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by shashank.ism » Wed Feb 10, 2010 9:29 am
There are three runners viz , Nishant , Deepak and Mohit who jog on the same path. Nishant goes jogging every two days. Deepak goes jogging every four days. Mohit goes jogging every seven days. If its the first day that they started this routine, what is the total number of days that each person will jog by himself in the next seven weeks?

A) 12
B) 13
C) 14
D) 15
E) 16
[spoiler]Correct Answer: B[/spoiler]
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by sanjayism » Wed Feb 10, 2010 9:43 am
deepak goes jogging every four days and Nishant goes every two days, so deepak meet with nishant on everry altermate jogging day. and they all meet each other when deepak comes for jogging, so the no of days they will meet for jogging is equal to no of days deepak arrive for jogging.

so total no days, deepak will come for jogging =13
so, answer= B

49=1+(n-1)*4
=> n=1+50/4
=> n=1+12
n=13
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by thephoenix » Wed Feb 10, 2010 9:44 am
Nishant-------->2
Deepak-------->4
Mohit----------->7
tot days=7*7=49
N & D will coincide in lcm of 2,4=4 , 49/4=12(approx) days out of 49 days
d&M will coincide in lcm of 4,7=28 ---> in 49/28=1 day out of 49 days
N & M--->lcm 2,7=14--->49/14=3 days

now N runs for 49/2=24 days so alone=24-12days with D-3 days with M=9 days alone

for D =49/4=12-12 days with N=0

for M =49/7=7- 1 day with D-3 day with N=3 days alone
tot=9+3=12

total=1 day at starting where each one is jogging alone +12 alone daye=13
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