# HCF and LCM Quiz Set 007

### Question 1

The sum of LCM and HCF of two numbers is 120 and LCM is 59 times the HCF. If one of the two numbers is 118, then the other number is?

A

2.

B

4.

C

3.

D

5.

Soln.
Ans: a

Let L be the LCM, and H the HCF. Then H + L = 120, and L = 59H. Solving these for H, we get H = \$120/{59 + 1}\$ = 2, and L = 118. The product of the two numbers is equal to the product of the lcm and hcf = 2 × 118 = 236. The other number is 236/118 = 2.

### Question 2

Which is the second smallest number, which, when divided by 12 and 60, leaves the remainder 7?

A

127.

B

67.

C

187.

D

53.

Soln.
Ans: a

The LCM of 12 and 60 = 60. All the numbers that leave remainder 7 upon being divided by either of the given two numbers are of the form 60k + 7. When k = 1, we get the smallest number, and k = 2 gives the next required number = 127.

### Question 3

Which is the smallest number, which, when divided by 24 and 85, leaves the remainder 8?

A

2048.

B

2050.

C

2047.

D

2051.

Soln.
Ans: a

The LCM of 24 and 85 = 2040, is the smallest number, which, when divided by either of them leaves the remainder 0. Now adding, 2040 + 8 = 2048 is the required number.

### Question 4

What is the LCM of \${3/7}\$, \${6/17}\$ and \$1{3/8}\$?

A

66.

B

67.

C

68.

D

69.

Soln.
Ans: a

The LCM of numerators 3, 6 and 11 is 66. The HCF of denominators 7, 17 and 8 is 1. So the required LCM of the given fractions is 66/1 = 66.

### Question 5

Three lights are flashing at regular intervals. They, respectively, flash after 6, 12 and 18 seconds. How many times do they together flash in 864 seconds?

A

25 times.

B

26 times.

C

24 times.

D

27 times.

Soln.
Ans: a

They will together flash after the LCM of their times. The LCM of 6, 12 and 18 = 36. So they will flash 1 + \$864/36\$ = 25 times.