Question

A choir is singing at a festival. On the first night $12$ choir members were absent so the choir stood in $5$ equal rows. On the second night only $1$ member was absent so the choir stood in $6$ equal rows. The same member of people stood in each row each night. How many members are in the choir?

Answer 1

To find the number of members in a choir for both nights let us assume their number as $C$ and equate (as the same number of people stood in both nights in each row) the number after subtracting with the members absent on both nights for each row. To find the members of choir we use the formula (as to equate the number of members standing in a row during first night to the number of members standing in a row during second night.):
$\dfrac{\text{C}-\text{Absente}{{\text{e}}_{\text{first night}}}}{\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}}=\dfrac{\text{C}-\text{Absente}{{\text{e}}_{\text{second night}}}}{\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}}$
where $\text{Absente}{{\text{e}}_{\text{first night}}}$ are the members absent in the first night of the festival, $\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}$ are the number of rows made after equating the absentees. Similarly, $\text{Absente}{{\text{e}}_{\text{second night}}}$ are the members absent in the second night of the festival, $\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}$ are the number of rows made after equating the absentees in the second night of the festival.

Complete step-by-step answer:
Now placing the values in the formula created:
$C$ are total members of choir for both the nights, $\text{Absente}{{\text{e}}_{\text{first night}}}=12$, $\text{Absente}{{\text{e}}_{\text{second night}}}=1$, $\text{Numbe}{{\text{r}}_{\text{rows in firstnight}}}=5$ and $\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}=6$
$\dfrac{\text{C}-\text{Absente}{{\text{e}}_{\text{first night}}}}{\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}}=\dfrac{\text{C}-\text{Absente}{{\text{e}}_{\text{second night}}}}{\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}}$
$\dfrac{\text{C}-\text{12}}{\text{5}}=\dfrac{\text{C}-\text{1}}{\text{6}}$
Cross multiplying the denominator from LHS to RHS and from RHS to LHS, we get:
$\Rightarrow \text{6}\left( \text{C}-\text{12} \right)=5\left( \text{C}-\text{1} \right)$
$\Rightarrow 6C-72=5C-5$
$\Rightarrow 6C-5C=72-5$
$\Rightarrow C=67$

$\therefore$ The total numbers of members in the choir for both nights are $67$

Note: Students may go wrong is they assume that the result they found is only for a single row, as the terms $\left( \dfrac{\text{C}-\text{Absente}{{\text{e}}_{\text{first night}}}}{\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}} \right),\left( \dfrac{\text{C}-\text{Absente}{{\text{e}}_{\text{second night}}}}{\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}} \right)$ are number of members in each row for two nights respectively. Now the value of $C$ is itself the total hence, there is no need to put it back into the above two terms and find the value out of it as if one puts the value in $\left( \dfrac{\text{C}-\text{Absente}{{\text{e}}_{\text{first night}}}}{\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}} \right),\left( \dfrac{\text{C}-\text{Absente}{{\text{e}}_{\text{second night}}}}{\text{Numbe}{{\text{r}}_{\text{rows in secondnight}}}} \right)$, they will only get the number of members in a single row for each night. Hence, the result $C=67$ is the final value.