How many mL of 20% soln must be added to 40% so....

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samsneaky

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Oct 13, 2007
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she had 500ML of a 40% love potion solution. she also has a love potion that is a 20% solution. How many ML of the 20% solution must she add to the 40% so that she can end up with a 25% solution that will help her finally get her man ?

thank you all that attempt the problem
 
Try using the standard set-up for "mixture" word problems:

. . .milliliters:
. . . . .40%: 500
. . . . .20%: x
. . . . .mix: 500 + x

. . .percent active ingredient:
. . . . .40%: 0.4
. . . . .20%: 0.2
. . . . .mix: [fill this in]

. . .milliliters active ingredient:
. . . . .40%: 0.4(500)
. . . . .20%: 0.2(x)
. . . . .mix: [fill this in]

Since the inputs (of the active ingredient) should match the output, sum the inputs and set equal to the output. Solve for the variable.

If you get stuck, please reply showing all of your work and reasoning. Thank you! :D

Eliz.
 
@stapel

thank you however i get the equation

(.4)(500) + (.2)(X)= .25(y)

i can not solve this i do not know what i am doing wrong, do i need two equations ???
 
samsneaky said:
she had 500ML of a 40% love potion solution. she also has a love potion that is a 20% solution. How many ML of the 20% solution must she add to the 40% so that she can end up with a 25% solution that will help her finally get her man ?

thank you all that attempt the problem
Click to expand...

Please show us your work, indicating exactly where you are stuck, so that we can find out where to begin to help you.
 
samsneaky said:
@stapel

thank you however i get the equation

(.4)(500) + (.2)(X)= .25(y)<--- what is y?

i can not solve this i do not know what i am doing wrong, do i need two equations ???
Click to expand...
 
@eliz

dont worry about it the mixture link helped i think i got the problem thank you very much for your help
 
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