SOLUTION: Milan needs 8 oz of a 20% alcohol solution for a chemistry lab. Unfortunately, his teacher only has a 1% alcohol solution and a 50% alcohol solution. How many ounces of the 50%

Click here to see ALL problems on Mixture Word Problems

Question 696763: Milan needs 8 oz of a 20% alcohol solution for a chemistry lab. Unfortunately, his teacher only has a 1% alcohol solution and a 50% alcohol solution.
How many ounces of the 50% alcohol solution should be mixed with the 1% alcohol solution to make 8 oz of the 20% alcohol solution Milan needs? Round to the nearest tenth of an ounce.
Found 3 solutions by MathLover1, checkley79, josmiceli:
Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!

let be the % alcohol solution and % alcohol solution

...substitute

.................ounces of % alcohol solution

............ounces of % alcohol solution

Answer by checkley79(3341)   (Show Source):
You can put this solution on YOUR website!
.01X+.50(8-X)=.20*8
.01X+4-.50X=1.6
-.49X=1.6-4
-.49X=-2.4
X=-2.4/-.49
X=4.9 OZ. OF 1% ALCOHOL IS USED.
8-4.9=3.1 OZ. OF 50% ALCOHOL IS USED.
PROOF:
.01*4.9+.50*3.1=.20*8
.049+1.1.55=1.6
1.6=1.6

Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website!
Let = ounces of 1% solution needed
Let = ounces of 50% solution needed
given:
(1)
(2)
--------------------------
(2)
(2)
(2)
Subtract (1) from (2)
(2)
(1)

and, since
(1)
(1)
(1)
(1)
4.898 ounces of 1% solution are needed
3.102 ounces of 50% solution are needed
check:
(2)
(2)
(2)
(2)
close enough- error due to rounding off