Question

A mixture contains five quarts of acid and water and is forty percent acid. If the mixture is to be weakened to thirty percent acid, how much water must be added?

Answer 1

Answer:  $1\dfrac{2}{3}\text{ quarts}$ of water must be added.

Step-by-step explanation:

Given : The total quantity of mixture = 5 quarts

The percentage of acid= 0.40

Then, total liters of acid is given by :-

$0.40\times5=2\text{ quarts}$

The amount of water in the given mixture : $5-2=3\text{ quarts}$

The percentage of acid in in weakened mixture = 0.30

Then , the percentage of water in weakened mixture = 0.70

Let 'x' amount of water is added to mixture.

Then According to the question ,we have the following equation :-

$\dfrac{\text{Water}}{\text{Acid}}=\dfrac{3+x}{2}=\dfrac{0.70}{0.30}\\\\\Rightarrow\ 3+x=\dfrac{7}{3}\times2\\\\\Righatrrow\ x=\dfrac{5}{3}=1\dfrac{2}{3}\text{ quarts}$

Hence, $1\dfrac{2}{3}\text{ quarts}$ of water must be added.

Answer 2

In five quarts, we have 2 quarts of acid and 3 quarts of water
total = (2 + 3) = 5 quarts
concentration = 2 / 5

We want a 30% concentration
total volume = 2 + x
.30 = 2 / (2+x)
.6 + .30 x  = 2
.30x = 1.4
x = 4.66666 gallons of water
(or we need to add 1.666666 gallons of water)

Double Check:
acid concentration = 2 / (2 + 4.666666)
acid concentration = .30