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CAT 2025 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 2025 Slot 2 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q47MCQQuadratic & Polynomial Equations

The equations 3x2−5x+p=03x^2 - 5x + p = 0 and 2x2−2x+q=02x^2 - 2x + q = 0 have one common root. The sum of the other roots of these two equations is
  1. 83+p+13q\frac{8}{3} + p + \frac{1}{3}q
  2. 23−2p+23q\frac{2}{3} - 2p + \frac{2}{3}q
  3. 83−p+32q\frac{8}{3} - p + \frac{3}{2}q
  4. 23−p+32q\frac{2}{3} - p + \frac{3}{2}q
Answer and solution

Answer: (C) 83−p+32q\frac{8}{3} - p + \frac{3}{2}q

Let the common root be α\alpha, the other root of 3x2−5x+p=03x^2 - 5x + p = 0 be β\beta, and the other root of 2x2−2x+q=02x^2 - 2x + q = 0 be γ\gamma. Sums of roots: α+β=53\alpha + \beta = \frac{5}{3} and α+γ=22=1\alpha + \gamma = \frac{2}{2} = 1. So β+γ=83−2α\beta + \gamma = \frac{8}{3} - 2\alpha. To find α\alpha, multiply the first equation by 2 and the second by 3: 6α2−10α+2p=06\alpha^2 - 10\alpha + 2p = 0 and 6α2−6α+3q=06\alpha^2 - 6\alpha + 3q = 0. Subtracting the second from the first: −4α+2p−3q=0-4\alpha + 2p - 3q = 0, so α=2p−3q4\alpha = \frac{2p - 3q}{4}. Then β+γ=83−2p−3q2=83−p+32q\beta + \gamma = \frac{8}{3} - \frac{2p - 3q}{2} = \frac{8}{3} - p + \frac{3}{2}q. Option D has the same pp and qq terms but the constant 23\frac{2}{3}, which comes from subtracting the two sums of roots (53−1\frac{5}{3} - 1) instead of adding them. Hence, option C (83−p+32q\frac{8}{3} - p + \frac{3}{2}q).

Q48MCQPolygons & Circles

Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is
  1. 6 : 19
  2. 5 : 24
  3. 7 : 24
  4. 6 : 25
Answer and solution

Answer: (B) 5 : 24

Let the side be ss. The hexagon's area is 332s2\frac{3\sqrt{3}}{2}s^2. In a regular hexagon the long diagonal AD has length 2s2s and is parallel to BC. So ABCD is a trapezium with parallel sides BC=sBC = s and AD=2sAD = 2s and height 32s\frac{\sqrt{3}}{2}s; it is half the hexagon. P and Q are the midpoints of its legs AB and CD, so PQ is its mid-line: parallel to BC, halfway between BC and AD, with PQ=s+2s2=3s2PQ = \frac{s + 2s}{2} = \frac{3s}{2}. The height of PBCQ is therefore 34s\frac{\sqrt{3}}{4}s. Area of PBCQ =12(s+3s2)⋅34s=5316s2= \frac{1}{2}\left(s + \frac{3s}{2}\right) \cdot \frac{\sqrt{3}}{4}s = \frac{5\sqrt{3}}{16}s^2. Ratio =5316÷332=516⋅23=524= \frac{5\sqrt{3}}{16} \div \frac{3\sqrt{3}}{2} = \frac{5}{16} \cdot \frac{2}{3} = \frac{5}{24}. The tempting 7 : 24 is the other part of the half-hexagon, APQD (1224−524\frac{12}{24} - \frac{5}{24}), not PBCQ. Hence, option B (5 : 24).

