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# 25+ Quant Questions for CAT with SOLUTIONS

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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

Q 8. How  many  prime  numbers  are  there  between  100  to  200?

(a)  21

(b)  20

(c)  16

(d)  11

Q 9. The  number  of  three  digit  numbers  which  are  multiples  of  9  are

(a)  100

(b)  99

(c)  98

(d)  101

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

Q 11. What is  the  sum  of  all  natural  numbers  from  1  to  100?

(a)  5050

(b)  6000

(c)  5000

(d)  5052

## 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

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

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

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

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.

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

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

Q 27. 7m –[3n–{8m–(4n–10m)}] simplifies to

(a)  11m  –  5n

(b)  11m  –  7n

(c)  11n  –  7m

(d)  13n  –  11m

## 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.

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