Question

A shopkeeper buys a certain no. of books for $Rs.960$.If the cost per book was $Rs.\,8$ less, the no. of books that could be bought for $Rs.960$ would be four more. Taking the original cost of each book as $Rs.\,x$, write an equation in $x$ and solve it.
$
  A) \,Rs.\,48 \\
  B) \,Rs.\,24 \\
  C) \,Rs.\,40 \\
  D) \,Rs.\,14 \\
 $

Answer 1

Hint:Let the initial shopkeeper buy $n$ number of books and each book cost $Rs.\,x$. So he buyed for $Rs.960$. Now if the cost of each book becomes $Rs.\,\left( {x - 8} \right)$ then the number of books increases to $n + 4$. For some amount of money. Generate two equations from this and simplify it.

Complete step-by-step answer:
According to the question shopkeeper buys certain number of books for $Rs.960$
So let the number of books bought be $n$ and the original cost of each book be $Rs.\,x$. So as we assumed that one book cost is $x$ and he bought $n$ books for $Rs.960$. So our equation becomes
$nx = 960\,\,\,\,\,\,\,\,\,\,\, \to \left( 1 \right)$
Now if the cost of each book is decreased by $Rs.\,8$ i.e. cost of each book becomes $Rs.\,\left( {x - 8} \right)$ and for$Rs.960$ shopkeeper can buy four more books this means total number of books $n + 4$
So here one book costs $Rs.\,\left( {x - 8} \right)$ and shopkeeper bought $n + 4$ books for$Rs.960$
So our equation becomes
$\left( {n + 4} \right)\left( {x - 8} \right) = 960\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 2 \right)$
Upon multiplication we get
$
  n\left( {x - 8} \right) + 4\left( {x - 8} \right) = 960 \\
  nx - 8n + 4x - 32 = 960 \\
 $
Substituting values from (1) we get
$
  {{960}} - 8n + 4x - 32 = {{960}} \\
  4x - 8n = 32$
Now from (1) we can write $n = \dfrac{{960}}{x}$
Hence putting this value we get
$
  4x - 8\left( {\dfrac{{960}}{x}} \right) = 32 \\
  {x^2} - 8x - 1920 = 0 \\
 $
Hence we can split this into
$
  {x^2} - 48x + 40x - 1920 = 0 \\
  x\left( {x - 48} \right) + 40\left( {x - 48} \right) = 0 \\
  \left( {x + 40} \right)\left( {x - 48} \right) = 0 \\
 $
So we get
$
  x + 40 = 0\,\,\,\,\,\& \,\,\,\,\,\,x - 48 = 0 \\
  x = - 40\,\,\,\,\,\,\,\,\,\,\& \,\,\,\,\,\,x = 48 \\
 $
Here $x$ represents the cost so it cannot be negative so we get $x = 48$

So, the correct answer is “Option A”.

Note:Here we got a quadratic equation of the form $a{x^2} + bx + c = 0$ . So it has solution or roots given by:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here $x$ has two values i.e. called the roots of the equation.
Here we are given equation ${x^2} - 8x - 1920 = 0$
So its roots can be given as
$
  x = \dfrac{{8 \pm \sqrt {{{\left( { - 8} \right)}^2} - 4\left( {1920} \right)} }}{2} \\
  x = \dfrac{{8 + \sqrt {7744} }}{2}\,\,or\,\,\,x = \dfrac{{8 - \sqrt {7744} }}{2} \\
  x = \dfrac{{8 + 88}}{2}\,\,or\,\,x = \dfrac{{8 - 88}}{2}\,\, \\
  x = 48\,\,or\,\,x = - 40\,\,$
As x cannot be negative so we get $x = 48\,$
We can use this method to solve complex quadratic equations also.