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- Jan 26, 2012

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- Thread starter Jameson
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- Thread starter
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- #1

- Jan 26, 2012

- 4,052

- Admin
- #2

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 $T=T$?

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

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

$\displaystyle x=40$