CAT 2023 Quant Practice Questions with Answers

CAT 2023 Quant Practice Questions with Answers: The majority of the questions in the CAT Quant section are difficult and tricky. Therefore, when it comes to answering these CAT questions, using a traditional and outdated textbook approach might not always be the preferred mode of preparation. In order to effectively answer CAT questions based on Quantitative Aptitude, candidates must possess not only a solid understanding of the CAT quantitative aptitude syllabus, and key concepts of the CAT 2023 important topics but also expertise in utilizing a variety of time-saving techniques and shortcuts to answer the questions rapidly. CAT quantitative aptitude questions covering important topics are compiled in this section along with comprehensive solutions. 

In order to accurately answer CAT questions on Quantitative Aptitude, try wrapping up the fundamentals first and then focusing on your conceptual clarity. After this, you can move on to the CAT Quant practice questions with answers. While solving the CAT questions on Quantitative Aptitude, recognize the correct formulas and learn how to use them strategically. After this, create a list of shortcuts that will help you solve the problem quickly and accurately. Never guess the answers because it might lead to a huge negative marking on the final result. Always keep in mind that there is a 40-minute sectional time limit, so in order to complete each question quickly, time management is the most crucial factor.

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Difficulty Level of CAT Quant Practice Questions with Answers

The level of difficulty of questions varies in the CAT quantitative aptitude section. Recently, the CAT 2023 exam has significantly lowered the overall difficulty level. But, it is impossible to predict with certainty how difficult the CAT quant questions will be. For this, candidates must be well-equipped to deal with modifications in the exam's difficulty level. Taking this into consideration, most of the questions in the Quant section of CAT fall into the range of moderate to challenging.

Why Solve CAT Quant Practice Questions with Answers?

The majority of the CAT Quant practice questions with answers range in difficulty from moderate to challenging. Consequently, it will be beneficial for your preparation to solve these questions only after you have mastered the fundamental concepts of every quant topic. It is suggested that the candidates solve the practice sets on the quant topics on which they finished understanding the concepts and have already practised elementary-level questions. Candidates aspiring to appear for the CAT 2023 exam must solve the CAT Quant practice questions with answers for the following reasons:

• CAT Quant practice questions with answers will help the candidates prepare effectively for a crucial part of the actual exam.
• It will help the test-takers improve their time management skills enabling them to complete the quant section within the allocated time.
• Practising CAT Quant practice questions regularly will improve the candidates' problem-solving skills and ability to solve complex mathematical problems.
• The understanding of mathematical concepts will be enhanced and reinforced, which is essential to answering a wide range of complex problems efficiently.
• Candidates will also be able to track their progress and identify areas for improvement. 

CAT Quant Practice Questions with Answers

CAT Quantitative Aptitude questions covering important topics are compiled in this section along with their detailed solutions. The high-quality sets of questions provided in this section aim to assess your conceptual understanding across a range of topics under quantitative aptitude.

1. For an odd number n, which is the highest number that always divides n × (n2 – 1)?

1. 12
2. 24
3. 48
4. 96

Answer: Option 2

Solution:

• We know, n × (n2 – 1) = (n – 1) × n × (n + 1), which is a product of three consecutive numbers.
• Since n is odd, the numbers (n – 1) and (n + 1) are both even. 
• One of these numbers will be a multiple of 2 and the other will be a multiple of 4 because they are two consecutive even numbers.
• Their product is therefore a multiple of 8. One of the three numbers will be a multiple of 3 since every three consecutive numbers is a multiple of 3.
• As a result, the product of three numbers will be a multiple of 8 × 3 = 24.
• Hence, the highest number that always divides n × (n2 – 1) is 24.

