How many ounces of pure alcohol should be added to 90 ounces of 4% alcohol solution to produce a solution that is 10%? Pesent your answer as an integer or a reduced fraction. Do not use decimals.

Question

How many ounces of pure alcohol should be added to

90 ounces of 4% alcohol solution to produce a solution

that is 10%?

Pesent your answer as an integer or a reduced fraction.

Do not use decimals.

Answer 1

Hi Rebecca!

Ok, let x = ounces of pure alcohol (100%)

let y = ounces of 4% alcohol = 90 (given in the problem)

(mixture of pure alcohol) + (mixture of 4%) = (mixture that is 10%)

100% = 1 4% = .04 10% = .10

1x + .04y = .10 (x + y)

But, we are given that y = 90 in the problem, so substitute

1x + .04 (90) = .10 (x + 90)

x + 3.6 = .10x + 9

-.10x -.10x subtract .10x

----------------------------------------------------------

.90x + 3.6 = 9

-3.6 -3.6 subtract 3.6

---------------------------------------------------------------

.90x = 5.4

multiply both sides by 100 to get rid of decimals

100 (.90x) = 100 (5.4)

90x = 540

------ ----- divide both sides by 90

90 90

x = 6oz (the amount of ounces of pure alcohol)

Answer 2

Hi Rebecca D

You want a final solution concentration of 10% and adding more liquid increases your overall volume

Let x = represent pure alcohol

Since you start with 90 ounces, your new volume will be

x + 90

Your increased concentration of alcohol will be raised from 4% to 10% by your increase in volume

x + 90(0.04) = x + 3.6

(x + 3.6)/(x + 90) = .10

.10(x + 90) = x + 3.6

Can you solve for x?