Question

$A$ and $B$ together can do a piece of work in $24$ days. $B$ takes $20$ days more than $A$ for doing the same work alone. In how many days $A$ and $B$ alone can do the same work?

Answer 1

Step 1: Determining the quadratic equation in $x$

Let the time taken by $A$ to finish the work alone be $x$ days.

So, the time taken by $B$ to finish the same work alone will be $(x+20)$ days.

In $1$ day, $B$ can finish $\frac{1}{x}$ of the work while $A$ can finish $\frac{1}{(x+20)}$ of the work.

It is given that $A$ and $B$ can finish the work together in $24$ days.

Therefore, in $1$ day both $A$ and $B$ can finish $\frac{1}{24}$ of the work.

$$\begin{gathered} \Rightarrow \frac{1}{x}+\frac{1}{x+20}=\frac{1}{24} \\ \Rightarrow \frac{x+20+x}{x(x+20)}=\frac{1}{24} \\ \Rightarrow 24(2x+20)=x(x+20) \\ \Rightarrow 48x+480=x^2+20x \\ \Rightarrow x^2-28x-480=0 \end{gathered}$$

Step 2: Finding the number of days taken by $A$ and $B$ to finish the work alone

By factorization method

$x^2-28x-480=0$

For splitting the middle i.e. $-28$, we need to find two numbers such that their sum is $-28$ and product is $480$.

Such numbers are $-40$ and $12$.

$$\begin{gathered} \Rightarrow x^2-40x+12x-480=0 \\ \Rightarrow x(x-40)+12(x-40)=0 \\ \Rightarrow (x+12)(x-40)=0 \\ \therefore x=-12,40 \end{gathered}$$

Since the value of $x$ cannot be negative,

$\therefore x=40$ and $(x+20)=40+20=60$

Therefore, $A$ takes $40$ days and $B$ takes $60$ days to finish the work alone.