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# Number System Questions For IBPS RRB Clerk

Download Top-20 IBPS RRB Clerk Number System Questions PDF. Number System questions based on asked questions in previous year exam papers very important for the IBPS RRB Assistant exam

Question 1: The product of two 2-digit numbers is 2160 and their H.C.F. is 12. The numbers are

a) (12, 60)

b) (72, 30)

c) (36, 60)

d) (60, 72)

Question 2: How many two digit numbers are divisible by 9?

a) 9

b) 8

c) 10

d) 11

Question 3: If 347P is divisible by 9, then what is the value of P?

a) 2

b) 3

c) 4

d) 7

Question 4: Which smallest number must be subtracted from 400, so that the resulting number is completely divisible by 7?

a) 6

b) 1

c) 2

d) 4

Question 5: Which smallest number to be subtracted from 300, so that the resulting number is completely divisible by 9?

a) 5

b) 6

c) 3

d) 1

Question 6: A number when divided by 18 leaves remainder 15. What is the remainder when the same number is divided by 6?

a) 3

b) 2

c) 1

d) 4

Question 7: If 123457Y is completely divisible by 8, then what will be the digit in place of Y?

a) 4

b) 5

c) 8

d) 6

Question 8: If a number is divided by 30 then it leaves 17 as a remainder. What will be the remainder when the same number is divided by 10?

a) 7

b) 3

c) 1

d) 2

Question 9: Which of the following number is divisible by 11?

a) 44433

b) 45332

c) 23581

d) 59609

Question 10: The LCM of two numbers is 162 and their HCF is 9. If one of the numbers is 18 then what is the other number?

a) 36

b) 81

c) 27

d) 162

Question 11: When the digits of a two natural number are interchanged, the number increases by 18. How many such two digit numbers exist?

a) 6

b) 8

c) 7

d) 9

Question 12: Find the number lying between 900 and 1000 which when divided by 38 and 57 leaves in each case a remainder 23.

a) 912

b) 926

c) 935

d) 962

Question 13: Each member of a society contributes as much rupees for a function as the number of members living in the society. However, for the event to be conducted, each member needs to contribute 10 rupees more. If the number of people in the society is 26 then what is the amount needed for conducting the function?

a) 936

b) 220

c) 660

d) 720

Question 14: On dividing a certain number by 378, we get 75 as the remainder. What will be the remainder when the same number is divided by 21?

a) 20

b) 0

c) 12

d) 1

Question 15: It is known that $0 < x < 1$. Which of the following relations is correct?

a) $x^2 > \sqrt{x} > x$

b) $\sqrt{x} < x < x^2$

c) $\sqrt{x} > x > x^2$

d) $x^2 < \sqrt{x} < x$

Question 16: A three digit number xyz is taken. The sum of the digits of this number is subtracted from the number. The resultant number is definitely divisible by ?

a) 11

b) 3 and 9

c) only 3

d) 3, 9 and 11

Question 17: A number on being divided by 36 leaves a remainder of 18. Which of the following is definitely true about this number?

a) The number is divisible by 18

b) The number is divisible by 12

c) The number is divisible by 24

d) More than one of the above

Question 18: What smallest number should be added to 1434 so that the sum is completely divisible by 17?

a) 13

b) 15

c) 9

d) 11

Question 19: The sum of all prime numbers between 45 and 63 is

a) 271

b) 220

c) 277

d) 224

Question 20: What digit should replace the blank for the number to be divisible by 11.
3859_572

a) 2

b) 5

c) 9

d) 4

Let the two numbers be $12x$ and $12y$ respectively where $x$ and $y$ are co-primes

Product of numbers = $(12x) \times (12y)=2160$

=> $xy=\frac{2160}{144}=15$

Possible pairs of $x$ and $y$ whose H.C.F. is 1 = $(3,5)$

$\therefore$ Required numbers = $(12 \times 3),(12 \times 5)$

= 36 , 60

=> Ans – (C)

2 digits numbers which are divisible by 9 are : 18,27,…..,99

Clearly, these numbers form an A.P. with first term, $a=18$, last term, $l=99$ and common difference, $d=9$

Let number of terms be $n$

=> Last term of an A.P. = $l=a+(n-1)d$

=> $99=18+(n-1)9$

=> $(n-1)9=99-18=81$

=> $(n-1)=\frac{81}{9}=9$

=> $n=9+1=10$

Thus, there are 10 two digit numbers that are divisible by 9.

=> Ans – (C)

If a number is divisible by 9, then the sum of its digits must also be divisible by 9.

