Question

A, B and C can do a work together in a certain number of days. If A leaves after half the work is done, then the work takes 4 more days for completion, but if B leaves after half the work is done, the work takes 5 more days for completion. If A takes 10 more days than B to do the work alone, then in how many days can C alone do the work?​

Answer 1

Answer:

C alone can do the work in 52.5 days

Step-by-step explanation:

Let B can do the work in x days

then, A can do the work = x + 10 days

In 1 day B does = 1/x work

In 1 day A does = 1/(x+10) work

Let C does work in y days

Then in 1 day C does = 1/y work

Let A, B, C can finish the work in n days

Then

$\frac{1}{x+10}+\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$   ............(a)

A, B, C can finish half work in = n/2 days

According to the question

B and C can do 1/2 work in 4 more days days

Thus

$(\frac{n}{2}+4)(\frac{1}{x}+\frac{1}{y})=\frac{1}{2}$

or, $\frac{1}{x}+\frac{1}{y}=\frac{1}{n+8}$  ......... (1)

Also

$(\frac{n}{2}+5)(\frac{1}{x+10}+\frac{1}{y})=\frac{1}{2}$

or, $\frac{1}{x+10}+\frac{1}{y}=\frac{1}{n+10}$   ............(2)

Subtracting eq (1) from eq (a) we get

$\frac{1}{x+10}=\frac{1}{n}-\frac{1}{n+8}$

$\implies \frac{1}{x+10}=\frac{8}{n(n+8)}$

$\implies x+10=\frac{n(n+8)}{8}$       ............ (i)

Subtracting eq (2) from eq (a) we get

$\frac{1}{x}=\frac{1}{n}-\frac{1}{n+10}$

$\implies \frac{1}{x}=\frac{10}{n(n+10)}$

$\implies x=\frac{n(n+10)}{10}$   .................(ii)

From (i) and (ii)

$\frac{n(n+10)}{10}+10=\frac{n(n+8)}{8}$

$\implies \frac{n(n+8)}{8}-\frac{n(n+10)}{10}=10$

$\implies \frac{n}{40}(5n+40-4n-40)=10$

$\implies n^2=400$

$\implies n=20$

Therefore,

$x=\frac{20(20+10)}{10}$

$\implies x=60$

Thus, from eq (1)

$\frac{1}{y}=\frac{1}{20+8}-\frac{1}{60}$

$\implies \frac{1}{y}=\frac{1}{28}-\frac{1}{60}$

$\implies \frac{1}{y}=\frac{60-28}{1680}$

$\implies \frac{1}{y}=\frac{32}{1680}$

$\implies y=\frac{1680}{32}$

$\implies y=\frac{1680}{32}$

$\implies y=52.5$

Thus, C alone can do the work in 52.5 days