Question

# $A$ takes $6$ days less than the time taken by $B$ to finish a piece of work. If both $A$and$B$ together can finish it in $4$ days, How many days will $B$ take to finish the work?

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Solution

## Step 1. Assuming that $B$ alone takes $x$ days to finish the work. Therefore $A$ alone can finish the work in $\left(x-6\right)$ daysStep 2. Now , $A\text{'}s$ one day’s work $+$ $B\text{'}s$ one day’s work $=\frac{1}{x}+\frac{1}{\left(x-6\right)}$Step 3. According to question , both $A$and$B$ together can finish it in $4$ days Therefore work done by $A$and$B$ together in one day$=\frac{1}{4}$ we have, $\frac{1}{x}+\frac{1}{\left(x-6\right)}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{x-6+x}{x\left(x-6\right)}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{2x-6}{{x}^{2}-6x}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒4\left(2x-6\right)={x}^{2}-6x\phantom{\rule{0ex}{0ex}}⇒8x-24={x}^{2}-6x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-6x-8x+24=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x+24=0\phantom{\rule{0ex}{0ex}}⇒x-\left(12+2\right)x+24=0\phantom{\rule{0ex}{0ex}}⇒x-12x-2x+24=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-12\right)-2\left(x-12\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-12\right)\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}⇒either\left(x-2\right)=0or\left(x-12\right)=0\phantom{\rule{0ex}{0ex}}⇒x=2orx=12$ Step 4. As $x$ can not be less than $6$. So, we have $x=12$Hence, $B$ will take $12$ days to finish the work.

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