Q49MCQLogarithms

If log⁡64(x2)+log⁡8(y)+3log⁡512(zy)=4\log_{64}(x^2) + \log_8(\sqrt{y}) + 3\log_{512}(z\sqrt{y}) = 4, where x, y and z are positive real numbers, then the minimum possible value of (x+y+z)(x + y + z) is
  1. 96
  2. 36
  3. 24
  4. 48
Answer and solution

Answer: (D) 48

Write every logarithm in base 2, using 64=2664 = 2^6, 8=238 = 2^3 and 512=29512 = 2^9: log⁡64(x2)=2log⁡2x6=13log⁡2x\log_{64}(x^2) = \frac{2\log_2 x}{6} = \frac{1}{3}\log_2 x log⁡8y=12log⁡2y3=16log⁡2y\log_8 \sqrt{y} = \frac{\frac{1}{2}\log_2 y}{3} = \frac{1}{6}\log_2 y 3log⁡512(zy)=39(log⁡2z+12log⁡2y)=13log⁡2z+16log⁡2y3\log_{512}(z\sqrt{y}) = \frac{3}{9}\left(\log_2 z + \frac{1}{2}\log_2 y\right) = \frac{1}{3}\log_2 z + \frac{1}{6}\log_2 y Adding: 13(log⁡2x+log⁡2y+log⁡2z)=4\frac{1}{3}(\log_2 x + \log_2 y + \log_2 z) = 4, so log⁡2(xyz)=12\log_2(xyz) = 12 and xyz=212xyz = 2^{12}. By AM–GM, x+y+z≥3xyz3=3⋅24=48x + y + z \ge 3\sqrt[3]{xyz} = 3 \cdot 2^4 = 48, with equality when x=y=z=16x = y = z = 16, which satisfies the equation. So 48 is reached and nothing smaller is possible: 24 and 36 lie below the AM–GM bound, and 96 is not the least value. Hence, option D (48).

Q50MCQInequalities & Modulus

The set of all real values of x for which (x2−∣x+9∣+x)>0(x^2 - |x + 9| + x) > 0, is
  1. (−∞,−3)∪(3,∞)(-\infty, -3) \cup (3, \infty)
  2. (−9,−3)∪(3,∞)(-9, -3) \cup (3, \infty)
  3. (−∞,−9)∪(9,∞)(-\infty, -9) \cup (9, \infty)
  4. (−∞,−9)∪(3,∞)(-\infty, -9) \cup (3, \infty)
Answer and solution

Answer: (A) (−∞,−3)∪(3,∞)(-\infty, -3) \cup (3, \infty)

Split on the sign of x+9x + 9. Case 1: x≥−9x \ge -9, so ∣x+9∣=x+9|x + 9| = x + 9. The inequality becomes x2−(x+9)+x>0x^2 - (x + 9) + x > 0, that is x2−9>0x^2 - 9 > 0, so x<−3x < -3 or x>3x > 3. With x≥−9x \ge -9 this gives [−9,−3)∪(3,∞)[-9, -3) \cup (3, \infty). (At x=−9x = -9: 81−0−9=72>081 - 0 - 9 = 72 > 0.) Case 2: x<−9x < -9, so ∣x+9∣=−(x+9)|x + 9| = -(x + 9). The inequality becomes x2+2x+9>0x^2 + 2x + 9 > 0. Its discriminant is 4−36=−32<04 - 36 = -32 < 0 and the leading coefficient is positive, so it holds for every xx. Case 2 gives (−∞,−9)(-\infty, -9). Union: (−∞,−9)∪[−9,−3)∪(3,∞)=(−∞,−3)∪(3,∞)(-\infty, -9) \cup [-9, -3) \cup (3, \infty) = (-\infty, -3) \cup (3, \infty). Option B is Case 1 without its endpoint x=−9x = -9: it misses x≤−9x \le -9, where the inequality also holds, e.g. x=−10x = -10 gives 100−1−10=89>0100 - 1 - 10 = 89 > 0. Hence, option A ((−∞,−3)∪(3,∞)(-\infty, -3) \cup (3, \infty)).

Q51MCQPercentages

A certain amount of money was divided among Pinu, Meena, Rinu and Seema. Pinu received 20% of the total amount and Meena received 40% of the remaining amount. If Seema received 20% less than Pinu, the ratio of the amounts received by Pinu and Rinu is
  1. 2 : 1
  2. 5 : 8
  3. 1 : 2
  4. 8 : 5
Answer and solution

Answer: (B) 5 : 8

Take the total amount as 100. Pinu gets 20% of 100 =20= 20, leaving 80. Meena gets 40% of 80 =32= 32, leaving 48 for Rinu and Seema. Seema gets 20% less than Pinu: 20−0.2×20=1620 - 0.2 \times 20 = 16. Rinu gets the rest: 48−16=3248 - 16 = 32. Pinu : Rinu =20:32=5:8= 20 : 32 = 5 : 8. Option D (8 : 5) is the same ratio turned round, Rinu : Pinu; the question asks for Pinu first, and Pinu received less. Hence, option B (5 : 8).

