A car covers 300 km at some speed. If its speed is increased by 10 kmph, it will cover the same distance in 1 hour less. Find the speed of the car.

Let the speed of the car be \( x \) km/h.

The time taken by the car to cover 300 km at a speed of \( x \) km/h is:

\[
\text{Time} = \frac{300}{x} \, \text{hours}
\]

If the speed is increased by 10 km/h, the new speed becomes \( x + 10 \) km/h. The time taken to cover the same distance at this new speed is:

\[
\text{New Time} = \frac{300}{x + 10} \, \text{hours}
\]

According to the problem, this time is 1 hour less than the original time. So, we set up the equation:

\[
\frac{300}{x} - \frac{300}{x + 10} = 1
\]

Now, solve this equation for \( x \).

First, find a common denominator:

\[
\frac{300(x + 10) - 300x}{x(x + 10)} = 1
\]

Simplifying the numerator:

\[
\frac{300x + 3000 - 300x}{x(x + 10)} = 1
\]

\[
\frac{3000}{x(x + 10)} = 1
\]

Now, multiply both sides by \( x(x + 10) \):

\[
3000 = x(x + 10)
\]

This simplifies to:

\[
3000 = x^2 + 10x
\]

Rearrange it into a standard quadratic equation:

\[
x^2 + 10x - 3000 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( x^2 + 10x - 3000 = 0 \), we have \( a = 1 \), \( b = 10 \), and \( c = -3000 \). Substituting these values into the quadratic formula:

\[
x = \frac{-10 \pm \sqrt{10^2 - 4(1)(-3000)}}{2(1)}
\]

\[
x = \frac{-10 \pm \sqrt{100 + 12000}}{2}
\]

\[
x = \frac{-10 \pm \sqrt{12100}}{2}
\]

\[
x = \frac{-10 \pm 110}{2}
\]

Thus, we have two solutions:

\[
x = \frac{-10 + 110}{2} = \frac{100}{2} = 50
\]
\[
x = \frac{-10 - 110}{2} = \frac{-120}{2} = -60
\]

Since speed cannot be negative, the speed of the car is \( x = 50 \) km/h.

### Final Answer:
The speed of the car is 50 km/h.

Would you like a more detailed explanation of any step?

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Here are 5 related questions:
1. What would the new time be at the increased speed?
2. How would the time change if the distance were 400 km instead of 300 km?
3. Can you solve for the speed if the increase was 20 km/h instead of 10 km/h?
4. How does the quadratic formula work in general?
5. How do we derive the equation for time and speed?

**Tip:** Always check the units when solving distance-speed-time problems to avoid confusion!

Learn how to calculate the speed of a car covering 300 km, given that increasing the speed by 10 km/h reduces travel time by 1 hour. This solution involves solving a quadratic equation and applying distance-time formulas. Suitable for students in Grades 9-10.

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