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9.5: Revenue problems

  • Page ID
    45131
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    Revenue problems are problems where a person buys a certain number of items for a certain price per item. If we multiply the number of items by the price per item we will get the total value. We can recall revenue problems in the word problems chapter. We used the formula \(AVT\):

    \[\text{Amount}\cdot\text{Value}=\text{Total}\nonumber\]

    We will continue to use the same formula, but rewrite it to model with rational equations as

    \[\text{Value}=\dfrac{\text{Total}}{\text{Amount}}\nonumber\]

    Example 9.5.1

    A man buys several fish for \($56\). After three fish die, he decides to sell the rest at a profit of \($5\) per fish. His total profit was \($4\). How many fish did he buy to begin with?

    Solution

    First, we can make a table to organize the given information and then create an equation. Let \(n\) represent the number of fish and \(p\) be the price of each fish.

    Table 9.5.1
    Amount Price Total Value
    Buy \(n\) \(p\) \($56\)
    Sell \(n-3\) \(p+5\) \($56+$4\)

    Let’s discuss the table for a moment. When the man purchased the fish, the total value of fish purchased was \($56\). Since the price of each fish and the quantity purchased is unknown, we leave it as \(p\) and \(n\), respectively. The man wants to sell the fish, but three died; hence, the amount left to sell is \(n − 3\). Since he wants to profit \($5\) per fish, then we take the price the man bought the fish for and add \($5\), \(p + 5\). It is given that his total profit was \($4\), so his total value from selling the fish was the original value, \($56\), plus the \($4\) profit; hence, a total of \($60\).

    Finally, let's set up the equations and solve:

    \[\begin{array}{ll}p_{Buy}=\dfrac{56}{n}&p_{Sell}+5=\dfrac{60}{(n-3)} \\ p_{Buy}=\dfrac{56}{n}&p_{Sell}=\dfrac{60}{(n-3)}-5\end{array}\nonumber\]

    Since we solved for \(p\) in each equation, we can set the \(p\)’s equal to each other:

    \[\begin{array}{rl}p_{Buy}=p_{Sell}&\text{Set }p\text{'s equal to each other} \\ \dfrac{56}{n}=\dfrac{60}{(n-3)}-5&\text{Multiply by the LCD} \\ \color{blue}{n(n-3)}\color{black}{}\cdot\dfrac{56}{n}=\color{blue}{n(n-3)}\color{black}{}\cdot\dfrac{60}{(n-3)}-\color{blue}{n(n-3)}\color{black}{}\cdot 5&\text{Clear denominators} \\ 56(n-3)=60n-5n(n-3)&\text{Distribute} \\ 56n-168=60n-5n^2+15n &\text{Combine like terms} \\ 56n-168=75n-5n^2&\text{Notice the }5n^2\text{ term; solve by factoring} \\ 5n^2-19n-168=0&\text{Factor} \\ (5n+21)(n-8)=0&\text{Apply zero product rule} \\ 5n+21=0\text{ or }n-8=0&\text{Isolate variable terms} \\ n=-\dfrac{21}{5}\text{ or }n=8&\text{Solutions}\end{array}\nonumber\]

    Since the quantity of fish is always positive, we omit the solution \(n = −\dfrac{21}{5}\). Thus, the man purchased \(8\) fish.

    Example 9.5.2

    A group of students bought a couch for their dorm that cost \($96\). However, \(2\) students failed to pay their share, so each student had to pay \($4\) more. How many students were in the original group?

    Solution

    First, we can make a table to organize the given information and then create an equation. Let \(n\) represent the number of students and \(p\) be the price of each share.

    Table 9.5.2
    Amount Price Total Value
    Original deal \(n\) \(p\) \($96\)
    Actual deal \(n-2\) \(p+4\) \($96\)

    Let’s discuss the table for a moment. The original deal was every student in the original group with \(n\) number of students were going to split the total value of the couch valued at \($96\). Since the price of each share and the number of students is unknown, we leave it as \(p\) and \(n\), respectively. When it came to actually paying for the couch, \(2\) students didn’t pay their share; hence, the number of students left to pay is \(n − 2\). Since this increases each share from the rest of the group, then we take the original share and add \($4\), \(p + 4\).

    Finally, let’s set up the equations and solve:

    \[\begin{array}{ll}p_{o}=\dfrac{96}{n}&p_{A}+4=\dfrac{96}{(n-2)} \\ p_{o}=\dfrac{96}{n}&p_{A}=\dfrac{96}{(n-2)}-4\end{array}\nonumber\]

    Since we solved for \(p\) in each equation, we can set the \(p\)’s equal to each other:

    \[\begin{array}{rl} p_{o}=p_{A}&\text{Set }p\text{'s equal to each other} \\ \dfrac{96}{n}=\dfrac{96}{(n-2)}-4&\text{Multiply by the LCD} \\ \color{blue}{n(n-2)}\color{black}{}\cdot\dfrac{96}{n}=\color{blue}{n(n-2)}\color{black}{}\cdot\dfrac{96}{(n-2)}-\color{blue}{n(n-2)}\color{black}{}\cdot 4&\text{Clear denominators} \\ 96(n-2)=96n-4n(n-2)&\text{Distribute} \\ 96n-192=96n-4n^2+8n&\text{Combine like terms} \\ 96n-192=104n-4n^2&\text{Notice the }4n^2\text{ term; solve by factoring} \\ 4n^2-8n-192=0&\text{Reduce all terms by a factor of }4 \\ n^2-2n-48=0&\text{Factor} \\ (n+6)(n-8)=0&\text{Apply zero product rule} \\ n+6=0\text{ or }n-8=0&\text{Isolate variable terms} \\ n=-6\text{ or }n=8&\text{Solutions}\end{array}\nonumber\]

    Since the quantity of students is always positive, we omit the solution \(n = −6\) and there were \(8\) students in the original group.

    Revenue Problems Homework

    Exercise 9.5.1

    A merchant bought some pieces of silk for \($900\). Had he bought \(3\) pieces more for the same money, he would have paid \($15\) less for each piece. Find the number of pieces purchased.

    Exercise 9.5.2

    A number of men subscribed a certain amount to make up a deficit of \($100\) but \(5\) men failed to pay and thus increased the share of the others by \($1\) each. Find the amount that each man paid.

    Exercise 9.5.3

    A merchant bought a number of barrels of apples for \($120\). He kept two barrels and sold the remainder at a profit of \($2\) per barrel making a total profit of \($34\). How many barrels did he originally buy?

    Exercise 9.5.4

    A dealer bought a number of sheep for \($440\). After \(5\) had died he sold the remainder at a profit of \($2\) each making a profit of \($60\) for the sheep. How many sheep did he originally purchase?

    Exercise 9.5.5

    A man bought a number of articles at equal cost for \($500\). He sold all but two for \($540\) at a profit of \($5\) for each item. How many articles did he buy?

    Exercise 9.5.6

    A clothier bought a lot of suits for \($750\). He sold all but \(3\) of them for \($864\) making a profit of \($7\) on each suit sold. How many suits did he buy?

    Exercise 9.5.7

    A group of boys bought a boat for \($450\). Five boys failed to pay their share, hence each remaining boys were compelled to pay \($4.50\) more. How many boys were in the original group and how much had each agreed to pay?

    Exercise 9.5.8

    The total expenses of a camping party were \($72\). If there had been \(3\) fewer persons in the party, it would have cost each person \($2\) more than it did. How many people were in the party and how much did it cost each one?


    This page titled 9.5: Revenue problems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz (ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.