Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 37 of Exercise 10E
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
A dealer sells an article for Rs. 75 and gains as much per cent as the cost price of the article. Find the cost price of the article.
Let the cost price of the article be x
Gain percent x%
According to the given condition,
(cost price + gain = selling price)
taking LCM
x2 + 100x = 7500 cross multiplying
x2 + 100x – 7500 = 0
Using the splitting middle term – the middle term of the general equation 
is divided in two such values that:
Product = a.c
For the given equation a = 1 b = 100 c = – 7500
= 1. – 7500 = – 7500
And either of their sum or difference = b
= 100
Thus the two terms are 150 and – 50
Difference = 150 – 50 = 100
Product = 150. – 50 = – 7500
x2 + 150x – 50x – 7500 = 0
x(x + 150) – 50(x + 150) = 0
(x + 150) (x – 50) = 0
(x + 150) = 0 or (x – 50) = 0
x = 50 (price cannot be negative)
Hence the cost price of the article is Rs 50
minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
