Fair-Asia is well known for its queer marketing practices. To lure its customers, out of the blue, it comes up with a very lucrative offer: In that offer it promises to sell and the air tickets, for a one-way trip to Bangkok that originates from Jakrata, in such a manner that the nth customer has to pay exactly Rs.n to buy an air ticket,but each prospective passenger has to bay a separate ticket for oneself. The airline claims that it had sold out all the seats at throm away prices in no time. However, at the last moment one of the customers had returned her ticket and got the full refund from the airline. Consequently, the average income of the airline from this particular flight is exactly rs 100 per passenger.
What could be the maximum number of passengers allowed to travel in this airline, if only one passenger occupies each seat and cabin crew or anyone else does not occupy any seat intended for the passengers?

Options

A. 320
B. 240
C. 200
D. none of these

Answer

To solve the problem, we need to determine the maximum number of passengers (n) that could have traveled on the airline flight, given the conditions stated in the problem. Let's break it down step by step.

### Step 1: Understand the Payment Structure
Each passenger pays an amount equal to their position number. Therefore:
- The 1st passenger pays Rs. 1
- The 2nd passenger pays Rs. 2
- The 3rd passenger pays Rs. 3
- ...
- The nth passenger pays Rs. n

### Step 2: Calculate Total Revenue from n Passengers
The total revenue (R) from n passengers can be calculated using the formula for the sum of the first n natural numbers:
\[ R = 1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2} \]

### Step 3: Adjust for One Passenger Returning Their Ticket
Since one passenger returned their ticket, the total number of passengers who actually paid is \( n - 1 \). The total revenue from these passengers is still \( R \), but the average income per passenger is given as Rs. 100.

### Step 4: Set Up the Average Income Equation
The average income per passenger after one passenger returned their ticket can be expressed as:
\[ \text{Average} = \frac{R}{n - 1} = 100 \]

Substituting the expression for R:
\[ \frac{\frac{n(n + 1)}{2}}{n - 1} = 100 \]

### Step 5: Simplify the Equation
To eliminate the fraction, we can multiply both sides by \( n - 1 \):
\[ \frac{n(n + 1)}{2} = 100(n - 1) \]

Now, multiply both sides by 2:
\[ n(n + 1) = 200(n - 1) \]

Expanding both sides:
\[ n^2 + n = 200n - 200 \]

### Step 6: Rearrange the Equation
Rearranging gives us:
\[ n^2 + n - 200n + 200 = 0 \]
\[ n^2 - 199n + 200 = 0 \]

### Step 7: Solve the Quadratic Equation
We can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 1, b = -199, c = 200 \):
\[ n = \frac{199 \pm \sqrt{(-199)^2 - 4 \cdot 1 \cdot 200}}{2 \cdot 1} \]
\[ n = \frac{199 \pm \sqrt{39601 - 800}}{2} \]
\[ n = \frac{199 \pm \sqrt{38801}}{2} \]

Calculating \( \sqrt{38801} \) gives approximately 197.0, so:
\[ n = \frac{199 \pm 197}{2} \]

Calculating the two possible values:
1. \( n = \frac{396}{2} = 198 \)
2. \( n = \frac{2}{2} = 1 \)

### Step 8: Determine the Maximum Number of Passengers
Since we are looking for the maximum number of passengers, we take \( n = 198 \).

### Step 9: Verify with the Average Calculation
If \( n = 198 \), then the total revenue is:
\[ R = \frac{198 \times 199}{2} = 19701 \]
The number of passengers who actually paid is \( 198 - 1 = 197 \).
The average income per passenger is:
\[ \text{Average} = \frac{19701}{197} = 100 \]

This confirms our solution.

### Final Answer
The maximum number of passengers allowed to travel in this airline is **198**.

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