Q52TITATime & Work

Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone, is

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

Answer and solution

Answer: 340

Let Chandan's efficiency be xx units/day. Then Bipin's is 2x2x and Ankita's is 4x4x.

Work in first 20 days (all three): 20×(4x+2x+x)=140x20 \times (4x + 2x + x) = 140x.

Remaining 40 days (Ankita and Chandan): 40×(4x+x)=200x40 \times (4x + x) = 200x.

Total work =340x= 340x.

Time for Chandan alone =340xx=340= \dfrac{340x}{x} = \mathbf{340} days.

Q53MCQSequences & Series

Let ana_n be the nnth term of a decreasing infinite geometric progression. If a1+a2+a3=52a_1 + a_2 + a_3 = 52 and a1a2+a2a3+a3a1=624a_1 a_2 + a_2 a_3 + a_3 a_1 = 624, then the sum of this geometric progression is
  1. 54
  2. 60
  3. 57
  4. 63
Answer and solution

Answer: (A) 54

Let the first term be aa and the common ratio rr. A decreasing infinite GP with a finite sum needs 0<r<10 < r < 1. a1+a2+a3=a(1+r+r2)=52a_1 + a_2 + a_3 = a(1 + r + r^2) = 52 a1a2+a2a3+a3a1=a2r+a2r3+a2r2=a2r(1+r+r2)=624a_1a_2 + a_2a_3 + a_3a_1 = a^2r + a^2r^3 + a^2r^2 = a^2r(1 + r + r^2) = 624 Dividing the second by the first: ar=12ar = 12, so a=12ra = \frac{12}{r}. Then 12r(1+r+r2)=52\frac{12}{r}(1 + r + r^2) = 52, so 12r2−40r+12=012r^2 - 40r + 12 = 0, that is 3r2−10r+3=03r^2 - 10r + 3 = 0, giving r=3r = 3 or r=13r = \frac{1}{3}. Decreasing means r=13r = \frac{1}{3}, so a=36a = 36. The terms are 36, 12, 4, … and S∞=361−13=54S_\infty = \frac{36}{1 - \frac{1}{3}} = 54. Check: the first three terms give 52 and the rest add only 4/31−1/3=2\frac{4/3}{1 - 1/3} = 2, so the sum is just above 52; 57, 60 and 63 are too large. Hence, option A (54).

Q54MCQFunctions & Graphs

Let f(x)=x2x−1f(x) = \dfrac{x}{2x - 1} and g(x)=xx−1g(x) = \dfrac{x}{x - 1}. Then, the domain of the function h(x)=f(g(x))+g(f(x))h(x) = f(g(x)) + g(f(x)) is all real numbers except
  1. 12,1\dfrac{1}{2}, 1 and 32\dfrac{3}{2}
  2. −12,12-\dfrac{1}{2}, \dfrac{1}{2} and 11
  3. −1,12-1, \dfrac{1}{2} and 11
  4. −12,1-\dfrac{1}{2}, 1 and 32\dfrac{3}{2}
Answer and solution

Answer: (C) −1,12-1, \dfrac{1}{2} and 11

f(x)=x2x−1f(x) = \frac{x}{2x - 1} needs x≠12x \ne \frac{1}{2}, and g(x)=xx−1g(x) = \frac{x}{x - 1} needs x≠1x \ne 1. f(g(x))f(g(x)) needs g(x)g(x) defined (x≠1x \ne 1) and g(x)≠12g(x) \ne \frac{1}{2}. Solving xx−1=12\frac{x}{x - 1} = \frac{1}{2} gives 2x=x−12x = x - 1, so x=−1x = -1 must be excluded. g(f(x))g(f(x)) needs f(x)f(x) defined (x≠12x \ne \frac{1}{2}) and f(x)≠1f(x) \ne 1. Solving x2x−1=1\frac{x}{2x - 1} = 1 gives x=1x = 1, already excluded. So the excluded values are −1-1, 12\frac{1}{2} and 11. Option B has −12-\frac{1}{2} in place of −1-1, but x=−12x = -\frac{1}{2} is allowed: g(−12)=13g(-\frac{1}{2}) = \frac{1}{3} and f(−12)=14f(-\frac{1}{2}) = \frac{1}{4}, and neither is a forbidden input for the outer function. Hence, option C (−1,12-1, \dfrac{1}{2} and 11).

