As we know that ,
Maximum number of students = The greatest common divisor = HCF of 1001 and 910

Correct Option: A

As we know that ,
Maximum number of students = The greatest common divisor = HCF of 1001 and 910 = 91

Required time = LCM of 48, 72 and 108 seconds

∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds = 7 minutes 12 second

Correct Option: A

Required time = LCM of 48, 72 and 108 seconds

∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds = 7 minutes 12 second
∴ Required time = 10 a.m. + 7 minutes 12 second = 10 : 07 : 12 hours

LCM of p and q = 161
∴ pq = 23 × 7
∴ p = 23; q = 7
∴ 3q – p = 3 × 7 – 23

Correct Option: A

LCM of p and q = 161
∴ pq = 23 × 7
∴ p = 23; q = 7
∴ 3q – p = 3 × 7 – 23
The value of (3q – p) = 21 – 23 = – 2

We find LCM of 60 ,

∴ 60 = 2 × 2 × 3 × 5
i.e., Numbers = 2, 3, 4 and 5

Correct Option: B

We find LCM of 60 ,

∴ 60 = 2 × 2 × 3 × 5
i.e., Numbers = 2, 3, 4 and 5
∴ Required sum = 2 + 3 + 4 + 5 = 14

Required answer = LCM of 36, 40 and 48 seconds

∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

Correct Option: B

Required answer = LCM of 36, 40 and 48 seconds
= 720 seconds