Dear reader here we are providing some question on H.C.F. and L.C.M which can be useful in your upcoming exam.
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
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
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
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
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
6. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
A. 101
B. 107
C. 111
D. 185
7. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
A. 40
B. 80
C. 120
D. 200
8. The G.C.D. of 1.08, 0.36 and 0.9 is:
A. 0.03
B. 0.9
C. 0.18
D. 0.108
9. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
A. 1
B. 2
C. 3
D. 4
10. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
A. 74
B. 94
C. 184
D. 364
ANSWERS WITH EXPLANATION:
1. (A)
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
2. (C)
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
3. (D)
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 (30/2) + 1 = 16 times.
4. (A)
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. (C)
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 - 399) = 9600.
6. (C)
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107,
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
7. (A)
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
8. (C)
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
H.C.F. of given numbers = 0.18.
9. (B)
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
10. (D)
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
