CAT 2024 Quant Topic-Wise Questions

CAT quant topic-wise questions are important for candidates who are preparing for the CAT 2024 exam and need more material to test their preparation. The CAT quantitative aptitude syllabus covers Arithmetic, Algebra, Geometry, Number Systems, and Modern Mathematics. Based on past years' data, the difficulty level of questions from this section is usually moderate to high.

A good way to prepare for CAT Quantitative Aptitude is to begin early, as it takes a lot of practice and focus to learn the formulas, tips & tricks for solving math problems in a shorter amount of time. These CAT Quant questions are designed to help you understand the different concepts and help you practice, identify your weak areas and work on them accordingly.



Also Read :

Tips for Solving Quant Questions in CAT 2024

Effective strategies during the CAT exam are essential for good performance. Here are some tips on how to answer CAT quantitative aptitude questions:

• Learn how to apply the exact formulas tactically.

• Identify shortcuts that will help you solve the problem more quickly and accurately.

• Do not guess the answers, if you are not sure skip the question since negative marking will impact the final score.

Important Topics for CAT 2024 Quant

Before we move on to the CAT topic-wise questions we must know which are the important topics for the CAT QA section. These areas and topics have a tendency to be asked in the CAT exam almost every year. Check out the list of CAT Quant topics mentioned below:

CAT 2024 Quant Topic-Wise Questions

We've listed the chapter-wise CAT quantitative aptitude questions for you. These questions will help you revise the concepts and improve your problem-solving skills. It'll also boost your confidence for the CAT quantitative aptitude test. Best of luck!

Before we begin, check out the formulas for CAT preparation:

CAT Time Speed Distance Questions

Question : When covering 30 km, Abhay takes two hours longer than Sameer. Abhay's time would be reduced by one hour if he doubled his speed. The speed of Abhay is:

5 kmph
6 kmph
6.25 kmph
7.5 kmph

Answer : Option A

Solution : Let Abhay's speed be x km/hr.
Then, 30/x - 30/2x = 3
6x = 30
x = 5 km/hr.

Question : In a room with a width of 30 m, Ram and Shyam are standing at the two ends. Starting at a speed of 2 m/s and 1 m/s, they walk towards each other along the width of the room. When Ram met Shyam for the third time, how far did he travel?

1. 110 m

2. 112 m

3. 120 m

4. 100 m

Answer (4) 100 m

Solution : Ram and Shyam would have covered a distance of 5d by the third time they met, i.e. 5x30m = 150m.

speed ratio of Shyam and Ram =  2:1,
the total distance travelled by them will be in the ratio 2:1
Since the time taken is constant so the distance travelled by Ram will be 2/3 x 150= 100 m

Question : The average speed of a person travelling from one place to another is 30 km/hour, and their return speed is 120 km/hour. Calculate the distance if the total time taken is 5 hours.

Answer : 120 km

Solution : Because the Distance is constant, the Time taken will be inversely proportional to the Speed. The ratio of Speed is given as 30:120, i.e. 1:4. Therefore, the ratio of time taken will be 4:1. Total Time taken = 5 hours; Time taken while going is 4 hours and returning is 1 hour. Hence, Distance = 30x 4 = 120 km.

CAT LCM and HCF

Question : What is the LCM of 8 & 6?

Answer : 24

Solution : To find: LCM (8, 6).

The multiples of 8 are 8, 16, 24, 32, 40, 48, ….
The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, …

The smallest common multiple is 24.
Therefore, the LCM of 8 & 6 = 24.

Question : What is the HCF of 408 and 1032?

Answer : 24

Solution :
To find: HCF (408, 1032).
1032 Prime factorisation = 2 × 2 × 2 × 3 × 43
408 Prime factorisation =  2 × 2 × 2 × 3 × 17
The product of the prime factors of 2 × 2 × 2 × 3 is 24

Hence, the HCF of 408 and 1032 is 24.

Question: The HCF and LCM of 2 positive numbers are 4 and 80 respectively. If one number is 4 more than the other number, find the smaller number.

