Question

How many gallons of a 90% antifreeze solution must be mixed with 70 gallons of 10% antifreeze to get a mixture that is 80% antifreeze?

Answer 1

Let x be the amount of 90% antifreeze.

After mixing we'll have a total of 70+x gallons.

The amount of "pure" antifreeze is 0.10(70) + 0.90x

We want the concentration to be 80%

$0.80 =\dfrac{ 0.10(70) + 0.90x}{70 + x}$

Let's clear the fractions and multiply both sides by 10.

$8(70+x)=70 + 9x$

$560 + 8x= 70 + 9x$

$490= x$

Answer: 490 gallons

Check:

(.1(70)+.9(490))/(70+490) = .8, good