CAT 2025 QA Slot 3 Question Paper With Detailed PDF Solutions

Correct Answer

4

Explanation

Let the digits be Thousands = a, Hundreds = b, Tens = c, Units = d. Given: a + b + c = 15, b + c + d = 16 and c = d + 6. So (c, d) can be (7, 1), (8, 2) or (9, 3).
The corresponding values of b can be 8, 6 and 4 respectively. Further corresponding values of a can be 0, 1, and 2 respectively.
But a must be ≥ 1.
Therefore, largest possible value will be = 2493 and smallest possible value will be = 1682.
Hence, required difference = 2493 – 1682 = 811.

Correct Answer

2

Explanation

Let average salary of engineers and managers be x and y respectively.
Total original salary = 30 × 60000 = Rs. 18,00,000 Managers get 20% increase; engineers' salaries remain unchanged.
Overall average increases by 5%, therefore, new average = Rs. 63,000.
New total salary: 30 × 63000 = Rs. 18,90,000 Increase in total salary = 1890000 – 1800000 = Rs. 90,000
Increase comes only from managers: 5 × 0.2y = y or, y = Rs. 90,000.
Original total salary is 5y + 25x = 18,00,000 Hence, 5 × (90000) + 25x = 1800000 or, x = Rs. 54,000.

Correct Answer

1

Explanation

Let marked price = M
In January, SP = 0.8M
In February, SP = 0.9M
In March, SP = 0.95M
In April, SP = M
Total profit = 120 × 0.8M + 135 × 0.9M + 150 × 0.95M + 165 × M – 570 × 240 = 138825
Simplifying: 525M – 136800 = 138825 or, 525M
= 275625 or, M = Rs. 525

Correct Answer

126

Explanation

Given: AB = AC = 50 cm, BC = 80 cm
Triangle is isosceles, therefore, the altitude drawn
from the vertex angle (between the two equal sides)
to the base acts as a perpendicular bisector.

Area of the triangle ABC
= 1/2 × 80 × 30
= 1200 cm₂.
Altitudes:
h_a = 2 × Area / BC = 30 cm
h_b = 2 × Area / AC = 2400 / 50 = 48 cm
h_c = 2 × Area / AB = 48 cm
Hence, the sum of altitudes = 30 + 48 + 48
= 126 cm.

Correct Answer

1

Explanation

As per the question.

Since AB = AC, therefore, ∠ABC = ∠ACB and its given ∠ACB = ∠AEB.
This implies that points C, B, E, A lie on a common circle (same angle subtended by chord AB). So A, B, C, E are concyclic.
Since D lies inside the circle and AD is extended to E: DA × DE = DB × DC (Power of Point property) But in an isosceles triangle with AB = AC, the altitude from A to BC also bisects BC. Hence D lies on the perpendicular bisector only if DB = DC.
So, DB = DC
Therefore, DB × DC = DB²
Thus, AD × DE = DB²
DE = AE – AD = AE – 8
So 8 × (AE – 8) = DB²
In triangle ABD, AB = 12 cm and AD = 8 cm.
Using Pythagoras:
DB² = AB² – AD² or, DB² = 12² – 8² = 80
Hence, 8(AE – 8) = 80 or, AE = 18 cm.

Correct Answer

3

Explanation

Correct Answer

1

Explanation

Correct Answer

4

Explanation

Let initial speed = v km/h and let total distance = D
Total scheduled time = 5 pm to 11 pm = 6 hours.
So, D = 6v
Case 1: He drives at speed v first, then stops for 20 minutes or 1/3 hours, then drives at speed (v + 3). Let time driven at speed v be 't' hours.
Then remaining driving time after stop = 6 – t – 1/3 = 17/3 – t
vt + (v + 3)(17/3 – t) = D …(1)
Substitute: D = 6v
⇒ vt + (v + 3)(17/3 – t) = 6v
⇒ –3t + 17v/3 + 17 = 6v
⇒ t = (51 – v)/9 …(A)
Case 2: He drives at speed v first, then stops for 30 minutes or 1/2 hours, then drives at speed (v + 5). Let time driven at speed v be 't' hours.
Time after stop: 6 – t – 1/2 = 11/2 – t
So vt + (v + 5)(11/2 – t) = 6v …(2)
⇒ vt – vt – 5t + 11v/2 + 55/2 = 6v
⇒ –5t + 11v/2 + 55/2 = 6v
⇒ t = (55 – v)/10 …(B)
Equate (A) and (B):
⇒ (51 – v)/9 = (55 – v)/10
⇒ v = 15 km/h

