A student walks from his house at 3 km/h and reaches his school 5 min late. If his speed had been 7 km/h he would have reached 10 min early. The distance of his school from his house is.

Options

A. 12.5 km
B. 12 km
C. `(21)/(16)` km
D. None of these

Answer

To solve the problem step by step, we will use the information given about the student's walking speeds and the time differences to find the distance from his house to his school.

### Step 1: Define Variables
Let the distance from the student's house to the school be \( d \) kilometers.

### Step 2: Calculate Time Taken at Different Speeds
- When the student walks at 3 km/h, the time taken to reach school is \( \frac{d}{3} \) hours.
- When the student walks at 7 km/h, the time taken to reach school is \( \frac{d}{7} \) hours.

### Step 3: Set Up the Time Difference Equation
According to the problem:
- The student is 5 minutes late when walking at 3 km/h.
- The student is 10 minutes early when walking at 7 km/h.

We convert these times into hours:
- 5 minutes = \( \frac{5}{60} = \frac{1}{12} \) hours
- 10 minutes = \( \frac{10}{60} = \frac{1}{6} \) hours

Thus, the equation for the time difference can be set up as:
\[ \frac{d}{3} - \frac{1}{12} = \frac{d}{7} + \frac{1}{6} \]

### Step 4: Simplify the Equation
To eliminate the fractions, we can find a common denominator. The least common multiple of 3, 7, 12, and 6 is 84. We will multiply the entire equation by 84:
\[ 84 \left( \frac{d}{3} \right) - 84 \left( \frac{1}{12} \right) = 84 \left( \frac{d}{7} \right) + 84 \left( \frac{1}{6} \right) \]

This simplifies to:
\[ 28d - 7 = 12d + 14 \]

### Step 5: Rearrange the Equation
Now, we will rearrange the equation to isolate \( d \):
\[ 28d - 12d = 14 + 7 \]
\[ 16d = 21 \]

### Step 6: Solve for \( d \)
Now, divide both sides by 16:
\[ d = \frac{21}{16} \text{ km} \]

### Conclusion
The distance from the student's house to his school is \( \frac{21}{16} \) kilometers.

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