# By increasing the list price of a book by Rs. 10, a person can buy 20 books less for Rs. 1200. The original price of the book was _____. A) Rs. 0 B) Rs. 20 C) Rs. 30 D) Rs. 24

## Question

By increasing the list price of a book by ₹10, a person can buy 20 books less for ₹1200. The original price of the book was _____.

A) ₹0

B) ₹20

C) ₹30

D) ₹24

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## Answer

The problem provides the scenario where increasing the list price of a book by ₹10 results in buying 20 fewer books for ₹1200. We are asked to find the **original price** of the book.

Let's denote the original price of the book as $ x $. Hence, after the price increase, the new price of the book becomes $ x + 10. $

The number of books that can be bought initially with ₹1200 is given by $ \frac{1200}{x},$ and the number of books that can be bought after the price increase is $ \frac{1200}{x + 10}. $

According to the problem, the difference in the number of books bought before and after the price increase is 20 books, leading to the equation: $$ \frac{1200}{x} - \frac{1200}{x + 10} = 20 $$

We can simplify and solve this equation by first clearing the denominators: $$ 1200(x + 10) - 1200x = 20x(x + 10) $$ $$ 12000 = 20x^2 + 200x $$

Dividing all terms by 20 to simplify: $$ 600 = x^2 + 10x $$

Rearranging into a standard quadratic equation form: $$ x^2 + 10x - 600 = 0 $$

Solving this quadratic equation using the quadratic formula where $ a = 1 $, $ b = 10 $, and $ c = -600 $: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ x = \frac{-10 \pm \sqrt{100 + 2400}}{2} $$ $$ x = \frac{-10 \pm \sqrt{2500}}{2} $$ $$ x = \frac{-10 \pm 50}{2} $$

We have two potential solutions for $ x $: $$ x = \frac{40}{2} = 20 \quad \text{and} \quad x = \frac{-60}{2} = -30 $$

Rejecting the negative value, $ x = -30 $, as a price cannot be negative, the original price of the book is therefore **₹20**. Hence, the correct answer is:

**B) ₹20**

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