**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.8

B.11

C.13

D.16

Answer

Option (d) is correct

**Explanation –**

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds,

i.e, 2 minutes.In 30 minutes,

they will toll together (30/2 + 1) = 16

**The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:**

A.68

B.98

C.180

D.364

Answer

Option (d) is correct

**Explanation –**

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.

**The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:**

A.11115

B.15110

C.15130

D.15310

Answer

Option (b) is correct

**Explanation –**

Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10.

Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10

= 15120 – 10 = 15110

**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:**

A.269

B.275

C.308

D.310

Answer

Option (c) is correct

**Explanation –**

Other number =[11 x 7700]/275 = 308

**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 meet again at the starting point?**

A.15 minutes 15 seconds

B.42 minutes 30 seconds

C.42 minutes

D.46 minutes 12 seconds

Answer

Option (d) is correct

**Explanation –**

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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