A new game show on TV has 100 boxes numbered 1, 2, . . . , 100 in a row, each containing a mystery prize. The prizes are items of different types, a, b, c, . . . , in decreasing order of value. The most expensive item is of type a, a diamond ring, and there is exactly one of these. You are told that the number of items at least doubles as you move to the next type. For example, there would be at least twice as many items of type b as of type a, at least twice as many items of type c as of type b and so on. There is no particular order in which the prizes are placed in the boxes.
Q.1 What is the minimum possible number of different types of prizes?
Explanation
- Minimum possible number is when there is 1 of type a and 99 of type b which is in accordance to the condition.
- Maximum possible number is when 1 of a, 2 of b, 4 of c, 8 of d, 16 of e, 32 of f,
Now left boxes woud be 100 – (1+2+…32) = 37
Now if one more type is to be added then we need at least 64 which is not available thus maximum possible is 6. - Lets try to prove the given options possible using easy numbers.
op1:never possible
op2: 1,30,69 is possible
op3: 1,2,4,18.75 is possible
op4: 1,9,30,60 is possible. - There have to be then at least 31 + `1 + 43 = 75 gifts of same type,
Thus maximum possible number of boxes = 5 when all types are lest 1,2,4,18,75
Q.2 What is the maximum possible number of different types of prizes?
Explanation
The minimum feasible number occurs when there is 1 item of type A and 99 items of type B, adhering to the specified conditions.
The maximum achievable quantity is when there is 1 item of type A, 2 of type B, 4 of type C, 8 of type D, 16 of type E, and 32 of type F. Subsequently, there would be 100 – (1+2+…+32) = 37 items remaining. If we were to introduce one more type, we would need a minimum of 64 items, which exceeds the available 37. Therefore, the maximum possible number of types is 6.
We can verify the provided options using manageable quantities:
Option 1: Not possible. Option 2: Possible with quantities 1, 30, and 69. Option 3: Possible with quantities 1, 2, 4, and 18.75. Option 4: Possible with quantities 1, 9, 30, and 60.
To meet the minimum requirement of at least 75 items of the same type, there must be at least 31 + 1 + 43 items. Consequently, the maximum feasible number of boxes is 5 when all types have quantities of 1, 2, 4, 18, and 75.
Q.3 Which of the following is not possible?
- There are exactly 45 items of type c.
- There are exactly 30 items of type b
- There are exactly 75 items of type e.
- There are exactly 60 items of type d.
Explanation
Anwer 1
The minimum feasible number occurs when there is 1 item of type A and 99 items of type B, adhering to the specified conditions.
The maximum achievable quantity is when there is 1 item of type A, 2 of type B, 4 of type C, 8 of type D, 16 of type E, and 32 of type F. Subsequently, there would be 100 – (1+2+…+32) = 37 items remaining. If we were to introduce one more type, we would need a minimum of 64 items, which exceeds the available 37. Therefore, the maximum possible number of types is 6.
We can verify the provided options using manageable quantities:
Option 1: Not possible. Option 2: Possible with quantities 1, 30, and 69. Option 3: Possible with quantities 1, 2, 4, and 18.75. Option 4: Possible with quantities 1, 9, 30, and 60.
To meet the minimum requirement of at least 75 items of the same type, there must be at least 31 + 1 + 43 items. Consequently, the maximum feasible number of boxes is 5 when all types have quantities of 1, 2, 4, 18, and 75.
Q.4 You ask for the type of item in box 45. Instead of being given a direct answer, you are told that there are 31 items of the same type as box 45 in boxes 1 to 44 and 43 items of the same type as box 45 in boxes 46 to 100.
What is the maximum possible number of different types of item
- 5
- 4
- 3
- 6
explanation
Answer 5
The minimum feasible number occurs when there is 1 item of type A and 99 items of type B, adhering to the specified conditions.
The maximum achievable quantity is when there is 1 item of type A, 2 of type B, 4 of type C, 8 of type D, 16 of type E, and 32 of type F. Subsequently, there would be 100 – (1+2+…+32) = 37 items remaining. If we were to introduce one more type, we would need a minimum of 64 items, which exceeds the available 37. Therefore, the maximum possible number of types is 6.
We can verify the provided options using manageable quantities:
Option 1: Not possible. Option 2: Possible with quantities 1, 30, and 69. Option 3: Possible with quantities 1, 2, 4, and 18.75. Option 4: Possible with quantities 1, 9, 30, and 60.
To meet the minimum requirement of at least 75 items of the same type, there must be at least 31 + 1 + 43 items. Consequently, the maximum feasible number of boxes is 5 when all types have quantities of 1, 2, 4, 18, and 75.
