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We know that the new solution is 3% vinegar and has a total of 62 ounces of liquid.uptowngirl92 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?
19.3%
17%
16.67%
15.5%
12.5%
Vena wrote:The resulted solution contains: 0.03(12+50)=1.86 ounce of vinegar.
Adding 50 ounce of water does not change the amount of vinegar in the original solution and the final one.
Hence, the %vinegar in the original solution is: (1.86/12)*100=15.5%
Choose D
Could anyone tell me if my approach is appropriate? Tnx a lot
To make the math easy, treat the given percentages as averages.uptowngirl92 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?
19.3%
17%
16.67%
15.5%
12.5%
