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Numerical Ability
Time Distance and Speed
two people one young and one old live together,and work in the same office it takes 20min and 30min respectively to walk from home to office when the young man catchup with the old man if the oldman starts at 10am and young at 10.05am?
Read Solution (Total 10)
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- 10.15 am is the answer.
- 11 years agoHelpfull: Yes(8) No(2)
- at 10.15 am...
- 11 years agoHelpfull: Yes(4) No(2)
- let x hours after he will meet.
so ,
distance travel by young boy in (5+x) min =distance travel by old man in x min.
let d be a total distance.
so
(5+x)d/20=x*d/30
x=15min
10:00 am +15 min
=10:1 am - 10 years agoHelpfull: Yes(3) No(0)
- let the distance be x..
now dis covered is same , let their speeds are y and o..
then y*20/60=o*30/60;
i.e. o=2y/3;
so relative speed is y-2y/3=y/3
with this relative velocity young has to cover (2y/3)*(5/60) m
i.e. y/180 m so it will take time =(y/180)/(y/3)=1/60 hr=1 min.
so he would catch at 10:06 am... - 11 years agoHelpfull: Yes(2) No(4)
- o yes 10:15 mistakenly done y/180 instead of y/18....
- 11 years agoHelpfull: Yes(1) No(0)
- 10;15am answer
- 11 years agoHelpfull: Yes(1) No(0)
- 10:15am is the answer
- 11 years agoHelpfull: Yes(1) No(0)
- above will v 10:15 am
- 10 years agoHelpfull: Yes(1) No(0)
- if young man starts 10 minute late ,he catches after 20 mintutes
10 min.........20 min
5 min............(20/10)*5=10 min late
ans...10:15 - 10 years agoHelpfull: Yes(1) No(0)
- let x hours after he will meet.
so ,
distance travel by young boy in (5+x) min =distance travel by old man in x min.
let d be a total distance.
so
(5+x)d/20=x*d/30
x=15min
10:00 am +15 min
=10:15am - 10 years agoHelpfull: Yes(0) No(0)
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