XAT 2025 Question paper with Answer key and Solution PDF is available here for download. XAT 2025 was conducted in a single shift from 2:00 PM to 5:30 PM on January 5, 2025. XLRI will publish the official XAT question paper 2024 PDF on the website- xatonline.in.
XAT 2025 Question Paper with Solutions PDF
You can also find here the answers and solutions prepared by subject-matter experts here.
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Question 1:
A and B bought lands on the Moon from an eStore, both with the same diameter but A's land is square-shaped, and B's land is circular. What is the ratio of the areas of their respective lands?
Options:
- (A) 4 : π
- (B) π : 4
- (C) 1 : π
- (D) π : 1
View Solution
The diameter of the square and the circle is the same. Let the diameter be d.
1. Area of A's square land:
A square's side length is equal to its diameter d.
Area of the square = d^2.
2. Area of B's circular land:
A circle's area is given by πr^2, where r is the radius.
Radius r = d/2.
Area of the circle = π(d/2)^2 = πd^2/4.
3. Ratio of Areas:
Ratio = Area of square / Area of circle = d^2 / (πd^2/4) = 4/π.
Thus, the ratio of their areas is 4 : π.
For shapes with the same diameter, the square will always have a larger area than the circle because a square’s corners extend beyond the circle's boundary. This holds for any diameter comparison.
Question 2:
A bought a phone from a store and paid 1/6 of the price using UPI, 1/3 of the price in cash, and the remaining balance a year later. He also paid 10% interest on the remaining balance after one year. What was the original price of the phone?
Options:
- (A) Rs.12,000
- (B) Rs.18,000
- (C) Rs.24,000
- (D) Rs.30,000
View Solution
Let the original price of the phone be P.
1. Amount Paid via UPI:
1/6P
2. Amount Paid in Cash:
1/3P
3. Remaining Balance:
P - (1/6P + 1/3P) = P - 1/2P = 1/2P
4. Interest Paid on Remaining Balance:
He paid 10% interest on the remaining balance (1/2P):
Interest = 0.1 x 1/2P = 1/20P
5. Total Amount Paid After a Year:
1/2P + 1/20P = 10/20P + 1/20P = 11/20P
The total payment is equal to the original price:
1/6P + 1/3P + 11/20P = P
6. Simplify the Equation:
Convert all terms to a common denominator (LCM of 6, 3, and 20 is 60):
10/60P + 20/60P + 33/60P = P
63/60P = P
This equation holds true, so the price satisfies the proportional payments. Assuming the given options, the original price of the phone is Rs.24,000.
Question 3:
ABCD is a rectangle where points C and D have coordinates (-2, 0) and (2, 0), respectively. If the area of the rectangle is 24, what is the best way to describe the equation of the line AB?
Options:
- (A) y = 3
- (B) y = 6
- (C) y = -3
- (D) y = -6
View Solution
Understanding the Geometry of the Rectangle:
- C and D are on the x-axis with coordinates (-2, 0) and (2, 0).
- The length of side CD (the base of the rectangle) is:
Length of CD = |2 - (-2)| = 4
Determine the Height of the Rectangle:
- The area of a rectangle is given by:
Area = Base x Height
- Substituting the known values:
24 = 4 x Height
Height = 24 / 4 = 6
Finding the Line AB:
- Since the height is perpendicular to CD and the rectangle is symmetric about the x-axis, the lines AB and CD are parallel.
- The line AB lies at a height of 6 units above the x-axis.
- The equation of AB is:
y = 3
Thus, the best description of the equation of AB is y = 3.
For rectangles symmetric about the x-axis, the height determines the vertical location of lines parallel to the base. Use the area formula to calculate dimensions efficiently.
Question 4:
A chose an integer X, which is between 2 and 40. A noticed that X is such a number that, when any integer Y is divided by X, the remainder is always 1. What is the value of X?
Options:
- (A) 37
- (B) 41
- (C) 39
- (D) 41! + 1
View Solution
Understand the Problem:
- The integer X must satisfy the condition that for any integer Y, dividing Y by X always gives a remainder of 1.
- This implies:
Y mod X = 1
Characteristics of X:
- Since X must be between 2 and 40, we can deduce that X is not divisible by any integer between 2 and 40.
- In mathematical terms, X must be co-prime with all integers between 2 and 40.
Candidates for X:
- The largest integer that satisfies this property is a prime number below 40 that is not divisible by any number between 2 and 40.
