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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.