Correct Answer

272

Explanation

Let initial number of coins in A and B be 17x and 7x respectively.
After shifting 108 coins, number of coins in A = 17x – 108 and in B = 7x + 108.
So (17x – 108)/(7x + 108) = 37/20
⇒ 20(17x – 108) = 37(7x + 108)
⇒ 340x – 2160 = 259x + 3996
⇒ 81x = 6156 ⇒ x = 76
So, after first shift:
A = 17 × 76 – 108 = 1184
B = 7 × 76 + 108 = 640
Difference = 1184 – 640 = 544
To make ratio 1:1, shift half the difference.
Required shift = 544/2 = 272

Correct Answer

397

Explanation

Given: p + q + r = 900 and r is a perfect square between 150 and 500.
Possible r values: 169, 196, 225, 256, 289, 324, 361, 400, 441, 484
We have 0.3q ≤ p ≤ 0.7q, or q + 0.3q ≤ p + q ≤ q + 0.7q or 1.3q ≤ p + q ≤ 1.7q.
Also, we can rewrite p + q = 900 – r. Therefore, 1.3q ≤ 900 – r ≤ 1.7q.
Since the extreme values (maximum and minimum) of p depend on the value of q (0.3q and 0.7q), we can maximise or minimise r to find the extreme values of p.
Since 1.3q ≤ 900 – r, the minimum possible value of q would be when 1.3q = 900 – r and when r is maximum. The maximum value of r is 484. We get 1.3q = 900 – 484, or q = 320.
This gives the minimum possible value of p as 0.3 × 320 = 96.
Since 900 – r ≤ 1.7q, the maximum possible value of q would be when 900 – r = 1.7q and when r is minimum. The minimum value of r is 169. We get 1.7q = 900 – 169 or q = 430.
This gives the maximum possible value of p as 0.7 × 430 = 301.
The sum of the maximum and minimum values of p is 96 + 301 = 397.

Correct Answer

1

Explanation

Given: AB || DC, AD ⊥ AB, AB = 3DC and Radius of incircle = 3 cm
Let DC = x ⇒ AB = 3x Since circle touches both parallel sides, Height = 2r = 6.
Area = 1/2 × (AB + DC) × height = 1/2 × (3x + x) × 6 = 12x
For a tangential trapezium: Sum of parallel sides
= sum of non-parallel sides
⇒ AB + DC = AD + BC
⇒ 4x = 6 + BC
⇒ BC = 4x – 6
Using right triangle: BC² = (AB – DC)² + AD²
⇒ (4x – 6)² = (2x)² + 36
⇒ 16x² – 48x + 36 = 4x² + 36
⇒ 12x² – 48x = 0
⇒ x = 4
Hence, the area = 12x = 48 sq. cm.

Correct Answer

16

Explanation

Let x litres be transferred from A to B.
After first transfer, A has 60 – x litres of alcohol and B has 'x' litres of alcohol and 60 litres of water.
Note: Fraction of alcohol in B = x / (60 + x) and, Fraction of water in B = 60 / (60 + x)
Again 'x' litres are taken from B and poured back into A.
Alcohol transferred back to A = x × x / (60 + x)
Water transferred back to A = x × 60 / (60 + x)
Therefore, alcohol in A = (60 – x) + x² / (60 + x) and water in A = 60x / (60 + x).
⇒ [(60 – x)(60 + x) + x²] / 60x = 15/4
⇒ (3600 – x² + x²) = 3600
⇒ 3600 / 60x = 15 / 4
⇒ 60 / x = 15 / 4
⇒ x = 16
Hence, initially 16 litres was transferred from
A to B.

