A car radiator needs a 60​% antifreeze solution. The radiator now holds 21 liters of a 40​% solution. How many liters of this should be drained and replaced with​ 100% antifreeze to get the desired​ strength?

Answer 1

Let x be the number of liters drained and replaced with 100% antifreeze.
The amount of pure antifreeze in the 40% solution that will remain after x liters are drained is 0.4 * (21 - x) liters.
The amount of pure antifreeze in the 100% antifreeze that will be added is 1 * x liters.
The total amount of pure antifreeze in the new solution will be 0.4 * (21 - x) + x liters.
The total amount of solution after x liters are drained and replaced is 21 liters.
The concentration of antifreeze in the new solution is (0.4 * (21 - x) + x) / 21 = 0.6.
0.4 * (21 - x) + x = 0.6 * 21.
8.4 - 0.4x + x = 12.6.
0.6x = 12.6 - 8.4.
0.6x = 4.2.
x = 4.2 / 0.6.
x = 7.
The number of liters drained and replaced with 100% antifreeze is 7. Answer: \boxed{7}.