104k views
1 vote
Chris needs to mix a 10% acid solution with a 60% acid solution to create 200 millileters of a 17.5% solution. How many millileters of each solution must Chris use?

1 Answer

3 votes

Answers:

170 mL of the 10% solution

30 mL of the 60% solution

=================================================

Work Shown

Label the two inputs as A and B

A: 10% acid

B: 60% acid

x = amount of A

200-x = amount of B

0.10x = amount of pure acid from A

0.60(200-x) = amount of pure acid from B

The goal is to have 200 mL of a 17.5% solution, so we want to have 200*0.175 = 35 mL of pure acid after mixing batches A and B.

---------

0.10x + 0.60(200-x) = 35

0.10x + 120 - 0.60x = 35

-0.50x + 120 = 35

-0.50x = 35-120

-0.50x = -85

x = -85/(-0.50)

x = 170 mL is the amount of the 10% acid solution

200-x = 200-170 = 30 mL is the amount of the 60% acid solution

---------

Check:

A: 10% of 170 = 0.10*170 = 17 mL of pure acid

B: 60% of 30 = 0.60*30 = 18 mL of pure acid

A+B = 17+18 = 35 mL of pure acid total, which matches the 35 mentioned earlier. We have confirmed the answers.

User Page
by
7.2k points