CAT 2025 QA Slot 2 Question Paper With Detailed PDF Solutions

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CAT 2025 QA Slot 2 Paper With Answers & Explanation

Q. 1 In a ΔABC, points D and E are on the sides BC and AC, respectively. BE and AD intersect at point T such that AD : AT = 4 : 3, and BE : BT = 5 : 4. Point F lies on AC such that DF is parallel to BE. Then, BD : CD is

Explanation
Correct Answer

4

Explanation

Q. 2

domain of the function h(x) = f(g(x)) + g(f(x)) is all
real numbers except

Explanation
Correct Answer

3

Explanation

Q. 3 The set of all real values of x for which (x2 – |x + 9| + x) > 0, is

Explanation
Correct Answer

3

Explanation

x2 – |x + 9| + x > 0
For x = 4 the expression becomes:
(4)2 – |4 + 9| + 4 = 16 – 13 + 4 = 8 > 0
So option (4) is discarded
For x = –10; (–10)2 – | –10 + 9| + (–10)
100 – 1 > 0
∴ Option (1) is discarded.
Now x = –4; (–4)2 – |– 4 + 9| + (–5)
16 – 5 – 5 > 0
∴ Option (2) is discarded.
Here answer is (-∞, -3) ∪ (3, ∞).
Hence, correct answer is option (3).

Q. 4 If 9x2 + 2x-3 -4(3x2+2x-2 ) + 27 = 0, then the product of all possible values of x is

Explanation
Correct Answer

3

Explanation

Q. 5 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

Explanation
Correct Answer

340

Explanation

Let Chandan’s efficiency = 1 unit/day.
Then Bipin’s efficiency = 2 units/day (twice Chandan)
Ankita’s efficiency = 4 units/day (twice Bipin)
So, combined efficiency of all three = 1 + 2 + 4 = 7 units/day
Work done in the first 20 days when all three are working = 20 × 7 = 140 units
After Bipin leaves, remaining workers are Ankita (4 units/day) and Chandan (1 unit/day)
Work done in next 40 days = 40 × 5 = 200 units Total work = 140 + 200 = 340 units Chandan’s efficiency is 1 unit per day.
Time taken by Chandan alone = 340/1 = 340 days.

Q. 6 If log64 x2 + log8 √y + 3 log512(√y z) = 4, where x, y and z are positive real numbers, then the minimum possible value of (x + y + z) is

Explanation
Correct Answer

3

Explanation

Q. 7 The number of divisors of (26 × 35 × 53 × 72), which are of the form (3r + 1), where r is a non-negative integer, is

Explanation
Correct Answer

3

Explanation

A number is of the form 3p + 1 if it leaves remainder 1 when divided by 3.
Since 35 is divisible by 3, any divisor containing 35 will be divisible by 3. Therefore, we consider only the case when the power of 3 is zero.
Even powers of 2: 20, 22, 24, 26 give remainder 1.
Even powers of 5: 50, 52 also give remainder 1.
So when both powers are even, the remainder is 1
So number of cases = 4 × 2 = 8
Odd power of 2: 21, 23, 25
Odd power of 5: 51, 53
So number of cases = 3 × 2 = 6
Total valid combinations of powers of 2 and 5 are 8 + 6 = 14
Now, powers of 7: 70, 71, 72 all are of the form (3p + 1).
So number of cases = 3
Hence, the number of divisors of the required form is 14 × 3 = 42.

Q. 8 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

Explanation
Correct Answer

49

Explanation

Total sales in first 7 days = 7 × 60 = 420
Total sales in first 8 days = 8 × 63 = 504
So, sales on the 8th day = 504 – 420 = 84
Sales on 9th day = 11 less than 8th day
⇒ 9th day sales = 84 – 11 = 73
From 2nd day to 9th day = 8 days ; Average = 66
⇒ Total sales from day 2 to day 9 = 8 × 66 = 528
Sales from day 2 to day 9 = (Total of first 8 days + 9th day) – 1st day
⇒ 528 = (504 + 73) – 1st day
⇒ 1st day = 577 – 528 = 49.

