Question

A shopkeeper buys a number of books for rupees 1200. If he had bought 10 more books for the same amount each book would have cost him rupees 20 less. How many books did he buy?

Answer 1

Hint: Let us assume that the shopkeeper buys ‘$x$’ number of books which cost rupees 1200. Then the cost of 1 book will be $\dfrac{1200}{x}$. Now, if he buys 10 more books for the same amount each book would cost him rupees 20 less. So, the cost of 1 book will be $\dfrac{1200}{x+10}+20$.

Complete step-by-step answer:
By equating the above equations we will find the number of books the shopkeeper has purchased.
It is given in the question that a shopkeeper buys a number of books for 1200 rupees. If he had bought 10 more books for the same amount each book would have cost him rupees 20 less. Then, we have to find the number of books the shopkeeper purchased.
Let us assume that the shopkeeper buys ‘$x$’ number of books which cost rupees 1200. Using unitary method we get cost of 1 book will be $\dfrac{1200}{x}$……….(i)
Also, given if a shopkeeper purchases 10 more books for the same amount each book would have cost him rupees 20 less. So, the cost of book will be $=\dfrac{1200}{x+10}+20$……….(ii)
By solving equation (i) and (ii), we get
$\begin{aligned} & \Rightarrow \dfrac{1200}{x}=\dfrac{1200}{(x+10)}+20 \\\ & \\\ \end{aligned}$
$\Rightarrow \dfrac{1200}{x}-\dfrac{1200}{(x+10)}=20$
Take 1200 common in L.H.S, we get
$\Rightarrow 1200\left[ \dfrac{1}{x}-\dfrac{1}{(x+10)} \right]=20$
$\Rightarrow \left[ \dfrac{1}{x}-\dfrac{1}{(x+10)} \right]=\dfrac{20}{1200}$
$\Rightarrow \left[ \dfrac{1}{x}-\dfrac{1}{(x+10)} \right]=\dfrac{1}{60}$
$\Rightarrow \dfrac{(x+10)-x}{x(x+10)}=\dfrac{1}{60}$
On cross multiplying, we get
$\Rightarrow 600={{x}^{2}}+10x$
$\Rightarrow {{x}^{2}}+10x-600=0$
On splitting the middle term, we get
$\Rightarrow {{x}^{2}}+30x-20x-600=0$
$\Rightarrow x(x+30)-20(x+30)=0$
$\Rightarrow (x-20)(x+30)=0$
$\Rightarrow x=20,x=-30$
$-30$ is not possible because the number of books will be a natural number. Therefore, we only consider $x=20$. So, the shopkeeper will purchase 20 books for rupees 1200.

Note: Students may take the second equation as $\dfrac{1200}{x+10}-20$ but, this will not give correct answer either we have to take $\dfrac{1200}{x+10}+20$ or we may take $\dfrac{1200}{x+10}$ and then first equation will become $\dfrac{1200}{x+10}+20$. On solving these 2 equations we will get the value of $x$.