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2 workers ,one old and one young,live together and work at the same office.the old man takes 30 mins where as the young man takes only 20 mins to reach the office.when will the young man catch up the old man ,if the old man starts at 10.00am and the young man starts at 10.05am?   ( 1 Marks )

 A. 10:25 AM B. 10:10 AM C. 10:05 AM D. 10:15 AM

Let distance = x

speed of young man = $\frac {x}{20}$

Speed of old man = $\frac {x}{30}$

relative speed $\frac {x}{20}-\frac{x}{30}=\frac{x}{60}$

Distance traveled by old man in 5 min =$\frac{x}{30}\times 5=\frac{x}{6}$

Time taken by the young man to catch up the old man  =$\frac{\frac{x}{6}}{\frac{x}{60}}$ =10 min

So,young man will catch up man the old man at 10:15 am.

Hence,Option(D)10:15am.
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+1 vote

let distance from there home to office be 60 km

OLD MAN

here old man takes 30 min to cover 60 km

so in 1 min old man covers 2km

YOUNG MAN

young man takes 20  min to cover 60 km

in one min it covers 3 km

old man step out at 10 am

young man step out 5 minutes late

so from 10 am to 10:05 am old man  covers  10 km

now suppoae time at clock is 10: 15 am so now old man covers 30 km and young also covers .30 km

so at this time young man catches the old man hence answer is

10:15 am

edited