Answer :
Problem:
A chemical company makes two brands of antifreeze the first brand is 20% pure antifreeze and the second brand is 70% pure antifreeze in order to obtain 90 gallons of a mixture that contains 60% pure antifreeze how many gallons of each brand of antifreeze must be used.
Solution:
Let x = amount of 20% pure antifreeze.
Let y = amount of 70% pure antifreeze.
Then, we have the following equations:
Equation 1:
x+y = 90 gallons (total of 90 gallons mixed)
Equation 2:
(0.20)x+(0.70)y = 0.60(x+y)
that is:
0.60x - 0.20x = 0.70y -0.60 y
that is :
0.40x = 0.10 y
Simplify and solve the system of equations. From equation 1, solve for x:
[tex]x\text{ = }90-y[/tex]replace this in equation 2:
[tex]0.40(90-y)\text{ = 0.10y}[/tex]that is:
[tex]36-0.40y\text{ = 0.10y}[/tex]that is:
[tex]36\text{= 0.10y}+\text{ 0.40y = 0.50y}[/tex]solve for y:
[tex]y\text{ = }\frac{36}{0.50}\text{ = 72 gallons }[/tex]
then we can conclude that we need 72 gallons of the brand with an amount of 70% pure antifreeze. Now, replace this value in equation 1, and solve for x :
[tex]x\text{ = }90-72\text{ = 18 gallons }[/tex]then we can conclude that we need 18 gallons of the brand with an amount of 20% pure antifreeze.
Then the solution is:
- we need 18 gallons of the brand with an amount of 20% pure antifreeze.
-we need 72 gallons of the brand with an amount of 70% pure antifreeze.