Question 212066: how many quarts of pure antifreeze must be added to 5 quarts of 10% antifreeze solution to obtain a 30% antifreeze solution
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
Let x=amount of pure antifreeze that needs to be added
Now we know that the amount of pure antifreeze added(x) plus the amount of pure antifreeze in the 5 qts of 10% antifreeze solution ((0.10)(5)) has to equal the amount of pure antifreeze in the final mixture ((0.30)(5+x)). So our equation to solve is:
x+0.10*5=0.30(5+x) get rid of parens
x+0.5=1.5+0.30x subtract 0.30x and also0.5 from each side
x-0.30x+0.5-0.5=1.5-0.5+0.30x-0.30x collect like terms
0.70x=1 divide each side by 0.70
x=1.429 qts--------------------------amount of pure antifreeze that needs to be added
ACTUALLY, IT MIGHT BE A LITTLE EASIER IN THIS CASE TO DEAL WITH THE WATER RATHER THAN THE ANTIFREEZE:
Amount of water before the mixture((0.90*5) equals amount of water after the mixture ((0.70*5+x), or:
0.90*5=0.70(5+x)
4.5=3.5+0.70x subtract 3,5 from each side
4.5-3.5=3.5-3.5+0.70x
0.70x=1------------------------same as before
CK
1.429 +0.5=0.30(5+1.429)
1.929+=1.5 + 0.429
1.929=1.929
Hope this helps---ptaylor
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