This problem involves calculating the average speed for a round trip when the speeds for the two trips are different. Let us solve it step by step.

Problem Breakdown:

1. Speed from P to Q = $40 \, \text{kmph}$.
2. Speed from Q to P = $40 + 50\%$ of $40$ = $40 + 20 = 60 \, \text{kmph}$.

The formula for average speed when the distance is the same for both trips is:

$\text{Average Speed} = \frac{2 \cdot S_1 \cdot S_2}{S_1 + S_2},$ where $S_1$ and $S_2$ are the speeds for the two trips.

Substituting the Values:

$\text{Average Speed} = \frac{2 \cdot 40 \cdot 60}{40 + 60}.$

1. Calculate the numerator: $2 \cdot 40 \cdot 60 = 4800.$

2. Calculate the denominator: $40 + 60 = 100.$

3. Divide the numerator by the denominator: $\text{Average Speed} = \frac{4800}{100} = 48 \, \text{kmph}.$

Final Answer:

The average speed for both trips is 48 kmph. The correct option is C.

Would you like a detailed explanation of the formula or help with a similar problem? Let me know!

Here are 5 related questions for practice:

1. If the speed for the return trip was increased by 100% instead of 50%, what would be the average speed?
2. How does the average speed formula change if the distances for the two trips are not the same?
3. Calculate the average speed for a round trip where the forward journey is 50 kmph and the return journey is 75 kmph.
4. Why is the average speed not equal to the arithmetic mean of the two speeds in this scenario?
5. How would the calculation change if the speeds were given in meters per second?

Tip: Always use the harmonic mean formula for average speed when the distances are the same but speeds vary.