# Number System Questions for RRB Group-D Set-3 PDF

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## Number System Questions for RRB Group-D Set-3 PDF

Download Top-10 RRB Group-D Number System Questions set – 3 PDF. RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: Sum of two positive integers is 15 and product of them is 50. What will be the L.C.M. of the numbers?

a) 50

b) 30

c) 40

d) 10

Question 2: If the product of the two numbers is 525 and L.C.M. of the numbers is 105, what will be the H.C.F. of the numbers?

a) 3

b) 4

c) 5

d) 6

Question 3: 14 is the HCF of two numbers ‘p’ and ‘q’, where 14 < p < q. What is the minimum values of (p,q)?

a) 42, 28

b) 28, 14

c) 14, 28

d) 28, 42

Question 4: Find the greatest number that divides 82, 147 and 199 so as to leave the same remainder in each case.

a) 7

b) 11

c) 13

d) 17

Question 5: Find the largest 4 digit number that is divisible by 78.

a) 9996

b) 9988

c) 9976

d) 9984

Question 6: Find the remainder when $79^{79}+79$ is divided by 80.

a) 78

b) 1

c) 79

d) 2

Question 7: The product of two numbers is 7920 and their HCF is 12. Find the number of pairs of such numbers possible.

a) 1

b) 2

c) 3

d) 4

Question 8: The sum of first 50 natural numbers is:

a) 1225

b) 1250

c) 1300

d) 1275

Question 9: The largest 5 digit number divisible by 99 is

a) 99999

b) 99990

c) 99991

d) 99980

Question 10: How many numbers between 1 and 1000 is divisible by 9?

a) 111

b) 110

c) 108

d) 114

Let’s say one of the number is x, hence another number will be 15-x
Hence, their product will be = $15x – x^2 = 50$
So after solving the above equation, we will get numbers as 10 and 5 and their L.C.M. will be 10

As we know that the product of the two numbers = (L.C.M. of the numbers)*(H.C.F. of the numbers)
Hence, 525 = 105*(H.C.F.)
H.C.F. = 5

Let the two numbers be 14x and 14y, where x and y are co-prime.
x, y > 1. The least possible values of x and y are 2 and 3 respectively.
So, the value of p is 14*2 = 28 and the value of q is 14*3 = 42

147 – 82 = 65
199 – 147 = 52
The HCF of these differences is the number that divides all the three numbers leaving the same remainder.
65 = 13 * 5
52 = 13 * 4
=> HCF = 13
Hence, 13 is the answer.

78*10 = 780
78 * 3 = 234
On adding the above, we get 78 * 13 = 1014
=> 78 * 130 = 10140
78 * 2 = 156
On subtracting the above, we get 78 * 128 = 9984
=> 9984 is the answer.

($79^{79}+79$ mod 80) can be written as ($(-1)^{79}+79$ mod 80)
= (-1 + 79) mod 80
= 78 mod 80
= 78

Let the two numbers be 12x and 12y
12x*12y = 7920
xy = 55 = 5*11 = 55*1
The two numbers can be [(12*5),(12*11)],[(12*55),(12*1)]

Sum of first n natural numbers = n(n+1)/2
Sum of first 50 natural numbers = 50*51/2 = 1275

Largest 5 digit number is 99999. This number on division by 99 gives 9 as a remainder.
The largest 5 digit number is 99999-9 = 99990