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How many gallons of a ​70% antifreeze solution must be mixed with gallons of 90​% antifreeze to get a mixture that is 30​% ​antifreeze? Use the​ six-step method.

Sagot :

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).

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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:

Hope this helps,

Have a great day

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MrRoyalidn

The relationship between each proportion of antifreeze can be used to form an equation.

The number of 90% antifreeze that must be added is 490 gallons.

The question is incomplete. So, I will use the following details to answer the question

[tex]Av ai la bl e = 70[/tex] gallons of 10% antifreeze

[tex]x \to[/tex] gallons of 90% antifreeze

[tex]Total = x + 70[/tex] --- after mixing the solution; to get a 80% solution

The equation is then represented as:

[tex]70 \times 10\% + x \times 90\% = (x + 70) \times 80\%[/tex]

[tex]7 + 0.9x = 0.8x + 56[/tex]

Collect like terms

[tex]0.9x - 0.8x = 56 - 7[/tex]

[tex]0.1x = 49[/tex]

Divide by 0.1

[tex]x = \frac{49}{0.1}[/tex]

[tex]x = 490[/tex]

Hence, 490 gallons of 90% antifreeze must be added.

Read more about equations at:

https://brainly.com/question/2264804

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