A 40% antifreeze solution is to be mixed with a 70% antifreeze
solution to get 240 liters of a 50% solution. How many liters of the
40% solution and how many liters of the 70% solution will be used?

of antifreeze solution and of antifreeze solutions will be used.

Given, two solutions namely antifreeze and antifreeze solutions.

Let litres of the antifreeze solution and litres of the antifreeze solutions will be used.

Total volume of the solution,

Now, of antifreeze solution is to be mixed with a of antifreeze

solution to get 240 liters of a solution,

From Equation (1)  , substitute the value of in Equation (2),we get

Putting the value of , we get

.

Hence of antifreeze solution and of antifreeze solutions will be used.

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160 liters of the

40% solution and 80 liters of the 70% solution will be used.

Step-by-step explanation:

Let x represent the number of liters of 40% antifreeze solution that should be used.

Let y represent the number of liters of 70% antifreeze solution that should be used.

The volume of the mixture to be mixed is 240 liters. It means that

x + y = 240

The 40% antifreeze solution is to be mixed with a 70% antifreeze

solution to get 240 liters of a 50% solution. This means that

0.4x + 0.7y = 0.5(240)

0.4x + 0.7y = 120 - - - - - - - - - - - -1

Substituting x = 240 - y into equation 1, it becomes

0.4(240 - y) + 0.7y = 120

96 - 0.4y + 0.7y = 120

- 0.4y + 0.7y = 120 - 96

0.3y = 24

y = 24/0.3

y = 80

x = 240 - y = 240 - 80

x = 160

step-by-step explanation:

107 units

4sqrt5

step-by-step explanation:

the base has dimensions of 8 x 4, so the diagonal is 64 + 16 square rooted or 4sqrt5.