Question

In a certain batch of guests in a museum, there are 50 guests; each guest buys either a 40 ticket or a 60 ticket, with at least one guest of each ticket type. The average (arithmetic mean) value of ticket-receipts from the batch is more than 50. If the average value of ticket-receipts is to be reduced to less than 50 by including few new guests with 40 tickets, what could definitely NOT be the number of new guests with $40 tickets that could be included?
Indicate all such numbers. 
[Note: Select one or more answer choices] 
 

• A. 1
• B. 2
• C. 3
• D. 4
• E.

Hint:

In problems with inequalities and phrases like "definitely," "at least," or "at most," it's crucial to identify the "worst-case" or most restrictive scenario. Here, to find a number of guests \(k\) that is definitely not enough, you must show that it's not enough even in the most favorable initial conditions for reducing the average (i.e., the lowest possible initial average, which corresponds to the lowest number of expensive tickets).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a problem involving inequalities and averages. We need to find the range of the initial average, and then see how adding new guests affects this average. The question asks what number of new guests could definitely not achieve the goal of reducing the average to below 50. This means we need to find the numbers of new guests for which it's possible to reduce the average to <50, and the answer will be the numbers for which it's impossible, regardless of the initial state. The word "definitely NOT" implies we must consider the most extreme initial case allowed by the problem statement.
Step 2: Key Formula or Approach:
1. Let x be the number of guests with 60 tickets and y be the number of guests with 40 tickets initially. We know x + y = 50, x ≥ 1, y ≥ 1.
2. The total initial revenue is R₁ = 60x + 40y.
3. The initial average is A₁ = R₁ / 50 > 50. Use this to find the possible range for x and y.
4. Let k be the number of new guests with 40 tickets.
5. The new total number of guests is 50 + k.
6. The new total revenue is R₂ = R₁ + 40k.
7. The new average is A₂ = (R₁ + 40k) / (50 + k) < 50.
8. We need to find values of k from the options for which this inequality can never be true, no matter what the initial valid values of x and y were. This means we should test the inequality against the 'worst-case' initial scenario, which is the lowest possible initial average A₁ that is still greater than 50.
Step 3: Detailed Explanation:
Analyze the initial state:
Let x be the number of 60 tickets and y be the number of 40 tickets. x + y = 50.
The initial average is A₁ = (60x + 40y) / 50 > 50.
Substitute y = 50 - x:
A₁ = (60x + 40(50 - x)) / 50 = (60x + 2000 - 40x) / 50 = (20x + 2000) / 50 = (2x / 5) + 40.
The condition is A₁ > 50:
(2x / 5) + 40 > 50
(2x / 5) > 10
2x > 50 → x > 25
Since x must be an integer, x ≥ 26. Also, we know x ≥ 1 and y ≥ 1 → x ≤ 49. So, the number of 60 tickets can be any integer from 26 to 49.
The initial total revenue is R₁ = 20x + 2000. Since x > 25, R₁ > 20(25) + 2000 = 2500. So, A₁ = R₁ / 50 > 50.
Analyze the final state:
Let k be the number of new guests with 40 tickets. The new average is A₂ = (R₁ + 40k) / (50 + k), and we want this to be less than 50.
(R₁ + 40k) / (50 + k) < 50
R₁ + 40k < 50(50 + k)
R₁ + 40k < 2500 + 50k
R₁ - 2500 < 10k
Substitute R₁ = 20x + 2000:
(20x + 2000) - 2500 < 10k
20x - 500 < 10k
2x - 50 < k
We know from the initial condition that x > 25, so 2x > 50, which means 2x - 50 > 0.
The condition for the average to drop below 50 is k > 2x - 50.
The question asks what could definitely not be the value of k. This means we are looking for a k such that the condition k > 2x - 50 can never be satisfied for any possible value of x.
Wait, this is the reverse. We want to find k for which it is always possible to satisfy the condition, and exclude those. The wording is tricky.
Let's re-read: "what could definitely NOT be the number of new guests ... that could be included?". This means, find k such that for any valid initial state, adding k guests is not enough.
The condition is k > 2x - 50. For this to fail, we need k ≤ 2x - 50.
We want the k for which k ≤ 2x - 50 is true for ALL possible initial states x.
The possible values for x are x ∈ {26, 27, ..., 49}.
The lowest value of x is 26.
