XAT 2025 Modern Math Questions

XAT 2025 Modern Math Questions: The XAT 2025 syllabus consists of topics from Verbal Ability and Logical Reasoning, Decision Making, Quantitative Aptitude & Data Interpretation, General Knowledge, and Essay Section. The Modern Math questions are covered under the Quantitative Aptitude section and contain questions on Probability, Permutation and Combinations and Set Theory. 

The  XAT 2025 Predicted Question Paper indicates that the number of expected questions for the Quantitative Ability & Data Interpretation will be 28. Of these, modern mathematics can have about 2-3 questions in the XAT 2025 exam. The list of important topics for the XAT 2025 Modern Math section is mentioned below. This article also contains the compilation of XAT 2025 Modern Math Solved Questions that you can read below.

Also Read | XAT 2025 Exam Dates

Important Topics Under XAT 2025 Modern Math

The XAT 2025 Syllabus is somewhat similar to CAT but is not officially released by XLRI. However, as per previous year trends, here are some topics that come under XAT Modern Math:

1. Probability
2. Permutation and Combinations
3. Set Theory
4. Time Speed Distance
5. Profit & Loss
6. Arithmetic Progression

Also Read | XAT 2025 QA and DI Important Topics with Weightage

XAT 2025 Modern Math Questions 

Here are some practice questions for you to understand what types of questions you can expect in the XAT 2025 Exam for topics such as Probability, Permutation and Combinations and Set Theory.

XAT 2025 Probability Questions 

Question 1 : One letter is chosen randomly from “ASSASSINATION”. What is the probability that the letter is a vowel? 

Answer: Vowel = a, a, i, a, i, o = 6 vowels
P (Vowel) = 6/ 13 

Question 2: If A and B are events such that P (A) = 1/ 4, P (B) = 1/2, P (A and B) = 1/8. Find the probability of getting no A and no B. 

Answer: P (A’ B’) = P (A B)’
= 1 - P (A B) { P (A B) = 1/ 4 + 1/ 2+ 1/ 8 = 5/ 8}
= 1 – 5/ 8 = 3/ 8.

Question 3 : The probability that a person will get an electrification contract is 2/ 5 and the probability of not getting a plumbing contractor is 4/ 7. If the probability that he will get not less than 1 contract is 2/ 3. Find the probability of the event that he will get both contracts. 

Answer: Let E be an event for getting an electrification contract and F be an event for getting a plumbing contract. 
P (E) = 2/ 5 and P (not F) = 4/ 7
P (F) = 1 – 4/ 7 = 3/ 7
P (E F) = 2/ 3 and P (E F) = P (E) + P (F) - P (E F)
= 2/ 5 + 3/ 7 – 2/ 3 = 17/ 105. 

Question 4: A book consists of 100 pages. One page is chosen randomly. Find the probability that the sum of the digit on the page is 9. 

Answer: Let the event be E.
E= {9, 18, 27, 36, 45, 54, 63, 72, 81, 90} and total number of pages is 100. 
P (E) = 10/100 = 1/10.

Question 5 : In a town, there are 6000 people of which 1200 are over 50 years old and 2000 are females. It is said that 30% of females are over 50 years. Find the probability that an individual chosen randomly from the town is either female or over 50 years. 

Answer: Let the event of person being a female is E1, and the event of person being over 50 years old is E2. 
n (E1) = 2,000, n (E2) = 1,200
n (E1 E2) = 30% of 2,000 = 30/100 2,000 = 600
n (E1 E2) = n (E1) + n (E2) - n (E1 E2)
= 2000 + 1200 – 600 = 2,600 
n (E1 E2) = 2600/ 6000 = 13/30. 

Question 6 : In a box carrying 1 dozen oranges 1/3 have become bad. If 3 oranges taken out from the box are random, what is the probability that at least one orange out of the 3 oranges picked up is good?

a) 2/55
b) 54/55
c) 40/55
d) 6/55
e) None of these

Answer: 54/55

Question 7: Two students Anil and Sheena appeared in an exam. The probability that Anil will clear the exam is 0.05 and that Sheena will clear the exam is 0. 10. The probability that both will clear the exam is 0.02. What is the probability that

(a) Both Anil and Sheena will clear the exam
(b) At least one of them will not clear the exam
(c) only one of them will clear the exam

Answer: Let the events A and B denote that Anil and Sheena will pass the exam

P (A) = 0. 05, P (B) = 0. 10, P (A B) = 0. 02

(a) P (Both Anil and Sheena will clear the exam)
P (A’ B’) = P (A B)’ = 1 - P (A B) = 1- [P (A) + P (B) - P (A B)]
= 1 – 0.13 = 0.87

(b) P (At least one of them will not clear the exam) = 
1 – P (both will clear) = 1- 0. 02 = 0. 98

(c) P (only one of them will clear the exam) = 
P (A ∩ B’) + P (A’ ∩ B) = P (A) - P (A ∩ B) + P (B) - P (A ∩ B)
= 0. 05 – 0.02 + 0. 10 – 0. 02
= 0. 11 

Question 8: In class XI of a school, 40% of students study mathematics and 30% study biology. 10% of the class study both mathematics and biology. If a student is selected at random from the class, what is the probability that she will be studying mathematics or biology? 

