CAT Questions based on Number System | Download [PDF]

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CAT Number System Quesitons
CAT Number System Quesitons

CAT Questions based on Number System
Download CAT 2020 CAT Number System Previous Year Questions PDF asked questions in CAT exam. Practice CAT 2020 Number System Questions PDF for CAT exam to understand the type and level of questions asked in the exam.

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Question 1: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct are there with 7 coming before 3?

Question 2: How many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ ?

a) 2018

b) 2019

c) 2017

d) 2020

Question 3: How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

a) 42

b) 41

c) 40

d) 43

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Question 4: Let N, x and y be positive integers such that $N=x+y,2<x<10$ and $14<y<23$. If $N>25$, then how many distinct values are possible for N?


Question 5: 
How many integers in the set {100, 101, 102, …, 999} have at least one digit repeated?


Question 6:
Let m and n be natural numbers such that n is even and $0.2<\frac{m}{20},\frac{n}{m},\frac{n}{11}<0.5$. Then $m-2n$ equals

a) 3

b) 1

c) 2

d) 4

Question 7: If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is

a) 49

b) 56

c) 59

d) 46

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Question 8: 
The mean of all 4-digit even natural numbers of the form ‘aabb’,where $a>0$, is

a) 4466

b) 5050

c) 4864

d) 5544

Question 9: How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Question 10: In a row at a bus stop, A is 7th from the left and B is 9th from the right. They both interchange their
positions. A becomes 11th from the left. How many people are there in the row?

a) 18

b) 19

c) 20

d) 21

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Question 11: 
The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

a) 58

b) 85

c) 50

d) 95

Question 12: In Roman Numerals, a number has been written as MMXVIII. In Arabic numbers it will be …………
(Note:- DO NOT include spaces in your answer)

Question 13: The number $37^{371}-26^{371}$ is divisible by:

a) 10

b) 11

c) 12

d) 15

Question 14: The average of 4 distinct prime numbers a, b, c, d is 35, where a < b < c < d. a and d are equidistant from 36 and b and c are equidistant from 34 and a, b are equidistant from 30 and c and d are equidistant from 40. The difference between a and d is:

a) 30

b) 14

c) 21

d) Cannot be determined

Question 15: A number G236G0 can be divided by 36 if G is:

a) 8

b) 6

c) 1

d) More than one values are possible.

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Answers & Solutions:

1) Answer: 315

Here there are two cases possible

Case 1: When 7 is at the left extreme

In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)

So total ways 3(8)(7)= 168

Case 2: When 7 is not at the extremes

Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left)

Hence in total 3(7)(7)=147 ways

Total ways 168+147=315 ways

2) Answer (A)

$ab\ =\ 4^{2017}=2^{4034}$

The total number of factors = 4035.

out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.

And since the given number is a perfect square we have one set of two equal factors.

.’. many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ = 2018.

3) Answer (B)

The number of multiples of 2 between 1 and 120 = 60

The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12

The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7

Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41

4) Answer: 6

Possible values of x = 3,4,5,6,7,8,9

When x = 3, there is no possible value of y

When x = 4, the possible values of y = 22

When x = 5, the possible values of y=21,22

When x = 6, the possible values of y = 20.21,22

When x = 7, the possible values of y = 19,20,21,22

When x = 8, the possible values of y=18,19,20,21,22

When x = 9, the possible values of y=17,18,19,20,21,22

The unique values of N = 26,27,28,29,30,31

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5) Answer: 252

Total number of numbers from 100 to 999 = 900

The number of three digits numbers with unique digits:

_ _ _

The hundredth’s place can be filled in 9 ways ( Number 0 cannot be selected)

Ten’s place can be filled in 9 ways

One’s place can be filled in 8 ways

Total number of numbers = 9*9*8 = 648

Number of integers in the set {100, 101, 102, …, 999} have at least one digit repeated = 900 – 648 = 252

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6) Answer (B)

$0.2<\frac{n}{11}<0.5$

=> 2.2<n<5.5

Since n is an even natural number, the value of n = 4

$0.2<\frac{m}{20}<0.5$  => 4< m<10. Possible values of m = 5,6,7,8,9

Since $0.2<\frac{n}{m}<0.5$, the only possible value of m is 9

Hence m-2n = 9-8 = 1

7) Answer (D)

Since $c<9$, we can have the following viable combinations for $b\times\ c\ =96$ (given our objective is to minimize the sum):

$48\times\ 2$ ; $32\times3$ ; $24\times\ 4$ ; $16\times6$ ; $12\times8$

Similarly, we can factorize $a\times\ b\ = 432$ into its factors. On close observation, we notice that $18\times24\ and\ 24\ \times\ 4\ $ corresponding to $a\times b\ and\ b\times\ c\ $ respectively together render us with the least value of the sum of $a+b\ +\ c\ \ =\ 18+24+4\ =46$

Hence, Option D is the correct answer.

