The correct answer is (C):
Given:
Let number of old visitors = \( x \). Then:
Therefore:
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
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.
Case 2: 32 platinum tickets
This satisfies all conditions: half to young, and equal number of the rest to middle-aged and old.
Answer: \( \boxed{32} \)
Given:
Let the number of old visitors = \( x \).
Therefore:
Let:
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:
Solving:
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:
Given directly:
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
Given:
Let number of old visitors = \( x \). Then:
So, \[ x + 2x + 4x = 7x = 140 \Rightarrow x = 20 \]
Therefore:
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:
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
The correct answer is (A):
Given:
Let’s denote:
So total visitors = \( x + 2x + 4x = 7x \Rightarrow 7x = 140 \Rightarrow x = 20 \)
Therefore:
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.
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