Answer the following question based on the information given below.A m...
We have the following two constraints:
1. The maximum number of tickets that a customer can buy is 5.
2. Tickets are sold to each customer in such a way that the total cost is a multiple of hundred.
Thus we can see that each of the 21 customers can buy tickets in the following ways only.
As no type C tickets were sold, we can eliminate cases 2, 3, 6, 9 and 10.
Also, as each of the customers had purchased at least one ticket of type B, we can eliminate case 7.
Hence possible combinations are,
Now, 21 people buy 26 tickets of type B. Hence no more than one person can buy the first combination of tickets.
Hence two cases are possible,
Case 1: No person buys the first combination.
Case 2: Exactly one person buys tickets as in the first combination.
Consider Case 1: As no person buys the first combination hence combinations
2, 3, 4 and 5 in the table are possible.
As 21 people buy 26 tickets hence the number of people buying 3rd and 5th combination is 5.
The remaining 16 people will buy the 2nd and 4th combination.
Let us assume that x tickets of type B and x tickets of type D were purchased in combinations 2 and 3.
Let y tickets of type B and 2y tickets of type A be purchased in combination 4.
Let z tickets of type B, z tickets of type A and z/2 tickets of type D be purchased in combination 5.
Hence, x + y+ z = 26, and 2x + 3y + 5z/2 = 60 Now,
2x + 3y + 5z/2 = 2(x + y + z) + y + z/2 = 52 + y + z/2 = 60
Hence, y + z/2 = 8 2y + z = 16 Now, number of A type tickets sold are,
2y + z
Hence number of A type tickets left = 30 - 16 = 14 Consider case 2: 1 person buys the first combination.
Hence remaining 20 persons buy tickets in the 2nd, 3rd, 4th or 5th combination.
Let us assume that x tickets of type B and x tickets of type D were purchased in combinations 2 and 3.
Let y tickets of type B and 2y tickets of type A be purchased in combination 4.
Let z tickets of type B, z tickets of type A and z/2 tickets of type D be purchased in combination 5.
Hence, x + y + z = 21, and 2x + 3y + 5z/2 = 55 Now,
2x + 3y + 5z/2 = 2(x + y + z) + y + z/2 = 42 + y + z/2 = 55
Hence, y + z/2 = 13 2y + z = 26 Now, number of tickets of type A that are sold is 2y + z = 26 Hence number of A type tickets left = 30 - 26 = 4 Either 4 or 14 tickets of type A were still available.
Answer: 14