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?

Answer 1

Step 1. Assuming that $B$ alone takes $x$ days to finish the work.

Therefore $A$ alone can finish the work in $(x-6)$ days

Step 2. Now ,

$A\text{'}s$ one day’s work $+$ $B\text{'}s$ one day’s work $=\frac{1}{x}+\frac{1}{(x-6)}$

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,

$$\begin{gathered} \frac{1}{x}+\frac{1}{(x-6)}=\frac{1}{4} \\ \Rightarrow \frac{x-6+x}{x(x-6)}=\frac{1}{4} \\ \Rightarrow \frac{2x-6}{x^2-6x}=\frac{1}{4} \\ \Rightarrow 4(2x-6)=x^2-6x \\ \Rightarrow 8x-24=x^2-6x \\ \Rightarrow x^2-6x-8x+24=0 \\ \Rightarrow x^2-14x+24=0 \\ \Rightarrow x-(12+2)x+24=0 \\ \Rightarrow x-12x-2x+24=0 \\ \Rightarrow x(x-12)-2(x-12)=0 \\ \Rightarrow (x-12)(x-2)=0 \\ \Rightarrow either(x-2)=0or(x-12)=0 \\ \Rightarrow x=2orx=12 \end{gathered}$$

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.