Question 896843: A 100% concentration is to be mixed with a mixture having a concentration of 40% to obtain 55 gallons of a mixture with a concentration of 75%. How much of the 100% concentration will be needed? (Round answers to nearest whole gallon.)
I've tried setting up the problem and this is what I've done-
100%con.+40%con=55 gal w/ 75% con. I'm really confused and I mainly need help setting up the problem and the best way to go about solving this.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 100% concentration is to be mixed with a mixture having a concentration of 40% to obtain 55 gallons of a mixture with a concentration of 75%. How much of the 100% concentration will be needed? (Round answers to nearest whole gallon.)
:
You can do it this way
Let x = amt of 100% concentration required
the total will be 55 gal, therefore:
(55-x) = amt of 40% mixture required
:
100x + 40(55-x) = 75(55)
100x + 2200 - 40x = 4125
100x - 40x = 4125 - 2200
60x = 1925
x = 1925/60
x = 32.0 gal of 100% stuff required
:
:
You can check this for yourself, find the amt of 40% stuff
55 - 32 = 23 gal
:
See if this comes out right
100(32) + 40(23) = 75(55) a slight error because we rounded down to 32 gal
;
Hopefully, this unconfused you a little?
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