SOLUTION: how many cups of 20% alcohol solution and 15% alcohol solution should be mixed in order to make 8 cups of 18% alcohol solution?
My try:
x+y=8, (x-x+y=8-x)= y+8-x
Let x=0.20
Le
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: how many cups of 20% alcohol solution and 15% alcohol solution should be mixed in order to make 8 cups of 18% alcohol solution?
My try:
x+y=8, (x-x+y=8-x)= y+8-x
Let x=0.20
Le
Log On
Question 573496: how many cups of 20% alcohol solution and 15% alcohol solution should be mixed in order to make 8 cups of 18% alcohol solution?
My try:
x+y=8, (x-x+y=8-x)= y+8-x
Let x=0.20
Let y=0.15
0.20x+0.15y= 0.18(8)
0.20x+0.15(8-x)=0.18(8)
0.20x+1.2-0.15x=1.44
0.35x+1.2=1.44
-1.2 -1.2
0.35x=0.24
x=0.68 Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! percent ----------------quantity
Alcohol type I 20.00% ----------------x cups
Alcohol type II 15.00% ------ 8-x cups
Mixture 18.00% ---------------- 8
20.00%x+15.00%(8-x)= 18.00% * 8
20x+15( 8-x)=144
20x+120 -15x=144
20x-15x=144-120
5 x = 24
/ 5
x= 4.8 cups 20.00% Alcohol type I
3.2 cups 15.00% Alcohol type II
m.ananth@hotmail.ca