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lukranit [14]
3 years ago
6

How many gallons of 60% antifreeze solution must be mixed with 70 gallons of 20% antifreeze to get a mixture that is 50% antifre

eze.
(Round to the nearest whole number)
Mathematics
1 answer:
Papessa [141]3 years ago
5 0

Answer:

Answer: Add 315 gallons of 70% antifreeze to 90 gallons of 25% antifreeze to make a 60% mixture.

You need to end up a mixture that is 60% antifreeze.

You have 90 gallons of 25% antifreeze, which means it has .25*90 = 22.5 gallons of pure antifreeze mixed with a solvent (probably water).

.

You will add 'x' gallons of 70% antifreeze to make the 60% mixture.

So, at the conclusion, you will have (90+x) gallons  60% 'pure' antifreeze. That can be shown with the equation:

.

.25*90 + .7*x = .6*(90+x)

.

Multiply by 100 to eliminate fractions.

.

25*90 + 70x = 60(90 +x)

.

2250 + 70x = 5400 + 60x

.

10x = 5400 -2250

.

10x = 3150

.

x = 315 gallons of 70% antifreeze.

.

How much will you end up having on hand?

90 + 315 = 405 gallons

.

If it is 60% antifreeze, how much 'pure' antifreeze will be in the mixture?

.6*405 = 243

.

In the original 90 gallons, you have 22.5 gallons of pure antifreeze.

In the additional 315 gallons, you have .7*315 = 220.5 gallons.

22.5+220.5 = 243 gallons

Step-by-step explanation:

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