My calculation:

20 x 15/100 = 3

20 x 20/100 = 4

1-gallon alcohol should be added to make a new solution that is 20% alcohol.

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My calculation:

20 x 15/100 = 3

20 x 20/100 = 4

1-gallon alcohol should be added to make a new solution that is 20% alcohol.

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- Jun 18, 2007

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- 21,308

Assume that we have to add '"A 20-gallon alcohol-water solution contains 15% pure alcohol. How much alcohol should be added to make a new solution that is 20% alcohol?"

My calculation:

20 x 15/100 = 3

20 x 20/100 = 4

1-gallon alcohol should be added to make a new solution that is 20% alcohol. .....................Incorrect

Volume of alcohol in original solution = 0.15 * 20 = 3 gallons

New volume of mixture = 20 + x

Volume of alcohol in final solution = 3 +x

given: new solution that is 20% alcohol

So:

(3 + x)/(20 + x) = 20% = 0.2

Solve for 'x'

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Could you explain

My calculation:

20 x 15/100 = 3

20 x 20/100 = 4

1-gallon alcohol should be added to make a new solution that is 20% alcohol.

If you add x gallons you will have 3+ x gallons of alcohol in 20+ x gallons of solution. You want \(\displaystyle \frac{3+ x}{20+ x}= 0.20\).

(3 + x)/(20 + x) = 20% = 0.2Assume that we have to add 'x' gallons of alcohol (to the 15% solution)

Volume of alcohol in original solution = 0.15 * 20 = 3 gallons

New volume of mixture = 20 + x

Volume of alcohol in final solution = 3 +x

given: new solution that is 20% alcohol

So:

(3 + x)/(20 + x) = 20% = 0.2

Solve for 'x'

3 + x = 0.2(20 + x)

3 + x = 4 + 0.2x

15 + 5x = 20 + x

-5 + 5x = x

-5 = - 4x

4x = 5

x = 1.25

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Almost good work!(3 + x)/(20 + x) = 20% = 0.2

3 + x = 0.2(20 + x)

3 + x = 4 + 0.2x

15 + 5x = 20 + x

-5 + 5x = x

-5 = - 4x

4x = 5

x = 1.25

x = 1.25 ....... number is correct - but what is the unit? ft