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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 
asked in Quantitative Aptitude by (456 points)   | 124 views

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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.
answered by (9.7k points)  
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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  

answered by (56 points)  
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