The distance of the college and the home of rajeev is 80 km. One day he was late by 1 hour than the normal time to leave for the college do he increased his speed by 4 km/h . And thus he reached to college at the normal time.What is the changed speed of rajeev?

Options

A. 28 km/h
B. 30 km/h
C. 40 km/h
D. 20 km/h

Answer

To solve the problem step by step, we will use the information provided in the question and apply the concepts of speed, time, and distance.

### Step 1: Define the Variables
Let:
- \( S \) = normal speed of Rajeev (in km/h)
- \( T \) = normal time taken to reach college (in hours)

### Step 2: Establish the Distance Equation
The distance from Rajeev's home to college is given as 80 km. Therefore, we can express the distance in terms of speed and time:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
So, we have:
\[ 80 = S \times T \]
This can be rearranged to:
\[ S = \frac{80}{T} \]

### Step 3: Analyze the Late Departure
Rajeev was late by 1 hour, so he left at \( T + 1 \) hours. He increased his speed by 4 km/h, making his new speed \( S + 4 \) km/h. He still covered the same distance of 80 km in the normal time \( T \):
\[ 80 = (S + 4) \times T \]

### Step 4: Set Up the Equation
Now we have two equations:
1. \( 80 = S \times T \)
2. \( 80 = (S + 4) \times T \)

### Step 5: Substitute and Simplify
From the first equation, we know \( S = \frac{80}{T} \). Substitute this into the second equation:
\[ 80 = \left(\frac{80}{T} + 4\right) \times T \]
Expanding this gives:
\[ 80 = 80 + 4T \]
Subtracting 80 from both sides leads to:
\[ 0 = 4T \]
This indicates that we need to re-arrange our equations properly.

### Step 6: Rearranging the Second Equation
From the second equation:
\[ 80 = (S + 4) \times T \]
Expanding gives:
\[ 80 = ST + 4T \]
Now, substituting \( S = \frac{80}{T} \) into this equation:
\[ 80 = \frac{80}{T} \times T + 4T \]
This simplifies to:
\[ 80 = 80 + 4T \]
Subtracting 80 from both sides gives:
\[ 0 = 4T \]
This means we need to find \( T \) in terms of \( S \).

### Step 7: Equate the Two Distances
From both equations:
1. \( S \times T = 80 \)
2. \( (S + 4) \times (T - 1) = 80 \)

Expanding the second equation:
\[ ST - S + 4T - 4 = 80 \]
Substituting \( ST = 80 \):
\[ 80 - S + 4T - 4 = 80 \]
This simplifies to:
\[ -S + 4T - 4 = 0 \]
Thus:
\[ S = 4T - 4 \]

### Step 8: Substitute Back
Now, substitute \( S = 4T - 4 \) into \( S \times T = 80 \):
\[ (4T - 4) \times T = 80 \]
Expanding gives:
\[ 4T^2 - 4T - 80 = 0 \]
Dividing the entire equation by 4:
\[ T^2 - T - 20 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -1, c = -20 \):
\[ T = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1} \]
\[ T = \frac{1 \pm \sqrt{1 + 80}}{2} \]
\[ T = \frac{1 \pm \sqrt{81}}{2} \]
\[ T = \frac{1 \pm 9}{2} \]
This gives us:
\[ T = 5 \quad \text{(since time cannot be negative)} \]

### Step 10: Find Normal Speed
Substituting \( T = 5 \) back into \( S = 4T - 4 \):
\[ S = 4(5) - 4 = 20 - 4 = 16 \text{ km/h} \]

### Step 11: Find Changed Speed
The changed speed is:
\[ S + 4 = 16 + 4 = 20 \text{ km/h} \]

### Final Answer
The changed speed of Rajeev is **20 km/h**.

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