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Last updated date: 02nd Dec 2023
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# 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.

$\Rightarrow n=15$
$\dfrac{60}{n}=\dfrac{60}{n+5}-1$ 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.