Q. 9 Suppose a, b, c are three distinct natural numbers, such that 3ac = 8(a + b). Then, the smallest possible value of 3a + 2b + c is

Explanation
Correct Answer

12

Explanation

Given, 3ac = 8(a + b), where a, b, c are distinct natural numbers, and we must minimize 3a + 2b + c.
Rearranging the given equation we get, 3ac = 8a + 8b ⇒ b = a(3c – 8) /8
For b to be a natural number, a(3c – 8) must be divisible by 8.
For a = 2 and c = 4 we get b = 1 and these are the smallest possible distinct values of a, b and c.
Also all are natural and distinct: (a, b, c) = (2, 1, 4)
Hence, 3a + 2b + c = 6 + 2 + 4 = 12.

Q. 10 Two tangents drawn from a point P touch a circle with centre 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°, then ∠APB, in degrees, equals

Explanation
Correct Answer

80

Explanation


∠AOB = 50°
∠AOQ = AOS = a
∠BOR = ∠BOS = b
∠QOR = ∠QOA + ∠AOS + ∠SOB + ∠BOR
= 2a + 2b
= 2 (a + b)
Now as given:
∠AOB = 50° = ∠AOS + ∠SOB = 2(a + b)
∴ QOR = 100°
In quadrilateral PQOR, Q and R are right angles.
Hence, ∠APB = 180° – 100° = 80°.

Q. 11 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

Explanation
Correct Answer

3

Explanation

The diagram of the hexagon can be drawn as follows:

In a regular hexagon area of each of the triangles AOB, BOC, COD, DOE, EOF and FOA is equal to one-sixth of the area of the hexagon.
Given that P and Q are midpoints of AB and CD, using midpoint theorem, we can say that the area enclosed by the trapezium will be equal to = 1/4 area of AOB + 3/4 area of BOC + 1/4 area of COD = 5/4 area of AOB.
Hence, required ratio = 5/4 : 6 = 5 : 24.

Q. 12 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

Explanation
Correct Answer

3

Explanation

Let P, R, M, and S represent the amount of Pinu,
Meena, Rinu and Seema, respectively.
Let P + R + M + S = 100

Q. 13 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

Explanation
Correct Answer

2

Explanation

Let the cost of 1 kg coffee be Rs.C and that of 1 kg cocoa be Rs.D.
From mixture 1: 0.16C + 0.84D = 240 …(i)
From mixture 2: 0.36C + 0.64D = 320 …(ii)
Subtract (i) from (ii) we get, 0.20C – 0.20D = 80
⇒ C – D = 400
From (1): 0.16C + 0.84(C – 400) = 240
⇒ C = Rs.576 and D = Rs.176
Let the fraction of coffee be x.
⇒ 576x + 176(1 – x) = 376 ⇒ x = 0.5
So, coffee content = 50%
Quantity of coffee in 10 kg = 10 × 0.5 = 5 kg.

Q. 14 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

Explanation
Correct Answer

2

Explanation

Let the annual rate of interest = r%
Since interest is compounded annually, the total value of the installments equals the loan amount.
530/(1 + r) + 594/(1 + r)2 = 1000
⇒ 530(1 + r) + 594 = 1000(1 + r2 + 2r)
⇒ 530r + 1124 = 1000 + 1000r2 + 2000r
⇒ 1000r2 + 1470r – 124 = 0
⇒ r = 0.08 or –1.55
We discard the negative value and take r = 0.08
Hence, rate = 8%

Q. 15 If m and n are integers such that (m + 2n)(2m + n) = 27, then the maximum possible value of 2m – 3n is

Explanation
Correct Answer

17

Explanation

Given (m + 2n)(2m + n) = 27 where m and n are integers.
Let (m + 2n) = a and (2m + n) = b
So n = (2a – b)/3 and m = (2b – a)/3
Also we know that, ab = 27
Possible values of (a, b) are {(1, 27), (3, 9), (9, 3), (27, 1), (–1, –27), (–3, –9), (–9, –3), (–27, –1)}
The only pairs that give integer values for m and n are (3, 9), (9, 3), (–3, –9) and (–9, –3).
The values of (2m – 3n) that are obtained from the respective pairs of (a, b) are 13, –17, –13 and 17.
Hence, the maximum possible value of (2m – 3n) is 17.

