# problems on hcf and lcm

1. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
 A. 4 B. 7 C. 9 D. 13

Explanation:

Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43)

= H.C.F. of 48, 92 and 140 = 4.

2. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
 A. 276 B. 299 C. 322 D. 345

Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).

Larger number = (23 x 14) = 322.

3. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
 A. 4 B. 10 C. 15 D. 16

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

 In 30 minutes, they will toll together 30 + 1 = 16 times. 2

4. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
 A. 4 B. 5 C. 6 D. 8

Explanation:

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

5. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
 A. 9000 B. 9400 C. 9600 D. 9800