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:

The phrase "definitely NOT" in a problem involving ranges and inequalities usually directs you to find a value that fails even under the most favorable conditions. Here, the most favorable condition for reducing the average is the initial average that is closest to $50 (which corresponds to the smallest possible number of $60 tickets, x=26). If a value of k doesn't work in this best-case scenario, it won't work in any other.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves analyzing how the average of a group changes when new members are added. The phrase "definitely NOT" is key; it means we are looking for a number of new guests, k, for which it is impossible to achieve the desired outcome (average < 50), regardless of the initial composition of the 50 guests (as long as it meets the initial criteria).
Step 2: Key Formula or Approach:
1. Define variables for the initial state. Let x be the number of guests with 60 tickets and y be the number with 40 tickets. We have x + y = 50.
2. Use the initial average condition (> 50) to find the possible range for x.
3. Define a variable k for the number of new guests with 40 tickets.
4. Set up an inequality for the new average to be less than 50.
5. The goal is to find the values of k from the options for which this inequality can never be satisfied, no matter the initial valid value of x. This means we test against the most 'favorable' initial condition for our goal. The easiest average to reduce is the one that is lowest to begin with.
Step 3: Detailed Explanation:
Initial State Analysis:
Let x be the number of 60 tickets. Then 50 - x is the number of 40 tickets.
The total initial revenue is R₁ = 60x + 40(50 - x) = 60x + 2000 - 40x = 20x + 2000.
The initial average A₁ = R₁ / 50 = (20x + 2000) / 50 = (2x / 5) + 40.
We are given A₁ > 50:
(2x / 5) + 40 > 50 → (2x / 5) > 10 → 2x > 50 → x > 25
Since x must be an integer, and there is at least one of each ticket type (x ≤ 49), the possible values for x are integers from 26 to 49.
Final State Analysis:
Let k be the number of new guests with 40 tickets.
New total guests = 50 + k.
New total revenue R₂ = R₁ + 40k = (20x + 2000) + 40k.
The new average A₂ = (20x + 2000 + 40k) / (50 + k) must be less than 50.
(20x + 2000 + 40k) / (50 + k) < 50
20x + 2000 + 40k < 50(50 + k) = 2500 + 50k
20x + 2000 - 2500 < 50k - 40k
20x - 500 < 10k
2x - 50 < k
So, to successfully reduce the average, the number of new guests k must be greater than 2x - 50.
The question asks for which k this is "definitely NOT" possible. This means for which k is the condition k > 2x - 50 impossible to satisfy for *any* of the allowed initial values of x (i.e., for any x ∈ {26, 27, ..., 49}).
The condition k > 2x - 50 is hardest to satisfy when the right side, 2x - 50, is as large as possible. It is easiest to satisfy when 2x - 50 is as small as possible. The smallest possible value for 2x - 50 occurs at the smallest possible x, which is x = 26.
Minimum value of 2x - 50 is 2(26) - 50 = 52 - 50 = 2.
So, for any valid initial scenario, the value of 2x - 50 will be at least 2.
The condition for success is k > 2x - 50.
If we choose k = 1, the condition becomes 1 > 2x - 50, which requires x < 25.5. This is not possible, as x must be at least 26. So k = 1 is definitely not enough.
If we choose k = 2, the condition becomes 2 > 2x - 50, which requires x < 26. This is also not possible, as x must be at least 26. So k = 2 is definitely not enough.
If we choose k = 3, the condition becomes 3 > 2x - 50, requiring x < 26.5. This is possible if the initial state was x = 26. In that specific case, adding 3 guests would work. Therefore, k = 3 is not a "definitely not" answer. The same logic applies to k = 4.
Step 4: Final Answer:
The values for the number of new guests that could definitely NOT be sufficient are 1 and 2.