◆ QA · Web-sourced PYQs

More CAT PYQs, web-sourced

Genuine CAT Quantitative Aptitude questions (2017-2019, all slots) that were not already in the rest of Percentyle. Each one was checked against every QA chapter page, the year papers and the sample papers, only the genuinely-missing ones are here. Every numeric/MCQ answer below was independently re-worked and confirmed; anything that couldn't be reproduced faithfully from the source was left out.

28new CAT PYQs

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What this page is. These are additional, web-sourced, independently-verified real CAT questions, gathered from the publicly published slot-wise CAT papers on 2IIM / Cracku and de-duplicated against the existing Percentyle. They are not coaching/practice questions, every item is a genuine CAT exam question from the year and slot shown. Percentyle's chapter pages already cover the overwhelming majority of CAT 2017-2025 QA; this page just picks up the real questions that weren't there yet.

Web-sourced CAT Previous-Year Questions

Real CAT questions, all verified. Difficulty: Easy Moderate Hard. Click any to reveal the worked solution.

CAT 2017

ModerateCAT 2017 · Slot 1

Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest is invested in stocks and the rest goes into his savings bank account. If the total amount he deposits in the bank (savings account plus fixed deposits) is Rs 59,500, then his total monthly savings (in Rs) is TITA

Show solution

70,000. Let savings = S. FD = 0.5S. Of the remaining 0.5S, stocks = 30% = 0.15S and savings account = 70% = 0.35S. Bank total = FD + savings account = 0.5S + 0.35S = 0.85S = 59,500 ⇒ S = 59,500/0.85 = 70,000.

HardCAT 2017 · Slot 1

A man travels by motor boat down a river to his office and back. With the river speed unchanged, if he doubles the speed of his motor boat, his total travel time reduces by 75%. The ratio of the original speed of the motor boat to the speed of the river is

(A) √6 : √2
(B) √7 : 2
(C) 2√5 : 3
(D) 3 : 2

Show solution

(B) √7 : 2. Let boat = b, river = r, each leg = d. T = d/(b+r) + d/(b−r). With boat 2b: T′ = d/(2b+r) + d/(2b−r) = T/4. Solving 1/(2b+r) + 1/(2b−r) = ¼[1/(b+r) + 1/(b−r)] gives 4b²−r² and b²−r² in the ratio that reduces to b/r = √7/2, i.e. √7 : 2.

ModerateCAT 2017 · Slot 1

A stall sells popcorn and chips in three sizes, large, super and jumbo. The numbers of large, super and jumbo packets are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets equals the total number of chips packets, then the numbers of jumbo popcorn and jumbo chips packets are in the ratio

(A) 1 : 1
(B) 8 : 7
(C) 4 : 3
(D) 6 : 5

Show solution

(A) 1 : 1. Popcorn total = (7+17+16)p = 40p; chips total = (6+15+14)c = 35c. Equal ⇒ 40p = 35c ⇒ p : c = 7 : 8. Jumbo popcorn = 16p, jumbo chips = 14c. Ratio = 16p : 14c = 16·7 : 14·8 = 112 : 112 = 1 : 1.

HardCAT 2017 · Slot 1

If Fatima sells 60 identical toys at a 40% discount on the printed price, she makes a 20% profit. Ten of these toys are destroyed in a fire. While selling the rest, what discount on the printed price should she give so that she makes the same amount of profit?

(A) 30%
(B) 25%
(C) 24%
(D) 28%

Show solution

(D) 28%. Let printed price = P. SP at 40% off = 0.6P. Revenue from 60 = 36P = 1.2·(cost of 60) ⇒ cost of 60 = 30P, so total cost = 30P and profit amount = 6P. She still paid for all 60 toys, so cost stays 30P. To earn profit 6P from 50 toys: revenue = 36P ⇒ SP = 36P/50 = 0.72P ⇒ discount = 28%.

ModerateCAT 2017 · Slot 1

In how many ways can 7 identical erasers be distributed among 4 kids so that each kid gets at least one eraser but nobody gets more than 3 erasers?

(A) 16
(B) 20
(C) 14
(D) 15

Show solution

(A) 16. Give each kid 1 first; distribute the remaining 3 with each ≤ 2. Non-negative solutions of y₁+y₂+y₃+y₄ = 3 = C(6,3) = 20. Subtract cases where some yᵢ ≥ 3 (i.e. yᵢ = 3): 4 such. 20 − 4 = 16.

