Solving a percent mixture problem using a system of linear equations

Question

A chemical company makes two brands of antifreeze.

The first brand is 55% pure antifreeze.

The second brand is 90% antifreeze.

In order to obtain 140 gallons of a mixture that contains 70% pure antifreeze - How many gallons of each antifreeze brand must be used?

Really need help with this.

Thank you,

Stephanie

Answer 1

Hi Stephanie!

Here is the solution:

Let X be the amount of first brand and Y be the amount for second brand.

There are 140 gallons total, that gives us: X+Y= 140 gallons

X gallons of first brand contains 55%*X antifreeze, which we can write as 0.55X gallons.

Y gallons of first brand contains 90%*X antifreeze, which we can write as 0.9Y gallons.

The mixture has 70% antifreeze in toal 140 gallons mixture, so there is 70%* 140 gallons= 98 gallons.

The 98 gallons antifreeze was the combination of frist brand and second brand, which is 0.55X+0.9Y= 98 gallons.

Now we need to solve two equations:

0.55X+0.9Y=98, X+Y= 140

after calculation, we get X=60 and Y= 80.

That is 60 gallons of first brand and 80 gallons of second brand.

Hope this helps!

Answer 2

Let xgalons of 55%, then 140 - x s 90%.

We get equation: 0.55x + 0.90(140 - x) = 0.70·140; 0.55x - 0.90x = 140(0.70 - 0.90); 0.35x = 28; x = 80 gal of 55%

and 140 - 80 = 60 gal 90%