Question

At a college football game, $\tfrac{4}{5}$ of the \emph{lower-deck} seats were sold. The lower deck contains $\tfrac{1}{4}$ of all stadium seats. Overall, $\tfrac{2}{3}$ of all stadium seats were sold. What fraction of the \emph{unsold} seats in the stadium lay in the lower deck?

• A. $\dfrac{3}{20}$
• B. $\dfrac{1}{6}$
• C. $\dfrac{1}{5}$
• D. $\dfrac{1}{3}$

Hint:

When questions ask "what fraction of the unsold (or sold) seats…", compute both \emph{the part} and \emph{the whole} as fractions of the \emph{same} total $T$ so that $T$ cancels cleanly in the final ratio.

Answer 1

The Correct Option is A

Step 1 (Name a convenient total).
Let the total number of seats be $T$. Then lower-deck seats $=\dfrac{1}{4}T$ and upper-deck seats $=\dfrac{3}{4}T$. \medskip Step 2 (Sold and unsold in the lower deck).
Given $\dfrac{4}{5}$ of the lower deck were sold:
Sold (lower) $=\dfrac{4}{5}\cdot\dfrac{1}{4}T=\dfrac{1}{5}T$.
Unsold (lower) $=$ (lower total) $-$ (lower sold) $=\dfrac{1}{4}T-\dfrac{1}{5}T=\left(\dfrac{5-4}{20}\right)T=\dfrac{1}{20}T$. \medskip Step 3 (Overall unsold in the whole stadium).
Overall sold $=\dfrac{2}{3}T \Rightarrow$ Overall \emph{unsold} $=T-\dfrac{2}{3}T=\dfrac{1}{3}T$. \medskip Step 4 (Required fraction: part of unsold that is in lower deck).
\[ \text{Fraction asked}=\frac{\text{Unsold in lower deck}}{\text{Total unsold in stadium}} =\frac{\frac{1}{20}T}{\frac{1}{3}T} =\frac{1}{20}\cdot 3 =\boxed{\dfrac{3}{20}}. \] \medskip (Sanity check) If $T=60$: lower deck $=15$; sold (lower) $=12$; unsold (lower) $=3$. Overall sold $=40$; overall unsold $=20$; fraction $=3/20$ — consistent.