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 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

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

$\frac{60}{100}\times 3.6 = 2.16 L$

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator  =50% of 3.6 L = $\frac{50}{100}\times 3.6=1.8 L$

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired  volume

2.16 L - 60% of( x) = 1.8 L

$2.16 L-0.6x=1.8 L$

$x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L$

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.