CAT 2025 Question Papers PDF FREE Download (Slot 1, 2 & 3), Memory-Based Questions With Answer Key & Solutions

CAT 2025 Question Paper PDF Download Links (Memory-Based Questions) - Slot 1, Slot 2 & Slot 3

This section gives you the memory-based CAT 2025 question paper PDFs for Slot 1, Slot 2 and Slot 3 that you can download right away. It also includes slot-wise analysis of CAT 2025 exam, recalled questions, and early percentile predictions to help you understand how you may have performed.

CAT 2025 Question Paper with Solutions (All Slots)

This section includes the memory-based CAT 2025 questions collected from students across all slots, along with simple solutions to help you understand the approach. It’s a quick way to revise the type of questions asked in the exam.

Question 1: Find the minimum number of integral solutions for

$x+y<14$

$x>y>=3$

Solution:

If y =3 , then x can be taken as 4 to 10 ( total 7 solutions)

If y = 4, then x can be taken from 5 to 9 ( total 5 solutions)

If y = 5, then x can be taken from 6 to 8 ( total 3 solutions)

If y = 6, then x can be taken as 7( total 1 solutions)

Total 16 solutions.

Question 2: The number of boys is 10 more than the number of girls. If 60% of the boys leave the class and 40% of the girls leave the class, the difference between the remaining boys and girls is 8. Find the number of boys in the class.

Solution:

Let number of Girls be G and the number of boys be B

According to the question

$\left|\frac{2}{5}(10+G)-\frac{3}{5} G\right|=8$

$20-G=40$ or $G-20=40$

$|20+G-3 G|=40$

$\begin{gathered}\text { So, } G=40 \\ B=50\end{gathered}$

Question 3: Alven, Alen, and Peter can do a work alone in 13 days, 19 days, and 20 days respectively. Alven was paid 2000 per day, Alen paid 1900 per day, and Peter was paid 1500 per day. What is the minimum amount to be paid so that work to be completed in 10 days or less.

Solution:

Alven, Alen, and Peter can do a work alone in 13 days, 19 days, and 20 days. Their daily wages are 2000, 1900, and 1500 respectively.

Work rates:
• Alven = 1/13 of work per day
• Alen = 1/19 of work per day
• Peter = 1/20 of work per day

Cost per unit work (wage divided by work rate):
• Alven: 2000 × 13 = 26000 per unit work
• Peter: 1500 × 20 = 30000 per unit work
• Alen: 1900 × 19 = 36100 per unit work

Alven is the cheapest per unit work, so we use him for the full 10 days.
Work done by Alven in 10 days = 10/13.
Remaining work = 1 – 10/13 = 3/13.

This remaining work should be given to the next cheapest worker, Peter.

Peter’s time to finish the remaining work = (3/13) ÷ (1/20) = 60/13 days.

Total cost = (10 × 2000) + (60/13 × 1500)
Total cost = 20000 + 90000/13
Total cost = 350000/13
Total cost ≈ 26923.08.

Minimum amount to be paid (with fractional working days allowed) is approximately 26,923.08.

If only full days are allowed, the minimum cost becomes 27,500.

Question 4: In a mixture of 200 liters, acid is 30% and the remaining is water. 20% mixture is replaced with water. And again 10% was replaced by acid. Find the percentage of acid in final solution.

Solution:

After 1st replacement:

Volume replaced = 20% of 200 = 40 litres

$\%$ of acid after $=\%$ of acid $\left[\frac{\text { vol. before addition }}{\text { replacement }}\right]$

$=30 \%\left(\frac{160}{200}\right)=24 \%$

After 2nd replacement

$\%$ of water after $=\%$ of water $\left[\frac{\text { vol. before addition }}{\text { replacement }}\right]$

$=76 \%\left(\frac{190}{200}\right)=73 \%$

So, the final volume acid after 2 replacements = 27%

Question 5: A rhombus has a side 36 cm and area 396 sq cm. Find the difference between the lengths of diagonals

Solution:

$\frac{1}{2} \times d_1 \times d_2 = 396$

$⇒ d_1 \times d_2 = 2 \times 396$

Also, we know that

$(d_1)^2 + (d_2)^2 = 4 a^2$ where a is the side of the rhombus

Now $(d_1 - d_2)^2 = (d_1)^2 + (d_2)^2 - 2d_1d_2$

$\begin{aligned}\left(d_1-d_2\right)^2 & =4(1296-396) \\ & =4 \times 900 \\ d_1-d_2 & =2 \times 30=60 \mathrm{~cm}\end{aligned}$

Question 6: A invests 100000 in stocks, bonds, and gold and earns an interest of 10%, 6%, 8% yearly. He invests 25% in bonds of the total gold value. His interest income of the year is 8200. Then calculate interest earned on bonds.

Solution:

1. S + B + G = 100000
   
   

2. 10% of S + 6% of B + 8% of G = 8200
   
   

3. B = 25% of G = 0.25G
   
   

From total investment:
S + 0.25G + G = 100000
S + 1.25G = 100000
So, S = 100000 − 1.25G

Substitute into interest equation:
0.10(100000 − 1.25G) + 0.06(0.25G) + 0.08G = 8200

This becomes:
10000 − 0.125G + 0.015G + 0.08G = 8200

Combine terms:
10000 − 0.03G = 8200

So, 0.03G = 1800
Thus, G = 60000

Then B = 0.25G = 15000

Interest on bonds = 6% of 15000 = 900

Final Answer: Interest earned on bonds = 900.

Question 7: $\{1,3,5,7, \ldots, 57\}$

If K is the number in this sequence such that the sum of numbers prior to it and after it are equal. Find K

Solution:

We have the sequence of odd numbers:
1, 3, 5, …, 57.

Let K be the term such that the sum of numbers before K equals the sum of numbers after K.

There are 29 terms in total because 57 is the 29th odd number.

The sum of the first n odd numbers is n².

Let K be the k-th term.
Then the sum before K = (k – 1)².
The value of K = 2k – 1.

Total sum of all numbers:
29² = 841.

Condition:
2 × (sum before K) + K = total sum
2(k – 1)² + (2k – 1) = 841.

Simplifying:
2k² – 2k + 1 = 841
k² – k – 420 = 0.

Solve the quadratic:
k = 21.

So K = 2 × 21 – 1 = 41.

Final Answer: 41.

Question 8: N is a 3-digit number with non-zero digits, no digit is a perfect square and only 1 of the digits is a prime number. Also, digits are distinct. Number of factors of the smallest such number possible?

Solution:

According to the given conditions, the possible smallest number = 268
$268 = 2^2 \times 67$
Number of factors = (2 + 1) × (1 + 1) = 6

Question 9:

$x^2 -5x + k = 0$, Find out the value of k, which is a non-negative integer, such that the roots are integers.

Solution:

Sum of the roots = 5

Possible roots are (1, 4), (0, 5), or (2, 3)

Product of roots = k

Possible values of k are $1 \times 4$ = 4 or $2 \times 3$ = 6 or $0 \times 5 = 0$

Question 10:

$\log _{1 / 4}\left(x^2-7 x+11\right)>0$, Find the number of natural number solutions in x.

Solution:

$\log _{1 / 4}\left(x^2-7 x+11\right)>0$
$⇒ x^2-7 x+11<1$
$⇒ (x-5)(x-2)<0$

So, $x$ must lie b/w 2 and 5.

But 3 and 4 do not satisfy the given condition of log.

So, the number of solutions = 0.

Question 11: A shopkeeper allows a discount of 22 percent on market price of each chair, gives 13 chair to a customer for a discounted price of 12 chairs, to earn a profit of 26 percent, the CP of each chair is 100. , calculate the market price of each chair

Solution:

Selling price of 13 chairs = 126% of cost price of 13 chairs = 126 $\times$ 13 = 1638

If the marked price of 1 chair be Rs $100x$

So, the discounted price of 1 chair = 78% of $100x$ = $78x$

the discounted price of 12 chair = $12 \times 78x = 936x$ = SP of 13 chairs

$936x=1638$

$x=1.75$

MP of 1 chair = Rs 175

Question 12: A sum amounts to 13920 in 3 years, and 18960 in 6 years and 6 months at SI. Find interest in 2 years if compounded half-yearly.