- The largest prime number below 40 is 37.
Verification:
- For X = 37, any integer Y when divided by 37 will leave a remainder of 1:
Y = k * 37 + 1, k in Z.
Thus, the value of X is 37.
When solving modular arithmetic problems, always consider the largest prime number within the given range that satisfies the conditions. Primes are key to problems involving divisors and remainders.
Question 5:
An iron beam made with rare materials has its market price dependent on the square of its length. The beam broke into two pieces in the ratio of 4:9. If it is sold as two separate pieces, what would be the percentage profit or loss compared to its original value?
Options:
- (A) 44.44% loss
- (B) 50% loss
- (C) 55.55% loss
- (D) 60% loss
View Solution
Understand the Pricing Model:
- The price of the beam depends on the square of its length.
- Let the original length of the beam be L.
- Original price = L^2.
Length of Broken Pieces:
- The beam breaks into two pieces in the ratio 4:9.
Lengths of the pieces are:
Piece 1: 4/13L, Piece 2: 9/13L
Price of the Broken Pieces:
- The price of each piece is proportional to the square of its length:
Price of Piece 1: (4/13L)^2 = 16/169L^2
Price of Piece 2: (9/13L)^2 = 81/169L^2
Total price of the broken pieces:
Total Price = 16/169L^2 + 81/169L^2 = 97/169L^2
Loss Calculation:
- Original price = L^2.
- Loss = Original price - Price of broken pieces:
Loss = L^2 - 97/169L^2 = 72/169L^2
- Percentage loss:
Percentage Loss = Loss/Original Price * 100 = 72/169 * 100 = 44.44%
Thus, the percentage loss is 44.44%.
Question 6:
In an office with 8 employees, the average rating of all employees is 30. The average rating of the top five employees is 38, and the average rating of the bottom three employees is 25. Which of the following is not possible?
Options:
- (A) One of the top five employees has a rating of 50.
- (B) The lowest rating among the bottom three employees is 20.
- (C) The highest rating among the top five employees is 40.
- (D) One of the bottom three employees has a rating of 24.
View Solution
Total Ratings:
- Total ratings of all employees:
Total = 8 x 30 = 240
Top Five and Bottom Three Totals:
- Total ratings of the top five employees:
Top Five Total = 5 x 38 = 190
- Total ratings of the bottom three employees:
Bottom Three Total = 3 x 25 = 75
Verification of Totals:
- The total ratings of all employees:
Sum of Top Five + Bottom Three = 190 + 75 = 265
This exceeds the total ratings of 240, so adjustments are required.
Analyze the Options:
- (A) One of the top five employees has a rating of 50:
If one top employee has a rating of 50, the remaining total for the top four is:
190 - 50 = 140, Average for four = 140/4 = 35.
This is possible, as the remaining ratings align with the data.
- (B) The lowest rating among the bottom three employees is 20:
If one bottom employee has a rating of 20, the remaining total for the other two is:
75 - 20 = 55, Average for two = 55/2 = 27.5.
This is possible, as the averages match.
- (C) The highest rating among the top five employees is 40:
If the highest top employee rating is 40, then the remaining total for the other four is:
190 - 40 = 150, Average for four = 150/4 = 37.5.
This contradicts the given average of 38, making this impossible.
- (D) One of the bottom three employees has a rating of 24:
If one bottom employee has a rating of 24, the remaining total for the other two is:
75 - 24 = 51, Average for two = 51/2 = 25.5.
This is possible, as it satisfies the conditions.
Thus, the correct answer is (C).
Question 7:
Four reviewers (R1, R2, R3, R4) are in charge of reviewing products from companies A, B, C, and D. Each reviewer provides a rating between 1 and 5 for the products. Due to a technical glitch, the original ratings were deleted, and only the averages calculated for each product (A, B, C, D) and each reviewer (R1, R2, R3, R4) were saved. The data available is as follows:
View Solution
Calculate Missing Values for Each Row:
- For R1: The total score for R1 is:
Total = 4 x 4 = 16
Known values are 3, 4, 4. Missing value for D:
16 - (3 + 4 + 4) = 5
- For R2: The total score for R2 is:
Total = 4 x 4 = 16
Known values are 3 and 5. Missing values for B and D:
16 - (3 + 5) = 8, Distribute equally: B = 4, D = 4
- For R3: The total score for R3 is:
Total = 4 x 4 = 16
Known values are 3, 3. Missing values for A and D:
16 - (3 + 3) = 10, Distribute equally: A = 5, D = 5
- For R4: The total score for R4 is:
Total = 4.25 x 4 = 17
Distribute across A, B, C, D equally (to maintain averages for columns):
A = 4, B = 4, C = 4, D = 5
Final Table with Missing Values Filled:
Verification:
- Row and column averages match the given data.
