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Maximum Profit for Movie Theatre

XAT 2024 · Quantitative Ability & Data Interpretation · Medium · Arithmetic

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The cost of running a movie theatre is Rs. 10,000 per day, plus additional Rs. 5000 per show. The theatre has 200 seats. A new movie released on Friday. There were three shows, where the ticket price was Rs. 250 each for the first two shows and Rs. 200 for the late-night show.

For all shows together, total occupancy was 80%. What was the maximum amount of profit possible?

  1. A.

    Rs. 1,20,000

  2. B.

    Rs. 87,000

  3. C.

    Rs. 95,000

  4. D.

    Rs. 91,000

  5. E.

    Rs. 1,16,000

Answer

D

Explanation

Total running cost =10000+3×5000=Rs. 25,000= 10000 + 3 \times 5000 = \text{Rs. } 25,000. Total seats available across 3 shows =3×200=600= 3 \times 200 = 600. Total occupancy =80% of 600=480= 80\% \text{ of } 600 = 480 seats filled (120 empty seats).

To maximize revenue, maximize sales of higher-priced tickets (Rs. 250) and put all 120 vacant seats in the late-night show (Rs. 200).

  • Show 1 (Rs. 250): 200 seats =200×250=50,000= 200 \times 250 = 50,000
  • Show 2 (Rs. 250): 200 seats =200×250=50,000= 200 \times 250 = 50,000
  • Show 3 (Rs. 200): 200−120=80200 - 120 = 80 seats =80×200=16,000= 80 \times 200 = 16,000

Total Revenue =50000+50000+16000=Rs. 1,16,000= 50000 + 50000 + 16000 = \text{Rs. } 1,16,000. Maximum Profit =1,16,000−25,000=Rs. 91,000= 1,16,000 - 25,000 = \text{Rs. } 91,000.

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