"In a musical class, the students either played piano or violin as head instrument. By a concert, the students got to choose whether they would do a solo or pair-performance. A piano player can only play with a violin player and a violin player with a piano player. It turned out that $2/3$ of the students with piano as head instrument and $3/5$ of the students with violin as head instrument chose to perform in pair. How many performed solo?"

Thoughts: I've looked up the answer, it's $7/19$. I know that I can do this:

$0.33$% + $0.4$% = $0.73$%

$0.73$/$2$ = $36.5$%

And $7$/$19$ may be close to 36.5%, but I still don't know how to get the answer in fractions. I've tried dividing and more, but I need to know the logic behind my actions :) (my brain is tired right now). Any help is appreciated!


The problem is coming from which group each fraction is referring to. The $\frac23$ refers to a fraction out of just the piano players, the $\frac35$ refers to a fraction out of just the violin players, and the answer you are looking for is a fraction out of all the students. Since the two original fractions are referring to different groups of students, you can't add them! It would be safest to work with the raw number of students.

Suppose $p$ is number of students that are piano players, and $v$ is the number students that are violin players.

Then the information says $\frac23 p$ piano players choose to play with $\frac35 v$ violin players. So we are left with $\frac13 p$ piano players who play solo and $\frac25 v$ violin players who play solo, for a total of $(\frac13 p + \frac25 v)$ students who play solo. We want to know what this is as a fraction of the total number of students.

Since each violin player plays with one piano player and vice versa, we know that $\frac23 p = \frac35 v$. We can multiply both sides of this equation by 3 and divide both sides by 2 to get $p = \frac9{10} v$.

Now the total number of solo students is $$ \begin{align} \frac13 p + \frac25v &= \frac13\cdot\frac9{10}v +\frac25v\\ &= \frac3{10}v+\frac25v \\ &= \frac3{10}v+\frac4{10}v\\ &= \frac7{10}v \end{align} $$ Also, the total number of students is $$ \begin{align} p+v &= \frac9{10}v + v \\ &= \frac9{10}v + \frac{10}{10}v \\ &= \frac{19}{10}v \end{align} $$

So the fraction of students playing solo is: $$ \begin{align} \frac{(\frac7{10}v)}{(\frac{19}{10}v)} &= \frac{(\frac7{10})}{(\frac{19}{10})}\\ &= \frac{7}{10}\cdot\frac{10}{19}\\ &= \frac{7}{19} \end{align} $$

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  • $\begingroup$ Ah yes, your answer makes sense and I'm thankful for your explanation. Really, thank you! $\endgroup$ – didnotcomeuptosomething Nov 20 '14 at 22:34

You can think of it this way:

$$ 0.33\%=\frac13,\qquad0.40\%=\frac25\\\\ $$

Then, we can add up these fractions:

$$ \begin{align} \frac13 + \frac25&=\frac{5}{15}+\frac{6}{15}\\\\ &=\frac{11}{15} \end{align} $$

Now, we can divide that fraction by two by multiplying with $\frac12$:

$$ \frac{11}{15}\cdot\frac12=\frac{11}{30} $$

But then, here we see that even with your method, our answer does not match the one in your textbook. But, $\frac{11}{30}$ is still very close to $\frac{7}{19}$. It might be that you have to try a different method, or the textbook is wrong. Either way, I hope this helps.

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  • 1
    $\begingroup$ This answer is wrong, sorry. $\endgroup$ – DavidButlerUofA Nov 20 '14 at 22:16
  • $\begingroup$ I tried the same thing you did! But it wasn't the textbook's answer so I don't know. $\endgroup$ – didnotcomeuptosomething Nov 20 '14 at 22:17
  • $\begingroup$ @didnotcomeuptosomething Look at DavidButler's answer. $\endgroup$ – Mark Fantini Nov 20 '14 at 22:18

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