The radiator in a car is filled with a solution of 75 per cent antifreeze and 25 per cent water. The...

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The radiator in a car is filled with a solution of 75 per cent antifreeze and 25 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the radiator is 4.7 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?

SnapAndSolve AI · Accepted Answer

Let x be the amount of coolant to be drained and replaced with water.

Initially, the radiator has 4.7 liters of 75% antifreeze, which is 0.75 * 4.7 = 3.525 liters of antifreeze.

After draining x liters and replacing with water, the total volume is still 4.7 liters, but the antifreeze amount is 3.525 - 0.75x liters.

We want the final concentration to be 50%, so:

(3.525 - 0.75x) / 4.7 = 0.5

Solving for x:

3.525 - 0.75x = 0.5 * 4.7
3.525 - 0.75x = 2.35
1.175 = 0.75x
x = 1.175 / 0.75
x ≈ 1.567 liters

Therefore, approximately 1.567 liters of coolant must be drained and replaced with pure water.

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The radiator in a car is filled with a solution of 75 per cent antifreeze and 25 per cent water. The...

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