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.

Given:

A takes 10 days less than the time taken by B to finish a piece of work.

Time taken to finish the work when A and B both work together$=12$ days.

To do:

We have to find the time taken by B to finish the work.

Solution:

Let the number of days taken by B to finish the work alone be $x$.

This implies,

The number of days A takes to finish the work$=x-10$.

Work done by A in a day$=\frac{1}{x-10}$

Work done by B in a day$=\frac{1}{x}$

Work done by A and B together in a day$=\frac{1}{12}$

Therefore,

$\frac{1}{x-10}+\frac{1}{x}=\frac{1}{12}$

$\frac{1(x)+1(x-10)}{(x-10)x}=\frac{1}{12}$

$\frac{x+x-10}{x^2-10x}=\frac{1}{12}$

$\frac{2x-10}{x^2-10x}=\frac{1}{12}$

$12(2x-10)=1(x^2-10x)$

$24x-120=x^2-10x$

$x^2-10x-24x+120=0$

$x^2-34x+120=0$

Solving for $x$ by factorization method, we get,

$x^2-30x-4x+120=0$

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

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

$x-30=0$ or $x-4=0$

$x=30$ or $x=4$

If $x=4$, $x-10=4-10=-6$, which is not possible.

Therefore, the value of $x=30$.

$x-10=30-10=20$

The time taken by B to finish the work alone is $30$ days.

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