A man travels three-fifths of a distance AB at a speed of 3a, and the remaining at a speed of 2b. If he goes from B to A and returns at a speed of 5c in the same time, then

Options

A. `(1)/(a) +(1)/(b) =(1)/(c )`
B. `a+b=c`
C. `3a+2b=5c`
D. `(1)/(a)+(1)/(b)=(2)/(c )`

Answer

To solve the problem step by step, we will analyze the information given and derive the necessary equations to find the relationship between the variables \( a \), \( b \), and \( c \).

### Step-by-Step Solution:

1. **Define the Total Distance**:
Let the total distance \( AB = x \).

2. **Calculate the Distances Covered**:
- The man travels \( \frac{3}{5} \) of the distance \( AB \) at a speed of \( 3a \).
- The remaining distance is \( x - \frac{3}{5}x = \frac{2}{5}x \), which he travels at a speed of \( 2b \).

3. **Calculate Time Taken for Each Segment**:
- Time taken to cover \( \frac{3}{5}x \) at speed \( 3a \):
\[ \text{Time}_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{\frac{3}{5}x}{3a} = \frac{x}{5a} \]
- Time taken to cover \( \frac{2}{5}x \) at speed \( 2b \):
\[ \text{Time}_2 = \frac{\frac{2}{5}x}{2b} = \frac{x}{5b} \]

4. **Calculate the Return Trip Time**:
- The man returns from \( B \) to \( A \) and back at a speed of \( 5c \) over a total distance of \( 2x \):
\[ \text{Time}_3 = \frac{2x}{5c} \]

5. **Set Up the Equation for Equal Times**:
According to the problem, the total time taken to travel from \( A \) to \( B \) and back is equal:
\[ \text{Time}_1 + \text{Time}_2 = \text{Time}_3 \]
Substituting the expressions we found:
\[ \frac{x}{5a} + \frac{x}{5b} = \frac{2x}{5c} \]

6. **Simplify the Equation**:
- We can cancel \( x \) from both sides (assuming \( x \neq 0 \)):
\[ \frac{1}{5a} + \frac{1}{5b} = \frac{2}{5c} \]
- Multiply through by \( 5 \):
\[ \frac{1}{a} + \frac{1}{b} = \frac{2}{c} \]

7. **Final Result**:
The final equation we derived is:
\[ \frac{1}{a} + \frac{1}{b} = \frac{2}{c} \]