SOLUTION: You need 325 mL of a 5% alcohol solution. On hand, you have a 25% alcohol mixture. How much of the 25% alcohol mixture and pure water will you need to obtain the desired solution?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: You need 325 mL of a 5% alcohol solution. On hand, you have a 25% alcohol mixture. How much of the 25% alcohol mixture and pure water will you need to obtain the desired solution?      Log On

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Question 788849: You need 325 mL of a 5% alcohol solution. On hand, you have a 25% alcohol mixture. How much of the 25% alcohol mixture and pure water will you need to obtain the desired solution?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
{pure alcohol content}/{total amount of solution}={Target Concentration}

u = amount of the water which is 0% alcohol
v = amount of the 25% alcohol

25%2Av%2F%28u%2Bv%29=5 and u%2Bv=325

v=325-u and substitute into the rational equation.
25%28325-u%29=5%28u%2B325-u%29
25%2A325-25u=5%2A325
25%2A325=5%2A325%2B25u
25%2A325-5%2A325=25u
325%2825-5%29=25u
20%2A325=25u
25u=20%2A13%2A25
u=20%2A13
highlight%28u=260%29 mL.

Steps can vary. Some people's steps may be more efficient than shown here. Using factors and making more steps was easier here to avoid using separate computations on paper for simplifications.