To find John's average speed for the entire trip, we will follow these steps:

Step 1: Total Distance

John rode 4 miles to the store and 4 miles back, so the total distance is:

$\text{Total Distance} = 4 \, \text{miles} + 4 \, \text{miles} = 8 \, \text{miles}$

Step 2: Total Time

The time it took John to get to the store is $\frac{3}{10}$ of an hour, and the time it took him to get back is $\frac{1}{2}$ of an hour. Adding these times together:

$\text{Total Time} = \frac{3}{10} \, \text{hours} + \frac{1}{2} \, \text{hours}$

We need to add these fractions. First, convert $\frac{1}{2}$ to have a denominator of 10:

$\frac{1}{2} = \frac{5}{10}$

Now, add the fractions:

$\text{Total Time} = \frac{3}{10} + \frac{5}{10} = \frac{8}{10} = 0.8 \, \text{hours}$

Step 3: Average Speed

The average speed is calculated by dividing the total distance by the total time:

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{8 \, \text{miles}}{0.8 \, \text{hours}} = 10 \, \text{miles per hour}$

So, John's average speed for the entire trip was 10 miles per hour.

Do you want further details or have any questions about this?

Related Questions:

1. What would his average speed be if he took 1 hour to return instead of 1/2 hour?
2. How would the average speed change if the distance to the store was 6 miles?
3. If John spent 15 minutes at the store, how would the total time for the trip be calculated?
4. How would you calculate the average speed if John had taken different routes to and from the store with different distances?
5. What would the average speed be if John spent twice as long going to the store than coming back?

Tip:

To handle fractional times more easily, try converting them to decimals before performing operations.