Question

At a college football game, \(\dfrac{4}{5}\) of the seats in the lower deck of the stadium were sold. If \(\dfrac{1}{4}\) of all the seating in the stadium is located in the lower deck, and if \(\dfrac{2}{3}\) of all the seats in the stadium were sold, then what fraction of the unsold seats in the stadium was in the lower deck?

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

Hint:

Work in terms of total seats x: find lower deck unsold and total unsold separately, then divide.

Answer 1

The Correct Option is A

Here is a second way, plugging in a convenient actual seat count instead of working with the variable x throughout.

1. Pick a convenient total. Since we divide by 4, 5 and 3, let the total number of seats be 60 (a common multiple of 4, 5, and 3).
2. Lower deck seats. $\dfrac{1}{4}$ of 60 is 15 seats in the lower deck; the upper deck has the remaining 45.
3. Sold and unsold in the lower deck. $\dfrac{4}{5}$ of 15 is 12 seats sold there, so $15-12=3$ seats are unsold in the lower deck.
4. Total sold and unsold in the stadium. $\dfrac{2}{3}$ of 60 is 40 seats sold overall, so $60-40=20$ seats are unsold overall.
5. Required fraction. $\dfrac{3}{20}$.

This numeric check gives the same fraction as the algebraic method, confirming option A. \[ \boxed{\dfrac{3}{20}} \]