ModerateCAT 2017 · Slot 2

Bottle 1 has milk and water in the ratio 7 : 2 and Bottle 2 has milk and water in the ratio 9 : 4. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to get a mixture of milk and water in the ratio 3 : 1?

(A) 27 : 14
(B) 27 : 13
(C) 27 : 16
(D) 27 : 18

Show solution

(B) 27 : 13. Milk fractions: 7/9 and 9/13; target 3/4. By alligation, ratio = (9/13 − 3/4) : (3/4 − 7/9) = (36−39)/52 : (27−28)/36, take magnitudes ⇒ (3/52) : (1/36). Cross-multiplying gives 27 : 13.

ModerateCAT 2017 · Slot 2

Arun drove from home to his hostel at 60 mph. Returning, he drove half the route at 25 mph, then took a bypass that increased the distance by 5 miles but let him drive at 50 mph. If the return took 30 minutes more than the onward trip, the total distance he travelled is

(A) 55 miles
(B) 60 miles
(C) 65 miles
(D) 70 miles

Show solution

(C) 65 miles. Onward distance D, time D/60. Return = (D/2)/25 + (D/2 + 5)/50, which is ½ hr more than D/60. Solving gives D = 30. Onward = 30, return = 35, total = 65 miles.

ModerateCAT 2017 · Slot 2

Three positive integers a, b, c are such that a : b = 3 : 4 and b : c = 2 : 1. Which of the following is a possible value of a + b + c?

(A) 201
(B) 205
(C) 207
(D) 210

Show solution

(C) 207. Make b common: a : b = 3 : 4, b : c = 4 : 2 ⇒ a : b : c = 3 : 4 : 2, sum = 9 parts. a + b + c must be a multiple of 9. 207 = 9 × 23 ✓ (others aren't). So 207.

HardCAT 2017 · Slot 2

A motorbike leaves A at 1 pm towards B at uniform speed. A car leaves B at 2 pm towards A at uniform speed double that of the motorbike. They meet at 3:40 pm at a point 168 km from A. The distance, in km, between A and B is

(A) 364
(B) 378
(C) 380
(D) 388

Show solution

(B) 378. Bike travels 1:00→3:40 = 8/3 hr to cover 168 km ⇒ speed = 63 kmph. Car travels 2:00→3:40 = 5/3 hr at 126 kmph ⇒ 210 km. AB = 168 + 210 = 378 km.

ModerateCAT 2017 · Slot 2

Three mixtures, the first with water and liquid A in ratio 1 : 2, the second with water and B in 1 : 3, the third with water and C in 1 : 4, are mixed in the proportion 4 : 3 : 2. The resulting mixture has

(A) the same amount of water and liquid B
(B) the same amount of liquids B and C
(C) more water than liquid B
(D) more water than liquid A

Show solution

(C) more water than liquid B. Water = 4(1/3) + 3(1/4) + 2(1/5) = 4/3 + 3/4 + 2/5 ≈ 2.483. A = 4(2/3) ≈ 2.667, B = 3(3/4) = 2.25, C = 2(4/5) = 1.6. Water (2.483) > B (2.25), so more water than liquid B.

HardCAT 2017 · Slot 2

A vertical pillar has uniform cross section in the shape of a trapezium whose parallel sides are 10 cm and 20 cm, with the other two sides equal and the perpendicular distance between the parallel sides 12 cm. If the height of the pillar is 20 cm, the total area, in sq cm, of all six surfaces is

(A) 1300
(B) 1340
(C) 1480
(D) 1520

Show solution

(C) 1480. Trapezium area = ½(10+20)·12 = 180; two ends = 360. Slant legs = √(5²+12²) = 13. Perimeter = 10+20+13+13 = 56; lateral = 56·20 = 1120. Total = 360 + 1120 = 1480.

ModerateCAT 2017 · Slot 2

If three sides of a rectangular park have total length 400 ft, then the area of the park is maximum when the length, in ft, of its longer side is TITA

Show solution

200. Let the single long side = x and the two equal sides = y, so x + 2y = 400. Area = xy = x(400−x)/2, maximised at x = 200 (vertex of the parabola). Longer side = 200 ft.

