Let x be the amount removed and then replaced with pure (100%) antifreeze
(10-x)*.3 +x =.5*10
3 -.3x + x = 5
.7x=2
x=20/7 qts
Andrew W.
asked 05/05/20A service station checks Mr. Gittleboro's radiator and finds it contains only 30% antifreeze. If the radiator holds 10 quarts and is full, how much must be drained off and replaced with pure antifreeze in order to bring it up to a required 50% antifreeze?
Let x be the amount removed and then replaced with pure (100%) antifreeze
(10-x)*.3 +x =.5*10
3 -.3x + x = 5
.7x=2
x=20/7 qts
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