A man is twice as fast as a woman and a woman is twice as fast as a boy. If all of them i.e. a man, a woman and a boy can finish a work together in 4 days, in how many days a boy shall do it alone?
A man is twice as fast as a woman and a woman is twice as fast as a boy. If all of them i.e. a man, a woman and a boy can finish a work together in 4 days, in how many days a boy shall do it alone?
1). 28 days
2). 21 days
3). 14 days
4). 7 days
1 answers
Assume the efficiency of a boy = B units/day
As a woman is twice as fast as a boy so the efficiency of a woman = 2B units/day
As a man is twice as fast as a woman so the efficiency of a man = 4B units/day
According to the question, all of them i.e. a man, a woman and a boy can finish a work together in 4 days.
In 4 days total work done by a man, a woman and a boy = 4 × (B + 2B + 4B) = 28B units.
∴ Total work = 28B units.
Time required for a boy to do the whole work alone = $(\frac{{Total\;work\;}}{{efficiency\;of\;a\;boy}}\; = \frac{{28B}}{B} = 28\;days)$
Time required for a boy to do the whole work alone = 28 days.Other Questions
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