The quantitative Aptitude section in the CAT exam is the most important. Through geometry questions, aspirants are tested on concepts related to circles, triangles, polygons, and other geometry-related topics. The difficulty level of the Geometry questions can be easy to moderate.
Tips to improve Quant Questions for CAT?
- Cover your basics and understand the topics, to get the concepts right and rectify your mistakes in the mock tests.
- Practice regularly to speed up your performance .
- Plan a daily study schedule that covers all the important topics in quantitative aptitude.
- Understand all the formulas, theorems, and concepts thoroughly, and practice problems based on them.
Quant Questions Tricks for the CAT Exam
- CAT Shortcut technique to find the square of 2 digit number mentioned below. Have a look below to know the trick:
- STEP 1: First, Deduct 25 from the number want to calculate square. Then first two digits of the Answer
- STEP 2: Then, Subtract 50 from the number form in the win which one wants to calculate the square. Then last two digits of the Answer and in case of carrying over. So add the number carry-over to the answer derived through the first step.
For Example:
Let us try with 63
Step 1: The difference from 25 is 38
Step 2: The difference from 50 is 13
Step 3: Square of 13 is 169
The answer to Step 1 (first two digits): = 38+1 = 39
The answer to Step 2 (last two digits): And the last 2 digits of 169= 69
Hence, the Final Answer is 3969.
Quant Questions Concepts for the CAT Exam
- Many questions in this section will come from Arithmetic, followed by Algebra and Geometry.
- Mensuration: Calculating areas, perimeters, volumes, and surface areas of various geometric shapes.
- Permutation & Combination, Probability, Set Theory, Functions & Graphs, Trigonometry: Advanced mathematical concepts and their applications.
- Data Interpretation: Analyzing and interpreting data from tables, graphs, charts, and caselets to derive meaningful insights.
Quantitative Aptitude Practice Questions for CAT exams
Q 1. In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 46, determine the dividend.
Solution. Remainder = 46 ; Divisor = 5 × 46 = 230 ; Quotient = =230 23.10
Dividend = Divisor × Quotient + Remainder = 230 × 23 + 46 = 5336.
Q 2. If three times the larger of the two numbers is divided by the smaller one, we get 4 as quotient and 3 as remainder. Also, if seven times the smaller number is divided by the larger one, we get 5 as quotient and 1 as remainder. Find the numbers.
Solution. Let the larger number be x and the smaller number be y.
Then, 3x = 4y + 3 ⇒ 3x – 4y = 3 ...(i)
And, 7y = 5x + 1 ⇒ – 5x + 7y = 1 ...(ii)
Multiplying (i) by 5 and (ii) by 3, we get:
15x – 20y = 15 ...(iii) and – 15x + 21y = 3 ...(iv)
Adding (iii) and (iv), we get: y = 18.
Putting y = 18 in (i), we get x = 25.
Hence, the numbers are 25 and 18.
Q 3. A number when divided by 6 leaves remainder 3. When the square of the same number is divided by 6, find the remainder.
Solution. On dividing the given number by 6, let k be the quotient and 3 the remainder.
Then, number = 6k + 3.
Square of the number = (6k + 3)2 = 36k2 + 9 + 36k = 36k2 + 36k + 6 + 3
= 6 (6k2 + 6k + 1) + 3, which gives a remainder 3 when divided by 6.
Q 4. Find the remainder when 96 + 7 is divided by 8.
Solution. (xn – an) is divisible by (x – a) for all values of n.
So, (96 – 1) is divisible by (9 – 1), i.e. 8 ⇒ (96 – 1) + 8 is divisible by 8 ⇒ (96 + 7) is divisible by 8.
Hence, required remainder = 0.
Q 5. Find the remainder when (397)3589 + 5 is divided by 398.
Solution. (xn + an) is divisible by (x + a) for all odd values of n.
So, [(397)3589 + 1] is divisible by (397 + 1), i.e. 398
⇒ [{(397)3589 + 1} + 4] gives remainder 4 when divided by 398
⇒ [(397)3589 + 5] gives remainder 4 when divided by 398.
Q 6. If 7126 is divided by 48, find the remainder.
Solution. 7126 = (72)63 = (49) 63.
Now, since (xn – an) is divisible by (x – a) for all values of n,
so [(49)63 – 1] or (7126 – 1) is divisible by (49 – 1) i.e. 48.
Remainder obtained when (7)126 is divided by 48 = 1.
What are the must-do Quant questions for the CAT exam?
Q 7. If (the place value of 5 in 15201) + (the place value of 6 in 2659) = 7 × ____, then the number of the blank space is:
(a) 800
(b) 80
(c) 90
(d) 900
Answer. Option (A)
Q 8. How many prime numbers are there between 100 to 200?
