# HCF and LCM Quiz Set 012

### Question 1

Which is the least number which must be added to 7347 so that it becomes divisible by 21, 75 and 98?

A

3.

B

4.

C

2.

D

5.

Soln.
Ans: a

The LCM of 21, 75 and 98 = 7350. The lcm when divided by either of these numbers leaves the remainder 0. The minimum number to be added is 7350 - 7347 = 3.

### Question 2

What is the HCF of (10 × 12), (12 × 16) and (16 × 10)?

A

8.

B

16.

C

4.

D

24.

Soln.
Ans: a

The required HCF is product of the three numbers divided by their LCM. The LCM of 10, 12 and 16 is 240. So the required HCF = \${10 × 12 × 16}/240\$ = 8. Please note that LCM(n1, n2, n3) × HCF(n1 × n2, n2 × n3, n3 × n1) = n1 × n2 × n3.

### Question 3

The HCF of two numbers is 13. The factors of their LCM are 13, 3 and 43. Which is the greater of the two numbers?

A

559.

B

1118.

C

279.

D

1677.

Soln.
Ans: a

The numbers are 13 × 3 = 39, and 13 × 43 = 559. The greater of them is 559.

### Question 4

Three cyclists are cycling in a circular track. They, respectively, take 3, 11 and 18 minutes to complete the circle once. After how many minutes will they all again meet at a single point?

A

198 minutes.

B

396 minutes.

C

99 minutes.

D

199 minutes.

Soln.
Ans: a

The answer lies in finding the LCM of their times. The LCM of 3, 11 and 18 = 198.

### Question 5

Which is the second smallest number, which, when divided by 18 and 90, leaves the remainder 2?

A

182.

B

92.

C

272.

D

88.

Soln.
Ans: a

The LCM of 18 and 90 = 90. All the numbers that leave remainder 2 upon being divided by either of the given two numbers are of the form 90k + 2. When k = 1, we get the smallest number, and k = 2 gives the next required number = 182.