SOLUTION: This is a teacher generated question.
How many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 40% solution.
I've tried the following
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How many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 40% solution.
I've tried the following
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Question 184199This question is from textbook
: This is a teacher generated question.
How many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 40% solution.
I've tried the following, but I'm not sure it's right at all.
a+b=80L
30a+80b=3200
Multiply by -30 to use the elimination format, this give you:
-30a - 30b = -2400L
30a + 80b = 3200L
50b = 800
b=16L, which makes a = 64L
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You can put this solution on YOUR website! How many liters of a 30% alcohol solution must be mixed with 80 liters
of a 80% solution to get a 40% solution.
:
I would do it this way:
Let x = amt of 30% solution:
:
Then when the two solutions are mixed together:
(x+80) = amt of 40% solution
:
a mixture equation using decimal equiv of per cent
:
30% amt + 80% amt = 40% amt
.3x + .8(80) = .4(x+80)
:
.3x + 64 = .4x + 32
:
.3x -.4x = 32 - 64
:
-.1x = -32
x =
x = +320 liters of 30 % solution required
You can put this solution on YOUR website! AT LEAST YOU GAVE IT A TRY BUT I CAN'T FOLLOW WHAT YOU DID BECAUSE I'M NOT SURE WHAT a AND b REPRESENT. ANYWAY, SEE IF YOU CAN FOLLOW WHAT I DID. NICE TRY!!!!!
Let x=number of liters of 30% solution needed
Now we know that the pure alcohol in the 30% solution (0.30x) plus the amount of pure alcohol in the 80% solution (0.80*80) has to equal the amount of pure alcohol in the final mixture (0.40(80+x)), so our equation to solve is:
0.30x+0.80*80=0.40(80+x) get rid of parens
0.30x+64=32+0.40x subtract 64 and also 0.40x from each side
0.30x-0.40x+64-64=32-64+0.40x-0.40x collect like terms
-0.10x=-32 divide each side by -0.10
x=320 liters----------------------ans
CK
0.30*320+0.80*80=0.40*400
96+64=160
160=160
Hope this helps---ptaylor