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Final CAT 2025 Slot 1 answer key and CAT 2025 Slot 1 Response sheet has been announced @iimcat.ac.in by IIM Kozhikode by December 17, 2025. This year, the CAT 2025 slot 1 was easy to moderate compared to last year. The candidate needs to have a basic understanding of the score vs percentile is useful for the candidates, as the MBA Admission 2026 will be based on the CAT 2025 percentile. In the CAT 2025 Slot 1 Score vs Percentile, 110+ marks in CAT 2025 would mean 99.95 percentile. While, 95 percentile in CAT 2025 would mean 56-58 marks. The CAT answer key consists of the CAT 2025 slot 1 DILR, CAT 2025 slot 1 VARC and CAT 2025 slot 1 QA answer keys. CAT 2025 exam was held on November 30, 2025
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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 Marks for 95 percentile | |
CAT 2025 Percentile | CAT 2025 Slot 1 Marks (Approximately) |
99.95 | 110+ |
99.9 | 100-104 |
99.5 | 90-92 |
99.0 | 78-82 |
95 | 56-58 |
90 | 45-48 |
85 | 40-42 |
80 | 35 -36 |
75 | 32-33 |
70 | 30-31 |
| 60 | 25-26 |
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:
S + B + G = 100000
10% of S + 6% of B + 8% of G = 8200
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:
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.
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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
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:
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
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.
| QUANT (22) | LRDI (22) | VARC (24 |
| Geometry - 3 | Pure DI - 5 | RC -16 |
| Arithmetic - 9 | - | VA - 8 Para Jumbling - 2 Para Completion - 2 Para Summary - 2 Odd one out - 2 |
| Algebra - 9 | - | - |
| PNC -1 | - | - |
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 Marks for 95 percentile | |
CAT 2025 Percentile | CAT 2025 Slot 1 Marks (Approximately) |
99.95 | 110+ |
99.9 | 100-104 |
99.5 | 90-92 |
99.0 | 78-82 |
95 | 56-58 |
90 | 45-48 |
85 | 40-42 |
80 | 35 -36 |
75 | 32-33 |
70 | 30-31 |
| 60 | 25-26 |
Those who appeared for the CAT 2025 can find the CAT 2025 Answer Key Slot 1 from the table below
Events | Date (Tentative)/ Link |
December 19-20, 2025 (Tentative) | |
CAT 2025 Slot 1 Response sheet | December 6-8, 2025 |
CAT 2025 Answer Key 2025 Link | OUT |
CAT 2025 Slot 1 Exam Objection Window 2025 Open Day | December 08, 2025 |
CAT 2025 Slot 1 Exam Objection Window 2025 Last Date | December 10, 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 Marking Scheme | MCQs | Non MCQs |
For every correct Answer | 3 marks will be awarded | 3 marks will be awarded |
For every incorrect answer | 1 mark will be deducted | 0 marks will be deducted |
For every attempted answer | 0 marks will be awarded or deducted | 0 marks will be awarded or deducted |
Check out the following CAT 2025 score vs percentile for Slot 1,2 and 3.
| CAT 2025 Slot 1 Score vs Percentile | Click on the Link |
| CAT 2025 Slot 2 Score vs Percentile | Click on the Link |
| CAT 2025 Slot 3 Score vs Percentile | Click on the Link |
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
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
Visit the CAT website - ‘iimcat.ac.in’
Log in under the ‘Candidate login’ tab and provide your CAT ID and Password.
Click on the ‘Submit’ button.
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
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 wise Question Papers | Download Link |
CAT 2025 Question Paper PDF Slot 1 (Memory-Based) | |
CAT 2025 Slot 2 Question Paper PDF (Memory-Based) | |
CAT 2025 Slot 3 Question Paper PDF (Memory-Based) | |
CAT 2025 Question Paper PDF Download (All Slots) |
Check out the sectional break-up along with the difficulty level of the CAT 2025 Slot 1 exam.
CAT 2025 Slot 1 section | Questions | Time | Number | Difficulty level |
CAT 2025 Slot 1 VARC | 24 | 40 | 72 | Easy to Moderate |
CAT 2025 Slot 1 DILR (Data Interpretation and Logical Reasoning) | 22 | 40 | 66 | Easy to Moderate |
CAT 2025 Slot 1 QA (Quantitative Aptitude) | 22 | 40 | 66 | Easy to Moderate |
Total | 68 | 120 | 204 | Easy - Moderate |
See also: CAT 2025 Section -Wise Score Expectations: What’s a Good score
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.
Visit CAT's official website ‘iimact.ac.in’.
Click on the ‘Candidate Login’ tab and enter your CAT login credentials.
Fill in the candidate's personal details page.
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.
Enter any remarks up to 500 words.
Pay the fees of Rs 1200 per question with other bank services and GST charges.
Submit the form.
Frequently Asked Questions (FAQs)
The CAT 2025 answer key is out on December 04, 2025
3 marks are awarded for each correct answer.
Yes. After IIM Calcutta releases the answer keys, the candidates will get a window of a couple of days to challenge the answer key.
The authorities will release the final CAT 2025 answer key alongside the CAT 2025 result.
Yes, CAT will release the answer key and the response sheet a few days after the completion of CAT 20024.
The CAT answer key can be downloaded by visiting the CAT official website- iimcat.ac.in.
Yes, the CAT 2025 Slot 1 answer key PDF download will be available from the official website of CAT. Candidates can get the CAT 2025 response sheet and the CAT 2025 slot 1 answer key by logging into the CAT 2025 Dashboard.
Candidates can check the CAT 2025 exam answer key by logging into and providing their CAT 2025 login credentials.
On Question asked by student community
Hello,
Here are some top colleges accepting XAT and CAT exams :
Top colleges accepting CAT:
IIMs (All Indian Institutes of Management)
FMS Delhi
SPJIMR Mumbai
MDI Gurgaon
IIT Bombay (SJMSOM)
IIT Delhi (DMS)
IIFT
IMT Ghaziabad
JBIMS Mumbai
Top colleges accepting XAT:
XLRI Jamshedpur
XIMB Bhubaneswar
IMT Ghaziabad
Great
Hi there,
A female candidate with a CAT percentile of 67.97 and low sectional scores should target private and tier-2/3 B-schools that accept overall CAT scores in the 60–70 percentile range and have flexible sectional criteria.
Some suitable options include AIMS Institute Bangalore, Doon Business School Dehradun, Christ Institute of
Hi there,
Yes, you are eligible for XISS Ranchi with a CAT percentile of 67.60.
According to recent admission trends, the CAT cutoff for the PGDM in Human Resource Management for the general category has been around 60 percentile. For other programs such as Marketing, Finance, and Rural Management, the
Hi there,
Careers360 offers a wide range of eBooks and study materials to assist with CAT preparation. You can access past CAT question papers with solutions to understand the exam pattern and difficulty level. Additionally, there are quantitative aptitude handbooks, cheat sheets, and section-specific practice sets for arithmetic, algebra, and
Hi there,
The minimum eligibility criteria for a general candidate to receive a call from IIM Sambalpur are as follows:
VARC: 65%ile
QUANTS: 65%ile
LRDI: 65%ile
Overall: 90%tile
Keep practising and aim to improve your score. You can also focus on other management exams where you may secure a strong
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