Q55MCQSimple & Compound Interest

A loan of Rs 1000 is fully repaid by two installments of Rs 530 and Rs 594, paid at the end of first and second year, respectively. If the interest is compounded annually, then the rate of interest, in percentage, is
  1. 8
  2. 9
  3. 10
  4. 11
Answer and solution

Answer: (A) 8

Let R=1+r100R = 1 + \frac{r}{100}. After one year the debt is 1000R1000R; after paying 530 it is 1000R−5301000R - 530. By the end of year 2 this grows to (1000R−530)R(1000R - 530)R, and the payment of 594 clears it: (1000R−530)R=594(1000R - 530)R = 594, so 1000R2−530R−594=01000R^2 - 530R - 594 = 0, that is 500R2−265R−297=0500R^2 - 265R - 297 = 0. R=265+2652+4⋅500⋅2971000=265+8151000=1.08R = \frac{265 + \sqrt{265^2 + 4 \cdot 500 \cdot 297}}{1000} = \frac{265 + 815}{1000} = 1.08 (the other root is negative). So r=8r = 8. Check: 1000×1.08−530=5501000 \times 1.08 - 530 = 550 and 550×1.08=594550 \times 1.08 = 594. At the next option, 9%, the balance after the first payment would be 1090−530=5601090 - 530 = 560, growing to 560×1.09=610.4560 \times 1.09 = 610.4, more than 594. Hence, option A (8).

Q56TITAPolygons & Circles

Two tangents drawn from a point P touch a circle with center O at points Q and R. Points A and B lie on PQ and PR, respectively, such that AB is also a tangent to the same circle. If ∠AOB=50∘\angle AOB = 50^\circ, then ∠APB\angle APB, in degrees, equals

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

Answer and solution

Answer: 80

Let ∠APB=P\angle APB = P, ∠PAB=A\angle PAB = A and ∠PBA=B\angle PBA = B, so A+B=180∘−PA + B = 180^\circ - P. A lies between P and Q, and B between P and R, so the circle lies beyond AB: it touches AB and the extensions of PA and PB. It is the excircle of △PAB\triangle PAB opposite P, not its incircle. Let AB touch the circle at T. From A the two tangents are AQ and AT, so OA bisects ∠QAT\angle QAT. Ray AQ points away from P, so ∠QAT=180∘−A\angle QAT = 180^\circ - A and ∠OAB=90∘−A2\angle OAB = 90^\circ - \frac{A}{2}. Likewise ∠OBA=90∘−B2\angle OBA = 90^\circ - \frac{B}{2}. In △AOB\triangle AOB: ∠AOB=180∘−(90∘−A2)−(90∘−B2)=A+B2=90∘−P2\angle AOB = 180^\circ - \left(90^\circ - \frac{A}{2}\right) - \left(90^\circ - \frac{B}{2}\right) = \frac{A + B}{2} = 90^\circ - \frac{P}{2}. So 50∘=90∘−P250^\circ = 90^\circ - \frac{P}{2}, which gives P=80∘P = 80^\circ. (Were the circle the incircle, ∠AOB\angle AOB would be 90∘+P290^\circ + \frac{P}{2}, more than 90∘90^\circ, so it could not be 50∘50^\circ.) The answer is 80.

