Question

A shopkeeper buys a number of books of Rs. 80. If he had bought 4 more books for the same amount, each book would have cost Rs. 1 less. How many books did he buy?

Answer 1

Hint: First, we assume the number of books. From the given total values of the books, we find the value of each book. Then we use the condition that on buying 4 more books for the same amount, each book would have cost Rs. 1 less. We get a new number of books which give new value for each book. We equate them to find a number of books.

Complete step by step answer:
Let’s assume that the shopkeeper bought n number of books. He paid Rs. 80 for those books.
Using the unitary system, we get that the price of every book is $ \dfrac{80}{n} $ Rs.
Now if he had bought 4 more books for the same amount, each book would have cost Rs. 1 less.
The number of books would have been $ \left( n+4 \right) $ and the total price would have been intact as Rs. 80.
In that case, the price of each book would have been $ \dfrac{80}{n+4} $ Rs.
The difference is Rs. 1.
The relation can be expressed in the form of mathematical equation as $ \dfrac{80}{n}-\dfrac{80}{n+4}=1 $ .
Now we solve the equation to find the value of n.
 $ \begin{align}
  & \dfrac{80}{n}-\dfrac{80}{n+4}=1 \\
 & \Rightarrow 80\left( \dfrac{n+4-n}{n\left( n+4 \right)} \right)=1 \\
\end{align} $
Now we find the quadratic equation as $ {{n}^{2}}+4n-320=0 $ .
We apply factorisation to find the value of n.
\[\begin{align}
  & {{n}^{2}}+4n-320=0 \\
 & \Rightarrow \left( n+20 \right)\left( n-16 \right)=0 \\
\end{align}\]
As n can’t be negative, value of n is 16.
Therefore, the shopkeeper bought 16 books.

Note:
We could have found the new value for each book as $ \left( \dfrac{80}{n}-1 \right) $ and multiplied with $ \left( n+4 \right) $ to equate that value with 80 as the total values of the books never changed. The equation would have been in the form of $ \left( \dfrac{80}{n}-1 \right)\left( n+4 \right)=80 $ . Solving the equation, we get same value of n.