(a) 21
(b) 20
(c) 16
(d) 11
Answer. Option (A)
Q 9. The number of three digit numbers which are multiples of 9 are
(a) 100
(b) 99
(c) 98
(d) 101
Answer. Option (A)
Q 10. Two consecutive even positive integers, sum of the squares of which is 1060, are
(a) 12 and 14
(b) 20 and 22
(c) 22 and 24
(d) 15 and 18
Answer. Option (C)
Q 11. What is the sum of all natural numbers from 1 to 100?
(a) 5050
(b) 6000
(c) 5000
(d) 5052
Answer. Option (A)
What are the most important Quant questions for the CAT 2024 exam?
Q 12. The traffic lights at three different signal points change after every 45 seconds, 75 seconds and 90 seconds respectively. If all change simultaneously at 7:20:15 hours, then they will change again simultaneously at
(a) 7:28:00 hours
(b) 7:27:45 hours
(c) 7:27:30 hours
(d) 7:27:50 hours
Answer. Option (B)
Q 13. A number x is divided by 7. When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of x is:
(a) 148
(b) 149
(c) 150
(d) 147
Answer. Option (D)
Q 14. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After
what time will they meet again at the starting point?
(a) 26 minutes 18 seconds
(b) 42 minutes 36 seconds
(c) 45 minutes
(d) 46 minutes 12 seconds
Answer. Option (D)
Q 15. Find the least multiple of 23, which when divided by 18, 21 and 24 leaves remainders 7, 10 and 13 respectively.
(a) 3002
(b) 3013
(c) 3024
(d) 3036
Answer. Option (D)
Q 16. Since 16 is not a factor of 136, it follows that there does not exist any pair of numbers with H.C.F. 16 and L.C.M.
Answer. No such pairs exist.
H.C.F. of two numbers divides their L.C.M. exactly. Since 16 is not a factor of 136, it states that there does not exist any pair of numbers with H,C.F. 16 and L.C.M. 136.
Q 17. Solve 2.5×4.8+7.2×1.5–1.2×14=?
(a) 1.2
(b) 6.5
(c) 4
(d) 4.8
(e) 6
Answer. Option (E)
Q 18. Simplify: b [b (a b) {b (b a b)} 2a].
Answer. Given exp. = b – [b – (a + b) – {b – (b – a + b)} + 2a]
= b – [b – a – b – {b – (2b – a)} + 2a]
= b – [– a – {b – 2b + a} + 2a]
= b – [– a – {– b + a} + 2a]
= b – [– a + b – a + 2a] = b – b = 0
Find some previous year Quant questions from the CAT 2023 exam
Q 19. Ram went to a shop to buy 50 kg of rice. He buys two varieties of rice which cost him ` 4.50 per kg and 5 per kg. He spends a total of 240. What was the quantity of rice bought which cost him 4.50 per kg?
Solution. Let the quantity of rice bought at ` 4.50 per kg be x kg.
Then, quantity of rice bought at ` 5 per kg = (50 – x) kg.
4.50x + 5(50 – x) = 240 ⇒ 250 – 0.5x = 240 ⇒ 0.5x = 10 ⇒ x = 20.
Hence, quantity of rice bought at ` 4.50 per kg = 20 kg.
Q 20. A boy was asked to multiply a certain number by 53. He multiplied it by 35 and got his answer less than the correct one by 1206. Find the number to be multiplied.
Solution. Let the required number be x.
Then, 53x – 35x = 1206 ⇔ 18x = 1206 ⇔ 1206 67.18x
Hence, number to be multiplied = 67.
Q 21. A man divides ` 8600 among 5 sons, 4 daughters, and 2 nephews. If each daughter receives four times as much as each nephew, and each son receives five times as much as each nephew, how much does each daughter receive?
Solution. Let the share of each nephew be x.
Then, share of each daughter = ` (4x); share of each son = (5x).
So, 5 × 5x + 4 × 4x + 2 × x = 8600 ⇔ 25x + 16x + 2x = 8600
⇔ 43x = 8600 ⇔ x = 200.
Share of each daughter = ` (4 × 200) = ` 800
Q 22. A zoo keeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo?
Solution. Let the number of pigeons be x and the number of horses be y.
Then, total number of heads = x + y.
total number of legs = 2x + 4y.
x + y = 80 ...(i)
And, 2x + 4y = 260 or x + 2y = 130 ...(ii)
Subtracting (i) from (ii), we get: y = 50.
Putting y = 50 in (i), we get: x = 30.
Hence, number of horses in the zoo = 50.
Q 23. In a caravan, in addition to 50 hens there are 45 goats and 8 camels with some keepers. If the total number of feet is 224 more than the number of heads, find the number of keepers.
Solution. Let the number of keepers be x. Then,
Total number of heads = (50 + 45 + 8 + x) = (103 + x).
