Distance Calculation: Bike and Walk Journey

The problem asks for the distance Sahil travelled one way, given his speed biking one way and walking back, and the total time for the round trip.

Understanding the Problem

• Speed biking (outward): $v_1 = 25$ kmph
• Speed walking (return): $v_2 = 4$ kmph
• Total time for journey: $T = 5$ hours $48$ minutes

Convert Total Time to Hours

First, convert the total time into hours.

$48 \text{ minutes} = \frac{48}{60} \text{ hours} = \frac{4}{5} \text{ hours} = 0.8 \text{ hours}$

Total time $T = 5 + 0.8 = 5.8$ hours.

Setting Up the Equation

Let the distance travelled one way be $d$ km.

• Time taken to travel by bike: $t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{d}{25}$ hours.
• Time taken to walk back: $t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{d}{4}$ hours.
• The total time is the sum of the time spent biking and walking: $t_1 + t_2 = T$.

Substitute the expressions for $t_1$, $t_2$, and $T$: $ \frac{d}{25} + \frac{d}{4} = 5.8 $

Solving for Distance

Combine the terms on the left side using a common denominator (100):

Simplify the equation:

Now, solve for $d$:

The distance Sahil travelled is 20 km.