1. 22

2. 16

3. 17

4. More than one of the above

Answer : 16

Solution :
HCF of two numbers = 4
LCM of two numbers = 80
One of the numbers is 4 more than the other

Let the smaller number be 4k
Then, the larger number = (4k + 4)
Now, the Product of two numbers = HCF × LCM

• 4k(4k + 4) = 4 × 80

• k(k + 1) = 20

• k = 4.

Smaller number = 4k
4(4) ⇒ 16
Therefore, The answer is 16.

Also Read : 15+ CAT 2024 Percentage Questions

CAT Profit, Loss, and Discount Questions

Question : Find the selling price of the bicycle of Rs. 1000 if the loss is Rs. 50.

Answer : C.P. of bicycle = Rs. 1000

Solution :
Loss (L) = Rs. 50
In the case of loss, C.P.>S.P.
Therefore, the formula of loss is
Loss = C.P. - S.P.
Let S.P. be 'Rs. x'
Therefore, 50 = 1000 - x
x = 1000 - 50
= Rs. 950
S.P. of a bicycle = Rs. 950
Now, C.P. = Rs. 1000

Question : Find the selling price of the bicycle of Rs. 1000 if %Profit is 50%.

Answer : The selling price of a bicycle is Rs. 1500.

Solution :
Let profit be ' x′
%P=50
formula: %profit = profit  CostPrice ×100
50=x1000×100
x=10×50
=Rs500

Hence, the profit is Rs. 500.
Now, S.P. = C.P. +P
=1000+500=1500
The selling price of a bicycle = Rs. 1500.

Question : If the list price of a book is 50, and a 10 discount is offered on the book, then what is the discount percentage?

Solution : Discount % = (Discount/marked Price) × 100

Marked Price = 50;discount = 10
Discount (%) = 1050×100
= 1005
= 20%
Therefore, the discount percentage is calculated as 20%.

Also Read : CAT 2024 Profit and Loss Questions and Answers

CAT 2024 Geometry Questions

Question: In the figure above, AB = BC = CD = DE = EF = FG = GA. Then ∠DAE is approximately

Options :

1. 15°

2. 20°

3. 30°

4. 25°

Answer : 25°

Question : The perimeter of an △ABC is 15. All sides have integral lengths. How many triangles can be made like that?

1. 7

2. 5

3. 6

4. 4

Answer : 7

Solution : Use the trial and error method to satisfy the condition that the sum of the two sides should be greater than the third side.

Let us assume a ≤ b ≤ c.
if a = 1, Possible sides of triangles 1, 7, 7
If a = 2, possible sides of triangles 2, 6, 7
if a = 3, possible sides of triangles 3, 6, 6 and 3, 5, 7
if a = 4, possible sides of triangles 4, 4, 7 and 4, 5, 6
if a = 5, possible sides of triangles 5, 5, 5
So a total of 7 triangles are possible.

CAT Sets Questions

Question : Write the set A = {1, 4, 9, 16, 25, . . . } in set-builder form.

Solution : There is a pattern here, the numbers are squares of natural numbers, such as:

12 = 1

22 = 4

32 = 9

42 = 16

And so on.

A = {x: x is the square of a natural number}

Or we can write; A = {x : x = n2 , where n ∈ N}

Question : Write the subsets of {1,2,3}.

Solution : Let A = {1, 2, 3}

The subsets of A are: φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}

Question : If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}. Find A ∩ (B ∪ C).

Solution : As we know, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

= {7, 9, 11} ∪ {11}

= {7, 9, 11}

Question : For sets A = {x | x is an integer, 1 ≤ x ≤ 6} and B = {x | x is an even integer, 2 ≤ x ≤ 8}, find the set A – B.

Solution :

A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8}.

A – B (difference) is the set of elements in A that are not in B.

A – B = {1, 3, 5}.

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CAT 2024 Probability Questions

Question : A pair of dice is thrown together. What is the probability a number obtained using one die is a multiple of a number obtained using the other die?

Solution : Total number of cases = 62 = 36
Considering a die's number must be multiple of another, it is possible to have the following outcome
(1, 1) (2, 2) (3, 3) ------ (6, 6) --- 6 ways
(2, 1) (1, 2) (1, 4) (4, 1) (1, 3) (3, 1) (1, 5) (5, 1) (6, 1) (1, 6) --- 10 ways
(2, 4) (4, 2) (2, 6) (6, 2) (3, 6) (6, 3) -- 6 ways
Favorable cases are = 6 + 10 + 6 = 22.

