Question

A chemical company makes two brands of antifreeze the first brand is 55% pure antifreeze and the second brand is 80% pure antifreeze in order to obtain 170 gallons of a mixture that contains 65% pure anti-freeze how many gallons of each brand of antifreeze must be used

Answer 1

102 gallons of 55 % pure antifreeze and 68 gallons of 80 % pure antifreeze is mixed to obtain 170 gallons of a mixture that contains 65% pure anti-freeze

Solution:

Let "x" be the gallons of 55 % pure antifreeze

Then, (170 - x) be the gallons of 80 % pure antifreeze

Then we can say, according to question,

"x"  gallons of 55 % pure antifreeze is mixed with (170 - x) gallons of 80 % pure antifreeze to obtain 170 gallons of a mixture that contains 65% pure anti-freeze

Thus, we frame a equation as:

$x \times 55 \% + (170-x) \times 80 \% = 170 \times 65 \%$

Solve the above expression for "x"

$x \times \frac{55}{100} + (170-x) \times \frac{80}{100} = 170 \times \frac{65}{100}\\\\\text{Simplify the above expression }\\\\0.55x + 0.80(170-x) = 0.65 \times 170\\\\0.55x + 136-0.8x = 110.5\\\\\text{Combine the like terms }\\\\0.25x = 136 - 110.5\\\\0.25x = 25.5\\\\\text{Divide both sides of equation by 0.25 }\\\\x = 102$

Thus, 102 gallons of 55 % pure antifreeze is used

Then, 170 - x = 170 - 102 = 68 gallons of 80 % pure antifreeze is used

Thus, 102 gallons of 55 % pure antifreeze and 68 gallons of 80 % pure antifreeze is mixed to obtain 170 gallons of a mixture that contains 65% pure anti-freeze