Question

A 5-gallon radiator containing a mixture of water and antifreeze was supposed to contain a 50% antifreeze solution. When tested, it was found to have only 40% antifreeze. How much must be drained out and replaced with pure antifreeze so that the radiator will then contain the desired 50% antifreeze solution?

Answer 1

Answer: drain 5/6 gallon from the radiator and replace with 5/6 gallon of antifreeze

Explanation:

Since we have 5gallon, for the radiator to contain 50% antifreeze, it must also contain 50% water.

Antifreeze = 50% of 5 gallon =0.5*5=2.5 gallon.

Similarly, amount of water is 2.5 gallons.

But analysis shows there's 60% water and 40% antifreeze

Antifreeze= 0.4*5= 2 gallons

Water = 0.6*5= 3 gallons.

So, we would need to drain out a solution containing approximately 0.5 gallons of water so that we can have quantity of water equals 2.5 gallon (3gallons - 0.5gallons=2.5).

Let us now work out amount of water to drain:

1 gallon -----> 0.6 gallon water (0.4 antifreeze)

x gallon-----> 0.5 gallon water (y antifreeze)

Cross multiply and work out x:

x = 0.5/0.6= 5/6 gallons

Amount of antifreeze to replace:

Hence we drain out 5/6 gallon and replace it with 5/6 gallon of antifreeze.