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Difference between revisions of "1999 AHSME Problems/Problem 14"

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==Solution==
 
==Solution==
  
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Alina and Tina must sing more than <math>4</math>, but less than <math>7</math>, songs.  Therefore, Alina sang <math>5</math> or <math>6</math> songs, and Tina sang <math>5</math> or <math>6</math> songs, with <math>4</math> possible combinations.
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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 <math>3</math>.  Thus, <math>7 + 4 + a + t</math> must be divisible by <math>3</math>, and <math>a+t - 1</math> must be divisible by <math>3</math>. 
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Since <math>10 \le a+t \le 12</math> by the bounds of <math>5 \le a, t \le 6</math>, we have <math>9 \le a + t - 1 \le 11</math>.  Because the middle number must be a multiple of <math>3</math>, we set <math>a + t - 1 = 9</math>, leading to <math>a + t = 10</math> and <math>a = t = 5</math>.
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So, the four girls sang <math>4, 5, 5</math> and <math>7</math> songs in trios.  That means there were <math>\frac{4 + 5 + 5 + 7}{3} = 7</math> songs sung, and the answer is <math>\boxed{A}</math>.
  
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1999|num-b=13|num-a=15}}
 
{{AHSME box|year=1999|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 13:34, 5 July 2013

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 (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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