SOLUTION: A 26-quart radiator contains 50% coolant solution. How much of the solution needs to be drained and replaced with pure coolant to obtain a 60% coolant solution?
Algebra ->
Length-and-distance
-> SOLUTION: A 26-quart radiator contains 50% coolant solution. How much of the solution needs to be drained and replaced with pure coolant to obtain a 60% coolant solution?
Log On
Question 1112665: A 26-quart radiator contains 50% coolant solution. How much of the solution needs to be drained and replaced with pure coolant to obtain a 60% coolant solution? Found 2 solutions by josgarithmetic, ikleyn:Answer by josgarithmetic(39616) (Show Source):
Let V be the volume to drain off from 26 quarts of antifreeze.
Step 1: Draining. After draining, you have 26-V quarts of the 50% antifreeze.
It contains 0.50*(26-V) of pure antifreeze.
Step 2: Replacing. Then you add V quarts of the pure antifreeze (the replacing step).
After the replacing, you have the same total liquid volume of 26 quarts.
It contains 0.50*(26-V) + V quarts of pure antifreeze.
So, the antifreeze concentration after replacement is .
It is the ratio of the pure antifreeze volume to the total volume.
Therefore, your "concentration equation" is
= 0.6. (1)
The setup is done and completed.
To solve the equation (1), multiply both sides by 26. You will get
0.50*(26-V) + V= 0.6*26,
13 - 0.5V + V= 15.6,
0.5V = 15.6 - 13 = 2.6 ====> V = = 5.2 liters.
Answer. 5.2 quarts of the 50% antifreeze must be drained and replaced by 5.2 quarts of pure antifreeze.
Check. = 0.6. ! Correct !