wittlocoodee had one solution that was 10% glycol and another that was 40% glycol. how much of each should she use to get 200 liters of soultion that was 19% glycol?
2007-04-02 14:59:25 · 4 answers · asked by argentina_mandy20 1 in Science & Mathematics ➔ Mathematics
Say X is volume if 10% glycol solution and Y is volume of 40% glycol solution.
X + Y = 200.
(0.1X + 0.4Y) / 200 = 0.19
Solve this system of equations and you'll get your answer
2007-04-02 15:06:20 · answer #1 · answered by Jack 3 · 0⤊ 0⤋
Suppose she uses x litres of 10% glycol, then she must use 200-x litres of 40% glycol to get a total of 200 litres.
The total amount of glycol is (10%Ãx + 40%Ã(200-x)), and we need this to be equal to 19%Ã200 = 38. So we get
(10 x + 40 (200-x)) / 100 = 38
<=> 10x + 8000 - 40x = 3800
<=> 30x = 4200
<=> x = 140
So she needs to use 140L of 10% glycol and 60L of 40% glycol.
2007-04-02 22:06:00 · answer #2 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
Let x = the liters of 40% glycol. Since the total is 200L, 200 - x = the liters of 10% glycol. Multiply % times amount of each part, add, make it equal % times amount of mixture.
.40x + .10(200-x) = .19(200)
.40x + 20 - .10x = 38
.30x + 20 = 38
Then solve this for x and go from there
2007-04-02 22:07:06 · answer #3 · answered by hayharbr 7 · 0⤊ 0⤋
x=10% glycol
y=40% glycol
Here's the system:
x+y=200
0.1x+0.4y=0.19(200)
Just solve for x and y.
2007-04-02 22:05:57 · answer #4 · answered by dcl 3 · 0⤊ 0⤋