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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?

User Oskarkv
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2 Answers

1 vote

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

User Mans
by
8.3k points
4 votes

SOLUTION:

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}}

User Amiri
by
8.2k points

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