1.

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

**Answer: Option 'C'**

**L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 - 37) = 23.**

2.

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:

**Answer: Option 'D'**

**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**

3.

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 'B'**

**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.**

4.

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?

**Answer: Option '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**

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.**

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