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If 12 ounces of a strong vinegar solution are diluted with 5
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02 Jul 2020, 09:31
02 Jul 2020, 09:31
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62% (02:21) correct
37% (01:32) wrong based on 67 sessions
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If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a three-percent vinegar solution, what was the concentration of the original solution?
A. 19.3%
B. 17%
C. 16.67%
D. 15.5%
E. 12.5%
A. 19.3%
B. 17%
C. 16.67%
D. 15.5%
E. 12.5%
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Re: If 12 ounces of a strong vinegar solution are diluted with 5
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02 Jul 2020, 09:37
02 Jul 2020, 09:37
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Carcass wrote:
If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a three-percent vinegar solution, what was the concentration of the original solution?
A. 19.3%
B. 17%
C. 16.67%
D. 15.5%
E. 12.5%
A. 19.3%
B. 17%
C. 16.67%
D. 15.5%
E. 12.5%
When 50 ounces of water is added to the 12 ounces of original solution, we get 62 ounces of a NEW solution.
The new solution is 3% vinegar.
3% of 62 = 1.86
So, the NEW solution contains 1.86 ounces of pure vinegar.
These 1.86 ounces of pure vinegar came from the ORIGINAL solution.
The ORIGINAL solution had a volume of 12 ounces. 1.86 ounces was pure vinegar.
1.86/12 = what percent?
Well, 1.8/12 = 15%, so 1.86/12 must be a little bit more than 15%.
Answer: D
Cheers,
Brent
Re: If 12 ounces of a strong vinegar solution are diluted with 5
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Updated on: 07 Jul 2020, 17:42
Updated on: 07 Jul 2020, 17:42
1
Let x+y=62
12+50=62
12x+50y=0.03(x+y)
12x+50y=0.03(62)
we are concerned with strong vinegar only so,
12x=1.86
x=15.5
Please correct me if i am wrong
12+50=62
12x+50y=0.03(x+y)
12x+50y=0.03(62)
we are concerned with strong vinegar only so,
12x=1.86
x=15.5
Please correct me if i am wrong
