#### Question

A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.

#### Solution

Let *B* alone takes *x *days to finish the work. Then,* B’*s one day’s work = 1/x.

Similarly, *A *alone can finish it in (x - 10) days to finish the work. Then,* A’*s one day’s work `=1/(x-10)`.

It is given that

*A’*s one day’s work + *B’*s one day’s work = (A + B)’s one day’s work

`1/x+1/(x-10)=1/12`

`(x-10+x)/(x(x-10))=1/12`

`(2x-10)/(x^2-10x)=1/12`

x^{2} - 10x = 12(2x - 10)

x^{2} - 10x = 24x - 120

x^{2} - 10x - 24x + 120 = 0

x^{2} - 34x + 120 = 0

x^{2} - 30x - 4x + 120 = 0

x(x - 30) - 4(x - 30) = 0

(x - 30)(x - 4) = 0

x - 30 = 0

x = 30

Or

x - 4 = 0

x = 4

But x = 4 is not correct.

therefore, x = 30 is correct

Hence, the time taken by *B *to finish the work in x = 30 days