2. If p = √8+√7√8−√78+78−7 and q = √8−√7√8+√78−78+7, what is the value of p2 + pq + q2?

1. 900
2. 901
3. 998
4. 899

Answer: Option 4

Solution:

p = √8+√7√8−√78+78−7 = (√8+√7)2(√8−√7)(√8+√7)(8+7)2(8−7)(8+7)= (√8+√7)2(8+7)2 = 15+2√5615+256

Similarly, q = (√8−√7)2(8−7)2 = 15−2√5615−256

p2 + pq + q2 = (15+2√56)2(15+256)2+ (15)2−(2√56)2(15)2−(256)2+ (15−2√56)2(15−256)2 = 675 + 224 = 899

3. Find the number of positive integers n in the range 12 ≤n ≤ 40 such that the product (n - 1) (n - 2) (n - 3)..3×2×1 is not divisible by n.

1. 0
2. 7
3. 13
4. None of these

Answer: Option 3

Solution:

The product is (n - 1) (n - 2) (n - 3)..3×2×1. Only when this product does not contain factors of n, meaning that n is a prime number, would the product not be divisible by n. 13, 17, 19, 23, 29, 31, 37 are the prime numbers in the range provided. A total of 7 numbers are there.

4. Three companies of soldiers, comprising 120, 192, and 144 soldiers, are to be divided into smaller groups in a way that each group has an equal number of soldiers and the soldiers from each group are from a single company. In total, how many groups have been formed the least in number?

Correct Answer: 19

Solution:

When each group has the number of soldiers equal to the HCF, the lowest number of groups will develop. For 120, 192, and 144, the HCF is 24. As a result, there will be 5, 8, and 6 groups, respectively, created for each of the three companies. Thus, the minimum number of groups that can be formed overall is 5 + 8 + 6 = 19.

5. What is the HCF of 2100– 1 and 2120– 1?

1. 220−1220−1
2. 210−1210−1
3. 240−1240−1
4. 1

Answer: Option 1

Solution:

2100– 1 = (220)5- 1 ⇒ divisible by 220 - 1 (an- bn can always be divided by a - b)

Similarly, 2120– 1 = (220)6- 1 ⇒ divisible by 220 - 1 (an- bn can always always be divided by a - b)

⇒ HCF = 220 - 1

6. Find the number of integer values of y that satisfy the given conditions for the given pair (x, y) of positive integers, such that 4x-17y=1 and x<1000.

1. 56
2. 57
3. 58
4. 59

Correct Answer: Option 4 (59)

Finding a solution for x and y is the first step. Once a solution is found, values of x would be in an AP with a common difference of 17 and values of y would be in an AP with a common difference of 4.

Possible Valid Solutions:

x = 13, y = 3

x = 30, y = 7

x = 47, y = 11

.

.

x = 999, y = 235

Number of terms = 999−13 +1= 58 + 1 = 59
                            17

7. Find the number of positive integral values of N less than 40 for which the equation 3a – Nb = 5 will have no integer solution.

1. 13
2. 14
3. 15
4. 12

Correct Answer: Option 1 (13)

Solution:
The RHS would not be divisible by 3 and the LHS could be divisible by 3 if N is a multiple of 3. Number of positive integral values less than 40 that are multiples of 3 = 13.

8. Exactly 2 years after Veeru invested Rs 10,000 at 5% simple interest anuually, Joy invested Rs 8,000 at 10% simple interest annually. Their balances—principal plus accumulated interest— will be equal after how many years of Veeru's investment?

Correct Answer: 10

Solution:

Let, both Veeru and Joy's investments amount to equal amounts after ‘n’ years.

Then, 1000(1+5×(n+2)) = 800(1+10×n) = 800(1+10×n)
                        100                      100                   100

i.e., 1.5 = 15n ⇒ n = 10
                100

Hence 10 years after Veeru’s investment, their balances will become equal.

9. Find the number of distinct positive integer-valued solutions that exist in the equation (x2−7x+11) (x2−13x+42) = 1.

1. 6
2. 8
3. 2
4. 4

Answer: Option 1 (6)

Solution:

(x²−7x+11) (x²−13x+42) = 1

We know, if ab = 1

⇒ a = 1 and b can be regarded as any number

or a = −1 and b is an even number

a>0 and b is 0

case 1: x²−13x+42 =0 ⇒ x=6,7

case 2: x²−7x+11 = 1 ⇒ x2−7x+10 = x = 2 or 5

case 3: x²−7x+11 = −1 ⇒ x²−7x+12 = 0

⇒ x = 4 or 3

Therefore, the number of solutions is 6.