Correct Answer

3

Explanation

Let rates of pipes A, B and C be 4k, 9k and 36k.
Since Pipe A alone fills the tank in 15 hours, so, rate of A = 1/15 tank per hour.
⇒ 4k = 1/15 ⇒ k = 1/60
Combined rate of all three pipes
A + B + C = (4k + 9k + 36k) = 49k = 49/60 tank per hour.
Time required = 1 ÷ (49/60) hours = 60/49 hours = (60/49) × 60 minutes ≈ 73.47 minutes
Hence, the nearest option = 73 minutes.

Correct Answer

1

Explanation

Given: f(x) = (x² + 3x)(x² + 3x + 2)
⇒ √(f(x) + 1) = 9701
⇒ f(x) + 1 = 9701²
⇒ f(x) = 9701² – 1
Let y = x² + 3x So y(y + 2) = 9700 × 9702
⇒ y² + 2y – 9700 × 9702 = 0
⇒ y = 9700, –9702
When y = 9700:
⇒ x² + 3x = 9700 or, x = 97, –100
When y = –9702:
⇒ x² + 3x = –9702 will give no real value of x, therefore, discarded.
Hence, the sum of real values of x = 97 – 100
= –3.

Correct Answer

1

Explanation

Number = 10⁵⁰ + 10²⁵ – 123 have 51 digits of which 51^st and 26^th digits are 1.
After subtracting 123, borrow comes from 26^th digit which is 1.
First three digits from right becomes 877 and digits
4^th to 25^th from right becomes 9. 26^th digits (initially 1) becomes 0.
Hence, the digit sum calculation = 1 + 9 × 22 + 8
+ 2 × 7 = 221.

Correct Answer

1

Explanation

Correct Answer

112

Explanation

Correct Answer

1

Explanation

As per the question, the rate of 1 member of A = 1/200, rate of 1 member of B = 1/400, And rate of 1 member of C = 1/40.
Initially, 2 members from A, 3 from B and 1 from C are working.
Work done per hour = 2(1/200) + 3(1/400) + 1/40 = 1/100 + 3/400 + 10/400 = 17/400
Hence, work done in 23 hours = 23 × 17/400 = 391/400
Remaining work = 1 – 391/400 = 9/400 To finish in the next 1 hour, required rate = 9/400 After member from C leaves, working rate
= 2(1/200) + (3 + x)(1/400) = 1/100 + (3 + x)/400
Therefore, 1/100 + (3 + x)/400 = 9/400
⇒ 4 + 3 + x = 9
⇒ x = 2.

Correct Answer

700

Explanation

Let science, arts and commerce students be S, A and C
Total students = S + A + C = 1500
Total fees = 1100S + 1000A + 800C = 1550000
⇒ 300S + 200A = 350000
⇒ 3S + 2A = 3500
Given A ⇒ S. To maximize S, take minimum A = S
Substitute and get: 3S + 2S = 3500 or, ⇒ S = 700.

Correct Answer

65

Explanation

4th + 7th + 10th terms = (a + 3d) + (a + 6d) + (a + 9d) = 3a + 18d = 99
⇒ a + 6d = 33 … eq(1)
Sum of first 14 terms = 14/2 (2a + 13d)
⇒ 2a + 13d = 71 …eq (2)
Subtracting equations (1) and (2) and get:
⇒ d = 5 and a = 3
Sum of first 5 terms = 5/2 (2a + 4d) = 5/2 (6 + 20) = 65

Correct Answer

444

Explanation

Let walking speed = v. Then, running speed = 1.4v.
Given: Time taken to walk B to C = time taken to run B to A = 3.5 hours
Therefore, BC = 3.5v and BA = 3.5 × 1.4v = 4.9v.
Required time = walking A to B + running B to C = 4.9v/v + 3.5v/1.4v
= 4.9 + 2.5 = 7.4 hours
Therefore, total time = 7.4 × 60 = 444 minutes.

Correct Answer

1

Explanation

We need to maximise a + d = 75 – 2x Let e = 4, then x = 20
Hence, a + d = 75 – 40 = 35 is the maximum
possible value.