EasyCAT 2017 · Slot 2

In how many ways can 8 identical pens be distributed among Amal, Bimal and Kamal so that Amal gets at least 1, Bimal at least 2 and Kamal at least 3 pens? TITA

Show solution

6. Pre-assign 1 + 2 + 3 = 6 pens, leaving 2 to distribute freely among 3 people: C(2+2, 2) = C(4,2) = 6.

CAT 2018

ModerateCAT 2018 · Slot 1

A trader sells 10 litres of a mixture of paints A and B, where the amount of B does not exceed that of A. Paint A costs Rs 8 more per litre than paint B. If he sells the whole mixture for Rs 264 at a profit of 10%, then the highest possible cost of paint B, in Rs per litre, is

(A) 20
(B) 16
(C) 22
(D) 26

Show solution

(A) 20. Total cost = 264/1.1 = 240. Let B = b litres of paint B (b ≤ 5) at price p, paint A = (10−b) litres at p+8. Cost = 10p + 8(10−b) = 240 ⇒ p = (160+8b)/10. p is largest at the largest allowed b = 5 ⇒ p = 200/10 = 20.

HardCAT 2018 · Slot 1

A wholesaler bought walnuts and peanuts, the walnut price per kg being thrice the peanut price. He sold 8 kg peanuts at 10% profit and 16 kg walnuts at 20% profit to a shopkeeper. The shopkeeper lost 5 kg walnuts and 3 kg peanuts in transit, then mixed the rest and sold the mixture at Rs 166/kg, making 25% overall profit. At what price, in Rs per kg, did the wholesaler buy the walnuts?

(A) 84
(B) 86
(C) 96
(D) 98

Show solution

(C) 96. Let peanut cost = p, walnut = 3p. Shopkeeper's cost = 8(1.1p) + 16(1.2·3p) = 8.8p + 57.6p = 66.4p. He sells 16 kg (11 walnut + 5 peanut) at Rs 166 ⇒ revenue = 2656 = 1.25 × 66.4p ⇒ p = 32. Walnut buy price = 3p = 96.

ModerateCAT 2018 · Slot 1

If u² + (u − 2v − 1)² = −4v(u + v), then the value of u + 3v is

(A) 1/4
(B) 1/2
(C) 0
(D) −1/4

Show solution

(D) −1/4. Moving everything to one side and expanding gives 2u² − 2u + 8v² + 4v + 1 = 0, i.e. 2(u − ½)² + 8(v + ¼)² = 0. Both squares must vanish ⇒ u = ½, v = −¼ ⇒ u + 3v = ½ − ¾ = −1/4.

HardCAT 2018 · Slot 2

Two drums of paint: Drum 1 has A and B in ratio 18 : 7. Liquids from Drum 1 and Drum 2 are mixed in ratio 3 : 4, giving a final mixture with A : B = 13 : 7. The ratio of A to B in Drum 2 is

(A) 251 : 163
(B) 239 : 161
(C) 220 : 149
(D) 229 : 141

Show solution

(B) 239 : 161. A-fraction in Drum 1 = 18/25; final A-fraction = 13/20. Let Drum 2 A-fraction = x: (3·18/25 + 4x)/7 = 13/20 ⇒ x = 239/400, so B-fraction = 161/400 ⇒ A : B = 239 : 161.

ModerateCAT 2018 · Slot 2

A 20% ethanol solution is mixed with another ethanol solution S in the ratio 1 : 3. This mixture is then mixed with an equal volume of the 20% solution to give a 31.25% solution. The strength of S is

(A) 50%
(B) 55%
(C) 48%
(D) 52%

Show solution

(A) 50%. First mixture M = (1·20 + 3·S)/4. Mixing M with an equal volume of 20%: (M + 20)/2 = 31.25 ⇒ M = 42.5 ⇒ (20 + 3S)/4 = 42.5 ⇒ 3S = 150 ⇒ S = 50%.

ModerateCAT 2018 · Slot 2

The ratio of the area of a rectangle to the square of its perimeter is 1 : 25. The ratio of the shorter to the longer side of the rectangle is

(A) 3 : 8
(B) 2 : 9
(C) 1 : 4
(D) 1 : 3

Show solution

(C) 1 : 4. lb / (2(l+b))² = 1/25 ⇒ 25lb = 4(l+b)². With r = l/b: 4r² − 17r + 4 = 0 ⇒ r = 4 or ¼. So shorter : longer = 1 : 4.

ModerateCAT 2018 · Slot 2

How many two-digit numbers with a non-zero units digit are more than thrice the number obtained by reversing their digits?

(A) 5
(B) 8
(C) 7
(D) 6

Show solution

(D) 6. Need 10a + b > 3(10b + a), i.e. 7a > 29b. Checking b = 1, 2: the qualifying numbers are 51, 61, 71, 81, 91, 92, exactly 6 of them.

