asked 151k views
5 votes
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?

2 Answers

2 votes
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


answered
User Do Good
by
7.8k points
4 votes

Answer:
1(2)/(3)\text{ quarts} of water must be added.

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*5=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 :-


\frac{\text{Water}}{\text{Acid}}=(3+x)/(2)=(0.70)/(0.30)\\\\\Rightarrow\ 3+x=(7)/(3)*2\\\\\Righatrrow\ x=(5)/(3)=1(2)/(3)\text{ quarts}

Hence,
1(2)/(3)\text{ quarts} of water must be added.

answered
User Superkhau
by
7.7k points
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