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

A. 7/8
B. 13/16
C. 31/40
D. 63/80
Explanation:
L.C.M of 8, 16, 40 and 80 = 80.
7/8 = 70/80; 13/16 = 65/80; 31/40 = 62/80
Since, 70/80 > 63/80 > 65/80 > 62/80,
So, 7/8 > 63/80 > 13/16 > 31/40
So, 7/8 is the largest.

A. 276
B. 299
C. 322
D. 345
Clearly, the numbers are (23 * 13) and (23 * 14). Larger number = (23 * 14) = 322.

A. 24
B. 48
C. 56
D. 60
Explanation:
H.C.F of two numbers divides their L.C.M exactly. Clearly, 8 is not a factor of 60.

A. 1
B. 117
C. equal to their H.C.F
D. Cannot be calculated
Explanation:
H.C.F of co-prime numbers is 1.
So, L.C.M = 117/1 = 117.

A. 55/601
B. 601/55
C. 11/120
D. 120/11
Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 * 120 = 600.
Required sum = 1/a + 1/b = (a + b)/ab = 55/600 = 11/120.

A. 10
B. 46
C. 70
D. 90
Explanation:
Let the numbers be x and (100 – x).
Then, x(100 – x) = 5 * 495
x2 – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10.

A. 279
B. 283
C. 308
D. 318
Explanation:
Other number = (11 * 7700)/275 = 308.

A. 12
B. 48
C. 84
D. 108
Explanation:
Let the numbers be x and 4x. Then, x * 4x = 84 * 21 x2 = (84 * 21)/4 = x = 21.
Hence, larger number = 4x = 84.

A. 40
B. 80
C. 120
D. 200
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M = 60x. So, 60x = 2400 or x = 40.
The numbers are (3 * 40), (4 * 40) and (5 * 40).
Hence, required H.C.F = 40.

A. 28
B. 32
C. 40
D. 64
Explanation:
Let the numbers be 2x and 3x.
Then, their L.C.M = 6x. So, 6x = 48 or x = 8.
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.