Question
Next door lives a man with his son. They both work in the same factory. I watch them going to work through my window. The father leaves for work 10 min earlier than his son. One day I asked him about it and he told me he takes 30 min to walk to his factory ,whereas his son is able to cover the distance in only 20 min. I wondered if the father were to leave the house 5 min earlier than his son, how soon would the son catch up with the father?
Answer: Option B
:
B
Let speed of son be s,and speed of father be f,then
sf=32 (inverse ratio of the time) ( If D is the distance ,then Df=30and Ds=20which implies sf=32)
Let speed of dad be 20 and son be 30m/s
Now in 5 mins, the father covers 100 m
And son covers 150 m
The son covers 100 m in 3.33 min,
Hence the son will be able to catch up with the father after 8.3 mins from the start. Option (b).
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:
B
Let speed of son be s,and speed of father be f,then
sf=32 (inverse ratio of the time) ( If D is the distance ,then Df=30and Ds=20which implies sf=32)
Let speed of dad be 20 and son be 30m/s
Now in 5 mins, the father covers 100 m
And son covers 150 m
The son covers 100 m in 3.33 min,
Hence the son will be able to catch up with the father after 8.3 mins from the start. Option (b).
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