Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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=$\$460$

share|cite|improve this question
up vote 1 down vote accepted

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 $$(6)(360)+4x=(10)(400).$$

share|cite|improve this answer

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

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

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

share|cite|improve this answer

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.

share|cite|improve this answer

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.