Question 181478
let x be the amount drained and added back to the radiator
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we know that the original mixture is 35% water so there is 1.05 liters of water
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we also know that as we drain this, whatever we drain has 35% water. so the drained portion of water(or water lost while draining), would be .35x. 
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we also know what ever we do drain we have to add back, that is x
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our end result needs to be 50% of capacity( which is 3 liters)
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so .35(3)-.35x + x =.5(3)
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1.05-.35x + x=1.5
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.65x=.45
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{{{highlight(x=.69)}}}liters