Question

A chemical company makes two brands of antifreeze. The first brand is 65% pure antifreeze, and the second brand is 90% pure antifreeze. In order to obtain 40 gallons of a mixture that contains 80% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

$\bf \begin{array}{lcccl} &\stackrel{solution}{gallons}&\stackrel{\textit{\% of }}{antifreeze}&\stackrel{\textit{gallons of }}{antifreeze}\\ \cline{2-4}&\\ \textit{1st brand}&x&0.65&0.65x\\ \textit{2nd brand}&y&0.90&0.9y\\ \cline{2-4}&\\ mixture&40&0.8&32 \end{array}~\hfill \to \begin{cases} x+y&=40\\ \boxed{x}=40-y\\ \cline{1-2} 0.65x+0.9y&=32 \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{substituting in the 2nd equation}}{0.65\left( \boxed{40-y} \right)+0.9y=32}\implies 26-0.65y+0.9y=32 \\\\\\ 26+0.25y=32\implies 0.25y=6\implies y=\cfrac{6}{0.25}\implies \blacktriangleright y=24 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that}}{x=40-y\implies }x=40-24\implies \blacktriangleright x=16 \blacktriangleleft$

Answer 2

Answer:

it should be used 16 gallons of 65% pue antifreeze and 24 gallons of 80% antifreeze.

Step-by-step explanation:

Let 'x' = amount of 65% antifreeze.

Let 'y' = amount of 90% antifreeze.

We need to obtain 40 gallons of mixture, then:

x + y = 40 gallons [1]

Also we know that the mixture should contain 80% pure antifreeze, then:

0.65x + 0.9y = 0.80(x+y) →  0.15x = 0.1y → y = 1.5x  [2]

Now, subtituting the value of 'y' into [1]:

x + y = 40 gallons → x + 1.5x = 40 gallons → 2.5x = 40 gallons

⇒ x = 16 gallons.

Then: y = 40 - 16 = 24 gallons.

Now, it should be used 16 gallons of 65% pue antifreeze and 24 gallons of 80% antifreeze.