A) 10: 30 AM.
B) 10 : 31 A.M.
C) 10:59 A.M.
D) 11 A.M.
Correct Answer: B
Solution :
H.C.P. of 60 and 62 = 2\[\times \]2\[\times \]15\[\times \]31 =1860 Hence, \[\frac{1860}{60}\,\min =31\min \] \[\therefore \] Hence, they would beep together at the earliest is 10:31 A.M.You need to login to perform this action.
You will be redirected in
3 sec