Answer : So, P (A) = 22/36 = 11/18

Question : If a card is drawn from a 56-card pack, determine the probability that the card will be numbered.

Solution : Total Cards = 52. Numbered Cards = (2, 3, 4, 5, 6, 7, 8, 9, 10) 9 from each suit 4 × 9 = 36. P (E) = 36/52 = 9/13

Question : 4 white balls, 5 red balls, and 6 blue balls are contained in a bag. Three balls are randomly selected from the bag. The probability that all of them are red is:

1. 1/22

2. 3/22

3. 2/91

4. 2/77

Answer : 2/91

Question : There are 15 boys and 10 girls in the class. Randomly, three students are selected. What is the probability that 1 girl and 2 boys are selected, is:

1. 21/46

2. 25/117

3. 1/50

4. 3/25

Answer : 21/46

Question : 5 red balls and some blue balls are in a bag. The probability of drawing a blue ball from the bag is four times greater than that of drawing a red ball, so there are four blue balls in the bag:

1. 10

2. 40

3. 20

4. More than one of the above

Answer : 20

CAT 2024 Time And Work Questions

Question : A and B can do a piece of work in 4 days, while C and D can do the same work in 12 days. In how many days will A, B, C and D do it together?

1. 12 days

2. 4 days

3. 3 days

4. 2 days

5. None of these

Answer : 3 Days

Solution : A, B, C and D will together take ¼ + 1/12 = 4/12 = 1/3 ⇒ 3 days to complete the work.

Question : A, B, C, and D can do a piece of work in 20 days. If A and B can do it together in 50 days, and C alone in 60 days, find the time in which D alone can do it.

1. 120 days

2. 200 days

3. 150 days

4. 90 days

5. 75 days

Answer : 75 Days

Solution : D alone will take 1/20 – 1/50 – 1/60 = 4/300 = 1/75

75 days to complete the work

​​​​​​ Question : A, B, and C can do a piece of work in 8 days. B and C together do it in 24 days. B alone can do it in 40 days. In what time will it be done by C working alone?

1. 25 days

2. 24 days

3. 60 days

4. 20 days

5. 30 days

Answer : 60 Days

Solution : B & C do this work in 24 days. B alone does this work in 40 days.  C alone will take 1/24 – 1/40 = 2/120=1/60 ⇒ 60 days to finish the work.

CAT 2024 Permutation and Combination Questions

Question : What is the maximum number o`f ways we can select a team of four students from a list of 15?

Solution : The number of possible selections = 15C4 = 15! / ((4 !) × (11 !))
The number of possible selections = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) = 1365

Question : With the letters from the word "DRIVER," how many words can you form in which all the vowels are never grouped?

Solution : All vowels are assumed to be single characters, e.g., "I,E" is a single character. We now have five characters in the word, namely D, R, V, R, and IE. However, R appears twice.
Therefore, number of possible arrangements = 5! / 2! = 60
Now, the two vowels can be arranged in 2! = 2 ways.
Number of words in which the vowels are always together = 60 x 2 = 120,
Total number of possible words = 6! / 2! = 720 / 2 = 360
In other words, there are 240 possible words in which the vowels are never together.

Question : How many ways, can we select a team of 4 students from a given choice of 15?

Solution : Number of possible ways of selection = 15C4 = 15 ! / ((4 !) × (11 !))
Number of possible ways of selection = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) = 1365 (Answer)

Question : How many words can be formed by using 3 letters from the word “DELHI”?

Solution : Here we will use the Permutations for this question. The formula for permutation is nPr, for this, we have, n = 5: Total 5 Letters, r = 3: Letters word we required:
nPr = n!/(n-r)! 5P3 = 5!/2! = 120/2 = 60.
Answer : So, in Total we can form 60 different permutations of words from the Letter Delhi.

Hopefully, the questions mentioned above will help you in your preparation for the CAT Quantitative Aptitude section! Check out the articles mentioned below to learn more!

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