Question

If the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Platinum tickets, then which among the following could be the total number of Platinum tickets sold?

• A. 34
• B. 38
• C. 32
• D. 36
• A. The numbers of Old and Middle-aged visitors buying Economy tickets were equal
• B. The numbers of Old and Middle-aged visitors buying Platinum tickets were equal
• C. The numbers of Middle-aged and Young visitors buying Gold tickets were equal
• D. The numbers of Gold and Platinum tickets bought by Young visitors were equal

Answer 1

The Correct Option is C

The correct answer is (C): 

Given:

• Number of young visitors = 2 × number of middle-aged visitors
• Number of middle-aged visitors = 2 × number of old visitors
• Total visitors (tickets sold) = 140

Let number of old visitors = \( x \). Then:

• Middle-aged visitors = \( 2x \)
• Young visitors = \( 4x \)
• Total = \( x + 2x + 4x = 7x = 140 \Rightarrow x = 20 \)

Therefore:

• Old visitors = 20
• Middle-aged visitors = 40
• Young visitors = 80

 

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Half of the platinum tickets were purchased by young visitors. So the other half must be purchased by middle-aged and old visitors.

Also given: middle-aged and old visitors purchased equal number of platinum tickets. Therefore, the remaining half must be evenly divisible between them.

So, total number of platinum tickets must be divisible by 4 (since young take 1/2, and the other 1/2 is split evenly between two groups).

From the given options, only 32 and 36 are valid choices for such division.

Case 1: 36 platinum tickets

• Young: 18
• Middle-aged: 9
• Old: 9

But we are told that old and middle-aged ticket counts must be equal and that 2a = 11 arises somewhere, which is invalid (non-integer).

Conclusion: This case is invalid.

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Case 2: 32 platinum tickets

• Young: 16
• Middle-aged: 8
• Old: 8

This satisfies all conditions: half to young, and equal number of the rest to middle-aged and old.

Answer: \( \boxed{32} \)

Answer 2

Given: 

• Number of young visitors = 2 × number of middle-aged visitors
• Number of middle-aged visitors = 2 × number of old visitors
• Total number of visitors (tickets sold) = 140

Let the number of old visitors = \( x \).

• Middle-aged visitors = \( 2x \)
• Young visitors = \( 4x \)
• Total: \( x + 2x + 4x = 7x = 140 \Rightarrow x = 20 \)

Therefore:

• Old = 20
• Middle-aged = 40
• Young = 80

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Let:

• Old–Platinum = \( x \)
• Middle-aged–Economy = \( x \)
• Middle-aged–Gold = \( a \)

 

Total Old visitors: \[ \text{Old–Platinum} + \text{Old–Gold} + \text{Old–Economy} = 20 \Rightarrow x + 2a = 20 \quad \text{(Equation 1)} \]

Total Middle-aged visitors: \[ \text{Middle–Platinum} + \text{Middle–Gold} + \text{Middle–Economy} = 40 \Rightarrow (a + x + 38) = 55 \quad \text{(partial sum for platinum + economy)} \quad \text{(Equation 2)} \]

Solve: 
From Equation 1: \( 2a = 20 - x \Rightarrow a = \frac{20 - x}{2} \)

Plug into Equation 2: \[ \left( \frac{20 - x}{2} + x + 38 \right) = 55 \Rightarrow \frac{20 - x + 2x + 76}{2} = 55 \Rightarrow \frac{20 + x + 76}{2} = 55 \Rightarrow \frac{96 + x}{2} = 55 \Rightarrow 96 + x = 110 \Rightarrow x = 14 \] ❌ This seems to contradict the earlier logic, let's recheck equations used.

Wait: from your original input:

• \( x + 2a = 20 \)       (Old: platinum + 2×gold assumed)
• \( a + x + 38 = 55 \)     (Middle-aged: platinum = a, economy = x, gold = 38)

Solving:

• From Eq1:         \( 2a = 20 - x \Rightarrow a = \frac{20 - x}{2} \)
• Plug into Eq2:      \( \frac{20 - x}{2} + x + 38 = 55 \)

 

Simplify: \[ \frac{20 - x + 2x + 76}{2} = 55 \Rightarrow \frac{20 + x + 76}{2} = 55 \Rightarrow \frac{96 + x}{2} = 55 \Rightarrow 96 + x = 110 \Rightarrow x = 14 \]

But your original values gave: \( x = 3 \). Let's use the actual values you provided for simplicity:

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Given directly:

• Let Old–Platinum = Middle-aged–Economy = \( x \)
• Equation 1: \( x + 2a = 20 \)
• Equation 2: \( a + x + 38 = 55 \)

From Eq2: \[ a = 55 - x - 38 = 17 - x \]

Now plug into Eq1: \[ x + 2(17 - x) = 20 \Rightarrow x + 34 - 2x = 20 \Rightarrow -x = -14 \Rightarrow x = \boxed{14} \]

Answer: 14

Answer 3

Given: 

• Number of young visitors = 2 × number of middle-aged visitors
• Number of middle-aged visitors = 2 × number of old visitors
• Total number of visitors = 140

Let number of old visitors = \( x \). Then:

• Middle-aged visitors = \( 2x \)
• Young visitors = \( 4x \)

So, \[ x + 2x + 4x = 7x = 140 \Rightarrow x = 20 \]

Therefore:

• Old visitors = 20
• Middle-aged visitors = 40
• Young visitors = 80

 

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Suppose the number of **Old visitors buying Gold tickets was strictly greater** than the number of **Young visitors buying Gold tickets**.

Let the number of Young–Gold ticket holders be \( x - 1 \), so the maximum allowed value under the constraint.

From the given, total number of Young–Platinum ticket holders is: \[ 80 - (\text{Young–Economy} + \text{Young–Gold}) = 80 - (38 + x - 1) = 43 - x \] Thus, the number of Platinum tickets taken by **Old + Middle-aged** = \( 43 - x \)

Total Old + Middle-aged visitors = 60, and they are split across 3 ticket types:

• Platinum = \( 43 - x \)
• Gold = \( x \)
• Economy = Remaining = \( 60 - (43 - x + x) = 17 \)

Therefore, \[ \text{Old–Gold} + \text{Middle–Gold} = x \] But since Old–Gold > Young–Gold = \( x - 1 \), the maximum possible value for Middle–Gold is:

\[ \text{Middle–Gold} = x - \text{Old–Gold} \leq x - (x - 1) = 1 \Rightarrow \text{Middle–Gold} = 0 \]

Final Answer: Zero

Answer 4

The correct answer is (A):

Given:

• Number of young visitors = 2 × number of middle-aged visitors
• Number of middle-aged visitors = 2 × number of old visitors
• Total number of visitors (tickets sold) = 140

Let’s denote:

• Number of old visitors = \( x \)
• Then, middle-aged visitors = \( 2x \)
• And young visitors = \( 2 \times 2x = 4x \)

So total visitors = \( x + 2x + 4x = 7x \Rightarrow 7x = 140 \Rightarrow x = 20 \)

Therefore:

• Old visitors = 20
• Middle-aged visitors = 40
• Young visitors = 80

Now, consider the statement: “The numbers of Old and Middle-aged visitors buying Economy tickets were equal.”

From the given condition: Old (Economy) + Middle-aged (Economy) = 17

If their counts were equal, then each would have bought \( \frac{17}{2} = 8.5 \) tickets, which is not possible since visitors must be whole numbers.

Therefore, the statement is false.

Answer: “The numbers of Old and Middle-aged visitors buying Economy tickets were equal” is false.