$B$ took $16$ days less than $A$ to do a piece of work. If both work together, they finish the work in $15$ days. How much time will $B$ alone takes to do the same work.

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Solution

Let $A$ alone can finish the work in $x$ days.
then $B$ alone can finish the work in $x - 16$ days.

Work done by $A$ alone in 1 day $= \frac{1}{x}$
Work done by $B$ alone in 1 day $= \frac{1}{x - 16}$

If both work together, they finish the work in $15$ days.

Work done by $A$ and $B$ together in 1 day is
$\frac{1}{x} + \frac{1}{x - 16}$

Work done by $A$ and $B$ together in 15 days is

$15 (\frac{1}{x} + \frac{1}{x - 16}) = 1$

$\frac{x - 16 + x}{x (x - 16)} = \frac{1}{15}$

$\frac{2 x - 16}{x (x - 16)} = \frac{1}{15}$

$15 (2 x - 16) = x (x - 16)$

$30 x - 240 = x^{2} - 16 x$

$x^{2} - 46 x + 240 = 0$

$x^{2} - 40 x - 6 x + 240 = 0$

$x (x - 40) - 6 (x - 40) = 0$

$(x - 40) (x - 6) = 0$

$x = 6$ or $x = 40$

Value of $x$ will be greater than $16$
So, $x = 40$

then $B$ alone can finish the work in $40 - 16 = 24$ days.

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