# Problems on H.C.F and L.C.M

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

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

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

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes). 4. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then the sum of the digits in N is:

A. 4
B. 5
C. 6
D. 8

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