Question

A can do a work in 7days.If b is 40% more efficient to a, then in how many days can b do the same work

Answer 1

Answer :-

B works 5 days to complete the same work.

Solution :-

Number of days A can do a work = 7 days

Work done by A in one day = 1/7

40 % of A's work = 40/100 * 1/7

= 4/10 * 1/7

= 4/70

Given

Work done by B = 40 % more efficient than A

= 1/7 + 40 % of A's work

= 1/7 + 4/70

= (10 + 4)/70

= 14/70

= 1/5

i.e, B's one day's work = 1/5

So B works 5 days to complete the same work.

Answer 2

• A can do a work in 7 days.

=> Work done by A in 1 day = $\dfrac{1}{7}$

• If B is 40% more efficient to A.

But before finding the work done by B we have to find the 40% work of A

i.e.

=> 40% of A

=> $\dfrac{40}{100}$ of A

=> $\dfrac{4}{10}$ × $\dfrac{1}{7}$

=> $\dfrac{4}{70}$ ______ (eq 1)

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Also, said in question work of B (work done by B) is 40% more efficient to A.

i.e.

=> Work done by B = Work done by A + 40% work of A

=> $\dfrac{1}{7}$ + $\dfrac{4}{70}$ [From (eq 1)]

Take LCM

=> $\dfrac{10\:+\:4}{70}$

=> $\dfrac{14}{70}$

=> $\dfrac{1}{5}$

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In 5 days, B can do the same work.

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