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A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days B alone can do the work

Answers (1)

A and B complete the work in 6 days.

$\rightarrow$ Let B do the work in $x$ days.

A do the work in $(x-5)$ days.

$\rightarrow$ Let the total work be W units.

$\rightarrow \text{A and B complete work in one day}=\frac{W}{6}$

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

$\rightarrow \text{A's one day work}=\frac{W}{x-5}$

$\Rightarrow \frac{W}{x}+\frac{W}{x-5}=\frac{W}{6}$

$\Rightarrow \frac{1}{x}+\frac{1}{x-5}=\frac{1}{6}$ (cancelling common W)

$\Rightarrow \frac{x-5+x}{x(x-5)}=\frac{1}{6}$ (Taking LCM )

$\Rightarrow (2x-5)6=x(x-5)$

$\Rightarrow 12x-30=x^2-5x$

$\Rightarrow x^2-5x-12x+30=0$

$\Rightarrow x(x-15)-2(x-15)=0$

$\Rightarrow (x-15)(x-2)=0$

$\Rightarrow x=2$ OR $x=15$

$x \neq 2$ because otherwise, A's work has no meaning

$\therefore x=15$

Hence B takes 15 days to complete the work.

Posted by

Safeer PP

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