Question

How many gallons of a 50​% antifreeze solution must be mixed with 70 gallons of ​10% antifreeze to get a mixture that is 40​% ​antifreeze?

Answer 1

Answer: 180 gallons needed

Explanation:

Zykeith,

Assume x gallons of 50% antifreeze is needed

Final mixture is x + 60 gallons

Amount of antifreeze in mixture is 0.4*(x+60)

Amount of antifreeze added is .5x + .1*60 = .5x + 6

so .5x + 6 = .4(x + 60)

.5x -.4x = 24 -6

.1x = 18

x = 180

Answer 2

Let x be the number of gallons of the 50% antifreeze solution needed. We know that the resulting mixture will be 70 + x gallons. To get a 40% antifreeze mixture, we can set up the following equation:

`{\implies 0.5x + 0.1(70) = 0.4(70 + x)}`

Simplifying the equation:

`\qquad\implies 0.5x + 7 = 28 + 0.4x`

`\qquad\quad\implies 0.1x = 21`

`\qquad\qquad\implies \bold{x = 210}`

`\therefore` We need 210 gallons of the 50% antifreeze solution to mix with 70 gallons of 10% antifreeze to get a mixture that is 40% antifreeze.

`\blue{\overline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}`