A car travels a distance of 840 km at a uniform speed if speed is 10 km/hr more it take two hour less original speed of car is

Options

A. A)45 km/hr
B. B)50km/hr
C. C)60km/hr
D. D)75 km/hr

Answer

To solve the problem, we will use the relationship between distance, speed, and time. The formula we will use is:

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

### Step-by-Step Solution:

1. **Define Variables:**
Let the original speed of the car be \( x \) km/hr.

2. **Calculate Time at Original Speed:**
The time taken to travel 840 km at the original speed \( x \) is given by:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{840}{x} \]

3. **Calculate Time at Increased Speed:**
If the speed is increased by 10 km/hr, the new speed becomes \( x + 10 \) km/hr. The time taken to travel the same distance at this new speed is:
\[ \text{Time} = \frac{840}{x + 10} \]

4. **Set Up the Equation:**
According to the problem, the time taken at the increased speed is 2 hours less than the time taken at the original speed. Therefore, we can set up the equation:
\[ \frac{840}{x} - \frac{840}{x + 10} = 2 \]

5. **Solve the Equation:**
To solve the equation, first find a common denominator:
\[ \frac{840(x + 10) - 840x}{x(x + 10)} = 2 \]
Simplifying the numerator:
\[ \frac{840x + 8400 - 840x}{x(x + 10)} = 2 \]
This simplifies to:
\[ \frac{8400}{x(x + 10)} = 2 \]

6. **Cross-Multiply:**
Cross-multiplying gives:
\[ 8400 = 2x(x + 10) \]

7. **Expand and Rearrange:**
Expanding the right side:
\[ 8400 = 2x^2 + 20x \]
Rearranging gives:
\[ 2x^2 + 20x - 8400 = 0 \]

8. **Divide by 2:**
To simplify, divide the entire equation by 2:
\[ x^2 + 10x - 4200 = 0 \]

9. **Use the Quadratic Formula:**
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 10, c = -4200 \):
\[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-4200)}}{2 \cdot 1} \]
\[ x = \frac{-10 \pm \sqrt{100 + 16800}}{2} \]
\[ x = \frac{-10 \pm \sqrt{16900}}{2} \]
\[ x = \frac{-10 \pm 130}{2} \]

10. **Calculate Possible Values:**
This gives us two possible solutions:
\[ x = \frac{120}{2} = 60 \quad \text{(valid speed)} \]
\[ x = \frac{-140}{2} = -70 \quad \text{(not valid)} \]

11. **Conclusion:**
The original speed of the car is \( \boxed{60} \) km/hr.