1999 AHSME Problems/Problem 14

Problem

Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?

$\textbf{(A)}\ 7 \qquad  \textbf{(B)}\ 8 \qquad  \textbf{(C)}\ 9 \qquad  \textbf{(D)}\ 10 \qquad  \textbf{(E)}\ 11$

Solution

Alina and Tina must sing more than $4$, but less than $7$, songs. Therefore, Alina sang $5$ or $6$ songs, and Tina sang $5$ or $6$ songs, with $4$ possible combinations.


However, since every song is a trio, if you add up all the numbers of songs a person sang for all four singers, it must be divisible by $3$. Thus, $7 + 4 + a + t$ must be divisible by $3$, and $a+t - 1$ must be divisible by $3$.


Since $10 \le a+t \le 12$ by the bounds of $5 \le a, t \le 6$, we have $9 \le a + t - 1 \le 11$. Because the middle number must be a multiple of $3$, we set $a + t - 1 = 9$, leading to $a + t = 10$ and $a = t = 5$.


So, the four girls sang $4, 5, 5$ and $7$ songs in trios. That means there were $\frac{4 + 5 + 5 + 7}{3} = 7$ songs sung, and the answer is $\boxed{A}$.

See Also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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