 60 + 42 = 112.

Q67TITAAverages, Mixtures & Alligations

In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is

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

Answer and solution

Answer: 700

Let SS, AA and CC be the numbers of science, arts and commerce students. S+A+C=1500S + A + C = 1500 and 1100S+1000A+800C=15500001100S + 1000A + 800C = 1550000, i.e. 11S+10A+8C=1550011S + 10A + 8C = 15500. Substitute C=1500−S−AC = 1500 - S - A: 11S+10A+12000−8S−8A=1550011S + 10A + 12000 - 8S - 8A = 15500, so 3S+2A=35003S + 2A = 3500 and A=3500−3S2A = \dfrac{3500 - 3S}{2}. The condition S≤AS \leq A gives S≤3500−3S2⇒5S≤3500⇒S≤700S \leq \dfrac{3500 - 3S}{2} \Rightarrow 5S \leq 3500 \Rightarrow S \leq 700. At S=700S = 700: A=3500−21002=700A = \dfrac{3500 - 2100}{2} = 700 and C=1500−1400=100C = 1500 - 1400 = 100, all whole and non-negative. Check the fees: 1100(700)+1000(700)+800(100)=770000+700000+80000=15500001100(700) + 1000(700) + 800(100) = 770000 + 700000 + 80000 = 1550000. So the maximum possible number of science students is 700. The answer is 700.

Q68MCQLogarithms

The sum of all possible real values of xx for which log⁡x−3(x2−9)=log⁡x−3(x+1)+2\log_{x-3}(x^2 - 9) = \log_{x-3}(x + 1) + 2 is
  1. −3-3
  2. 33\sqrt{33}
  3. 3+332\dfrac{3 + \sqrt{33}}{2}
  4. 33
Answer and solution

Answer: (C) 3+332\dfrac{3 + \sqrt{33}}{2}

For the logarithms to be defined, the base needs x−3>0x - 3 > 0 and x−3≠1x - 3 \neq 1, so x>3x > 3 and x≠4x \neq 4. Then x2−9>0x^2 - 9 > 0 and x+1>0x + 1 > 0 as well. Write 2=log⁡x−3((x−3)2)2 = \log_{x-3}\left((x - 3)^2\right). The equation becomes log⁡x−3(x2−9)=log⁡x−3((x+1)(x−3)2)\log_{x-3}(x^2 - 9) = \log_{x-3}\left((x + 1)(x - 3)^2\right), so (x−3)(x+3)=(x+1)(x−3)2(x - 3)(x + 3) = (x + 1)(x - 3)^2. Since x≠3x \neq 3, divide by x−3x - 3: x+3=(x+1)(x−3)=x2−2x−3x + 3 = (x + 1)(x - 3) = x^2 - 2x - 3, so x2−3x−6=0x^2 - 3x - 6 = 0. x=3±9+242=3±332x = \dfrac{3 \pm \sqrt{9 + 24}}{2} = \dfrac{3 \pm \sqrt{33}}{2}. Only 3+332≈4.37\dfrac{3 + \sqrt{33}}{2} \approx 4.37 is greater than 3 (and it is not 4); the other root is negative. So the sum of all valid values is 3+332\dfrac{3 + \sqrt{33}}{2}. Option D (3) is the sum of both roots of the quadratic, but it wrongly includes the negative root, where the base x−3x - 3 is negative. Hence, option C (3+332\dfrac{3 + \sqrt{33}}{2}).