Hi swerve,

We're told that a chemical supply company has 60 liters of a 40% HNO3 solution. We're asked for the number of liters of pure undiluted (meaning 100%) HNO3 the chemists must add so that the resultant solution is a 50% solution. This question is essentially a 'Weighted Average' question, but there's a math 'shortcut' that you can use to simplify the calculations involved. We can TEST THE ANSWERS to take advantage of the shortcut.

To start, since we're mixing in a pure (100%) solution, it won't take much of that solution - relatively speaking - to raise a 40% solution to an overall 50% solution. Thus, the correct answer is likely one of the smaller answers. Both Answer A (12) and Answer B (15) are factors of 60, so those numbers would form a simple ratio (1:5 and 1:4, respectively) with the 60 liters that are already there. This means that we can calculate the average using the ratio - and not the larger numbers involved.

Answer A: 12 liters --> forms a ratio of 1:5 with the original 60 liter mixture

[(1)(100%) + (5)(40%)]/(1+5) = 300%/6 = 50%
This is an exact match for what we were told, so this must be the answer

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

swerve 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

The OA is 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