Q. 16 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

Explanation
Correct Answer

8

Explanation

Q. 17 Let an be the nth term of a decreasing infinite geometric progression. If a1 + a2 + a3 = 52 and a1a2 + a2a3 + a3a1 = 624, then the sum of this geometric progression is

Explanation
Correct Answer

4

Explanation

Q. 18 If a, b, c and d are integers such that their sum is 46, then minimum possible value of (a – b)2 + (a – c)2 + (a – d)2 is

Explanation
Correct Answer

2

Explanation

a + b + c + d = 46
To find the minimum possible value of
(a – b)2 + (a – c)2 + (a – d)2
If all are equal, minimum value will be 0. But 46 is
not divisible by 4.
So take a = 11, b = 11, c = 12, d = 12
Hence, the minimum possible value = 0 + 1 + 1 = 2.

Q. 19 The sum of digits of the number (625)65 × (128)36, is

Explanation
Correct Answer

25

Explanation

N = (625)65 × (128)36
= (54)65 × (27)36
= 5260 × 2252
= (10)252 × 58
= (390625) × 10252
Hence, the sum of digits = 3 + 9 + 0 + 6 + 2 + 5 = 25.

Q. 20 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

Explanation
Correct Answer

4

Explanation

Given Cost Price = Rs.1,650 and Profit = 20% of 1650 = Rs.330
Selling Price = 1650 + 330 = Rs.1,980
Let the original discount be d%.
1980 = MP(1 – d/100) …(i)
According to the question, if the discount is doubled (i.e. 2d%), profit becomes Rs.110.
New Selling Price = 1650 + 110 = Rs.1,760
1760 = MP(1 – 2d/100) …(ii)
From equation (i) and (ii) we get, 1760/1980 = (100 – 2d)/(100 – d)
⇒ 800 – 8d = 900 – 18d
⇒ 10d = 100 ⇒ d = 10%
From equation (i), we get, MP = 1980/0.9
= Rs. 2,200
Let new discount = x%
Now the profit percentage has to be equal to the
discount percentage.
Selling price at x% discount = 2200(1 – x/100)
Profit percentage = (SP – 1650)/1650 × 100 = (2200 – 22x – 1650)/1650 × 100
⇒ (550 – 22x)/1650 × 100 = x ⇒ 165x = 5500 – 220x ⇒ 385x = 5500
⇒ x = 5500/385 = 14.286%
Hence, the correct answer is 14%.

Q. 21 The equations 3x2 – 5x + p = 0 and 2x2 – 2x + q = 0 have one common root. The sum of the other roots of these two equations is

Explanation
Correct Answer

1

Explanation

Q. 22 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

Explanation
Correct Answer

2

Explanation

Given: Expenditure ratio of Lakshmi : Meenakshi = 2 : 3.
Let Lakshmi’s expenditure = 2x and Meenakshi’s expenditure = 3x
Given: Income of Lakshmi : Expenditure of Meenakshi = 6 : 7
⇒ Income of Lakshmi = 6/7 × (3x) = 18x/7
Now, Savings = Income – Expenditure Lakshmi’s savings = 18x/7 – 2x = 4x/7
Let Meenakshi’s income = y ; Meenakshi’s savings = y – 3x
Given: Savings of Lakshmi : Meenakshi = 4 : 9
⇒ (4x/7) : (y – 3x) = 4 : 9
⇒ y = 30x/7
Lakshmi’s income =18x/7 ; Meenakshi’s income = 30x/7
Hence, the ratio of incomes = 18x/7 : 30x/7 = 3 : 5.