Some notebooks were bought for a total of 720 rupees. If the price of each notebook had been 6 rupees less, 4 more notebooks could have been bought for the same amount. What was the original price of each notebook?

36 rupees

Let the original price of each notebook be x rupees. Then the number of notebooks bought is \(\frac{720}{x}\), while at a price reduced by 6 rupees, the number would be \(\frac{720}{x-6}\). Hence, \(\frac{720}{x-6}-\frac{720}{x}=4\). On simplifying, \(x^2-6x-1080=0\), or \((x-36)(x+30)=0\). Since a price must be positive, \(x=36\) rupees. Thus, option C is correct. Exam tip: In price-based word problems, express the number of items as total cost divided by the price per item, and reject any negative root.

This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Word Problems and Applications.