Q57MCQAverages, Mixtures & Alligations

A mixture of coffee and cocoa, 16% of which is coffee, costs Rs 240 per kg. Another mixture of coffee and cocoa, of which 36% is coffee, costs Rs 320 per kg. If a new mixture of coffee and cocoa costs Rs 376 per kg, then the quantity, in kg, of coffee in 10 kg of this new mixture is
  1. 6
  2. 2.5
  3. 4
  4. 5
Answer and solution

Answer: (D) 5

Let coffee cost CC and cocoa KK rupees per kg. First mixture: 0.16C+0.84K=2400.16C + 0.84K = 240. Second mixture: 0.36C+0.64K=3200.36C + 0.64K = 320. Subtracting the first from the second: 0.20C−0.20K=800.20C - 0.20K = 80, so C−K=400C - K = 400. Put C=K+400C = K + 400 into the first: 0.16K+64+0.84K=2400.16K + 64 + 0.84K = 240, so K=176K = 176 and C=576C = 576. In the new mixture let the coffee fraction be ff: 576f+176(1−f)=376576f + 176(1 - f) = 376, so 400f=200400f = 200 and f=0.5f = 0.5. Coffee in 10 kg =0.5×10=5= 0.5 \times 10 = 5 kg. Option A (6 kg) would make the price 0.6×576+0.4×176=4160.6 \times 576 + 0.4 \times 176 = 416 rupees per kg, not 376. Hence, option D (5).

Q58TITAInequalities & Modulus

If a, b, c and d are integers such that their sum is 46, then the minimum possible value of (a−b)2+(a−c)2+(a−d)2(a - b)^2 + (a - c)^2 + (a - d)^2 is

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

Answer and solution

Answer: 2

Given a+b+c+d=46a + b + c + d = 46 with integers a,b,c,da, b, c, d.

To minimize (a−b)2+(a−c)2+(a−d)2(a - b)^2 + (a - c)^2 + (a - d)^2, keep the values as close to each other as possible.

If all four were equal: 4a=46⇒a=11.54a = 46 \Rightarrow a = 11.5, which is not an integer. So perfect equality is impossible.

The closest integer distribution with sum 46 is: 12,12,11,1112, 12, 11, 11 (sum =46= 46).

Take a=12, b=12, c=11, d=11a = 12,\ b = 12,\ c = 11,\ d = 11:

(a−b)2+(a−c)2+(a−d)2=0+1+1=2(a - b)^2 + (a - c)^2 + (a - d)^2 = 0 + 1 + 1 = 2.

S=0S = 0 or 11 would require all four equal or three equal with one off by 1 — both impossible with sum 46. Therefore, the minimum possible value is 2\mathbf{2}.

Q59TITAAverages, Mixtures & Alligations

The average number of copies of a book sold per day by a shopkeeper is 60 in the initial seven days and 63 in the initial eight days, after the book launch. On the ninth day, she sells 11 copies less than the eighth day, and the average number of copies sold per day from second day to ninth day becomes 66. The number of copies sold on the first day of the book launch is

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

Answer and solution

Answer: 49

Total copies in 7 days =60×7=420= 60 \times 7 = 420.

Total copies in 8 days =63×8=504= 63 \times 8 = 504.

Copies on 8th day =504−420=84= 504 - 420 = 84.

Copies on 9th day =84−11=73= 84 - 11 = 73.

Total copies in 9 days =504+73=577= 504 + 73 = 577.

Total copies from day 2 to day 9 =66×8=528= 66 \times 8 = 528.

Copies on day 1 =577−528=49= 577 - 528 = \mathbf{49}.

Q60MCQQuadratic & Polynomial Equations

If 9x2+2x−3−4⋅3x2+2x−2+27=09^{x^2 + 2x - 3} - 4 \cdot 3^{x^2 + 2x - 2} + 27 = 0, then the product of all possible values of x is
  1. 15
  2. 20
  3. 5
  4. 30
Answer and solution

Answer: (B) 20

Let y=3x2+2x−2y = 3^{x^2 + 2x - 2}.

Then 9x2+2x−3=32(x2+2x−3)=19⋅32(x2+2x−2)=y299^{x^2 + 2x - 3} = 3^{2(x^2 + 2x - 3)} = \dfrac{1}{9} \cdot 3^{2(x^2 + 2x - 2)} = \dfrac{y^2}{9}.

The equation becomes: y29−4y+27=0⇒y2−36y+243=0\dfrac{y^2}{9} - 4y + 27 = 0 \Rightarrow y^2 - 36y + 243 = 0.

Factoring: (y−27)(y−9)=0(y - 27)(y - 9) = 0, so y=27y = 27 or y=9y = 9.

