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A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it ?

Answers (1)

Let B alone take $x$ days to finish work 'W' then A alone takes $(x-6)$ days to finish work 'W'.

Given : Both A and B finish the work in 4 days

B's one day work $=\frac{W}{x}$

A's one day work $=\frac{W}{x-6}$

A and B both one day work $=\frac{W}{4}$

Now,

$\frac{W}{x}+\frac{W}{x-6}=\frac{W}{4}$

$\Rightarrow \frac{1}{x}+\frac{1}{x-6}=\frac{1}{4}$ (cancelling common W)

$\Rightarrow \frac{(x-6)}{x(x-6)}+\frac{x}{(x-6)x}=\frac{1}{4}$ (Taking LCM)

$\Rightarrow \frac{2x-6}{x(x-6)}=\frac{1}{4}$

$\Rightarrow 4(2x-6)=x(x-6)$ (cross multiplication)

$\Rightarrow 8x-24=x^2-6x$

$\Rightarrow x^2-6x-8x+24=0$

$\Rightarrow x^2-14x+24=0$

$\Rightarrow x^2-(12+2)x+24=0$ $(\because 12\times 2=24)$

$\Rightarrow x^2-12x-2x+24=0$

$\Rightarrow x(x-12)-2(x-12)=0$

$\Rightarrow (x-12)(x-2)=0$

$\Rightarrow x=12$ OR $x=2$

$x \neq 2$ because otherwise A's work will bw negative and will have no value.

Hence $x =12$

$\Rightarrow$ B takes 12 days to complete work and A takes $(12-6)=6$ days to complete the work.

Posted by

Ravindra Pindel

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