Calculating Average Speed for a Round Trip

The question asks us to find the average speed of a man who travels from point A to point B and then returns from B to A. We are given the speed for the trip from A to B and the percentage increase in speed for the return trip.

Step 1: Understand the Given Information

• Speed from A to B = 50 km/h
• Speed from B to A is increased by 40% compared to the speed from A to B.

Step 2: Calculate the Speed for the Return Trip (B to A)

The speed for the return trip is the initial speed plus a 40% increase.

• Increase in speed = 40% of 50 km/h
• Increase in speed = $\frac{40}{100} \times 50$ km/h
• Increase in speed = $0.40 \times 50$ km/h
• Increase in speed = 20 km/h

So, the speed for the return trip is:

• Return speed = Speed (A to B) + Increase in speed
• Return speed = 50 km/h + 20 km/h
• Return speed = 70 km/h

Step 3: Determine the Formula for Average Speed

For a trip where the distance covered in both directions (A to B and B to A) is the same, the average speed is not simply the arithmetic mean of the two speeds. Instead, we use the harmonic mean formula for average speed:

$\text{Average Speed} = \frac{2 \times \text{Speed}_1 \times \text{Speed}_2}{\text{Speed}_1 + \text{Speed}_2}$

Where:

• $\text{Speed}_1$ is the speed in one direction (A to B).
• $\text{Speed}_2$ is the speed in the opposite direction (B to A).

Step 4: Apply the Formula and Calculate Average Speed

Using the speeds we have:

• $\text{Speed}_1 = 50$ km/h
• $\text{Speed}_2 = 70$ km/h

Substitute these values into the average speed formula:

$\text{Average Speed} = \frac{2 \times 50 \times 70}{50 + 70}$

$\text{Average Speed} = \frac{2 \times 3500}{120}$

$\text{Average Speed} = \frac{7000}{120}$

$\text{Average Speed} = \frac{700}{12}$

$\text{Average Speed} = \frac{350}{6}$

$\text{Average Speed} = \frac{175}{3}$

Now, we perform the division:

$\text{Average Speed} \approx 58.3333...$ km/h

Step 5: Round the Result to Two Decimal Places

The question asks for the average speed to 2 decimal places. Rounding 58.3333... to two decimal places gives us 58.33.

Therefore, the average speed for both trips is 58.33 km/h.

Revision Table: Key Concepts Reviewed

Additional Information: Why Not Use Arithmetic Mean?

It's a common mistake to think that the average speed is the simple arithmetic mean of the two speeds ($\frac{50+70}{2} = 60$ km/h). This is incorrect because the man spends different amounts of time traveling at each speed. Average speed is calculated based on total distance and total time.

Let's assume the distance from A to B is 'd' km.

• Time taken from A to B = $\frac{\text{Distance}}{\text{Speed}} = \frac{d}{50}$ hours
• Time taken from B to A = $\frac{\text{Distance}}{\text{Speed}} = \frac{d}{70}$ hours
• Total Distance = $d + d = 2d$ km
• Total Time = $\frac{d}{50} + \frac{d}{70} = d \left( \frac{1}{50} + \frac{1}{70} \right) = d \left( \frac{70+50}{50 \times 70} \right) = d \left( \frac{120}{3500} \right) = \frac{120d}{3500} = \frac{12d}{350}$ hours

Now calculate the average speed:

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{\frac{12d}{350}} = 2d \times \frac{350}{12d} = \frac{2 \times 350}{12} = \frac{700}{12} = \frac{175}{3}$ km/h

$\text{Average Speed} \approx 58.33$ km/h

This confirms that the harmonic mean formula is correct for this type of problem where distances are equal but speeds differ.