SOLUTION: What amount of each mixture, one 95% alcohol and the other 15% alcohol, must be used to make 10 liters of a mixture which is 45% alcohol? what equation or equations will express th
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Question 107937: What amount of each mixture, one 95% alcohol and the other 15% alcohol, must be used to make 10 liters of a mixture which is 45% alcohol? what equation or equations will express this problem? I was thinking it would be this equation: 0.95x+0.15x=0.45 Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! AT LEAST YOU GAVE IT A SHOT BUT YOU ARE GENERALLY FAR BETTER OFF IF YOU LAY OUT THE WORD PROBLEMS (WHAT EQUALS WHAT) AS I HAVE DONE BELOW.
Let x=the amount of 95% alcohol mixture
Then 10-x=amount of 15% alcohol mixture
Now we know that the amount of pure alcohol in the 95% mixture (0.95x) plus the amount of pure alcohol in the 15% mixture (0.15(10-x)) has to equal the amount of pure alcohol in the 45% mixture (0.45*10). So our equation to solve is:
0.95x+0.15(10-x)=0.45*10 get rid of parens (distributive law)
0.95x+1.5-0.15x=4.5 subtract 1.5 from both sides
0.95x+1.5-1.5-0.15x=4.5-1.5 collect like terms
0.80x=3 divide both sides by 0.80
x=3.75 liters-----------amount of 95% alcohol mixture
10-x=10-3.75= 6.25 liters-----------------amount of 15% alcohol mixture
CK
0.95*3.75+0.15*6.25=0.45*10
3.5625+0.9375=4.5
4.5=4.5
ANOTHER WAY
Let x=amount of 95% alcohol mixture
And let y=amount of 15% alcohol mixture
Now we are told that:
x+y=10----------------------eq1
We also know that:
0.95x+0.15y=0.45*10------------------eq2
From eq1, y=10-x. Substitute this into eq2 and we get:
0.95x+0.15(10-x)=4.5-----------------same equation as before.
