The cooling system on a truck contains 5 gallons of collant that is 40% antifreeze mixture. How much must be withdrawn and replaced with pure 100% antifreeze to bring the coolant in the system to a 50% antifreeze mixture?

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The cooling system on a truck contains 5 gallons of collant that is 40% antifreeze mixture. How much must be withdrawn and replaced with pure 100% antifreeze to bring the coolant in the system to a 50% antifreeze mixture?
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Equation:
anti -anti + anti = anti
0.40*5 - 0.40x + x = 0.50(5)
Multiply thru by 100 to get:
40*5 - 40x + 100x = 50*5
60x = 10*5
x = (1/6)5 = 5/6 gallon (amt. that should be replaced in the mixture)
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Cheers,
stan H.
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