If y=27y = 27: 3x2+2x−2=33⇒x2+2x−2=3⇒x2+2x−5=03^{x^2 + 2x - 2} = 3^3 \Rightarrow x^2 + 2x - 2 = 3 \Rightarrow x^2 + 2x - 5 = 0. Product of roots =−5= - 5.

If y=9y = 9: 3x2+2x−2=32⇒x2+2x−2=2⇒x2+2x−4=03^{x^2 + 2x - 2} = 3^2 \Rightarrow x^2 + 2x - 2 = 2 \Rightarrow x^2 + 2x - 4 = 0. Product of roots =−4= - 4.

Product of all possible values of x=(−5)×(−4)=20x = ( - 5) \times ( - 4) = \mathbf{20}.

Q61TITAProperties of Numbers

Suppose a, b, c are three distinct natural numbers, such that 3ac=8(a+b)3ac = 8(a + b). Then, the smallest possible value of 3a+2b+c3a + 2b + c is

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

Answer and solution

Answer: 12

From 3ac=8(a+b)3ac = 8(a + b): 8b=a(3c−8)8b = a(3c - 8), so b=a(3c−8)8b = \frac{a(3c - 8)}{8}. Since b≥1b \ge 1, 3c>83c > 8, so c≥3c \ge 3. a=1a = 1: b=3c−88b = \frac{3c - 8}{8} needs 3c3c to be a multiple of 8, so c≥8c \ge 8. c=8c = 8 gives b=2b = 2 and 3a+2b+c=3+4+8=153a + 2b + c = 3 + 4 + 8 = 15. a=2a = 2: b=3c−84b = \frac{3c - 8}{4}. c=3c = 3 gives b=14b = \frac{1}{4}; c=4c = 4 gives b=1b = 1. So (a,b,c)=(2,1,4)(a, b, c) = (2, 1, 4), all distinct, and 3a+2b+c=6+2+4=123a + 2b + c = 6 + 2 + 4 = 12. Check: 3⋅2⋅4=24=8(2+1)3 \cdot 2 \cdot 4 = 24 = 8(2 + 1). A larger cc raises both bb and cc, so the total only grows. a≥3a \ge 3: 3a≥93a \ge 9, and distinct b,c≥1b, c \ge 1 give 2b+c≥2⋅1+2=42b + c \ge 2 \cdot 1 + 2 = 4, so the total is at least 13. So the smallest value is 12. The answer is 12.

Q62MCQTriangles & Lines

In a △ABC\triangle ABC, points D and E are on the sides BC and AC, respectively. BE and AD intersect at point T such that AD:AT=4:3AD : AT = 4 : 3, and BE:BT=5:4BE : BT = 5 : 4. Point F lies on AC such that DF is parallel to BE. Then, BD:CDBD : CD is
  1. 9 : 4
  2. 7 : 4
  3. 15 : 4
  4. 11 : 4
Answer and solution

Answer: (D) 11 : 4

From AD:AT=4:3AD : AT = 4 : 3, AT:TD=3:1AT : TD = 3 : 1. From BE:BT=5:4BE : BT = 5 : 4, BT:TE=4:1BT : TE = 4 : 1, so BE=5 TEBE = 5\,TE. F is on AC with DF parallel to BE, and TE lies along BE. In △ADF\triangle ADF, TE is parallel to DF with T on AD and E on AF, so TEDF=ATAD=34\frac{TE}{DF} = \frac{AT}{AD} = \frac{3}{4}, giving DF=43TEDF = \frac{4}{3}TE. In △CBE\triangle CBE, D is on CB, F is on CE and DF is parallel to BE, so CDCB=DFBE=43TE5 TE=415\frac{CD}{CB} = \frac{DF}{BE} = \frac{\frac{4}{3}TE}{5\,TE} = \frac{4}{15}. So CD=415BCCD = \frac{4}{15}BC and BD=1115BCBD = \frac{11}{15}BC, giving BD:CD=11:4BD : CD = 11 : 4. Option C (15 : 4) is BC:CDBC : CD; it uses the whole side BC in place of BD. Hence, option D (11 : 4).

