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A certain solution that is 30 percent acid is obtained by mixing x
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13 Feb 2024, 21:44
13 Feb 2024, 21:44
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A certain solution that is 30 percent acid is obtained by mixing x gallons of a 20 percent acid solution with y gallons of a 60 percent acid solution, where the percents are according to volume. How many gallons of the 60 percent acid solution must be used to obtain 10 gallons of the 30 percent acid solution?
A. 0.75
B. 2.5
C. 3.3
D. 4.0
E. 5.0
A. 0.75
B. 2.5
C. 3.3
D. 4.0
E. 5.0
Quant Chat Moderator
Joined: 22 Dec 2016
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Re: A certain solution that is 30 percent acid is obtained by mixing x
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13 Feb 2024, 22:49
13 Feb 2024, 22:49
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DerekLin wrote:
A certain solution that is 30 percent acid is obtained by mixing x gallons of a 20 percent acid solution with y gallons of a 60 percent acid solution, where the percents are according to volume. How many gallons of the 60 percent acid solution must be used to obtain 10 gallons of the 30 percent acid solution?
A. 0.75
B. 2.5
C. 3.3
D. 4.0
E. 5.0
A. 0.75
B. 2.5
C. 3.3
D. 4.0
E. 5.0
Volume of the solution that is 30% acid = \(x + y\)
Amount of acid in that solution = \(\frac{30}{100}(x+y) = 0.3x + 0.3y\)
The amount of acid in the solution that is 30% acid = Volume of acid in the solution that is 20% acid + Volume of acid in the solution that is 60% acid
\(0.3x + 0.3y = 0.2x + 0.6y\)
\(0.1x = 0.3y\)
\(\frac{x}{y} = \frac{3}{1}\)
Hence, for every 1 part of 'y' we have 3 parts of 'x'.
The number of gallons of the 60 percent acid solution must be used to obtain 10 gallons of the 30 percent acid solution = \(\frac{1}{4} * 10 = 2.5\)
Option B
Re: A certain solution that is 30 percent acid is obtained by mixing x
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02 May 2024, 06:12
02 May 2024, 06:12
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A certain solution that is 30 percent acid is obtained by mixing x gallons of a 20 percent acid solution with y gallons of a 60 percent acid solution, where the percents are according to volume. How many gallons of the 60 percent acid solution must be used to obtain 10 gallons of the 30 percent acid solution?
A. 0.75
B. 2.5
C. 3.3
D. 4.0
E. 5.0
Here's the teeter-totter method: If you have 2 percentages of the additive solution and the final solution percentage, you set up up a line with with the two additive solutions with the final solution in the middle 20-----30------60, you find the difference to the final solution, 10 and 30, then reduce, 1:3, whichever is closest to the final solution (20) you give that the favor of the larger amount, so its 1:3 (1/4) 60 solution and 3:1(3/4) 20 solution, multiply 1/4 by 10 gallons to find the answer. 2.5
Hope that helps
A. 0.75
B. 2.5
C. 3.3
D. 4.0
E. 5.0
Here's the teeter-totter method: If you have 2 percentages of the additive solution and the final solution percentage, you set up up a line with with the two additive solutions with the final solution in the middle 20-----30------60, you find the difference to the final solution, 10 and 30, then reduce, 1:3, whichever is closest to the final solution (20) you give that the favor of the larger amount, so its 1:3 (1/4) 60 solution and 3:1(3/4) 20 solution, multiply 1/4 by 10 gallons to find the answer. 2.5
Hope that helps