Total number of feet = (45 + 8) × 4 + (50 + x) × 2 = (312 + 2x).
(312 + 2x) – (103 + x) = 224 ⇔ x = 15.
Hence, number of keepers = 15.
What are the important Quant practice questions for CAT 2024?
Q 24. Arun and Sajal are friends. Each has some money. If Arun gives ` 30 to Sajal, then Sajal will have twice the money left with Arun. But, if Sajal gives ` 10 to Arun, then Arun will have thrice as much as is left with Sajal. How much money does each have?
Solution. Suppose Arun has ` x and Sajal has y. Then,
2(x – 30) = y + 30 ⇒ 2x – y = 90 ...(i)
and x + 10 = 3(y – 10) ⇒ x – 3y = – 40 ...(ii)
Solving (i) and (ii), we get: x = 62 and y = 34.
Arun has 62 and Sajal has 34.
Q 25. An executive goes on a business trip. His daily schedule has a definite pattern. If he is busy with a meeting in the morning, he is free in the afternoon. When he returns, he realises that he attended 15 meetings altogether.
There were 12 free mornings and 13 free afternoons. What was the duration of his trip?
Solution. Let M and A represent the number of busy mornings and busy afternoons respectively.
Then, total number of mornings in the trip = 12 + M.
And, total number of afternoons in the trip = 13 + A.
Clearly, 12 + M = 13 + A or M – A = 1 ...(i)
Total number of meetings = M + A
So, M + A = 15 ...(ii)
Adding (i) and (ii), we get: 2M = 16 or M = 8.
Hence, duration of the trip = (12 + 8) = 20 days.
Q 26. If a = 7, b = 5, then the value of a3–b3+3a2b is
(a) 218
(b) 307
(c) 735
(d) 953
Answer. Option: (D)
Q 27. 7m –[3n–{8m–(4n–10m)}] simplifies to
(a) 11m – 5n
(b) 11m – 7n
(c) 11n – 7m
(d) 13n – 11m
Answer. Option: (B)
Which are the most repeated Quant questions for the CAT?
Q 28. What value will replace the question mark in each of the following questions?
(i)? – 1936248 = 1635773
(ii) 9587 –? = 7429 – 4358
Solution. (i) Let x – 1936248 = 1635773. Then, x = 1635773 + 1936248 = 3572021.
(ii) Let 9587 – x = 7429 – 4358.
Then, 9587 – x = 3071 ⇒ x = 9587 – 3071 = 6516.
Q 29. What could be the maximum value of Q in the following equation?
5P9 + 3R7 + 2Q8 = 1114
Solution. We may analyse the given equation as shown:
Clearly, 2 + P + R + Q = 11.
So, the maximum value of Q can be (11 – 2), i.e. 9 (when P = 0, R = 0).
Q 30. Which of the following numbers is divisible by 4?
(i) 67920594
(ii) 618703572
Solution. (i) The number formed by the last two digits in the given number is 94, which is not divisible by 4.
Hence, 67920594 is not divisible by 4.
(ii) The number formed by the last two digits in the given number is 72, which is divisible by 4.
Hence, 618703572 is divisible by 4
Q 31. Is 52563744 divisible by 24?
Solution. 24 = 3 × 8, where 3 and 8 are co-primes.
The sum of the digits in the given number is 36, which is divisible by 3. So, the given number is divisible by 3.
The number formed by the last 3 digits of the given number is 744, which is divisible by 8. So, the given number is divisible by 8. Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes.
So, it is divisible by 3 × 8, i.e. 24.
Q 32. What are the values of M and N respectively if M39048458N is divisible by both 8 and 11, where M and N are single-digit integers?
Solution. Since the given number is divisible by 8, it is obvious that the number formed by the last three digits, i.e.
58N is divisible by 8, which is possible only when N = 4.
Now, (sum of digits at even places) – (sum of digits at odd places)
= (8 + 4 + 4 + 9 + M) – (4 + 5 + 8 + 0 + 3)
= (25 + M) – 20 = M + 5, which must be divisible by 11.
So, M = 6.
Hence, M = 6, N = 4.
Q 33. Find the number of digits in the smallest number which is made up of digits 1 and 0 only and is divisible by 225.
Solution. 225 = 9 × 25, where 9 and 25 are co-primes.
Clearly, a number is divisible by 225 if it is divisible by both 9 and 25.
Now, a number is divisible by 9 if the sum of its digits is divisible by 9 and a number is divisible by 25 if the number formed by the last two digits is divisible by 25.
The smallest number which is made up of digits 1 and 0 and divisible by 225 = 11111111100.
Hence, the number of digits = 11.
Stay informed, Stay ahead and Stay inspired with MBA Rendezvous.