BTGmoderatorDC wrote:A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemists add so that the resultant solution is a 50% solution?

A) 12
B) 15
C) 20
D) 24
E) 30

OA A

Source: Magoosh

We can lert n = the amount of pure undiluted HNO3 to be added and create the equation:

(0.4 x 60 + n)/(60 + n) = 1/2

2(24 + n) = 60 + n

48 + 2n = 60 + n

n = 12

Alternate Solution:

We have 60 liters of 40% solution and add x liters of 100% solution to get (60 + x) liters of 50% solution. Converting the percents to decimals and putting the information into an equation, we have:

60(0.40) + x(1.0) = (60 + x)(0.50)

24 + x = 30 + 0.5x

0.5x = 6

5x = 60

x = 12

Answer: A