1.
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Answer: Option 'C'
Answer: Option 'C'
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 × 4) + 4 = 364.
2.
The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
Answer: Option 'C'
Other number = (11 x 7700/275) = 308.
3.
Four prime numbers are arranged in ascending order according to their magnitude. Product of first Three is 20,677 and the product of last three 33,263
Answer: Option 'D'
37
4.
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
Answer: Option 'C'
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.
5.
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point?
Answer: Option 'B'
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.