If we choose the "worst case" (the initial average that is hardest to bring down), we should choose the highest possible initial average, which corresponds to the largest x, i.e., x = 49.
If x = 49, the condition becomes k > 2(49) - 50 = 98 - 50 = 48. In this case, we would need to add at least 49 guests.
If we choose the "best case" (the initial average that is easiest to bring down), we should choose the lowest possible initial average, which corresponds to the smallest x, i.e., x = 26.
If x = 26, the condition becomes k > 2(26) - 50 = 52 - 50 = 2. In this case, we need to add at least 3 guests.
The question asks for which k could definitely not bring the average down. This means, find a k such that, no matter what the initial composition of the group was (as long as it fits the criteria), adding k guests is insufficient.
This means we are looking for values of k such that the inequality k > 2x - 50 is FALSE for at least one possible value of x.
The phrase "definitely not" means it's impossible. So we are looking for values of k for which the goal is impossible to achieve.
The goal is possible if there exists at least one valid initial state x for which adding k guests works. The goal is impossible (k is definitely not the number) if for ALL valid initial states x, adding k guests is not enough.
So we want to find k such that for all x ∈ {26, ..., 49}, the condition A₂ < 50 is FALSE.
This means for all x ∈ {26, ..., 49}, we have A₂ ≥ 50, which is equivalent to k ≤ 2x - 50.
For k to be less than or equal to 2x - 50 for all possible x, it must be less than or equal to the minimum possible value of 2x - 50.
The minimum value of 2x - 50 occurs at the minimum value of x, which is x = 26.
Min value of 2x - 50 is 2(26) - 50 = 52 - 50 = 2.
So, if k ≤ 2, the goal of getting the average below 50 is definitely not achievable. For k = 1 or k = 2, there is no value of x that allows the condition k > 2x - 50 to be met, because the smallest value 2x - 50 can take is 2.
Let's test the options.
If k = 1, is it possible to have A₂ < 50? We need 1 > 2x - 50. This requires 51 > 2x or x < 25.5. But we know from the problem that x must be at least 26. So it's impossible. Thus k = 1 is a number that could definitely NOT be the number of new guests.
If k = 2, is it possible? We need 2 > 2x - 50. This requires 52 > 2x or x < 26. But x must be at least 26. So it's impossible. Thus k = 2 is also a correct answer.
If k = 3, is it possible? We need 3 > 2x - 50. This requires 53 > 2x or x < 26. If the initial state had x = 26, then adding 3 guests would work. A₂ < 50. The question asks what could definitely not be the number. Since for k = 3, we found a scenario where it is possible, k = 3 is not a "definitely not" answer.
Let me re-read the question again. It is very subtle.
"what could definitely NOT be the number of new guests ... that could be included?"
This can be interpreted as: find k such that there exists some initial condition for which adding k guests is not enough.
Let's re-evaluate the question's logic.
The goal is to reduce the average to less than 50.
Let's see for which k values it is always possible.
It is always possible if k > 2x - 50 for all x ∈ {26, ..., 49}. For this to be true, k must be greater than the maximum value of 2x - 50.
Max value of 2x - 50 is 2(49) - 50 = 48. So if we add k = 49 new guests, it is always possible to bring the average down.
Let's take the other interpretation. A number k is a "definitely not" number if it's impossible to achieve the goal with k new guests, for at least one of the valid initial scenarios.
No, "definitely not" means impossibility across all scenarios.
My first interpretation was correct. A value k is a "definitely not" value if, regardless of the initial number of 60 ticket holders (as long as it's between 26 and 49), adding k new guests is insufficient to bring the average below 50.
This requires k ≤ 2x - 50 for all x ∈ {26, ..., 49}.
This means k ≤ min(2x - 50) = 2(26) - 50 = 2.
So k = 1, 2 are the answers.
Step 4: Final Answer:
The values of k for which it is impossible to reduce the average ticket price to less than 50, regardless of the initial distribution of guests, are k = 1 and k = 2. Therefore, options (A) and (B) are the correct answers. There might be an error in the question or the expected answer if C and D are also considered correct. Based on a rigorous mathematical analysis of the problem as stated, only A and B fit the criteria.