a car radiator needs a 60% antifreeze solution. The radiator now holds 35 L of a 50% solution. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?

7 L should be drained off and replaced with 7L 100% antifreeze.

Step-by-step explanation:

The first step is to figure out the number of Liters of antifreeze are present.

50% of 35L is antifreeze.

50% * 35 L = 50/100 * 35L = 17.5 L are antifreeze.

Now you have to figure out how many L of antifreeze you need for 60% antifreeze 60*35 = 21.00 L of antifreeze are needed for a 60% solution.

The difference is 21 - 17.5 = 3.5 L of antifreeze are needed.

Since you are using 100% antifreeze, all you need do is drain off 3.5 L of what is in there and replace it with 3.5 L of 100% antifreeze. The problem is not quite that simple because the 3.5 L of antifreeze that you drained off contains 1.75 L of antifreeze. So you have to refine your answer.

Let the number of Liters you have to drain off = x

Let the number of 100% antifreeze you need to put in of 100% also be x.

50% (35 - x) + 100%*x = 60%35L Remove the brackets on the left.

17.5 - 50%x + 100%x = 60%*35 L Expand the right side.

17.5 - 50%x + 100%x = (60/100) * 35L Find the number of Liters on the right

17.5 - 50%x + 100%x = 2100/100 = 21L of antifreeze in a 60% solution. Now combine the percents on the left.

17.5 + 50%x = 21L Subtract 17.5 from both sides.

17.5 - 17.5 + 50%x = 21L - 17.5

50%x = 3.5L Translate 50% to a decimal

(50/100) x = 3.5L Multiply both sides by (100/50)

(50/100) x * (100/50) = 3.5 * (100/50)

x = 7 L

What you do is take out 7 L of 50% antifreeze and put in 7L of 100% antifreeze in the 7Ls place.