Mohan covers a distance of 187 km at the speed of S km/hr. If Mohan increases his speed by 6 km/hr, then he takes 6 hours less. What is the value of S?

Options

A. 12 km/hr
B. 14 km/hr
C. 11 km/hr
D. 13 km/hr

Answer

To solve the problem step by step, we need to find the speed \( S \) at which Mohan covers a distance of 187 km. The steps are as follows:

### Step 1: Define the Variables
Let:
- Distance \( D = 187 \) km
- Speed \( S = s \) km/hr
- Increased Speed \( S + 6 = s + 6 \) km/hr

### Step 2: Write the Time Equations
The time taken to cover the distance at speed \( S \) is given by:
\[ t = \frac{D}{S} = \frac{187}{s} \]
When the speed is increased by 6 km/hr, the time taken becomes:
\[ t - 6 = \frac{D}{S + 6} = \frac{187}{s + 6} \]

### Step 3: Set Up the Equation
From the above, we can set up the equation:
\[ \frac{187}{s} - 6 = \frac{187}{s + 6} \]

### Step 4: Clear the Fractions
To eliminate the fractions, multiply through by \( s(s + 6) \):
\[ 187(s + 6) - 6s(s + 6) = 187s \]

### Step 5: Expand and Simplify
Expanding both sides gives:
\[ 187s + 1122 - 6s^2 - 36s = 187s \]
Now, simplify:
\[ 1122 - 6s^2 - 36s = 0 \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ 6s^2 + 36s - 1122 = 0 \]

### Step 7: Divide by 6
To simplify, divide the entire equation by 6:
\[ s^2 + 6s - 187 = 0 \]

### Step 8: Factor the Quadratic Equation
Now we need to factor the quadratic equation. We look for two numbers that multiply to \(-187\) and add to \(6\):
\[ (s - 11)(s + 17) = 0 \]

### Step 9: Solve for \( s \)
Setting each factor to zero gives:
1. \( s - 11 = 0 \) → \( s = 11 \)
2. \( s + 17 = 0 \) → \( s = -17 \) (not valid since speed cannot be negative)

### Conclusion
Thus, the only valid solution is:
\[ s = 11 \text{ km/hr} \]