Discussion

Explanation:

All of them buy at least 1 ticket.

Let number of people buying tickets for exactly
1 movie = a
2 movies = b
3 movies = c

⇒ a + b + c = 1000   ...(1)

Also, when we add number of tickets bought for Avengers (400), Interstellar (500) and Eat Pray & Love (600) we would add 'a' once, 'b' twice and 'c' thrice.
⇒ a + 2b + 3c = 400 + 500 + 600
⇒ a + 2b + 3c = 1500   ...(2)

(2) - (1)
⇒ b + 2c = 500

The highest possible value of 'c' here is 250.

In this case b = 0 [from (2)] and a = 750 [from (1)]

This case is as follows:

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Hence, option (d).

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