How many gallons of 60% antifreeze solution must be mixed with 70 gallons of 20% antifreeze to get a mixture that is 50% antifreeze. (Round to the nearest whole number)

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).

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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:

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

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Multiply by 100 to eliminate fractions.

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

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2250 + 70x = 5400 + 60x

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10x = 5400 -2250

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10x = 3150

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x = 315 gallons of 70% antifreeze.

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How much will you end up having on hand?

90 + 315 = 405 gallons

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

.6*405 = 243

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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:

21/58.

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step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

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