The ratio between the rates of travelling of A and B is 2:3 and therefore A takes 10 min more than the time taken by B to reach a destination if A had walked at double the speed he would have covered the distance is

Options

A. 30min
B. 25min
C. 15min
D. 20min

Answer

To solve the problem, let's break it down step by step.

### Step 1: Understand the given information
We know that the ratio of the speeds of A and B is 2:3. This means if A's speed is 2x, then B's speed is 3x for some value of x.

### Step 2: Set up the time relationship
Since A takes 10 minutes more than B to reach the destination, we can express the time taken by A and B in terms of distance (D) and speed (S).

- Time taken by A = Distance / Speed of A = D / (2x)
- Time taken by B = Distance / Speed of B = D / (3x)

From the problem, we know:
\[ \frac{D}{2x} = \frac{D}{3x} + 10 \]

### Step 3: Solve for D
To eliminate D from the equation, we can multiply through by \(6x\) (the least common multiple of the denominators):
\[ 6x \cdot \frac{D}{2x} = 6x \cdot \frac{D}{3x} + 6x \cdot 10 \]
This simplifies to:
\[ 3D = 2D + 60x \]
Now, subtract \(2D\) from both sides:
\[ 3D - 2D = 60x \]
\[ D = 60x \]

### Step 4: Find the time taken by B
Now that we have D, we can find the time taken by B:
\[ \text{Time taken by B} = \frac{D}{3x} = \frac{60x}{3x} = 20 \text{ minutes} \]

### Step 5: Find the time taken by A
Since A takes 10 minutes more than B:
\[ \text{Time taken by A} = 20 + 10 = 30 \text{ minutes} \]

### Step 6: Calculate the distance if A walked at double speed
If A had walked at double the speed, his speed would be \(4x\). The time taken by A at double speed would be:
\[ \text{Time} = \frac{D}{4x} = \frac{60x}{4x} = 15 \text{ minutes} \]

### Step 7: Conclusion
Thus, the distance covered by A (and B) is:
\[ D = 60x \]
Since we don't have the value of x, we can conclude that the distance is 60 times whatever x is. However, we can express the distance in terms of time and speed.

### Final Answer
The distance is \(60x\) where \(x\) is a unit of speed.