Working together A, B and C can complete a certain work in 40 days. B and C together can complete thesame work in 60 days. B is 25% more efficient than A. In how many days will C alone complete 40% ofthe same work?

Correct option is B

Given :

A + B + C can complete the work in 40 days

B + C can complete the work in 60 days

B is 25% more efficient than A

Find time taken by C alone to complete 40% work

Formula Used :
Work = Rate × Time

Solution :

Let total work = 1 unit

A + B + C =$$\frac{1}{40}$$​​

B + C =$$\frac{1}{60}$$​​

Subtracting,

A = $$\frac{1}{40} - \frac{1}{60}$$​​

​$$= \frac{3 - 2}{120}$$​​
=$$\frac{1}{120}$$​​

So, efficiency of A =$$\frac{1}{120}$$​​

B is 25% more efficient than A:

B = 1.25$$A = \frac{5}{4} \times \frac{1}{120} = \frac{1}{96}$$​​

Now,

B + C =$$\frac{1}{60}$$​​

C =$$\frac{1}{60} - \frac{1}{96}$$​​

C =$$\frac{8 - 5}{480} = \frac{3}{480} = \frac{1}{160}$$​​

So, C alone can complete full work in 160 days

Time taken by C to complete 40% work:

= 16$$0 \times 0.4 = 64$$ days
C alone will complete 40% of the work in 64 days.

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