How many liters of a 30% alcohol solution should be mixed with 10 liters of pure alcohol to produce a 40% solution?
2006-07-19 16:13:56 · 9 answers · asked by Anonymous in Education & Reference ➔ Homework Help
Let x = liters of 30% solution
You know that 10 = liters of pure alcohol (which is 100%)
You mixture is
liters of 30% + liters of pure, so
liters of 40% mixture = x + 10
30(x) + 100(10) = 40(x+10)
30x + 1000 = 40x + 400
Subtract 30x from each side
1000 = 10x + 400
Subtract 400 from each side
600 = 10x
Divide each side by 10
60 = x
2006-07-19 16:32:46 · answer #1 · answered by MsMath 7 · 0⤊ 0⤋
y = 0.3x + 10 (where x = liters of solution added and y = liters of total alcohol)
y = 0.4(x + 10) (40% of the solution is alcohol)
Where do these two lines intersect? That will give you the answer.
0.3x + 10 = 0.4(x + 10)
0.3x + 10 = 0.4 x + 4
6 = 0.1x
60 = x
You have to add 60 L of 30% solution to 10 L pure alcohol to produce 70 L of a 40% solution.
2006-07-19 23:39:39 · answer #2 · answered by jimbob 6 · 0⤊ 0⤋
y = 0.3x + 10
y = 0.4(x + 10)
0.3x + 10 = 0.4(x + 10)
0.3x + 10 = 0.4 x + 4
6 = 0.1x
60 = x
Add 60 L of 30% solution to 10 L pure alcohol to make 70 L of a 40% solution.
2006-07-20 01:05:57 · answer #3 · answered by choirgirl1987 2 · 0⤊ 0⤋
10 liters i think . haven't done albebra in years and only first year. but you are 10% off .so you would need to add 10% more to get to the 40% level.
2006-07-19 23:23:57 · answer #4 · answered by Anonymous · 0⤊ 0⤋
10*1+.3*x = .4(x+10)
3x=.4x+4
2.6x=4
x = 1.5385 Liters
2006-07-19 23:19:01 · answer #5 · answered by dudeabides 2 · 0⤊ 0⤋
x30 + 10(100)=40
2006-07-19 23:18:07 · answer #6 · answered by Mr. Sly 4 · 0⤊ 0⤋
2.
You start with 10L of the good stuff, 100%, and you dillute it with 2L of 30% for an over all cut of 60% - leaving you with 40% left over.
2006-07-19 23:17:35 · answer #7 · answered by Mongoose 2 · 0⤊ 0⤋
100 liters.
2006-07-19 23:17:39 · answer #8 · answered by AlexS 2 · 0⤊ 0⤋
I HAVE NO IDEA, but I love that every expert that answered you has a different answer.
I wouldn't trust any of them. Go to a school help site (use google to find one for algebra.)
Good luck.
2006-07-19 23:48:34 · answer #9 · answered by ? 3 · 0⤊ 0⤋