A person goes from point S to T and comes back. His average speed for the whole journey is 70 km/hr. If his speed while coming back from T to S is 80 km/hr, then what will be the speed of the person (in km/hr) while going from S to T?

Options

A. 58.13
B. 62.22
C. 60
D. 65

Answer

To find the speed of the person while going from S to T, we can use the formula for average speed and the relationship between distance, speed, and time.

### Step-by-Step Solution:

1. **Define Variables**:
Let the speed from S to T be \( s \) km/hr. The speed from T to S is given as 80 km/hr.

2. **Distance Calculation**:
Let the distance from S to T be \( d \) km. Therefore, the total distance for the round trip (S to T and back) is \( 2d \) km.

3. **Time Calculation**:
The time taken to go from S to T is \( \frac{d}{s} \) hours, and the time taken to return from T to S is \( \frac{d}{80} \) hours.

4. **Total Time**:
The total time for the journey is:
\[ \text{Total Time} = \frac{d}{s} + \frac{d}{80} \]

5. **Average Speed Formula**:
The average speed for the whole journey is given by:
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{\frac{d}{s} + \frac{d}{80}} \]
Since the average speed is 70 km/hr, we set up the equation:
\[ 70 = \frac{2d}{\frac{d}{s} + \frac{d}{80}} \]

6. **Simplifying the Equation**:
We can simplify the equation by canceling \( d \) (assuming \( d \neq 0 \)):
\[ 70 = \frac{2}{\frac{1}{s} + \frac{1}{80}} \]
This can be rewritten as:
\[ \frac{1}{s} + \frac{1}{80} = \frac{2}{70} \]
Simplifying \( \frac{2}{70} \) gives:
\[ \frac{1}{s} + \frac{1}{80} = \frac{1}{35} \]

7. **Finding Common Denominator**:
To solve for \( \frac{1}{s} \), we can rearrange:
\[ \frac{1}{s} = \frac{1}{35} - \frac{1}{80} \]
To combine these fractions, we need a common denominator. The least common multiple of 35 and 80 is 280.

Converting each fraction:
\[ \frac{1}{35} = \frac{8}{280}, \quad \frac{1}{80} = \frac{3.5}{280} \]
Therefore:
\[ \frac{1}{s} = \frac{8}{280} - \frac{3.5}{280} = \frac{4.5}{280} = \frac{9}{560} \]

8. **Calculating \( s \)**:
Taking the reciprocal gives:
\[ s = \frac{560}{9} \]
Performing the division:
\[ s \approx 62.22 \text{ km/hr} \]

### Final Answer:
The speed of the person while going from S to T is approximately **62.22 km/hr**.