ModerateCAT 2018 · Slot 2

If p³ = q⁴ = r⁵ = s⁶, then log_s(pqr) equals

(A) 24/5
(B) 1
(C) 47/10
(D) 16/5

Show solution

(C) 47/10. Let the common value be k. Then p = k^1/3, q = k^1/4, r = k^1/5, s = k^1/6. pqr = k^{(1/3+1/4+1/5)} = k^47/60. log_s(pqr) = (47/60)/(1/6) = 47/10.

CAT 2019

ModerateCAT 2019 · Slot 1

If (5.55)^x = (0.555)^y = 1000, then the value of 1/x − 1/y is

(A) 1
(B) 1/3
(C) 2/3
(D) 3

Show solution

(B) 1/3. 1/x = log₁₀₀₀5.55 and 1/y = log₁₀₀₀0.555. 1/x − 1/y = log₁₀₀₀(5.55/0.555) = log₁₀₀₀10 = 1/3 (since 1000 = 10³). So 1/3.

ModerateCAT 2019 · Slot 1

At their usual efficiencies, A and B together finish a task in 12 days. If A had worked at half her usual efficiency and B at thrice his usual efficiency, the task would finish in 9 days. How many days would A take alone at her usual efficiency?

(A) 36
(B) 24
(C) 18
(D) 12

Show solution

(C) 18. Let A, B do a, b per day. a + b = 1/12 and ½a + 3b = 1/9. Solving: a = 1/18, b = 1/36. A alone needs 1/a = 18 days.

ModerateCAT 2019 · Slot 1

A chemist mixes two liquids, 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half a litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, by volume, is

(A) 70
(B) 85
(C) 80
(D) 75

Show solution

(C) 80. Half a litre weighs 480 gm ⇒ 1 litre weighs 960 gm. If fraction v is liquid 1: 1000v + 800(1−v) = 960 ⇒ 200v = 160 ⇒ v = 0.8 = 80%.

ModerateCAT 2019 · Slot 2

In 2010 a library had 11,500 books (fiction and non-fiction). By 2015 it had 12,760 books, the fiction having increased by 10% and the non-fiction by 12%. How many fiction books did it have in 2015?

(A) 6600
(B) 6160
(C) 6000
(D) 5500

Show solution

(A) 6600. f + n = 11,500 and 1.1f + 1.12n = 12,760. Subtracting 1.1×(first): 0.02n = 12,760 − 12,650 = 110 ⇒ n = 5,500, f = 6,000. Fiction in 2015 = 1.1 × 6,000 = 6,600.

HardCAT 2019 · Slot 2

Vessels A, B and C each contain 500 ml of solution of strengths 10%, 22% and 32% respectively. 100 ml is transferred from A to B, then 100 ml from B to C, then 100 ml from C to A. The strength, as a percentage, of the resulting solution in A is

(A) 15
(B) 12
(C) 13
(D) 14

Show solution

(D) 14. A→B: A loses 10 units solute (now 40 in 400 ml); B has 110+10 = 120 in 600. B→C: B's conc = 20%, send 20 units; C now 160+20 = 180 in 600 ⇒ 30%. C→A: send 30 units to A. A now 40 + 30 = 70 solute in 400 + 100 = 500 ml ⇒ 70/500 = 14%.

HardCAT 2019 · Slot 2

If 5^x − 3^y = 13438 and 5^x−1 + 3^y+1 = 9686, then x + y equals TITA

Show solution

13. Try powers: 5⁶ = 15625, 15625 − 13438 = 2187 = 3⁷, so x = 6, y = 7. Check: 5⁵ + 3⁸ = 3125 + 6561 = 9686 ✓. So x + y = 6 + 7 = 13.

ModerateCAT 2019 · Slot 2

Ramesh, Ganesh and Rajesh had salaries in the ratio 6 : 5 : 7 in 2010, and 3 : 4 : 3 in 2015. If Ramesh's salary rose by 25%, the percentage increase in Rajesh's salary is closest to

(A) 7
(B) 8
(C) 9
(D) 10

Show solution

(A) 7. 2010: 6k, 5k, 7k; 2015: 3m, 4m, 3m. Ramesh: 3m = 1.25·6k ⇒ m = 2.5k. Rajesh: 2010 = 7k, 2015 = 3m = 7.5k. Increase = 0.5k/7k ≈ 7.1% ⇒ closest to 7.