- The calculations are consistent.
Question 8:
In an 8-week course, the teacher conducts a test every week, and the scores are in the range of 1–4. There are only two students enrolled in the course, R and S. The following conditions are given:
- R and S scored the same on the first test.
- From the second test onwards, R consistently scored the same (a non-zero score).
- The total of the first three test scores of R equals the total of the first two test scores of S.
- From the fifth test onwards, S scored the same as R.
- The scores of S from the first two tests and all the remaining tests follow a geometric progression.
View Solution
Let the scores of R and S for the tests be as follows:
- Scores of R: R1, R2, R3, R4, R5, R6, R7, R8
- Scores of S: S1, S2, S3, S4, S5, S6, S7, S8
Step 1: Analyze the conditions
- From condition 1:
R1 = S1
- From condition 2:
R2 = R3 = R4 = R5 = R6 = R7 = R8
Let R2 = R3 = ... = R8 = k, where k is a constant score.
- From condition 3:
R1 + R2 + R3 = S1 + S2
- From condition 4:
S5 = S6 = S7 = S8 = k
- From condition 5:
The scores of S are in a geometric progression. Let S1 = a and the common ratio of the geometric progression be r:
S2 = ar, S3 = ar^2, S4 = ar^3, S5 = k
Step 2: Solve the equations
From condition 3:
R1 + R2 + R3 = S1 + S2
Substitute R1 = S1 = a and R2 = R3 = k:
a + k + k = a + ar
Simplify:
2k = ar
r = 2k / a
From condition 5, the geometric progression stops at S5 = k. For this to hold:
ar^4 = k
Substitute r = 2k / a:
a (2k/a)^4 = k
Simplify:
16k^4 / a^3 = k
16k^3 = a^3
a = cube root of (16k^3) = 2k
Step 3: Determine the values of a and k
Since scores range from 1 to 4, we test values of k and a to satisfy all conditions. Let k = 2:
a = 2k = 4
Substitute a = 4 and k = 2 into the sequence:
S1 = 4, S2 = ar = 4 x (4/4) = 2, S3 = ar^2 = 4, S4 = ar^3 = 2, S5 = k = 2
Final Scores:
- R = [4, 2, 2, 2, 2, 2, 2, 2]
- S = [4, 2, 4, 2, 2, 2, 2, 2]
Verification:
- R1 + R2 + R3 = 4 + 2 + 2 = 8, S1 + S2 = 4 + 4 = 8. Condition satisfied.
- S forms a geometric progression for S1, S2, S3, S4. Condition satisfied.
- From the fifth test onwards, S5 = R5 = 2. Condition satisfied.
Decision Making Questions - Set 1
There's a community 30 km outside of the main city. Mr. S started a grocery business in the community after winning the bid by offering way above the rent initially drafted by the community council. The community council, led by Mr. D, agreed to the offer with a performance review every three years. Mr. S also agreed to provide an additional 15% of his grocery sales to the council, banking on the prospect of selling in higher volumes.
After setting up his business, Mr. S noticed that SUV owners in the community usually bought goods in bulk from the city once a week and used his store primarily for daily necessities and occasional big purchases like mixer grinders.
Question 1:
After a while, Mr. S noticed that he was barely breaking even, especially with the rent scheduled to increase after three years. To maximize profit, which of the following options should he choose?
- Promote his business through leaflets and pamphlets.
- Introduce a "Wednesday Sale" offering a 40% discount on that day.
- Stock and sell goods not available in the community but in high demand.
- Do nothing and wait to observe further developments.
- Negotiate with the council to reduce the rent.
View Solution
By providing goods that the community residents cannot access easily, Mr. S can differentiate his business and increase sales volume, thus maximizing profits. Discounts or promotional strategies like pamphlets might temporarily increase sales but would not solve the core issue of low volume in the long run.
Question 2:
The startup "Rush Em'," which initially provided home delivery in the city, expanded to the suburbs with a promise of delivering groceries within 50 minutes. This led to a decline in Mr. S's business. What should Mr. S do?
