The radiator needs water

Question

The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests that, for the summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 3.6 L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?

Answer 1

Given:

Total volume = 3.6L

current mix = 40% water 60% antifreeze

Required:

How much solution should be drained? And

How much water is needed to get 50/50 solution?

Solution:

Current water = 0.4x3.6 = 1.44L

Ccurrent antifreeze = 0.6x3.6 = 2.16L

Final solution should be 1.8L water and 1.8L antifreeze.

Let y = the amount of current solution needed to get antifreeze to be 1.8L.

0.6y= 1.8

Y= 3L

So 3.6 - 3.0 = 0.6L

Answer:

0.6L of solution needs to be drained and replaced with water.