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2 men can complete the job in 24 days. If the faster one can alone complete the job 20 days faster than the second man, find the time taken by faster man to complete the work.

A) 68 days

B) 40 days

C) 20 days

D) 60 days

Let the time taken by slower worker = x days and faster = (x + 20) days

Now, let the rate of slower worker complete the job in = $\frac{1}{x}days$

and faster worker complete the job in = $\frac{1}{x+20}days$

They together complete the work in = $\frac{1}{x}+\frac{1}{x+20}=\frac{x+20+x}{x\left(x+20\right)}=\frac{2x+20}{x\left(x+20\right)}days$

Therefore, ATQ,

$\frac{1}{{\displaystyle \frac{2x+20}{x\left(x+20\right)}}}=24\phantom{\rule{0ex}{0ex}}\frac{x\left(x+20\right)}{2x+20}=24\phantom{\rule{0ex}{0ex}}{x}^{2}+20x=24(2x+20)\phantom{\rule{0ex}{0ex}}{x}^{2}+20x=48x+480$

x

^{2 }- 28x - 480 = 0

x

^{2 }- 40x + 12x - 480 = 0

x(x - 40) + 12(x - 40) = 0

(x - 40)(x + 12) = 0

Either x - 40 = 0 or x + 12 = 0

x = 40 or x = - 12

Neglect the negative term, we get,

x = 40 days = time taken by slower worker

and

x + 20 = 40 + 20 = 60 days = time taken by faster worker

Hence, D) option is correct.

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