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? 2/3 quarts 1/2/3 quarts 2/2/3 quarts

Answer 1

Answer: Second option is correct.

Step-by-step explanation:

Since we have given that

Amount of quarts of acid and water = 5

Percentage of acid = 40%

So, Amount of acid is given by

$\frac{40}{100}\times 5\\\\=2\ quarts$

Amount of water is given by

$5-2=3\ quarts$

So, we need to find the amount of water must be added to the mixture to get weakened to 30% acid.

Let x amount of water is added to mixture, so that water gets 70% of mixture and acid becomes 30% of mixture.

According to question,

$\frac{3+x}{2}=\frac{70}{30}\\\\\frac{3+x}{2}=\frac{7}{3}\\\\3(3+x)=2\times 7\\\\9+3x=14\\\\3x=14-9\\\\3x=5\\\\x=\frac{5}{3}\\\\x=1\frac{2}{3}\ quarts$

Hence, Second option is correct.

Answer 2

.4 * 5 = 2 quarts of acid and 3 quarts of water

We need a 30% concentration so
.30 * (2 + x) = 2 quarts of acid
.6 + .3x = 2
.3x = 1.4
x = 4.6666666667 (total amount of water so we have to add 1.66666 quarts)
or 1 (2/3) quarts