Question

A and B can do a work in 20 days. When A works at 60% capacity, B has to work at 150% capacity to finish the work. Find in how many days the faster one will finish the work alone.

Answer 1

Let x be the number of days the faster one (let's assume A is faster) will finish the work alone.

When A works at 60% capacity and B works at 150% capacity, their combined work rate is \(\frac{3}{5}+\frac{3}{2}\)​ of the normal capacity.

The equation \((\frac{3}{5}+\frac{3}{2}).\frac{1}{x}=\frac{1}{20}\)​ is derived from the fact that A and B together can finish the work in 20 days.

Solving the equation, we find x=36.

Therefore, the faster one (A) will finish the work alone in 36 days.

Answer 2

Let's assume W be the total amount of work.
And a and b be the efficiencies of A and B respectively.
According to the question :
⇒ a + b = \(\frac{W}{20}\) (1 day work)     ….. (i)
And given that A is doing only 60% : \(\frac{3a}{5}\)
B is doing 150% : \(\frac{3b}{2}\)
Now , using this ,we get :
⇒ \(\frac{3a}{5}+\frac{3b}{2}=\frac{W}{20}\) (1 day work)

⇒ a + b = \(\frac{3a}{5}+\frac{3b}{2}\)

⇒ \(\frac{a}{b}=\frac{4}{5}\)
This implies that A is more efficient person.
By using equation (i) , we get :
⇒ \(a+\frac{4b}{5}=\frac{W}{20}\)

⇒ \(\frac{9}{5}a=\frac{W}{20}\)

⇒ \(a=\frac{W}{36}\)

Therefore, A takes 36 days to finish the given work.