RRB Practice Questions | Mathematics – Day-06

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Mathematics is a complex web. So only most of the applicant’s thought is RRB Mathematics questions are tough to work out. It’s really not, to score the maximum mark in this section you need of regular practice on RRB Maths every day.

Start to practice Quiz for RRB Mathematics section and reveal the answers with detailed explanation. We have taken much care on the explanation part that to understand the RRB Mathematics questions easily. Take RRB Mathematics quiz regularly to perform better on  RRB Maths section without any hurdle.

RRB Exam Mathematics Day – 06

maximum of 10 points
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1) Two numbers are in the ratio of 11:17. The product of their H.C.F and L.C.M is 26928.The sum of the numbers is

a) 336

b) 204

c) 132

d) 156

2) H.C.F and L.C.M of three numbers 3600, 800 and p are 120 and 25 × 52 × 32 × 7 respectively. Find p.

a) 23 x 32 x 50 x 7

b) 23 x 3 x 52 x 7

c) 23 x 3 x 5 x 7

d) 23 x 32 x 5 x 73

3) L.C.M of two numbers is 432 times their H.C.F. The sum of H.C.F and L.C.M is 1732. If one number is 64 find the other number.

a) 99

b) 108

c) 120

d) 180

4) Find the least number which when divided by 2, 6, 9 and 12 leaves the reminder 2 in each case.

a) 36

b) 34

c) 74

d) 38

5) Out of three numbers the product of the first two is 3168 and that of the last two is 1683. The H.C.F of all three numbers is 1. The sum of the three numbers is

a) 148

b) 180

c) 30

d) 150

6) Three numbers are in the ratio 14:17:11 and the L.C.M is 47124. Their H.C.F is

a) 25

b) 16

c) 18

d) 12

7) The least number which should be added to 2867 so that the sum is exactly divisible by 2, 3, 6 and 5 is

a) 14

b) 17

c) 13

d) 21

8) Ram, Lakshman and Bharath start at the same time in the same direction to run around a circular park. Ram completes a around in 64 sec, Lakshman in 80 sec and Bharath in 72 sec, all starting at the same point. After what time will they meet again at the starting point?

a) 52min

b) 48min

c) 45min 24sec

d) 52 min 28sec

9) Let the lest number of five digits, which divided by 4,6 and 10 leaves in each case a remainder of 1, be z . The sum of the digits in z is

a) 4

b) 5

c) 3

d) 2

10) The ratio of two numbers is 2 : 7 and their H.C.F is 4. Their L.C.M is

a) 64

b) 58

c) 84

d) 56

Answers :

1) Answer: a)

First number × second number = H.C.F × L.C.M

11x × 17x = 26928

187x2 = 26928

x2= 144

x = 12

Sum of the numbers = 11×12 + 17×12 = 336

2) Answer: c)

3600 =24 x 32 x 52

800 = 25 x 52

L.C.M = 25 x 52 x 32 x 7

H.C.F =120

P = 23 × 3 ×5 × 7

3) Answer: b)

Let the H.C.F be h and L.C.M be 432h

i.e., 432h + h = 1732

h= 1732/433= 4

L.C.M = 432 x 4 = 1728

The required number = (4 x 1728) / 64 = 108

4) Answer: d)

Required number = L.C.M (2, 6, 9, 12) +2

= 36 + 2 = 38

5) Answer: a)

Let the three numbers be a, b and c

a × b = 3168 and b × c = 1683

Since b is common and so it is given by

H.C.F of 3168 and 1683 = 99

i.e., a = 3168/99 = 32 and c =1683/99 = 17

The three numbers are 32, 99, 17

a + b + c = 148

6) Answer: c)

The numbers are 14x, 17x and 11x.

Where x is the H.C.F

L.C.M = (14×17×11)x = 2618x = 47124

x = 18

7) Answer: c)

L.C.M of 2, 3, 6, 5 = 30

On dividing 2867 by 30 the reminder is 17

Therefore, the number to be added = 30 – 17 =13

8) Answer: b)

L.C.M of 64, 80 and 74 is 2880 sec

i.e., 48 min

9) Answer: a)

The L.C.M of 4, 6 and 10 = 60

When 10000 is divided by 60 remainder is 40.

Least number when divided by 60 = 10000 + (60 – 40) = 10020

i.e., z = 10020 + 1 = 10021

Sum of the digits = 4

10) Answer: d)

Since every number is a multiple of its H.C.F, the two numbers are 2 x 4 and 7 x 4 i.e., 8 and 28.

L.C.M of 8 and 28 is 56.

 

 

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