X takes 3 hours more than Y to walk 30 km. But if X doubles his pace, he is ahead of Y by `1(1)/(2)` hours. The speed of X is

Options

A. 5km/hr
B. 9km/hr
C. 9/7 km/hr
D. 10/3 km/hr

Answer

To solve the problem, we will define the variables and set up equations based on the information given in the question.

### Step 1: Define Variables
Let:
- \( v_x \) = speed of X (in km/h)
- \( v_y \) = speed of Y (in km/h)

### Step 2: Set Up the First Equation
According to the problem, X takes 3 hours more than Y to walk 30 km. The time taken by X to walk 30 km is given by:
\[ \text{Time taken by X} = \frac{30}{v_x} \]
The time taken by Y to walk 30 km is:
\[ \text{Time taken by Y} = \frac{30}{v_y} \]
From the information given, we can write:
\[ \frac{30}{v_x} = \frac{30}{v_y} + 3 \]

### Step 3: Set Up the Second Equation
If X doubles his pace, his new speed becomes \( 2v_x \). The problem states that at this new speed, X is ahead of Y by \( 1 \frac{1}{2} \) hours, which is \( \frac{3}{2} \) hours. Thus, we can write:
\[ \frac{30}{2v_x} = \frac{30}{v_y} - \frac{3}{2} \]

### Step 4: Solve the First Equation
Rearranging the first equation:
\[ \frac{30}{v_x} - \frac{30}{v_y} = 3 \]
Multiplying through by \( v_x v_y \):
\[ 30v_y - 30v_x = 3v_x v_y \]
This simplifies to:
\[ 3v_x v_y = 30(v_y - v_x) \]
Dividing through by 3:
\[ v_x v_y = 10(v_y - v_x) \quad \text{(1)} \]

### Step 5: Solve the Second Equation
Rearranging the second equation:
\[ \frac{30}{2v_x} + \frac{3}{2} = \frac{30}{v_y} \]
Multiplying through by \( 2v_x v_y \):
\[ 30v_y + 3v_x v_y = 60v_x \]
This simplifies to:
\[ 3v_x v_y = 60v_x - 30v_y \]
Dividing through by 3:
\[ v_x v_y = 20v_x - 10v_y \quad \text{(2)} \]

### Step 6: Solve the System of Equations
Now we have two equations:
1. \( v_x v_y = 10(v_y - v_x) \)
2. \( v_x v_y = 20v_x - 10v_y \)

Setting the right-hand sides equal to each other:
\[ 10(v_y - v_x) = 20v_x - 10v_y \]
Expanding and rearranging gives:
\[ 10v_y - 10v_x = 20v_x - 10v_y \]
Combining like terms:
\[ 20v_y = 30v_x \]
Thus:
\[ v_y = \frac{3}{2}v_x \quad \text{(3)} \]

### Step 7: Substitute Equation (3) into Equation (1)
Substituting \( v_y \) from equation (3) into equation (1):
\[ v_x \left(\frac{3}{2}v_x\right) = 10\left(\frac{3}{2}v_x - v_x\right) \]
This simplifies to:
\[ \frac{3}{2}v_x^2 = 10\left(\frac{1}{2}v_x\right) \]
\[ \frac{3}{2}v_x^2 = 5v_x \]
Dividing both sides by \( v_x \) (assuming \( v_x \neq 0 \)):
\[ \frac{3}{2}v_x = 5 \]
Thus:
\[ v_x = \frac{5 \times 2}{3} = \frac{10}{3} \text{ km/h} \]

### Conclusion
The speed of X is \( \frac{10}{3} \) km/h.