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

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Math Expert
Joined: 02 Sep 2009
Posts: 52938
A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of  [#permalink]

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19 Jan 2017, 06:58
1
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00:00

Difficulty:

15% (low)

Question Stats:

84% (01:37) correct 16% (01:36) wrong based on 141 sessions

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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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Manager
Joined: 17 May 2015
Posts: 249
Re: A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of  [#permalink]

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19 Jan 2017, 08:52
1
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.
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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Joined: 07 Dec 2014
Posts: 1156
A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of  [#permalink]

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21 Jan 2017, 14:21
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]

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21 Jan 2019, 17:55
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%.

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Manager
Joined: 22 Sep 2018
Posts: 243
Re: A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of  [#permalink]

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22 Jan 2019, 14:25
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.
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?
Manager
Joined: 30 Oct 2018
Posts: 59
Re: A chemist mixes a 12-ounce solution that is 45% acid with 8 ounces of  [#permalink]

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27 Jan 2019, 13:07
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, 13:07
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