Answer: Let M be mathematics and B be biology. 

P (M) = 40/ 100 and P (B) = 30/ 100
P (M B) = 10/ 100
P (M B) = P (M) + P (B) – P (M B)
= 40/ 100 + 30/ 100 – 10/ 100 = 0.6

Question 9: Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral, equals (express your answer in decimal).

Answer: (0.1
There are 6 vertices in a hexagon. Using 3 vertices out of 6 vertices we can form 6C3 triangles. But there can be only triangles out of 6C3 equilateral triangles

(i) ∆ACE (ii) ∆BDF.

Required Probability = (2/6C3) =(2/20) = 0.1

Question 10: Five persons entered the lift cabin on the ground floor of a seven-storied building. Suppose that each of them independently and with equal probability, can leave the cabin on any floor beginning with the first. What will be the probability of all the five persons leaving on different floors?

1. 0.03
2. 0.15
3. 0.35
4. 0.32

Answer: (B)

There are seven levels above the main floor.75 is the total number of ways each of the five people can exit the cabin on any of the seven floors.And the optimal number of ways, i.e. the number of ways in which the five individuals evacuate the building via different floors, is 7P5.
Required Probability = (7P5/75) = 0.15

Permutation and Combinations Questions XAT 2025

Question 1: How many numbers of up to 5 digits can be created using the digits 1, 2, 3 and 5 each at least once such that they are a multiple of 15?

1. 20
2. 13
3. 16
4. 12

Answer: 12 

Question 2: How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

1. 5
2. 10
3. 15
4. 20

Answer: 20

Explanation: Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it. The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.Therefore, Required number of numbers = (1 x 5 x 4) = 20.

Question 3: How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are always together? 

Answer: 120

Solution: In these types of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character. So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times. 

Number of possible arrangements = 5! / 2! = 60
Now, the two vowels can be arranged in 2! = 2 ways.
Total number of possible words such that the vowels are always together = 60 × 2 = 120 

Question 4: How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are never together

Answer: 240

Solution: We assume all the vowels to be a single character, i.e., “IE” is a single character.
So, now we have 5 characters in the word, namely, D, R, V, R, and IE.
But, R occurs 2 times.
Number of possible arrangements = 5! / 2! = 60
Now, the two vowels can be arranged in 2! = 2 ways.
Total number of possible words such that the vowels are always together = 60 x 2 = 120 ,
Total number of possible words = 6! / 2! = 720 / 2 = 360
Therefore, the total number of possible words such that the vowels are never together 240

XAT 2025 Set Theory Questions 

Question 1: In a survey of 60 people, 25 liked tea, 30 liked coffee, and 10 liked both. How many people liked only tea?

Answer: 15 people liked only tea.

Solution: Number of people who liked only tea = Number who liked tea – Number who liked both. = 25 – 10 = 15

Question 2: A class has 60 students. 35 students play football, 40 play cricket, and 15 play both. How many students play only one sport?

Answer: 45 Students

Solution: Number of students playing only one sport = Number playing football only + Number playing cricket only.
= (Number playing football – Number playing both) + (Number playing cricket – Number playing both).
= (35 – 15) + (40 – 15) = 20 + 25 = 45.

Question 3: For sets P = {a, b, c, d} and Q = {c, d, e, f, g}, find P ∪ Q and P – Q.

Answer: P ∪ Q (union) is the set of elements in P, or Q, or both.
P ∪ Q = {a, b, c, d, e, f, g}.
P – Q (difference) is the set of elements in P that are not in Q.
P – Q = {a, b}.

Question 4: Set X contains all even numbers between 1 and 20, and Set Y contains all multiples of 3 between 1 and 20. Find X ∩ Y.

Answer: X = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}, Y = {3, 6, 9, 12, 15, 18}.
X ∩ Y (intersection) is the set of elements common to both X and Y.
X ∩ Y = {6, 12, 18}.

Also Read | XAT 2025 Study Plan - One Year Study Plan for XAT

XAT 2025 Modern Maths Books

The books for XAT preparation are designed to enhance students' capacity to analyze intricate scenarios and form reliable conclusions. Check out the following books that can assist you in your preparation:

1. Analytical Reasoning by MK Pandey
2. Modern Approach to Verbal & Non-Verbal Reasoning by R.S. Aggarwal
3. Quantitative Aptitude by Nishit K Sinha
4. How to prepare for Quantitative for the CAT by Arun Sharma, published by MGH Education
5. Quantitative Aptitude Quantum by Sarvesh K Verma

Candidates who focus on XAT 2025 Modern Maths Important Topics will have an advantage in the Quantitative Ability (QA) and Data Interpretation (DI) sections of the XAT Exam. To master this section you must practice regularly, understand mathematical concepts, and prepare strategically and effectively for the exam. 

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