Q63MCQProperties of Numbers

The number of divisors of (26×35×53×72)(2^6 \times 3^5 \times 5^3 \times 7^2), which are of the form (3r+1)(3r + 1), where r is a non-negative integer, is
  1. 36
  2. 56
  3. 24
  4. 42
Answer and solution

Answer: (D) 42

A divisor has the form 2a⋅3b⋅5c⋅7e2^a \cdot 3^b \cdot 5^c \cdot 7^e with 0≤a≤60 \le a \le 6, 0≤b≤50 \le b \le 5, 0≤c≤30 \le c \le 3, 0≤e≤20 \le e \le 2. To leave remainder 1 on division by 3 it cannot contain the factor 3, so b=0b = 0. Modulo 3, 2≡−12 \equiv -1, 5≡−15 \equiv -1 and 7≡17 \equiv 1, so the divisor is ≡(−1)a+c\equiv (-1)^{a + c}. It is ≡1\equiv 1 exactly when a+ca + c is even. aa has 4 even values (0, 2, 4, 6) and 3 odd; cc has 2 even (0, 2) and 2 odd. Pairs with a+ca + c even: both even 4×2=84 \times 2 = 8, both odd 3×2=63 \times 2 = 6, total 14. ee is free: 3 choices. Count =14×3=42= 14 \times 3 = 42. Check: the 7×4×3=847 \times 4 \times 3 = 84 divisors not divisible by 3 split evenly, 42 of the form 3r+13r + 1 and 42 of the form 3r+23r + 2, so 36, 24 and 56 cannot be right. Hence, option D (42).

Q64TITATime, Speed & Distance

Rita and Sneha can row a boat at 5 km/h and 6 km/h in still water, respectively. In a river flowing with a constant velocity, Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream. If Rita starts from a certain location in the river, and returns downstream to the same location, taking a total of 100 minutes, then the total distance, in km, Rita will cover is

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

Answer and solution

Answer: 8

Let the river flow at vv km/h. Sneha rows at 6 km/h in still water, so her upstream and downstream speeds are 6−v6 - v and 6+v6 + v. 48 minutes =45= \frac{4}{5} h, so 146−v−146+v=45\frac{14}{6 - v} - \frac{14}{6 + v} = \frac{4}{5}. The left side is 14⋅2v36−v2=28v36−v2\frac{14 \cdot 2v}{36 - v^2} = \frac{28v}{36 - v^2}, so 140v=144−4v2140v = 144 - 4v^2, that is v2+35v−36=0v^2 + 35v - 36 = 0, or (v+36)(v−1)=0(v + 36)(v - 1) = 0. Hence v=1v = 1 km/h. Rita rows at 5 km/h in still water: 4 km/h upstream and 6 km/h downstream. If she goes xx km upstream and comes back, 100 minutes =53= \frac{5}{3} h gives x4+x6=5x12=53\frac{x}{4} + \frac{x}{6} = \frac{5x}{12} = \frac{5}{3}, so x=4x = 4. Total distance =2x=8= 2x = 8 km. The answer is 8.

Q65MCQProfit, Loss & Discount

An item with a cost price of Rs 1650 is sold at a certain discount on a fixed marked price to earn a profit of 20% on the cost price. If the discount was doubled, the profit would have been Rs 110. The rate of discount, in percentage, at which the profit percentage would be equal to the rate of discount, is nearest to
  1. 18
  2. 12
  3. 16
  4. 14
Answer and solution

Answer: (D) 14

CP =1650= 1650. A 20% profit means SP =1980= 1980. With marked price MM and discount dd rupees: M−d=1980M - d = 1980. With the discount doubled the profit is 110, so SP =1760= 1760: M−2d=1760M - 2d = 1760. Subtracting: d=220d = 220, so M=2200M = 2200. Now let the selling price be xx, with discount % equal to profit %: 2200−x2200=x−16501650\frac{2200 - x}{2200} = \frac{x - 1650}{1650}. Multiplying by 6600: 3(2200−x)=4(x−1650)3(2200 - x) = 4(x - 1650), so 7x=132007x = 13200 and x=132007≈1885.71x = \frac{13200}{7} \approx 1885.71. Discount % =2200−1885.712200×100=1007≈14.29= \frac{2200 - 1885.71}{2200} \times 100 = \frac{100}{7} \approx 14.29, and profit % =235.711650×100≈14.29= \frac{235.71}{1650} \times 100 \approx 14.29 as well. The nearest option is 14; the next closest, 12 and 16, are each about 2 away. Hence, option D (14).

