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# A shopkeeper buys some books for Rs. 80. If he had bought 4 more books for the same amount, each book would have cost him Rs. 1 less. Determine the number of books the shopkeeper bought.

Hint: In this question, the shopkeeper buys some books for Rs. 80. Suppose he buys $n$ number of books and the price of each book is Rs. $x$. Then we will form an equation $nx=80$. Now if the shopkeeper buys 4 more books for the same amount, then each book would have cost him Rs. 1 less. Then the equation becomes $\left( n+4 \right)\left( x-1 \right)=80$. On solving both the equations, we will get the desired value for the number of books that the shopkeeper bought.

Suppose that a shopkeeper buys some books for Rs. 80.
Let the number of books that the shopkeeper bought is given by $n$.
Now let if possible the cost of each book that the shopkeeper bought is Rs. $x$.
Then we have the equation for the total cost of $n$ books, which is given by
$nx=80............(1)$
Now if the shopkeeper buys 4 more books for the same amount, then each book would have cost him Rs. 1 less.
That is if the shopkeeper would have bought $n+4$ books, then the price of each book would have cost him Rs. 1 less which is given by
$x-1$
Then we can form another equation for the total cost of $n+4$ books, which is given by
$\left( n+4 \right)\left( x-1 \right)=80............(2)$
On simplifying equation (2), we get
$nx-n+4x-4=80$
Now by using equation (1) in equation (2), we get
$80-n+4x-4=80$
Now subtracting 80 from both side of the above equation and simplifying the same, we get
$-n+4x-4=0$
$\Rightarrow n-4x+4=0.............(3)$
Now from equation (1) we get,
$x=\dfrac{80}{n}$
On substituting the value of $x$ in equation (3), we will get
\begin{align} & n-4\left( \dfrac{80}{n} \right)+4=0 \\ & \Rightarrow {{n}^{2}}-320+4n=0 \\ & \Rightarrow {{n}^{2}}+4n-320=0 \\ \end{align}

Now in order to find the value of $n$, we have to solve the quadratic equation ${{n}^{2}}+4n-320=0$.
On solving the quadratic equation ${{n}^{2}}+4n-320=0$ by splitting the middle term, we will have
\begin{align} & {{n}^{2}}+20n-16n-320=0 \\ & \Rightarrow n\left( n+20 \right)-16\left( n+20 \right)=0 \\ & \Rightarrow \left( n-16 \right)\left( n+20 \right)=0 \\ \end{align}
This implies either $n=16$ or $n=-20$.
Now since the number of books that the shopkeeper bought cannot be negative, hence $n=16$.
Therefore we get the shopkeeper both 16 books for Rs. 80.

Note: In this problem, please do not consider both the values $n=16$ and $n=-20$. Since the total number of books that the shopkeeper bought can never be negative. So discard that possibility.