Question

A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 Liters of a 64% solution. How many liters of

the 40% and how many liters of the 70% solution will be used?

Answer 1

Let x be the number of liters of the 40% antifreeze solution to be used, and y be the number of liters of the 70% antifreeze solution to be used.

We can start by setting up two equations based on the information given:

Equation 1: x + y = 240 (total volume of the mixture is 240 liters)

Equation 2: 0.4x + 0.7y = 0.64(240) (total amount of antifreeze in the mixture is 64% of 240 liters)

Simplifying Equation 2, we get:

0.4x + 0.7y = 153.6

Now we can use Equation 1 to solve for one of the variables in terms of the other. Let's solve for y:

y = 240 - x

Substituting this into Equation 2, we get:

0.4x + 0.7(240 - x) = 153.6

Simplifying and solving for x, we get:

0.4x + 168 - 0.7x = 153.6

-0.3x = -14.4

x = 48

So we need 48 liters of the 40% antifreeze solution. Using Equation 1 to find y, we get:

y = 240 - x = 240 - 48 = 192

So we need 192 liters of the 70% antifreeze solution.

Therefore, 48 liters of the 40% antifreeze solution and 192 liters of the 70% antifreeze solution should be mixed to obtain 240 liters of a 64% antifreeze solution.