Q66TITAIndices & Surds

The sum of digits of the number (625)65×(128)36(625)^{65} \times (128)^{36}, is

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

Answer and solution

Answer: 25

(625)65=(54)65=5260(625)^{65} = (5^4)^{65} = 5^{260}.

(128)36=(27)36=2252(128)^{36} = (2^7)^{36} = 2^{252}.

Product =2252×5260=2252×5252×58=(2×5)252×58=10252×390625= 2^{252} \times 5^{260} = 2^{252} \times 5^{252} \times 5^8 = (2 \times 5)^{252} \times 5^8 = 10^{252} \times 390625.

This is the number 390625 followed by 252 trailing zeros.

Sum of digits =3+9+0+6+2+5=25= 3 + 9 + 0 + 6 + 2 + 5 = \mathbf{25}.

Q67MCQRatios, Proportions & Partnership

The ratio of expenditures of Lakshmi and Meenakshi is 2 : 3, and the ratio of income of Lakshmi to expenditure of Meenakshi is 6 : 7. If excess of income over expenditure is saved by Lakshmi and Meenakshi, and the ratio of their savings is 4 : 9, then the ratio of their incomes is
  1. 3 : 5
  2. 5 : 6
  3. 7 : 8
  4. 2 : 1
Answer and solution

Answer: (A) 3 : 5

Let the expenditures be EL=2xE_L = 2x and EM=3xE_M = 3x. Lakshmi's income : Meenakshi's expenditure =6:7= 6 : 7, so IL=67×3x=18x7I_L = \frac{6}{7} \times 3x = \frac{18x}{7}. Lakshmi saves SL=18x7−2x=4x7S_L = \frac{18x}{7} - 2x = \frac{4x}{7}. Savings are in the ratio 4:94 : 9, so SM=94×4x7=9x7S_M = \frac{9}{4} \times \frac{4x}{7} = \frac{9x}{7}. Meenakshi's income is IM=3x+9x7=30x7I_M = 3x + \frac{9x}{7} = \frac{30x}{7}. IL:IM=18:30=3:5I_L : I_M = 18 : 30 = 3 : 5. Option B (5 : 6) would give Meenakshi an income of 65×18x7=21.6x7\frac{6}{5} \times \frac{18x}{7} = \frac{21.6x}{7} and savings of only 0.6x7\frac{0.6x}{7}, not 9x7\frac{9x}{7}. Hence, option A (3 : 5).

Q68TITAProperties of Numbers

If m and n are integers such that (m+2n)(2m+n)=27(m + 2n)(2m + n) = 27, then the maximum possible value of 2m−3n2m - 3n is

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

Answer and solution

Answer: 17

Let x=m+2nx = m + 2n and y=2m+ny = 2m + n, so xy=27xy = 27.

From these: m=2y−x3m = \dfrac{2y - x}{3} and n=2x−y3n = \dfrac{2x - y}{3} (integers require x+y≡0(mod3)x + y \equiv 0 \pmod{3}).

Integer factor pairs of 27 with x+y≡0(mod3)x + y \equiv 0 \pmod{3}: (3,9), (9,3), (−3,−9), (−9,−3)(3,9),\ (9,3),\ ( - 3, - 9),\ ( - 9, - 3).

Computing 2m−3n2m - 3n:

(3,9)⇒m=5,n=−1⇒2(5)−3(−1)=13(3,9) \Rightarrow m = 5, n = - 1 \Rightarrow 2(5) - 3( - 1) = 13

(9,3)⇒m=−1,n=5⇒2(−1)−3(5)=−17(9,3) \Rightarrow m = - 1, n = 5 \Rightarrow 2( - 1) - 3(5) = - 17

(−3,−9)⇒m=−5,n=1⇒2(−5)−3(1)=−13( - 3, - 9) \Rightarrow m = - 5, n = 1 \Rightarrow 2( - 5) - 3(1) = - 13

(−9,−3)⇒m=1,n=−5⇒2(1)−3(−5)=17( - 9, - 3) \Rightarrow m = 1, n = - 5 \Rightarrow 2(1) - 3( - 5) = 17

Maximum =17= \mathbf{17}.