Manoj takes twice the time to cover a distance 'D' km than time taken by Shreya to cover 2D km. Manoj started from his home & after 30 min, Shreya started from his house but she catched him after travelling for 20/3 km. Find speed of Shreya. (speed is considered in kmph)

Options

A. A)40 kmph
B. B)13.33 kmph
C. C)28.5 kmph
D. D)17.77 kmph

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: Understanding the relationship between Manoj and Shreya's speeds
Manoj takes twice the time to cover a distance 'D' km compared to the time taken by Shreya to cover 2D km.

Let:
- Speed of Manoj = \( V_m \)
- Speed of Shreya = \( V_s \)

From the information given, we can set up the relationship:
\[ \text{Time taken by Manoj} = \frac{D}{V_m} \]
\[ \text{Time taken by Shreya} = \frac{2D}{V_s} \]
According to the problem, Manoj takes twice the time of Shreya:
\[ \frac{D}{V_m} = 2 \times \frac{2D}{V_s} \]
This simplifies to:
\[ \frac{D}{V_m} = \frac{4D}{V_s} \]
Cross-multiplying gives us:
\[ V_s = 4V_m \]

### Step 2: Setting up the time equations
Manoj starts from home and after 30 minutes (which is \( \frac{1}{2} \) hours), Shreya starts. Let \( t \) be the time (in hours) that Shreya travels until she catches Manoj.

During the first 30 minutes, Manoj travels:
\[ \text{Distance covered by Manoj in 30 minutes} = V_m \times \frac{1}{2} \]

After 30 minutes, Manoj continues to travel for an additional \( t \) hours, covering:
\[ \text{Additional distance by Manoj} = V_m \times t \]

So, the total distance covered by Manoj when Shreya catches him is:
\[ \text{Total distance by Manoj} = V_m \times \frac{1}{2} + V_m \times t = V_m \left( \frac{1}{2} + t \right) \]

### Step 3: Distance covered by Shreya
Shreya travels a distance of \( \frac{20}{3} \) km in time \( t \):
\[ \text{Distance covered by Shreya} = V_s \times t \]

### Step 4: Setting the distances equal
Since Shreya catches Manoj, the distances they cover are equal:
\[ V_m \left( \frac{1}{2} + t \right) = V_s \times t \]
Substituting \( V_s = 4V_m \):
\[ V_m \left( \frac{1}{2} + t \right) = 4V_m \times t \]

### Step 5: Simplifying the equation
Dividing both sides by \( V_m \) (assuming \( V_m \neq 0 \)):
\[ \frac{1}{2} + t = 4t \]
Rearranging gives:
\[ \frac{1}{2} = 4t - t \]
\[ \frac{1}{2} = 3t \]
Thus:
\[ t = \frac{1}{6} \text{ hours} \]

### Step 6: Finding Shreya's speed
Now we can find Shreya's speed using the distance she covered:
\[ \text{Distance} = \frac{20}{3} \text{ km} \]
Using the formula for speed:
\[ V_s = \frac{\text{Distance}}{\text{Time}} = \frac{\frac{20}{3}}{\frac{1}{6}} = \frac{20}{3} \times 6 = 40 \text{ km/h} \]

### Final Answer
The speed of Shreya is **40 km/h**.

---