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I can't seem to solve the following problem any suggestions would be appreciated..

In a small snack shop the average revenue was $\$400$ a day over a $10$ day period. During this period, if the average daily revenue was $\$360$ for the first 6 days, what was the average daily revenue for the last 4 days? Ans=$\$460$

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4 Answers

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The total revenue over the $10$ days was $(10)(400)$.

The total revenue over the first $6$ days was $(6)(360)$.

So if the average daily revenue over the last $4$ days was $x$, then $$(6)(360)+4x=(10)(400).$$

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Let $D_6$, $D_4$ be the average revenues over the first 6 and last 4 days, respectively. We then have

$\frac{6 \times D_6 + 4 \times D_4}{10}=400$.

Solving for $D_4$ and using the fact that $D_6=360$ gives us

$4000-6\times D_6=4\times D_4 \Rightarrow 4000-2160=4\times D_4$

$\Rightarrow \frac{1840}{4}=460=D_4$.

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I am not an algebra nor calculus major, but I work in healthcare finance. I have to look at things simply. Here goes: $360x6 = $2160. $400 x 10 = $4000. $4000-2160 = $1840. $1840/4 = $460.

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What is the total revenue of all $10$ days?

What is the total revenue of the first $6$ days?

What is the remaining total revenue for the remaining $4$ days?

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