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Cammie's question from Facebook (finding a mixture through systems of equations)

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Jameson

Administrator
Staff member
Jan 26, 2012
4,042
A girl scout troop has 20 pounds of candy worth 80 cents per pound. the troop wishes to mix it with candy worth 50 cents per pound so that the total mixture can be sold at 60 cents per pound without any gain or loss. how much of the 50 cent candy must be used? Solve by elimination.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Cammie,

Let's let $T$ be the total value of the two candies in dollars, and $x$ be the amount in pounds of the 50-cent candy that must be used to meet the stated goal.

Before mixing, we may state the combined value of the two candies as:

$T=20\cdot0.80+x\cdot0.50$

and we may arrange this as:

(1) $2T-x=32$

After mixing, we may state the total value of the mixture as:

$T=0.60(20+x)$

which we may arrange as:

(2) $5T-3x=60$

To eliminate $T$, we may multiply (1) by 5 and (2) by -2 and add them to get:

$x=40$

Thus, we have found the troop needs to add 40 lbs. of the 50-cent candy to the 20 lbs. of 80-cent candy to get a mixture that can be sold for 60 cents per pound without any gain or loss.

You may be interested to know that what we have done is equivalent to using a weighted average to solve the problem:

$\displaystyle \frac{20\cdot0.80+x\cdot0.50}{20+x}=0.60$

$\displaystyle 20\cdot0.80+x\cdot0.50=0.60(20+x)$

Do you see that this is $T=T$?

$\displaystyle 16+\frac{1}{2}x=12+\frac{3}{5}x$

$\displaystyle 4=\left(\frac{3}{5}-\frac{1}{2} \right)x=\frac{1}{10}x$

$\displaystyle x=40$