Question

A alone can finish i piece of work in 42 eays. B is 20% more efficient than a and c is 40% more efficient 5hsn



b. In how many days b and c working together can finish the same piece of work.

Answer 1

● A alone can finish i piece of work in 42 days.

In 1 day, A can do = 1/42 of the work

● B is 20% more efficient than A.

So B can work 20% faster than A.

In order to complete the work, B will take = 42 ÷ 120/100 = 42 × 100/120 = 35 days.

In 1 day, B can do = 1/35 of the work

● C is 40% more efficient than B

So C can work 40% faster than B

In order to complete the work, C will take = 35 ÷ 140/100 = 35 × 100/140 = 25 days.

In 1 day, C can do = 1/25 of the work

● Together, B and C can do in 1 day

$\frac{1}{35} + \frac{1}{25} = \frac{5+7}{175} = \frac{12}{175}$

Days required to finish the work is:

$1 \div \frac{12}{175} = \frac{175}{12} = 14 \: \frac{7}{12} \: days$

Hence, they can finish it in 14 7/12 days.

Answer 2

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