8) Answer (D)

The four digit even numbers will be of form:

1100,1122,1144…1188,2200,2222,2244….9900,9922,9944,9966,9988

Their sum ‘S’ will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+…)….+(9900+9900+22+9900+44+9900+66+9900+88)

=> S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)….+9900*5+(22+44+66+88)

=> S=5*1100(1+2+3+…9)+9(22+44+66+88)

=>S=5*1100*9*10/2 + 9*11*20

Total number of numbers are 9*5=45

.’. Mean will be S/45 = 5*1100+44=5544.

Option D

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9) Answer: 21

Let the number be ‘abc’. Then, $2<a\times\ b\times\ c<7$. The product can be 3,4,5,6.

We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12 numbers fulfilling the criteria.

We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3 more distinct numbers.

We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6 additional distinct numbers.

Thus a total of 12 + 3 + 6 = 21 such numbers can be formed.

10) Answer (B)

After interchanging the positions, A becomes 11th from the left. So B must have been 11th from the left and 9th from the right before the interchange. Hence the total number of people = 11+9-1=19

11) Answer (C)

Assume the numbers are a and b, then ab=616

We have, $\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$ = $\ \frac{\ 157}{3}$

=> $\ 3\left(a^3-b^3\right)\ =\ 157\left(a^3-b^3+3ab\left(b-a\right)\right)$

=> $154\left(a^3-b^3\right)+3*157*ab\left(b-a\right)$ = 0

=> $154\left(a^3-b^3\right)+3*616*157\left(b-a\right)$ = 0        (ab=616)

=>$a^3-b^3+\left(3\times\ 4\times\ 157\left(b-a\right)\right)$    (154*4=616)

=> $\left(a-b\right)\left(a^2+b^2+ab\right)\ =\ 3\times\ 4\times\ 157\left(a-b\right)$

=> $a^2+b^2+ab\ =\ 3\times\ 4\times\ 157$

Adding ab=616 on both sides, we get

$a^2+b^2+ab\ +ab=\ 3\times\ 4\times\ 157+616$

=> $\left(a+b\right)^2=\ 3\times\ 4\times\ 157+616$ = 2500

=> a+b=50

12) Answer: 2018

In Roman numerals,

The values of M = 1000, D = 500, C = 100, L=50, X = 10, V = 5, 1 = 1.

The value of MMXVIII = 1000+1000+10+5+1+1+1 = 2018

2018 is the correct answer.

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13) Answer (B)

Option A:

$37^{371}-26^{371}$ mod 10

=$7^{371}-6^{371}$

Since 7 and 10 are co-prime to each other

E(10) = 10*($\left(1-\ \frac{\ 1}{2}\right)\left(1-\ \frac{\ 1}{5}\right)$

=4

$7^{4}$ mod 10 = 1

$7^{4*92}*7^{3}$ mod 10 = $1*7^{3}$ mod 10 =3

$6^{371}$ mod 10 = 0

Hence $37^{371}-26^{371}$ is not divisible by 10

Option B:

$37^{371}-26^{371}$ mod 11

$4^{371}-4^{371}$ mod 11

Hence $37^{371}-26^{371}$ is divisible by 11

Option C:

$37^{371}-26^{371}$ mod 12

$1^{371}-2^{371}$ mod 12

Hence $37^{371}-26^{371}$ is not divisible by 12

Option D:

$37^{371}-26^{371}$ mod 15

$7^{371}-11^{371}$ mod 15

$37^{371}-26^{371}$ is not divisible by 15

B is the correct answer.

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14) Answer (B)

Given,

The average of the four prime numbers = 35.

a + b + c + d = 35 * 4 = 140.

Since a and d are equidistant from 36.

a + d = 72 — Eq (1)

b + c = 68 — Eq (2)

a + b = 60 — Eq (3) and c + d = 80 — Eq (4)

Using equation (3) let us look for the prime values of a and b and the corresponding values of c and d using Eq 2 and 1.

Also given that a < b < c < d.

(a, b, c, d) = 29, 31, 37, 43

d – a = 43 – 29 = 14

B is the correct answer.

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15) Answer (A)

36 = 9*4 Since 9 and 4 are co-prime to each other, we can say that for a number to be divisible by 36 it must be divisible by both 9 and 4.

For G236G0, to be divisible by 4, last two digits should be divisible by 4.

Hence G0 should be a multiple of 4.

Possible values of G are 2, 4, 6, 8, 0

For the number to be divisible by 9, the sum of the digits should be a multiple of 9

G+2+3+6+G+0 = 11+2G should be a multiple of 9

If G = 0, 11+2G is not a multiple of 9

If G = 2, 11+2G is not a multiple of 9

If G = 4, 11+2G is not a multiple of 9

If G = 6, 11+2G is not a multiple of 9

If G = 8, 11+2G is a multiple of 9.

Hence 8 is the correct answer.

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