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A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of

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Bunuel
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A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of [#permalink] New post   19 Jan 2017, 07:58
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A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of solution that is 70% acid. What percent of acid is in the mixture?

A. 11
B. 20
C. 37.5
D. 55
E. 57.5
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D
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ganand
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Re: A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of [#permalink] New post   19 Jan 2017, 09:52
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Bunuel wrote:
A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of solution that is 70% acid. What percent of acid is in the mixture?

A. 11
B. 20
C. 37.5
D. 55
E. 57.5


Method 1:
Amount of acid = 12*45/100 + 8*70/100 = 11 ounces. Total volume = 20 ounces.

% of acid = 11*100/20 = 55 %.

Method 2:

Apply alligation method.
Let percent of acid in the mixture is x.
Please refer the attached diagram.
We have following relation:
\(\frac{70 - x}{x - 45} = \frac{12}{8} \Rightarrow 5x = 275 \Rightarrow x = 55\)

% of acid = 55%

Thanks.
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gracie
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A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of [#permalink] New post   21 Jan 2017, 15:21
Bunuel wrote:
A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of solution that is 70% acid. What percent of acid is in the mixture?

A. 11
B. 20
C. 37.5
D. 55
E. 57.5


let x=% of acid in mixture
.45*12+.7*8=20x
x=55%
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A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of [#permalink] New post   21 Jan 2019, 18:55
Expert Reply
Bunuel wrote:
A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of solution that is 70% acid. What percent of acid is in the mixture?

A. 11
B. 20
C. 37.5
D. 55
E. 57.5


The percent of acid in the mixture is:

[0.45(12) + 0.7(8)]/(12 + 8)

=(5.4 + 5.6)/20 x 100

= 11/20 x 100

= 55

Alternate Solution:

We combine 12 ounces of 45% solution with 8 ounces of 70% solution, getting 20 ounces of x% solution. We can express this in an equation as:

12(0.45) + 8(0.7) = 20x

5.4 + 5.6 = 20x

11 = 20x

11/20 = x

x = .55, or 55%.

Answer: D
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Re: A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of [#permalink] New post   22 Jan 2019, 15:25
ganand wrote:
Bunuel wrote:
A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of solution that is 70% acid. What percent of acid is in the mixture?

A. 11
B. 20
C. 37.5
D. 55
E. 57.5


Method 1:
Amount of acid = 12*45/100 + 8*70/100 = 11 ounces. Total volume = 20 ounces.

% of acid = 11*100/20 = 55 %.

Method 2:

Apply alligation method.
Let percent of acid in the mixture is x.
Please refer the attached diagram.
We have following relation:
\(\frac{70 - x}{x - 45} = \frac{12}{8} \Rightarrow 5x = 275 \Rightarrow x = 55\)

% of acid = 55%

Thanks.


Hi in your allegation method you wrote x -45, but in the diagram it says 45-x. What did you do to change the signs in this case?
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Re: A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of [#permalink] New post   27 Jan 2019, 14:07
70−x/x−45=12/8
⇒5x=275
⇒x=55 (D)
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Re: A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of [#permalink]
27 Jan 2019, 14:07
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