Question

# A shopkeeper buys a number of books for Rs. 80. If he had bought 4 more books for the same amount, each book would have cost Re. 1 less. How many books did he buy? a.8 b.16 c.24 d.28

Hint: Here, we can assume that the total number of books in n. Then, we will find the price of each book for both the given cases. We will use the information that the difference between the price of each book in both the cases is Re.1 to form an equation and then we can find the value of n.

Let us consider that the total number of books that he bought be = n
The total price of the books in both the cases = Rs. 80
Since, price of n books is = Rs. 80
So, the price of one book = Rs. $\dfrac{80}{n}$
Now, in the second case also the total price of the books is = Rs. 80
But, in this case it is given that the number of books is 4 more than the first case.
So, the total number of books in the second case is = (n+4)
Therefore, price of each book in the second case is = Rs. $\dfrac{80}{n+4}$
Since, the difference between the price of each book in both the cases is Re.1. So, we can write:
$\dfrac{80}{n}-\dfrac{80}{n+4}=1$
We can solve this equation to get the value of n:
\begin{align} & \dfrac{80\left( n+4 \right)-80n}{n\left( n+4 \right)}=1 \\ & \Rightarrow \dfrac{80n+320-80n}{{{n}^{2}}+4n}=1 \\ & \Rightarrow {{n}^{2}}+4n-320=0 \\ & \Rightarrow {{n}^{2}}+20n-16n-320=0 \\ & \Rightarrow n\left( n+20 \right)-16\left( n+20 \right)=0 \\ & \Rightarrow \left( n+20 \right)\left( n-16 \right)=0 \\ & \\ \end{align}
From, this we get two values of n as:
Either n = 16 or n=-20
But here negative value of n is not possible.
Therefore, the value of n is 16 and also the number of books that the shopkeeper bought is 16.
Hence, option (b) is the correct answer.

Note: Students should note here that we get two values of n because we obtained a quadratic equation in n. But one of the values of n is negative and since, the number of books can’t be negative, so we neglected the negative value.