Solution:

1. We are told the amount becomes 13920 in 3 years and 18960 in 6.5 years at simple interest. Let P be the principal and r the simple interest rate.

Using simple interest:
A = P (1 + r t)

Equation 1:
P (1 + 3r) = 13920

Equation 2:
P (1 + 6.5r) = 18960

Divide equation 2 by equation 1:
(1 + 6.5r) / (1 + 3r) = 18960 / 13920 = 1.36207

Solve for r:
1 + 6.5r = 1.36207 (1 + 3r)
1 + 6.5r = 1.36207 + 4.08621r
6.5r - 4.08621r = 1.36207 - 1
2.41379r = 0.36207
r = 0.15 or 15 percent per year

Now find principal:
P (1 + 3 × 0.15) = 13920
P × 1.45 = 13920
P = 9600

2. Now calculate compound interest for 2 years, compounded half-yearly.
   Annual simple interest rate is 15 percent.
   Half-yearly rate is 15 percent divided by 2 = 7.5 percent or 0.075.
   Number of half-year periods in 2 years = 4.
   
   

Amount after 2 years = $P × (1.075)^4$
Amount = 9600 × 1.33547 = 12820.5
Interest for 2 years = 12820.5 - 9600 = 3220.5

Final answer: 3220.5

Question 13: a,b,c are distinct natural number, such that , 3ac=8(a+b),find minimum possible value of 3a+2b+c

Solution:

Given:

3ac = 8a + 8b

b = a·k (k is a natural number)

Substitute:

3a·c = 8a + 8(a·k)
3a·c = 8a(1 + k)

3c = 8(1 + k)

c = 8(1 + k) / 3

For c to be a natural number, the numerator must be divisible by 3.

Smallest possible k:
k = 2, 5, 8, …

We want to minimise the expression:

3a + 2b + c
= 3a + 2(a·k) + 8(1 + k)/3

To minimize, start with a = 1.

Then:
b = k
c = 8(1 + k)/3

Try smallest k satisfying k ≡ 2 mod 3:

k = 2

b = 2
c = 8(1 + 2)/3 = 8×3/3 = 8

All are natural AND distinct:
a = 1, b = 2, c = 8 → valid.

Compute expression:

3a + 2b + c
= 3(1) + 2(2) + 8
= 3 + 4 + 8
= 15

CAT 2025 CAT 2025 Analysis - Slot 1, 2 and 3

This section shares the detailed exam analysis for all three slots, along with the YouTube video by our CAT expert, Mr. Ark Srinivas. It helps you understand the difficulty level, question trends, and overall student experience.

CAT 2025 Official Question Paper PDF Download - Slot 1, Slot 2 & Slot 3

The CAT 2025 question paper PDFs for Slot 1, Slot 2 and Slot 3 will be available for download once the official papers are released. This section gives you the direct download details, along with what each slot-wise paper will include.

CAT 2025 Exam Overview: Slots, Pattern

This section gives you a quick look at how the CAT 2025 exam was held across all three slots, along with the overall paper pattern. It helps you understand the structure, number of questions, and what pattern students experienced in each slot.

How to Download the CAT 2025 Question Paper?

IIMs do not release a separate official question paper PDF, but you can view all the questions from your slot through the response sheet. Once the response sheets are live, students can also access questions from all three slots online. Here’s the easy process:

1. Visit the official CAT website: iimcat.ac.in.

2. Click on the Login button on the homepage.

3. Enter your CAT registration number and password.

4. After logging in, open the Response Sheet and Answer Key section.

5. Download your response sheet - it contains every question that appeared in your slot.

6. Save the PDF to your device for later review or analysis.

CAT 2025 Expected Score vs Percentile: First Cut – All Slots

This section gives the first rough estimate of score vs percentile for all three slots. It includes a clear table showing the predicted percentiles based on expected scores to help you understand where you might stand.

CAT 2025 Score vs Percentile - Slot 1, 2 and 3

This part explains the expected score vs percentile for each slot, supported by insights from the YouTube analysis by Mr. Ark Srinivas. It gives you a simple idea of how your raw score may convert into a percentile.