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AcademicGround >> Help with a easy math problem


10/30/04 9:58 PM
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quamrh
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Edited: 30-Oct-04
Member Since: 01/01/2001
Posts: 195
 
This is probably easy but it has me stumped. I know the answer but can't put it into a good formula for future reference. If I need a 40% antifreeze mixture and have 20L of solution at 20% how much do I need to drain and add 100% to in order to get the 20L @ 40%?
10/31/04 9:21 AM
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Andrew Yao
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Edited: 31-Oct-04 09:24 AM
Member Since: 01/01/2001
Posts: 2426
20 L at 20% implies that there is 20 * .2 = 4 L of pure in your bottle. You want to have 20 * .4 = 8 L of pure in the bottle to get 40%. let x = amount of solution to drain and y = amount of pure to add. 20L solution * (.2 pure/solution) - xL solution*(.2 pure/solution) + yL solution(1 pure/solution) = 20L solution * .4 pure/solution. 4 - 1/5x + y = 8 -1/5 x + y = 4 y = 1/5x + 4 This formula does not take into account that we want to have 20L of solution at the end. We need to solve for x and y such that x = y, so that the end volume is the same as the start volume. y = 1/5x + 4
-y = -x
-----------------
0 = -4/5x + 4
4/5x = 4
x = 5
So drain 5 liters and add 5 liters of pure. verification: start with 4 L pure in 20 L solution. Drain 5 liters, resulting in 4 - 5 * .2 = 3 L pure in 15 L of solution. Add 5 L pure resulting in 8 L pure in 20 L solution. High school algebra in the hizouse!
10/31/04 8:24 PM
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quamrh
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Edited: 31-Oct-04
Member Since: 01/01/2001
Posts: 198
Thanks man. I am trying to help my kid with this stuff and it has been a long time since I have taken a math class.
12/26/04 7:01 PM
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ValeTudo020
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Edited: 26-Dec-04
Member Since: 07/30/2002
Posts: 903
6, seriously

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