Radha and Shyam start travelling from Delhi to Goa at the speed of 15 km/h and 12 km/h respectively. After half an hour Meera also leaves from Delhi to Goa. After same time Meera crosses Shyam and 90 minutes further on Meera crosses Radha. Find the speed of Meera.

Answer

To solve the problem step-by-step, we will break down the information given and use the relationships between distance, speed, and time.

### Step 1: Understand the Problem
Radha travels at 15 km/h, Shyam at 12 km/h, and Meera starts half an hour later. We need to find Meera's speed.

### Step 2: Define Variables
Let the speed of Meera be \( x \) km/h.

### Step 3: Calculate Distances at Meeting Points
1. **When Meera meets Shyam:**
- Shyam travels for \( t \) hours.
- Distance covered by Shyam: \( \text{Distance} = \text{Speed} \times \text{Time} = 12t \) km.
- Meera starts half an hour later, so she travels for \( t - 0.5 \) hours.
- Distance covered by Meera: \( \text{Distance} = x(t - 0.5) \) km.
- At the meeting point: \( 12t = x(t - 0.5) \).

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ 12t = xt - 0.5x \]
\[ xt - 12t = 0.5x \]
\[ t(x - 12) = 0.5x \]
\[ t = \frac{0.5x}{x - 12} \quad \text{(1)} \]

### Step 5: When Meera meets Radha
2. **When Meera meets Radha:**
- After meeting Shyam, Meera travels for another 90 minutes (1.5 hours).
- Total time for Radha until they meet: \( t + 1.5 \) hours.
- Distance covered by Radha: \( \text{Distance} = 15(t + 1.5) \) km.
- Distance covered by Meera: \( x(t - 0.5 + 1.5) = x(t + 1) \) km.
- At this meeting point: \( 15(t + 1.5) = x(t + 1) \).

### Step 6: Rearranging the Second Equation
Rearranging gives:
\[ 15t + 22.5 = xt + x \]
\[ xt - 15t = 22.5 - x \]
\[ t(x - 15) = 22.5 - x \]
\[ t = \frac{22.5 - x}{x - 15} \quad \text{(2)} \]

### Step 7: Equate the Two Expressions for t
From equations (1) and (2):
\[ \frac{0.5x}{x - 12} = \frac{22.5 - x}{x - 15} \]

### Step 8: Cross-Multiply and Solve
Cross-multiplying gives:
\[ 0.5x(x - 15) = (22.5 - x)(x - 12) \]
Expanding both sides:
\[ 0.5x^2 - 7.5x = 22.5x - 270 - x^2 + 12x \]
Combining like terms:
\[ 0.5x^2 - 7.5x + x^2 - 34.5x + 270 = 0 \]
\[ 1.5x^2 - 42x + 270 = 0 \]

### Step 9: Simplify the Quadratic Equation
Dividing the entire equation by 1.5:
\[ x^2 - 28x + 180 = 0 \]

### Step 10: Solve the Quadratic Equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -28, c = 180 \):
\[ x = \frac{28 \pm \sqrt{(-28)^2 - 4 \cdot 1 \cdot 180}}{2 \cdot 1} \]
\[ x = \frac{28 \pm \sqrt{784 - 720}}{2} \]
\[ x = \frac{28 \pm \sqrt{64}}{2} \]
\[ x = \frac{28 \pm 8}{2} \]
Thus, \( x = 18 \) or \( x = 10 \).

### Step 11: Determine the Valid Speed
Since Meera must be faster than both Radha and Shyam, we take \( x = 18 \) km/h as the valid speed.

### Final Answer
The speed of Meera is **18 km/h**.