A car travels a distance of 300 km at a uniform speed. If the speed of the car is 5 km/h more, it takes 2 h less to cover the same distance. Find the original speed of the car.

Correct option is B

Given:

Distance traveled = 300 km

If the speed increases by 5 km/h, the time taken to cover the same distance decreases by 2 hours.

Formula Used:

Time =$$\frac{\text{Distance}}{\text{Speed}}$$​

Solution:

Let the original speed of the car be x km/h.

For the original speed x, the time taken is:

​$$\text{Time}_{\text{original}} = \frac{300}{x}$$​

For the increased speed x + 5, the time taken is:

​$$\text{Time}_{\text{new}} = \frac{300}{x + 5}$$​

We are told that the difference in time is 2 hours:

​$$\text{Time}_{\text{original}} - \text{Time}_{\text{new}} = 2$$​​

Using the formula for time, substitute the values into the equation:

​$$\frac{300}{x} - \frac{300}{x + 5} = 2$$​

​$$\frac{300(x + 5) - 300x}{x(x + 5)} = 2$$​​

​$$\frac{300x + 1500 - 300x}{x(x + 5)} = 2$$​

​$$\frac{1500}{x(x + 5)} = 2$$​

​$$1500 = 2x^2 + 10x$$​

​$$2x^2 + 10x - 1500 = 0$$​

​$$x^2 + 5x - 750 = 0$$​

x =$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$​

Here, a = 1, b = 5, and c = -750. Substituting the values:

​$$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-750)}}{2(1)}$$​​

​$$x = \frac{-5 \pm \sqrt{25 + 3000}}{2}$$​

​$$x = \frac{-5 \pm \sqrt{3025}}{2}$$​

​$$x = \frac{-5 \pm 55}{2}$$​

​$$x = \frac{-5 + 55}{2} = \frac{50}{2} = 25$$​

​$$x = \frac{-5 - 55}{2} = \frac{-60}{2} = -30$$​

Since speed cannot be negative, the original speed is x = 25 km/h.

Alternate Method:

D = $$\frac{S_1S_2}{S_1-S_2}\times t$$ 

300 = $$\frac{x(x+5)}{5}\times 2$$

750 = x (x + 5)

Above equation satisfied only x = 25 km/h.