10.

Answer: Option 2

11. Given below is a figure in which A B C D is a rectangle and BCE and CDF are equilateral triangles.

∠AEB +∠BFC =

1. 60^∘
2. 30^∘
3. 45^∘
4. Cannot be determined 

Correct Answer: Option 2 (30^∘) 

Solution:

From the information provided, it is known that the triangles BEC and CDF are equilateral triangles. Therefore, we obtain,

∠ABE=∠BCF=90^∘+60^∘=150^∘

AB = CF

BC = BE

AB = CF

BC = BE, and 

∠ABE = ∠BCF

⇒Triangles ABE and CFB are found to be congruent triangles.

⇒∠BFC=∠BAE

In ΔABE, ∠ABE + ∠AEB + ∠BAE = 180^∘

150^∘ + ∠AEB+ ∠BFC = 180^∘

⇒∠AEB + ∠BFC=30^∘

12. Australia and India compete with each other in a one-day international cricket series until one team wins all four matches. There was never a draw observed in a match. In how many ways is it possible to win the series?

1. 35
2. 70
3. 105
4. 140

Correct Answer: Option 2 (70)

Hence, it is possible for India to win the series in 1+4+10+20 = 35 ways. 

Similarly, it is possible for Australia to win the series in 35 ways. 

Hence, the total number of ways in which both India and Australia can win is 35+35=70 ways. 

13. A pair of fair dice are rolled together until the sum of them comes out to be either 5 or 7. What is the probability that the sum 5 comes before sum 7?

1. 0.45
2. 0.4
3. 0.5
4. 0.6

Answer: Option 2 (0.4)

Solution:

We can obtain 5 by getting any of [(4, 1), (3, 2). (2. 3). (1. 4)]

We can obtain 7 by getting any of [(6, 1). (5, 2). (4. 3). (3. 4). (2. ‘5). (1. 6)]

Probability that the sum 5 comes before sum 7 = 4/10 = 0.4

14. In the ratio 10 : 9 : 8 : 7 : 6, a student obtains an aggregate of 60% marks in five subjects. In how many subjects did he pass the exam if the passing marks are equal to 50% of the maximum marks and the maximum marks are the same for each subject?

1. 2
2. 3
3. 4
4. 5

Correct Answer: Option 3 (4)

Solution:

For each of the five subjects, let his marks be 10, 9, 8, 7, and 6. As a result, he scored 40 points in total which is just 60% of the total marks. The maximum score for all five subjects is therefore 40/0.6 = 66.7 or 67 approximately.

The maximum mark in each subject will be approximately 67/5 = 13, as all subjects have the same total number of marks. The passing mark is 50%. To state differently, a student must obtain 6.5 marks in order to pass a subject. He only received a low score in one subject, which is subject number 6. So he passed in 4 subjects.

15. There are x number of zeros n! and (x + 3) zeroes in (n + 1)! at the end. Now, 1≤n≤1000. Find the number of solutions are possible for ‘n’.

1. 8
2. 7
3. 1
4. 4

Correct Answer: Option 2 (7)

Solution:
It is observed that by adding 1 to the natural number, we are accumulating 3 more powers of 5. As a result, (n + 1) is a multiple of 125 rather than 625, since doing so would produce 4 additional powers of 5. As a result, every multiple of 125 minus 625 equals (n + 1).

There are 8 multiples of 125 less than 1000 

There is 1 multiple of 625 less than 1000 

Therefore, the required answer is (8 – 1) = 7

It is extremely crucial to solve CAT 2023 Quant practice questions with answers if one desires to achieve a competitive advantage in the exam. Your mathematical skills will be enhanced and your chances of success in the CAT 2023 exam will be increased greatly. 

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