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Bunuel
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.)

A. 14
B. 11
C. 2
D. 8
E. 5

Are You Up For the Challenge: 700 Level Questions
Let the Tootal number of tickets sold be n.

The number of $11 tickets sold be “a”.

The number of $7 tickets sold be “n-a” .

Given further that n/4 people bought refreshments for $4.

Total amount spent on tickets and refreshments = $124

11a + 7(n-a) + (n/4)* $4 = 124

Solving it we get , a + 2n =31

Possible values of (n, a, n-a) are
(15,1,14)
(14,3,11)
(13,5,8)
(12,7,5)
(11,9,2)

Only value of n divisible by 4 is 12.

7*$11 + 5*$7 + (12/4)*4
= 77 + 35 + 12
=$124

Number of $7 tickets = 5
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11X+7Y+4((X+Y)/4)=124 so

3X+2Y=31 and X+Y=4K.

Substituting

12K-Y=31

Y=14 ? 12K=45 NO
Y=11 ? 12K=42 NO
Y=2 ? 12K=33 NO
Y=8 ? 12K=39 NO

Y=5 ? 12K=36 YES
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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.)

Let the number of $7 and $11 tickets sold be x & y respectively

7x + 11y + (x+y)/4*4 = 124

8x + 12y = 124

2x + 3y = 31

Since x & y are integers
x = 2; y= 9; x+y=11; Not divisible by 4; not feasible
x = 5; y= 7; x+y=12; divisible by 4; feasible

The number of $7 tickets sold = x = 5

IMO E
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