A man starts moving from a place A and reaches the place B in 26 hours. He covers `1//3^(rd)` of the distance at the speed of 4 km/hr and covers the remaining distance at the speed of 5 km/hr. What is the distance (in km) between A and B?

Options

A. 45
B. 80
C. 120
D. 90

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Define the total distance
Let the total distance between A and B be \( x \) km.

### Step 2: Calculate the distance covered at each speed
The man covers \( \frac{1}{3} \) of the distance at a speed of 4 km/hr. Therefore, the distance covered at this speed is:
\[ \text{Distance}_1 = \frac{x}{3} \text{ km} \]

The remaining distance, which is \( \frac{2}{3} \) of the total distance, is covered at a speed of 5 km/hr. Thus, the distance covered at this speed is:
\[ \text{Distance}_2 = \frac{2x}{3} \text{ km} \]

### Step 3: Calculate the time taken for each part of the journey
Using the formula for time, \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \):

1. For the first part of the journey:
\[ t_1 = \frac{\text{Distance}_1}{\text{Speed}_1} = \frac{\frac{x}{3}}{4} = \frac{x}{12} \text{ hours} \]

2. For the second part of the journey:
\[ t_2 = \frac{\text{Distance}_2}{\text{Speed}_2} = \frac{\frac{2x}{3}}{5} = \frac{2x}{15} \text{ hours} \]

### Step 4: Set up the equation for total time
The total time taken for the journey is given as 26 hours. Therefore, we can set up the equation:
\[ t_1 + t_2 = 26 \]
Substituting the values of \( t_1 \) and \( t_2 \):
\[ \frac{x}{12} + \frac{2x}{15} = 26 \]

### Step 5: Find a common denominator and solve for \( x \)
The least common multiple (LCM) of 12 and 15 is 60. We can rewrite the equation:
\[ \frac{5x}{60} + \frac{8x}{60} = 26 \]
Combining the fractions:
\[ \frac{13x}{60} = 26 \]

### Step 6: Solve for \( x \)
Multiplying both sides by 60:
\[ 13x = 1560 \]
Now, divide both sides by 13:
\[ x = \frac{1560}{13} = 120 \text{ km} \]

### Conclusion
The distance between A and B is \( \boxed{120} \) km.