# Aniruddha bought some books for Rs. 60. Had he bought 5 more books for the same amount, each book would have cost him 1 rupee less. Find the number of books bought by Aniruddha.

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Hint: Assume that the number of books bought by Aniruddha equals ‘n’. He bought ‘n’ books for Rs 60. So, the price of each book is $\dfrac{60}{n}$. It is given that if he had bought 5 more books i.e. ‘n+5’ books for the same price i.e. Rs 60, each book would have cost him Rs. $\dfrac{60}{n+5}$.

Using this data and the information given in the question, solve for ‘n’.

Let us assume that Aniruddha has bought n books. In the question, it is given that he bought this book for Rs. 60. So, the price of each book will be equal to Rs $\dfrac{60}{n}........\left( 1 \right)$.

If he would have bought 5 more books i.e. n+5 books for the same price i.e. Rs 60, the price of each book will be equal to $\dfrac{60}{n+5}...........\left( 2 \right)$

Also, it is given in the question that if he would have 5 more books i.e. n+5 books for the same price i.e. Rs 60, each book would have cost him 1 rupee less.

Using equation $\left( 1 \right)$ and $\left( 2 \right)$ and the information mention in the above paragraph, we get,

$\begin{align}

& \dfrac{60}{n}-1=\dfrac{60}{n+5} \\

& \Rightarrow \dfrac{60-n}{n}=\dfrac{60}{n+5} \\

& \Rightarrow \left( 60-n \right)\left( n+5 \right)=60n \\

& \Rightarrow 60n+300-{{n}^{2}}-5n=60n \\

& \Rightarrow {{n}^{2}}+5n-300=0 \\

& \Rightarrow {{n}^{2}}-15n+20n-300=0 \\

& \Rightarrow n\left( n-15 \right)+20\left( n-15 \right)=0 \\

& \Rightarrow \left( n-15 \right)\left( n+20 \right)=0 \\

& \Rightarrow n=15,n=-20 \\

\end{align}$

Since n represents the number of books, so, it must be positive.

$\Rightarrow n=15$

Hence, the answer is 15.

Note: There is a possibility that one may commit a mistake by writing the equation as

$\dfrac{60}{n}=\dfrac{60}{n+5}-1$ instead of $\dfrac{60}{n}-1=\dfrac{60}{n+5}$. So, one should read this question very carefully in order to obtain the correct equation and the correct answer.

Using this data and the information given in the question, solve for ‘n’.

Let us assume that Aniruddha has bought n books. In the question, it is given that he bought this book for Rs. 60. So, the price of each book will be equal to Rs $\dfrac{60}{n}........\left( 1 \right)$.

If he would have bought 5 more books i.e. n+5 books for the same price i.e. Rs 60, the price of each book will be equal to $\dfrac{60}{n+5}...........\left( 2 \right)$

Also, it is given in the question that if he would have 5 more books i.e. n+5 books for the same price i.e. Rs 60, each book would have cost him 1 rupee less.

Using equation $\left( 1 \right)$ and $\left( 2 \right)$ and the information mention in the above paragraph, we get,

$\begin{align}

& \dfrac{60}{n}-1=\dfrac{60}{n+5} \\

& \Rightarrow \dfrac{60-n}{n}=\dfrac{60}{n+5} \\

& \Rightarrow \left( 60-n \right)\left( n+5 \right)=60n \\

& \Rightarrow 60n+300-{{n}^{2}}-5n=60n \\

& \Rightarrow {{n}^{2}}+5n-300=0 \\

& \Rightarrow {{n}^{2}}-15n+20n-300=0 \\

& \Rightarrow n\left( n-15 \right)+20\left( n-15 \right)=0 \\

& \Rightarrow \left( n-15 \right)\left( n+20 \right)=0 \\

& \Rightarrow n=15,n=-20 \\

\end{align}$

Since n represents the number of books, so, it must be positive.

$\Rightarrow n=15$

Hence, the answer is 15.

Note: There is a possibility that one may commit a mistake by writing the equation as

$\dfrac{60}{n}=\dfrac{60}{n+5}-1$ instead of $\dfrac{60}{n}-1=\dfrac{60}{n+5}$. So, one should read this question very carefully in order to obtain the correct equation and the correct answer.

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