Mr. Chaalu while travelling by Ferry Queen has travelled the distance one kilometre more, than the fare he paid per km. Initially he had total amount of Rs. 350 in his wallet. Now he is only left with the minimum sum of (if all the distance travelled by him is in integers) :

Options

A. a. Rs. 26
B. b. Rs. 8
C. c. Rs. 19
D. d. Can't be determined

Answer

To solve the problem step by step, let's break down the information given and derive the solution systematically.

### Step 1: Understand the relationship between distance and fare
Let the fare paid per kilometer be \( n \) (in Rs). According to the problem, Mr. Chaalu traveled a distance of \( n + 1 \) kilometers.

### Step 2: Calculate total fare
The total fare paid by Mr. Chaalu can be expressed as:
\[ \text{Total Fare} = \text{Fare per km} \times \text{Distance} = n \times (n + 1) \]

### Step 3: Set up the equation for total amount
Initially, Mr. Chaalu had Rs. 350. After paying the fare, the amount left with him is:
\[ 350 - n(n + 1) \]
We need to find the minimum amount left with him, which means we need to maximize the total fare paid.

### Step 4: Find the maximum fare paid
To find the maximum fare, we need to ensure that \( n(n + 1) \) is less than or equal to 350. Thus, we can set up the inequality:
\[ n(n + 1) \leq 350 \]

### Step 5: Solve the inequality
This is a quadratic inequality:
\[ n^2 + n - 350 \leq 0 \]
To find the roots of the equation \( n^2 + n - 350 = 0 \), we can use the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = -350 \).

Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \times 1 \times (-350) = 1 + 1400 = 1401 \]

Now calculating \( n \):
\[ n = \frac{-1 \pm \sqrt{1401}}{2} \]
Calculating \( \sqrt{1401} \) gives approximately \( 37.45 \):
\[ n = \frac{-1 + 37.45}{2} \approx 18.225 \]
Since \( n \) must be an integer, we can take \( n = 18 \) or \( n = 19 \).

### Step 6: Calculate total fare for \( n = 18 \) and \( n = 19 \)
1. For \( n = 18 \):
\[ \text{Total Fare} = 18 \times (18 + 1) = 18 \times 19 = 342 \]
Amount left:
\[ 350 - 342 = 8 \]

2. For \( n = 19 \):
\[ \text{Total Fare} = 19 \times (19 + 1) = 19 \times 20 = 380 \]
This exceeds 350, so it is not valid.

### Step 7: Conclusion
The minimum amount left with Mr. Chaalu, when he pays the maximum possible fare, is Rs. 8.

### Final Answer
The minimum sum left with Mr. Chaalu is **Rs. 8**.

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