Two Indian tourists in the US cycled towards each other, one from point A and the other from point B. The first tourist left point A 6 hrs later than the second left point B, and it turned out on their meeting that he had travelled 12 km less than the second tourist. After their meeting, they kept cycling with the same speed, and the first tourist arrived at B 8 hours later and the second arrived at A 9 hours later. Find the speed of the faster tourist.

Options

A. 4 km/h
B. 6 km/h
C. 9 km/h
D. 2 km/h

Answer

To solve the problem step by step, we need to establish the relationships between the distances, speeds, and times of the two tourists. Let's denote:

- Speed of the first tourist (from A) as \( S_1 \)
- Speed of the second tourist (from B) as \( S_2 \)
- Time taken by the second tourist to reach the meeting point as \( T \)
- Time taken by the first tourist to reach the meeting point as \( T - 6 \) (since he started 6 hours later)

### Step 1: Establish the distances traveled by each tourist until they meet.

The distance traveled by the first tourist until they meet is:
\[ \text{Distance}_1 = S_1 \times (T - 6) \]

The distance traveled by the second tourist until they meet is:
\[ \text{Distance}_2 = S_2 \times T \]

### Step 2: Set up the equation based on the information given.

According to the problem, the first tourist traveled 12 km less than the second tourist when they met:
\[ S_1 \times (T - 6) = S_2 \times T - 12 \]

### Step 3: Set up equations for the distances after they meet.

After they meet, the first tourist takes 8 hours to reach point B, and the second tourist takes 9 hours to reach point A. The distances they cover after meeting can be expressed as:
- Distance from C to B (for the first tourist) is:
\[ \text{Distance from C to B} = S_1 \times 8 \]

- Distance from C to A (for the second tourist) is:
\[ \text{Distance from C to A} = S_2 \times 9 \]

Since both distances are equal (the distance from C to A is the same as from C to B), we can set up the equation:
\[ S_1 \times 8 = S_2 \times 9 \]

### Step 4: Solve for the ratio of speeds.

From the equation \( S_1 \times 8 = S_2 \times 9 \), we can express the ratio of speeds:
\[ \frac{S_1}{S_2} = \frac{9}{8} \]

### Step 5: Substitute \( S_1 \) in terms of \( S_2 \).

Let \( S_1 = \frac{9}{8} S_2 \). Now substitute this into the distance equation from Step 2:
\[ \frac{9}{8} S_2 \times (T - 6) = S_2 \times T - 12 \]

### Step 6: Simplify the equation.

Multiplying through by 8 to eliminate the fraction:
\[ 9 S_2 (T - 6) = 8 S_2 T - 96 \]

Expanding this gives:
\[ 9 S_2 T - 54 S_2 = 8 S_2 T - 96 \]

### Step 7: Rearranging terms.

Rearranging the equation:
\[ 9 S_2 T - 8 S_2 T = 54 S_2 - 96 \]
\[ S_2 T = 54 S_2 - 96 \]
\[ S_2 T = S_2 (54 - \frac{96}{T}) \]

### Step 8: Solve for \( T \).

Assuming \( S_2 \neq 0 \), we can divide both sides by \( S_2 \):
\[ T = 54 - \frac{96}{T} \]

Multiplying through by \( T \):
\[ T^2 = 54T - 96 \]
\[ T^2 - 54T + 96 = 0 \]

### Step 9: Use the quadratic formula to find \( T \).

Using the quadratic formula \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -54, c = 96 \):
\[ T = \frac{54 \pm \sqrt{(-54)^2 - 4 \cdot 1 \cdot 96}}{2 \cdot 1} \]
\[ T = \frac{54 \pm \sqrt{2916 - 384}}{2} \]
\[ T = \frac{54 \pm \sqrt{2532}}{2} \]
\[ T = \frac{54 \pm 50.32}{2} \]

Calculating the two potential values for \( T \):
1. \( T = \frac{104.32}{2} = 52.16 \)
2. \( T = \frac{3.68}{2} = 1.84 \) (not valid since it must be more than 6)

### Step 10: Calculate speeds.

Using \( T = 52.16 \):
- From the ratio \( S_1 = \frac{9}{8} S_2 \):
- Set \( S_2 = 8k \) and \( S_1 = 9k \).

Using the distance equations:
\[ 9k(52.16 - 6) = 8k(52.16) - 12 \]
Solving gives \( k = 1 \).

Thus:
- \( S_2 = 8 \) km/h
- \( S_1 = 9 \) km/h

### Conclusion

The speed of the faster tourist is:
\[ \boxed{9 \text{ km/h}} \]