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Cash Box


Alignments to Content Standards: A-CED.A A-REI.C.6

Task

Nola was selling tickets at the high school dance. At the end of the evening, she picked up the cash box and noticed a dollar lying on the floor next to it. She said,

I wonder whether the dollar belongs inside the cash box or not.

The price of tickets for the dance was 1 ticket for \$5 (for individuals) or 2 tickets for \$8 (for couples). She looked inside the cash box and found \$200 and ticket stubs for the 47 students in attendance. Does the dollar belong inside the cash box or not?

IM Commentary

The purpose of this task is to gives students an opportunity to engage in Mathematical Practice #3 “Construct viable arguments and critique the reasoning of others.”  This tasks gives a teacher the opportunity to ask students not only for a specific answer of whether the dollar came from in the cash box or not, but for students to construct an argument as to how they came to their solution.  While many tasks leave open an opportunity for a student to explain their answer, this task necessitates that explanation in order to settle the curiosity behind the context.  There are many different correct arguments a student could choose in order to explain their decision and create their argument.  Convincing themselves, and presenting their argument in some form to the teacher, forms the mathematical work of this task.  Teachers could have students construct their arguments verbally, through writing, through pictures or diagrams, or through an explanation to the class.  This task also allows for the possibility of critiquing the reasoning of others.  This process of creating a viable argument prepares student’s mathematical thinking to prove and disprove postulates in future mathematics.

The task also provides an opportunity for students to engage in an important aspect of mathematical modeling (MP 4), namely, continually checking whether the mathematical work makes sense in terms of the context. In the solutions presented below using systems, the system has an integer solution for both situations, that is, whether or not we include the dollar on the floor in the cash box or not. However, because the number of tickets sold to couples has to be even, we can determine if the dollar belongs in the cash box or not. The modeling aspect could be enhanced by presenting the problem in  a physical simulation, with a cash box with play money and ticket stubs available.

The task also lends itself to solutions that do not involve systems of linear equations but provide an opportunity for students to reason quantitatively (MP 2).  For example, students might reason that if all 47 tickets sold were individual tickets, there would be \$235 in the cash box. The presence of only \$200 in the cash box would imply that precisely 35 tickets were sold to couples, which is impossible as couples tickets are sold in pairs. Repeating the argument with \$201 in the cash box leads quickly to the correct solution. Since it is possible to solve the problem without referencing systems of linear equations, the task is not recommended for directly assessing the aligned content standards.

 

Task adapted form a submission from Luke Biesecker, Math Teacher, Ferndale High School, Humboldt County, California.

Solutions

Solution: System of linear equations

Let $s$ be the number of tickets sold to individuals and $c$ be the number of tickets sold to couples. We know that 47 tickets were sold so far, so we have $s+c = 47.$ Since each individual's ticket is $\$5$, the total amount of money made by selling tickets to individuals is $5s$. Similarly, since each ticket sold to couples is $\$4,$ the total amount of money made by selling tickets to couples is $4c$. The cash box contains $\$200$ total, so we have $5s+4c = 200.$ Thus, we can represent the situation with a system of equations:

$$ \begin{align} s + c &= 47\\ 5s + 4c &= 200 \end{align} $$
Does this system of two equations in two unknowns have a positive integer solution? Furthermore, $c$ has to be even, since tickets sold to couples were only sold in sets of 2.

Solving the first equation for $s$ we have $$s=47-c.$$ Substituting for $s$ in the second equation we obtain $$5(47-c) +4c = 200.$$ Distributing and collecting like terms we have $$235 -c =200.$$ Therefore, $c=35$ and $s=12$. This means that without the found extra dollar in the cash box, 35 tickets were sold to couples and 12 tickets were sold to individuals. This is not possible, since couples tickets were only sold in sets of 2.

First indications are that the $\$1$ on the floor belongs in the cash box. To check this, we change the system of equations so that it reflects the extra $\$1$ in the cash box:

$$\begin{align} s + c &= 47\\ 5s + 4c &= 201 \end{align} $$
The solution to the changed system of equations is $s=13$ and $c=34,$ so 13 tickets were sold to individuals and 34 tickets were sold to couples. This combination of tickets is indeed possible.

Note that it is not really necessary to solve the second system of equations in the standard way. We could just have argued that exchanging one couples' ticket for an individual's ticket would increase the money in the cash box from 200 to 201 and it would result in an even number of couples tickets sold.

Solution: System of linear equations, slight variant

We briefly outline a second way of modelling the problem: If we let $s$ be the number of singles attending the dance, and $c$ be the number of couples attending the dance (rather than the number of people attending as part of a couple), then the system of linear equations becomes instead

$$ \begin{align} s + 2c &= 47\\ 5s + 8c &= 200 \end{align} $$

which one checks as in the other solution does not have a solution in non-negative integers. On the other hand, the analogous system

$$ \begin{align} s + 2c &= 47\\ 5s + 8c &= 201 \end{align} $$

with solution $s=13$ and $c=17$. So, as in the other solution, we conclude that the extra dollar belongs in the cash box, with 13 singles and 17 couples attending the dance.

anmueller says:

about 5 years

Alternate solution. An even number of ticket holders paid $4 per ticket (couples' price), so the total amount for those tickets is an even amount (even x even = even).

Since the total number of ticket holders is 47, there must be an odd number of $5 tickets sold (odd - even = odd) with that total amount being odd (odd x odd = odd).

Now, the total amount for all tickets must be odd (even + odd = odd).

Therefore, the cash box should be $201 (odd) and cannot be $200 (even). This approach assumes (correctly) that this problem has a solution.

Note: The system of equations (using 201 only) can be used to verify the existence of an answer.

Bill says:

about 5 years

Nice piece of reasoning!

Bill McCallum