A student walks from his house at ` 2 (1)/(2)` km /h and reaches his school late by 6 min . Next day increase his speed by 1 km/h and reaches 6 min before school time reaches 6 min before school time . How far is the school form his house ?

Options

A. `5/4` km
B. `7/4`km
C. `9/4`km
D. `(11)/(4) `km

Answer

To solve the problem step by step, we will use the information given about the student's speed and the time he is late or early.

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

### Step 2: Calculate Time Taken on First Day
The speed of the student on the first day is \( 2.5 \) km/h. The time taken to reach school can be calculated using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{d}{2.5} \text{ hours} \]
Since he reaches late by 6 minutes, we convert 6 minutes to hours:
\[ 6 \text{ minutes} = \frac{6}{60} = \frac{1}{10} \text{ hours} \]
Thus, the effective time he should have taken to reach on time is:
\[ \frac{d}{2.5} - \frac{1}{10} \text{ hours} \]

### Step 3: Calculate Time Taken on Second Day
On the second day, the speed of the student is increased to \( 2.5 + 1 = 3.5 \) km/h. The time taken on the second day is:
\[ \text{Time} = \frac{d}{3.5} \text{ hours} \]
He reaches 6 minutes early, which is also \( \frac{1}{10} \) hours. Therefore, the effective time he should have taken is:
\[ \frac{d}{3.5} + \frac{1}{10} \text{ hours} \]

### Step 4: Set Up the Equation
Since both expressions represent the same effective time to reach school, we can set them equal to each other:
\[ \frac{d}{2.5} - \frac{1}{10} = \frac{d}{3.5} + \frac{1}{10} \]

### Step 5: Clear the Fractions
To eliminate the fractions, we can multiply the entire equation by \( 70 \) (the least common multiple of 10, 2.5, and 3.5):
\[ 70 \left( \frac{d}{2.5} - \frac{1}{10} \right) = 70 \left( \frac{d}{3.5} + \frac{1}{10} \right) \]
This simplifies to:
\[ 28d - 7 = 20d + 7 \]

### Step 6: Solve for \( d \)
Now, we can isolate \( d \):
\[ 28d - 20d = 7 + 7 \]
\[ 8d = 14 \]
\[ d = \frac{14}{8} = \frac{7}{4} \text{ km} \]

### Final Answer
The distance from the student's house to the school is \( \frac{7}{4} \) km or \( 1.75 \) km.

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