the solution in the large container was 60% alcohol, and the solution in the small container was only 30% alcohol. how much of each should be used to get 300 ml that is 42% alcohol?
2007-04-23 13:02:01 · 4 answers · asked by argentina_mandy20 1 in Science & Mathematics ➔ Mathematics
You have a 60% solution (x) and a 30% solution (y), that together need to combine to a 42% solution (z).
0.6x+0.3y=0.42z
However, you know you'll end up with 300 mL of z, and since we assume no evaporation, you must put a total of 300 mL in. so y=300-x, and z=300
0.6x+0.3(300-x)=0.42(300)
0.6x+90-0.3x=126
0.3x+90=126
0.3x=36
x=120
y=300-120
y=180
120mL of the 60% solution, 180mL of the 30% solution
2007-04-23 13:34:31 · answer #1 · answered by apachetear 2 · 0⤊ 0⤋
Let x be the number of mL of 60% alcohol; since the total is 300 mL, 300 - x is the amount of 30%. Then write an equation based on percents:
60x + 30 (300 - x) = 42 (300)
60x + 9000 - 30x = 12600
Solve for x
2007-04-23 13:09:20 · answer #2 · answered by hayharbr 7 · 0⤊ 0⤋
120 ml of 60%
180 ml of 30%
2007-04-23 13:13:55 · answer #3 · answered by kennyk 4 · 0⤊ 0⤋
.60x+.40y=300*.=0.42*300
x+y=300
solve for x and y
2007-04-23 13:06:35 · answer #4 · answered by leo 6 · 0⤊ 0⤋