Question 1145968: By combining water with pure alcohol, louisse wants to obtain 2 liters of a solution that is 20% alcohol. On his first attempt at arbitrarily mixing the two together . louisse obtain 2 liters of a solution that test out 36% alcohol. On his second attempt , he obtains 2 liter of a solution that is 16% alcohol. Louisse now has two solutions (neither of which is correct)no more pure alcohol, and no idea of what he is doing. His sister dhenna tells him that by using algebra and combining a certain amounts of the 36% solution with 16% solution, she can solve his problem. How much of each solution does dhenna combine to get louisse his 2 liters of 20% alcohol?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x equal the number of liters of the 36% solution.
let y equal the number of liters of the 16% solution.
your equations are:
.36 * x + .16 * y = .2 * 2
x + y = 2
simplify the first equation and leave the second equation as is to get:
.36 * x + .16 * y = .4
x + y = 2
leave the first equation as is and multiply both sides of the second equation by .36 to get:
.36 * x + .16 * y = .4
.36 * x + .36 * y = .72
subtract the first equation from the second equation to get:
.2 * y = .32
solve for y to get:
y = .32 / .2 = 1.6
since x + y = 2, then x must be equal to .4.
you get:
.36 * x + .16 * y = .4 becomes .36 * .4 + .16 * 1.6 = .4 which becomes .144 + .256 = .4 which becomes .4 = .4.
x + y = 2
.36 * x + .16 * y = .4
.4 / 2 = .2 = 20%
your solution is that you combine .4 liters of 36% solution with 1.6 liters of 16% solution to get 2 liters of 20% solution.
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