Question

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?

Answer 1

Answer: 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

Answer 2

$160 \ litres$ of $40 \%$ antifreeze solution and $80 \ litres$ of $70 \%$ antifreeze solutions will be used.

Given, two solutions namely $40 \%$ antifreeze and $70 \%$ antifreeze solutions.

Let $x$ litres of the $40 \%$ antifreeze solution and $y$ litres of the $70 \%$ antifreeze solutions will be used.

Total volume of the solution,

$x+y=240..........(1)$

Now, $40\%$ of antifreeze solution is to be mixed with a $70 \%$ of antifreeze

solution to get 240 liters of a $50 \%$ solution,

$0.4x+0.7y=240\times 0.5$

$0.4x+0.7y=120........(2)$

From Equation (1)  $y=240-x$, substitute the value of $y$ in Equation (2),we get

$0.4x+0.7(240-x)=120\\0.4x+168-0.7x=120\\-0.3x=120-168\\-0.3x=-48\\x=160$

Putting the value of $x=160$, we get

$y=240-160$

$y=80$.

Hence $160 \ litres$ of $40 \%$ antifreeze solution and $80 \ litres$ of $70 \%$ antifreeze solutions will be used.

For more details on solving equations in two variables follow the link:

https://brainly.com/question/1836867