SOLUTION: Problem Page
In the lab, Heather has two solutions that contain alcohol and is mixing them with each other. She uses
500
milliliters less of Solution A than Solution B. Soluti
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Question 1047306: Problem Page
In the lab, Heather has two solutions that contain alcohol and is mixing them with each other. She uses
500
milliliters less of Solution A than Solution B. Solution A is
17%
alcohol and Solution B is
11%
alcohol. How many milliliters of Solution A does she use, if the resulting mixture has
167
milliliters of pure alcohol?
Found 2 solutions by ankor@dixie-net.com, stanbon:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
In the lab, Heather has two solutions that contain alcohol and is mixing them with each other.
She uses 500 milliliters less of Solution A than Solution B.
Solution A is 17% alcohol and
Solution B is 11% alcohol.
How many milliliters of Solution A does she use, if the resulting mixture has
167 milliliters of pure alcohol?
:
a = amt of A solution
b = amt of B solution
"500 milliliters less of Solution A than Solution B."
a = b - 500
:
A mixture equation
.17a + .11b = 167
Replace a with (b-500)
.17(b-500) + .11b = 167
.17b - 85 + .11b = 167
.17b + .11b = 167 + 85
.28b = 252
b = 252/,28
b = 900 ml of B solution used
therefore
900 - 500 = 400 ml of A solution
:
;
Check by finding the actual amt of alcohol in each
.17(400) = 68 ml
.11(900) = 99
------------------------
total alcohol: 167 mo
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Heather has two solutions that contain alcohol and is mixing them with each other. She uses 500 milliliters less of Solution A than Solution B.
Solution A is 17% alcohol and Solution B is 11% alcohol.
How many milliliters of Solution A does she use, if the resulting mixture has
167 milliliters of pure alcohol?
------------
B - A = 500
0.11B + 0.17A = 1*167
----------------------
11B - 11A = 11*500
11B + 17A = 100*167
-----------------
6A = 22200
A = 3700 ml
------------
Cheers,
Stan H.
------------
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