Problem of the Day

A high school sells tickets to a basketball game. Adult tickets cost $9 and student tickets cost $5. If n tickets are sold for a total revenue of $927, how many possible values are there for n?

A. 18
B. 19
C. 20
D. 21
E. 22

Reveal Answer


D. 21

See the Solution



From the given information we can set up the following  equation to model the revenue, where $x$ represents the number of student tickets sold and $y$ represents the number of adult tickets sold: $5x+9y=927$. First, note that $x$ and $y$ are both non-negative integers. Next, we can see that 927 is a multiple of 9 because its digits add to 18, and $9y$ is clearly a multiple of 9. This implies that $5x$ is also a multiple of 9, and since 5 does not contain any factors of 9, $x$ itself must be a multiple of 9. That is, all possible values of $x$ are of the form $x=9t$ for some non-negative integer, $t$. Plugging this in to the original equation: $5(9t)+9y=927$. Solve this for $y$:




Values of $y$ must be non-negative integers, so it must be true that $103-5t >=0$. Solving this inequality gives us $t<=20.6$. We initially defined $t$ to be a non-negative integer, so $0<= t <= 20$, which means there are 21 possible values for $t$. To show that this means that there are 21 possible values of $n$, note that $n=x+y$, so $n=9t+103-5t$ or $n=103+4t$. Clearly each of the possible 21 values for $t$ will produce a distinct value for $n$.

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