John rode his bicycle to town at the rate of 15 milles per hour. He left the bicycle in town for minor repairs and walked home along the same route at the rate 3 miles per hour. Excluding the time John spent in taking the bike into the repair shop, the trip took 3 hours. How many hours did John take to walk back?

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Define Variables
Let:
- \( d \) = distance from home to town (in miles)
- \( t_1 \) = time taken to ride to town (in hours)
- \( t_2 \) = time taken to walk back home (in hours)

### Step 2: Write the Distance Equations
Using the formula for distance, which is:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]

For the bicycle ride to town:
\[ d = 15 \times t_1 \] (since speed = 15 miles/hour)

For the walk back home:
\[ d = 3 \times t_2 \] (since speed = 3 miles/hour)

### Step 3: Set the Equations Equal
Since both expressions equal \( d \), we can set them equal to each other:
\[ 15t_1 = 3t_2 \]

### Step 4: Express \( t_1 \) in terms of \( t_2 \)
Rearranging the equation gives:
\[ t_1 = \frac{3t_2}{15} \]
\[ t_1 = \frac{t_2}{5} \]

### Step 5: Use the Total Time Condition
According to the problem, the total time for the trip (biking to town and walking back) is 3 hours:
\[ t_1 + t_2 = 3 \]

Substituting \( t_1 \) from the previous step:
\[ \frac{t_2}{5} + t_2 = 3 \]

### Step 6: Combine Terms
To combine the terms, express \( t_2 \) with a common denominator:
\[ \frac{t_2}{5} + \frac{5t_2}{5} = 3 \]
\[ \frac{6t_2}{5} = 3 \]

### Step 7: Solve for \( t_2 \)
Multiply both sides by 5 to eliminate the fraction:
\[ 6t_2 = 15 \]
Now, divide by 6:
\[ t_2 = \frac{15}{6} = \frac{5}{2} \]

### Step 8: Convert to Hours and Minutes
Converting \( \frac{5}{2} \) hours gives us:
\[ \frac{5}{2} = 2 \text{ hours and } 30 \text{ minutes} \]

### Final Answer
John took **2 hours and 30 minutes** to walk back home.

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