CAT 2025 Slot 1 Answer Key & Response Sheet Released: Download PDF Link, Dates

CAT 2025 Slot 1 Score vs Percentile (Expected)

The CAT 2025 score vs percentile helps candidates to understand what percentile they can get based on their marks, depending on the CAT 2025 answer key of that slot. The CAT 2025 percentile reflects the percentage of candidates who scored less than the highest CAT 2025 score. A CAT percentile is not a direct representation of marks. are calculated after applying scaled scores and slot-wise normalisation. The competition level and exam difficulty affect score-to-percentile conversion, which varies each year. Look at the table below for the CAT 2025 Slot 1 Score vs Percentile.

CAT 2025 Slot 1 Question Paper with Solutions

In this section, we have included several memory-based CAT 2025 questions, which were acquired from students, along with simple solutions to help you understand the approach. With the help of these questions and answers, one will get a rough idea of how the CAT 2025 Slot 1 paper was.

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

|25(10+G)−35G|=8

20−G=40 or G−20=40

|20+G−3G|=40

So, G=40B=50

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 [ vol. before addition replacement ]

=30%(160200)=24%

After 2nd replacement

% of water after =% of water [ vol. before addition replacement ]

=76%(190200)=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:

12×d1×d2=396

⇒d1×d2=2×396

Also, we know that

(d1)2+(d2)2=4a2 where a is the side of the rhombus

Now (d1−d2)2=(d1)2+(d2)2−2d1d2

(d1−d2)2=4(1296−396)=4×900d1−d2=2×30=60 cm

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,…,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=22×67
Number of factors = (2 + 1) × (1 + 1) = 6

Question 9:

x2−5x+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 1×4 = 4 or 2×3 = 6 or 0×5=0

Question 10:

log1/4⁡(x2−7x+11)>0, Find the number of natural number solutions in x.

Solution:

log1/4⁡(x2−7x+11)>0
⇒x2−7x+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 × 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 = 12×78x=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 Slot 1 Answer Key 2025: CAT 2025 Slot 1 Exam Pattern

The CAT 2025 slot 1 didn't go through major changes. The CAT 2025 slot 1 exam pattern had 68 questions divided between the three sections -VARC (24), LRDI (22) and QUANT (22). Apart from that in the CAT 2025 Slot 1, LRDI section had 11 TITA questions and 11 MCQ. On the other hand CAT 2025 Slot 1 question there were 8 TITA questions and 14 MCQ in the LRDI section.

CAT 2025 Slot 1 Score vs Percentile (Expected)

The CAT 2025 score vs percentile helps candidates to understand what percentile they can get based on their marks, depending on the CAT 2025 answer key of that slot. The CAT 2025 percentile reflects the percentage of candidates who scored less than the highest CAT 2025 score. A CAT percentile is not a direct representation of marks. are calculated after applying scaled scores and slot-wise normalisation. The competition level and exam difficulty affect score-to-percentile conversion, which varies each year. Look at the table below for the CAT 2025 Slot 1 Score vs Percentile.

CAT 2025 Slot 1 Response Sheet: Links and Dates

Those who appeared for the CAT 2025 can find the CAT 2025 Answer Key Slot 1 from the table below

CAT 2025 Slot 1 Answer Key 2025: CAT Marking Scheme 2025

With the help of the CAT 2025 marking scheme, those who appeared for the CAT 2025 can calculate their probable marks after they have completed the CAT 2025 answer key PDF download. Here is the CAT 2025 Slot 1 the marking scheme is -

CAT 2025 Score vs Percentile: Slot 1,2,3

Check out the following CAT 2025 score vs percentile for Slot 1,2 and 3.

CAT 2025 Slot 1 Answer Key 2025: How to Calculate CAT Score 2025

The candidates who have appeared for the CAT 2025 Slot 1 can calculate their marks, following these methods. They can verify their answer by matching it with the CAT 2025 answer key for slot 1 Answer Key and the response sheet released by IIM Kozhikode.

CAT 2025 Slot 1 MCQ CAT Raw Score = 3* (Number of the correct answers) - 1*(wrong answer)
CAT 2025 Slot 1 For Non MCQ CAT Raw Score = 3 * (Number of Correct answers)
CAT 2025 Raw Score = Raw CAT score for MCQ questions Raw CAT Score for
Non-MCQ Questions

See Also: Calculate CAT 2025 Percentile from CAT Raw score

CAT 2025 Slot 1 Answer Key: How to download CAT 2025 Slot 1 Response Sheet

Candidates who have appeared for the CAT 2025 and were allotted Slot 1 are eligible for the CAT 2025 Slot 1 Answer Key by visiting the CAT official website- iimcat.ac.in. Candidates can download the CAT 2025 answer key and the CAT 2025 slot 1 response sheet by following the instructions given below

1. Visit the CAT website - ‘iimcat.ac.in’

2. Log in under the ‘Candidate login’ tab and provide your CAT ID and Password.

3. Click on the ‘Submit’ button.

4. Click on the ‘Download PDF’ and download the CAT 2025 Answer Key along with the response sheet.

ALSO CHECK: CAT 2025 Slot 1 Score vs Percentile | CAT 2025 Slot 2 Score vs Percentile | CAT 2025 Score vs Percentile vs Colleges

CAT 2025 Slot 1 Response Sheet : Detailed Mentioned

Upon downloading the CAT 2025 Slot 1 Answer Key and the response sheet, one needs to check the following details if they are presented properly or not.

• CAT 2025 Registration Number

• Candidates Name

• Date of Exam

• Time of the Test as well as the Slot 3

• Exam centre details

• Questions

• Type of Questions

• Status

• Right Option/Answer

CAT 2025 Slot 1 VARC vs CAT 2025 Slot 1 DILR vs CAT 2025 Slot 1 QA: Analysis

Check out the sectional break-up along with the difficulty level of the CAT 2025 Slot 1 exam.

See also: CAT 2025 Section -Wise Score Expectations: What’s a Good score

CAT 2025 Slot 1 Answer Key: How to raise objections to CAT 2025 Slot 1 Answer Key ?

The objection-raising window for the CAT 2025 Slot 1 will be opened a couple of days after the release of the CAT 2025 answer keys. Candidates facing any objections regarding inaccuracy in the answer key can follow these steps to raise an objection against the CAT 2025 Slot 1 Answer Key.

1. Visit CAT's official website ‘iimact.ac.in’.

2. Click on the ‘Candidate Login’ tab and enter your CAT login credentials.

3. Fill in the candidate's personal details page.

4. Click on the ‘Objection Form’ tab and choose the section and question number. They also need to choose the type of objection from these three options: None of the options is the correct answer/ More than one option is correct/ The answer key is incorrect.

5. Enter any remarks up to 500 words.

6. Pay the fees of Rs 1200 per question with other bank services and GST charges.

7. Submit the form.