 QUESTION

# If the list price of a book is reduced by Rs. 5, a person can buy 5 more books for Rs. 300. Then the original cost of the book is-$A{\text{ R}}{\text{s}}{\text{.15}} \\ {\text{B R}}{\text{s}}{\text{. 20}} \\ {\text{C R}}{\text{s}}{\text{. 25}} \\ {\text{D R}}{\text{s}}{\text{. 30}} \\$

Hint: Here we will proceed by assuming the price of one book as Rs. y and number of books purchased be x. Then we will equate the given total cost with the relative price such that it will form a quadratic equation and further we will use factorization method to solve the equation and get the right answer.

Let the price of one book be Rs. y.
Let the number of books purchased be x.
So, total cost $= x \times y$
Also we are given that the total cost is Rs. 300.
So, $xy = 300$ ----(1)
Also that a book is reduced by Rs. 5 a person can buy 5 more books.
So, the price of one book = y – 5
The number of books purchased = x + 5
So, Total cost = (x + 5) (y – 5)
Since we are given that the total cost is Rs. 300.
So, (x + 5) (y – 5) = 300
$\Rightarrow xy - 5x + 5y - 25 = 300$
$\Rightarrow 300 - 5x + 5y - 25 = 300$ [Using equation (1)]
$\Rightarrow - 5x + 5y = 25$
$\Rightarrow - x + y = 5$
$\Rightarrow - \dfrac{{300}}{y} + y = 5$ [Using equation (1)]
$\Rightarrow - \dfrac{{300 + {y^2}}}{y} = 5$
$\Rightarrow {y^2} - 5y - 300 = 0$
Now we will factorise the quadratic equation i.e. ${y^2} - 5y - 300 = 0$
$\Rightarrow {y^2} - 20y + 15y - 300 = 0$
$\Rightarrow y\left( {y - 20} \right) + 15\left( {y - 20} \right) = 0$
$\Rightarrow \left( {y - 20} \right)\left( {y + 15} \right) = 0$
$\therefore y = 20,y = - 15$
Since the price cannot be negative.
So, the original cost price of the book is Rs. 20.

Note: While solving this type of question, one can get confused in solving the quadratic equation by factorization. So we can also use square roots, completing the squares or quadratic formula. Also we must concentrate and always accept the positive value of the solved equation because price cannot be negative.