The given question is GMAT 700+ Math problem solving question combining concepts in number properties and linear equations. The question is a GMAT hard math word problem.

Question 16: A movie hall sold tickets to one of its shows in two denominations, $11 and $7. A fourth of all those who bought a ticket also spent $4 each on refreshments at the movie hall. If the total collections from tickets and refreshments for the show was $124, how many $7 tickets were sold? Note: The number of $11 tickets sold is different from the number of $7 tickets sold.

- 14
- 11
- 2
- 8
- 5

INR

Let ‘x’ and ‘y’ be the number of tickets sold at $11 and $7 respectively.

Then, total number of tickets sold = x + y

Collection by selling 'x' $11 tickets = 11x

Collection by selling 'y' $7 tickets = 7y

Total collection by selling (x + y) tickets = 11x + 7y

**A fourth of all those who bought a ticket also spent $4 each on refreshments** at the movie hall.

i.e., \\frac{1}{4})(x + y) spent $4 on refreshments.

∴ Collections from sale of refreshments = \\frac{1}{4})(x + y) × 4 = (x + y)

**Total collection from tickets and refreshments** = $124

i.e., 11x + 7y + (x + y) = 124

Or 12x + 8y = 124

__Divide the equation by 4__: 3x + 2y = 31

**Fact 1**: x and y are positive integers.

**Fact 2**: \\frac{1}{4}) (x + y) is an integer because a fourth of total tickets sold should be an integer. i.e., **(x + y) should be a multiple of 4**.

Let us list down the different possibilities that satisfy the equation and fact 1

x | y | 3x + 2Y | x + y |
---|---|---|---|

9 | 2 | 27 + 4 = 31 | 11 |

7 | 5 | 21 + 10 = 31 | 12 |

5 | 8 | 15 + 16 = 31 | 13 |

3 | 11 | 9 + 22 = 31 | 14 |

1 | 14 | 3 + 28 = 31 | 15 |

Of all the different positive integer values of x and y that satisfy the equation, the only combination (x, y) = (7, 5) is divisible by 4

So, the movie hall sold __five__ $7 tickets.

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