Question

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.

Answer 1

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.

Complete step-by-step answer:
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.