Number : 347P

Sum of digits = $3+4+7+P=(14+P)$

Now, for above value to be divisible by 9, it should be equal to 18 (next highest multiple of 9)

=> $14+P=18$

=> $P=18-14=4$

=> Ans – (C)

On dividing 400 by 7, we get : $400=7\times57+1$

Thus, the smallest number which should be subtracted from 400 = 1

Also, $400-1=399$ is completely divisible by 7.

=> Ans – (B)

On dividing 300 by 9, we get : $300=9\times33+3$

Thus, the smallest number which should be subtracted from 300 = 3

Also, $300-3=297$ is completely divisible by 9.

=> Ans – (C)

The number when divided by 18 leaves remainder 15, => Number is of the form = $N=18k+15$, where $k$ is a whole number.

Now, when $N$ is divided by 6, we get : $\frac{18k+15}{6}$

= $(\frac{18k}{6})+(\frac{15}{6})$

$18k$ is completely divisible by 6, hence the second term will determine the remainder.

=> Remainder when 15 is divided by 6 is $15\%6=3$

=> Ans – (A)

Number : 123457Y

If a number is completely divisible by 8, then the last three digits of the number must also be divisible by 8.

=> $57Y$ must be divisible by 8 and the only three digit number starting with ’57’ which is divisible by 8 is = 576

=> $Y=6$

=> Ans – (D)

The number when divided by 30 leaves remainder 17, => Number is of the form = $N=30k+17$, where $k$ is a whole number.

Now, when $N$ is divided by 10, we get : $\frac{30k+17}{10}$

= $(\frac{30k}{10})+(\frac{17}{10})$

$\because30k$ is completely divisible by 10, hence the second term will determine the remainder.

=> Remainder when 17 is divided by 10 is $17\%10=7$

=> Ans – (A)

If the positive difference between the sum of even digits and odd digits (starting from unit’s place) is divisible by 11, then the number is also divisible by 11.

(A) : 44433 = $(3+4+4)-(3+4)=11-7=4$

(B) : 45332 = $(2+3+4)-(3+5)=9-8=1$

(C) : 23581 = $(1+5+2)-(8+3)=8-11=3$

(D) : 59609 = $(9+6+5)-(0+9)=20-9=11$

In the above numbers, only in the last option, 11 is divisible by 11, hence 59609 is divisible by 11.

=> Ans – (D)

We know that LCM*HCF = product of two numbers
=> 162*9 = 18*x
=> x = 162*9/18 = 9*9 = 81
Hence, the required number is 81.

Let the number be 10x + y
=> We have been given that
10y + x – 10x – y = 18
=> 9 (y – x) = 18
=> y – x = 2
Since, both x and y are natural numbers so, y can range from 3 to 9. Thus, there are 7 such natural numbers.

L.C.M. (38,57) = 114

Now, multiple of 114 between 900 and 1000 = 912

Now, the number which leaves remainder 23 = $912+23=935$

=> Ans – (C)

Let there are ‘x’ members in the society. So the amount needed for the function = x(x + 10) = $x^2 + 10x$
Here x = 26. So the required amount will be 26*36 = 936. Thus, option A is correct.

We have been given that a number on division by 378 leaves a remainder of 75. So the number can be expressed as
N = 378*k + 75
We can write this as
N = 18*k + 21*k + 75
Hence, we can see that remainder with 21 will be nothing but the remainder when 75 is divided by 21. This will be equal to 12. Hence, 12 is the correct answer.

We know that when the number is between 0 and 1, then higher the power smaller will be the number. Consider an example to understand this.
Let x = .25
So $\sqrt{x} = .5$
=> $x^2 = .0625$
We can see that the correct order is $\sqrt{x} > x > x^2$
Hence, option C is the correct answer.

We can express the given number as
100x + 10y + z
We have been given that x + y + z is subtracted from this number. So we have
100x + 10y + z – x – y – z = 99x – 9z = 9(11x – z)
Thus, it must be divisible by 3 and 9.

The number on division by 36 leaves a remainder of 18. So the number can be expressed as

N = 36k + 18

This can be re-written as

N = 18(2k + 1)

Thus, we can see that N must be divisible by 18. Hence, option A is definitely true.

We know that 17*85 = 1445
1445 is the smallest number above 1434 and divisible by 17.
So 1445 – 1434 = 11 must be added to 1434 to get a number which is divisible by 17.
Hence, option D is the correct option.

Number between 45 and 63 which are prime are – 47,53,59,61.
Required sum = 47 + 53 + 59 + 61 =220
Hence, option B is the correct option.