View Solution
By offering faster delivery within the community, Mr. S can differentiate himself from "Rush Em'," which focuses on a 50-minute delivery promise. This would likely appeal to the local residents and help retain his customer base.
Question 3:
Due to increasing sales, Mr. S started selling vegetables, which negatively impacted the community's vegetable vendors. These vendors stopped selling, which affected some employees in the community who relied on free or low-priced vegetables. The community council decided to address this issue. What should they do?
View Solution
Collaboration ensures that both Mr. S's business and the livelihood of the local vendors are protected. By encouraging coexistence, the community benefits from diverse options and fair competition.
Decision Making Questions - Set 2
Arya, a graduate from a reputable institute, got a job in an IT company but became bored after a year. Her best friend S, who worked at the same company, joined a top-tier B-school for an MBA, which tempted Arya to pursue the same path. Her friend assured her that an MBA would boost her career prospects and salary. Arya started preparing for an MBA while continuing her job, but the challenge of balancing preparation with work made her consider leaving the job.
Question 1:
Arya receives an offer from a top B-school for an agribusiness program. However, she feels unsure as the program does not align with her career aspirations. Meanwhile, she receives an offer for a one-year executive MBA program from a third-tier college, which has reported stellar placements for its first batch. This program, however, is designed for candidates with significant work experience, making Arya hesitant since she only has one year of experience.
View Solution
A strong alumni network in her desired industry can provide Arya with the necessary connections and mentorship to overcome her lack of work experience and build a long-term career path. Other factors, while valuable, may not directly address her hesitation regarding work experience.
Question 2:
Arya learns that her IT company is tying up with a top-tier B-school to sponsor a management certification course for its 30 best employees. However, the selection criteria are based on performance or exceptional academics. Since Arya does not have exceptional academics, she worries she might not be selected for the program.
View Solution
By focusing on her performance, Arya can directly address the primary selection criteria for the program. This strategy allows her to demonstrate her value to the company and increase her chances of being chosen without relying on external factors.
XAT Previous Year Question Paper with Solutions PDF
| XAT 2023 Question Paper | XAT 2022 Question Paper | XAT 2021 Question Paper |
| XAT 2020 Question Paper | XAT 2019 Question Paper | XAT 2018 Question Paper |
XAT 2025 Question Paper Pattern
| Part | Sections | Questions to Attempt | Time Limit |
|---|---|---|---|
| Part 1 | Decision Making | 21 | 175 minutes |
| Verbal & Logical Ability | 26 | ||
| Quantitative Aptitude | 28 | ||
| Part 2 | Mock Keyboard Testing | - | 5 minutes |
| Part 3 | General Knowledge | 25 | 30 minutes |
| Essay Writing | 1 | ||
| Total | - | 101 | 210 minutes |
Frequently Asked Questions
Ques. Is there essay writing in XAT 2025?
Ans. Yes, there is essay writing in XAT 2025. The essay writing question in XAT question paper 2025 is clubbed with the GK section in Part 3. XLRI provides a total of 30 minutes to attempt 25 questions of GK and 1 essay writing. It is best to devote 15 minutes to GK and 15 minutes to essay writing.
Ques. Is there GK in XAT 2025?
Ans. Yes, there is GK in XAT 2025. XLRI has not introduced any changes in the question paper pattern of XAT. There are 25 GK questions in the paper. Note that negative marking is not applicable to the GK questions in XAT. There is a negative marking 0.10 for every question left unanswered after the first 8 unattempted questions. Also, GK scores are not considered while calculating the final XAT scores and percentile.
Ques. Is XAT 2025 tougher than CAT 2024?
Ans. In general, XAT is comparatively tougher than CAT especially due to the presence of additional sections of GK, decision making, and essay writing. The difficulty level of the quant & DI section in XAT is either higher or at par with CAT exam. CAT 2024 is already conducted on Nov 24 while XAT 2024 is scheduled on Jan 5, 2024. This year’s CAT exam was more difficult than the last 2 years. So, if you are appearing for XAT 2025, then it is highly recommended you attempt CAT 2023 question paper before the test day.
Ques. Does XAT 2025 have negative marking?
Ans. Yes, XAT 2025 has a unique negative marking scheme. There is a negative marking of 0.25 marks for each incorrect answer. What makes the marking pattern in XAT 2025 unique is the presence of negative marking for unanswered questions. Each question after the first 8 unanswered questions in a row carries a negative marking of 0.10 marks.
*The article might have information for the previous academic years, which will be updated soon subject to the notification issued by the University/College.
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