An automobile radiator contains 16 liters of antifreeze andwater. This mixture i

Question

An automobile radiator contains 16 liters of antifreeze andwater. This mixture is 30% antifreeze. How much of this mixtureshould be drained and replaced with pure antifreeze so that therewill be 50% antifreeze? Please define variables and write a system of equations inyour answer. Thanks! An automobile radiator contains 16 liters of antifreeze andwater. This mixture is 30% antifreeze. How much of this mixtureshould be drained and replaced with pure antifreeze so that therewill be 50% antifreeze? Please define variables and write a system of equations inyour answer. Thanks!

Explanation / Answer

You know that 30 percent of the 16 L is antifreeze so thatmeans there's 4.8 L of antifreeze and 11.2 L of water in theradiator. I'm assuming that when you drain the mixture, anequal amount of water and antifreeze are being drained at the sametime and since you need equal parts water and antifreeze, thats atotal of 8 L each. You would need to drain 3.2 L of antifreeze and 3.2 L of water to make the amount of water equal 8 L. So that's a total of 6.4 L of mixture.
The way I thought of it: 16 x 0.3 = 4.8 L of antifreeze
4.8 L of antifreeze - 3.2 L = 1.6 L (add 6.4 L of antifreezeto make it 50%) 11.2 L of water (need to make this 8 L) - 3.2 L = 8 L 3.2 + 3.2 = 6.4 L of mixture

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