If a student run with the speed of 5 km./hr. then he reach his school 10 min late but if he increases his speed by 1 km./hr., he reach school 5 min late. Find his original speed to reach school on time ?

Options

A. 7.5 km./hr.
B. 5 km./hr.
C. 10 km./hr.
D. 6 km./hr.

Answer

To solve the problem step by step, let's analyze the information given and derive the required speed.

### Step 1: Define Variables
Let the distance to the school be \( D \) km. The original speed of the student is \( S \) km/hr.

### Step 2: Establish the Time Taken at Different Speeds
1. When the student runs at 5 km/hr, he is 10 minutes late.
- Time taken = \( \frac{D}{5} \) hours
- Time should have been = \( \frac{D}{S} \) hours
- Therefore, we can write the equation:
\[ \frac{D}{5} = \frac{D}{S} + \frac{10}{60} \]
- Simplifying gives:
\[ \frac{D}{5} = \frac{D}{S} + \frac{1}{6} \]

2. When the student runs at 6 km/hr, he is 5 minutes late.
- Time taken = \( \frac{D}{6} \) hours
- Therefore, we can write the equation:
\[ \frac{D}{6} = \frac{D}{S} + \frac{5}{60} \]
- Simplifying gives:
\[ \frac{D}{6} = \frac{D}{S} + \frac{1}{12} \]

### Step 3: Set Up the Equations
Now we have two equations:
1. \( \frac{D}{5} - \frac{D}{S} = \frac{1}{6} \) (Equation 1)
2. \( \frac{D}{6} - \frac{D}{S} = \frac{1}{12} \) (Equation 2)

### Step 4: Solve the Equations
From Equation 1:
\[ \frac{D}{5} - \frac{D}{S} = \frac{1}{6} \]
Rearranging gives:
\[ \frac{D}{S} = \frac{D}{5} - \frac{1}{6} \]
Multiplying through by \( 30S \) (to eliminate the denominators):
\[ 6D - 5D = 5S \cdot \frac{1}{6} \cdot 30 \]
\[ D = 5S \]

From Equation 2:
\[ \frac{D}{6} - \frac{D}{S} = \frac{1}{12} \]
Rearranging gives:
\[ \frac{D}{S} = \frac{D}{6} - \frac{1}{12} \]
Multiplying through by \( 12S \):
\[ 2D - 2D = S \cdot \frac{1}{12} \cdot 12 \]
\[ D = 2S \]

### Step 5: Equate the Two Expressions for D
From the two expressions we derived:
1. \( D = 5S \)
2. \( D = 2S \)

Setting them equal:
\[ 5S = 2S \]
This gives us:
\[ 3S = 0 \implies S = 0 \]

### Step 6: Substitute Back to Find the Original Speed
We need to find the speed \( S \) that allows the student to reach on time.

Using \( D = 5S \) in \( \frac{D}{S} = \frac{D}{5} - \frac{1}{6} \):
Substituting \( D = 5S \):
\[ \frac{5S}{S} = \frac{5S}{5} - \frac{1}{6} \]
\[ 5 = S - \frac{1}{6} \]
Multiplying through by 6:
\[ 30 = 6S - 1 \]
\[ 6S = 31 \implies S = \frac{31}{6} \approx 5.17 \text{ km/hr} \]

### Final Answer
The original speed to reach school on time is approximately **5.17 km/hr**.