Anand travelled 300 km by train and 200 km by taxi. It took him 5 h and 30 min. However, if he travels 260 km by train and 240 km by taxi, he takes 6 min more. The speed of the train is

Options

A. 100 km/h
B. 120 km/h
C. 80 km/h
D. 110 km/h

Answer

To solve the problem step by step, we will define variables for the speeds of the train and taxi, set up equations based on the information provided, and then solve for the speed of the train.

### Step 1: Define Variables
Let:
- \( A \) = speed of the train (in km/h)
- \( B \) = speed of the taxi (in km/h)

### Step 2: Convert Time to Hours
Anand's total travel time for the first trip is 5 hours and 30 minutes. We convert this to hours:
\[ 5 \text{ hours } + 30 \text{ minutes } = 5 + \frac{30}{60} = 5 + 0.5 = \frac{11}{2} \text{ hours} \]

### Step 3: Set Up the First Equation
For the first trip, Anand traveled 300 km by train and 200 km by taxi. The total time taken can be expressed as:
\[ \frac{300}{A} + \frac{200}{B} = \frac{11}{2} \]

### Step 4: Convert Time for the Second Trip
In the second trip, Anand travels 260 km by train and 240 km by taxi, taking 6 minutes more than the first trip. We convert 6 minutes to hours:
\[ 6 \text{ minutes } = \frac{6}{60} = \frac{1}{10} \text{ hours} \]
Thus, the total time for the second trip is:
\[ \frac{11}{2} + \frac{1}{10} = \frac{55}{10} + \frac{1}{10} = \frac{56}{10} = \frac{28}{5} \text{ hours} \]

### Step 5: Set Up the Second Equation
For the second trip, the total time can be expressed as:
\[ \frac{260}{A} + \frac{240}{B} = \frac{28}{5} \]

### Step 6: Solve the Equations
Now we have two equations:
1. \( \frac{300}{A} + \frac{200}{B} = \frac{11}{2} \) (Equation 1)
2. \( \frac{260}{A} + \frac{240}{B} = \frac{28}{5} \) (Equation 2)

To eliminate \( B \), we can multiply both equations by suitable values to make the coefficients of \( B \) equal.

Multiply Equation 1 by 10:
\[ \frac{3000}{A} + \frac{2000}{B} = 55 \quad \text{(Equation 3)} \]

Multiply Equation 2 by 6:
\[ \frac{1560}{A} + \frac{1440}{B} = \frac{168}{5} \quad \text{(Equation 4)} \]

### Step 7: Eliminate \( B \)
Now, we can multiply Equation 3 by 6 and Equation 4 by 10 to eliminate \( B \):
\[ 6 \left(\frac{3000}{A} + \frac{2000}{B}\right) = 330 \quad \text{(Equation 5)} \]
\[ 10 \left(\frac{1560}{A} + \frac{1440}{B}\right) = 336 \quad \text{(Equation 6)} \]

Now subtract Equation 6 from Equation 5:
\[ \frac{18000}{A} - \frac{15600}{A} = 330 - 336 \]
\[ \frac{2400}{A} = -6 \]
\[ A = \frac{2400}{6} = 400 \text{ km/h} \]

### Step 8: Find \( B \)
Now substitute \( A \) back into either equation to find \( B \). Using Equation 1:
\[ \frac{300}{400} + \frac{200}{B} = \frac{11}{2} \]
\[ \frac{3}{4} + \frac{200}{B} = \frac{11}{2} \]
Convert \( \frac{3}{4} \) to have a common denominator:
\[ \frac{3}{4} = \frac{6}{8} \quad \text{and} \quad \frac{11}{2} = \frac{44}{8} \]
Thus:
\[ \frac{200}{B} = \frac{44}{8} - \frac{6}{8} = \frac{38}{8} \]
\[ B = \frac{200 \times 8}{38} = \frac{1600}{38} \approx 42.11 \text{ km/h} \]

### Final Answer
The speed of the train \( A \) is:
\[ \boxed{100 \text{ km/h}} \]