Two buses leave a station at the same time and travel in opposite directions. One bus travels 113 mi/h faster than the other. If the two buses are 333 miles apart after 3 hours, what is the rate of each bus?

Let's denote the speed of the slower bus as \( x \) miles per hour. Therefore, the speed of the faster bus would be \( x + 113 \) miles per hour, since it travels 113 miles per hour faster than the slower bus.

Since the buses are traveling in opposite directions, their relative speed is the sum of their individual speeds. After 3 hours, the distance between them is 333 miles, so we can set up the following equation:

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

The distance between the two buses after 3 hours is given by:

\[
333 = 3(x + (x + 113))
\]

Simplifying the equation:

\[
333 = 3(2x + 113)
\]

\[
333 = 6x + 339
\]

Now, subtract 339 from both sides:

\[
333 - 339 = 6x
\]

\[
-6 = 6x
\]

Finally, divide both sides by 6:

\[
x = -1
\]

However, this negative result suggests that there was a mistake in interpreting the problem or setting up the equation. Let's review the equation setup carefully:

The correct interpretation should be:

The buses together cover 333 miles in 3 hours, so the sum of their speeds multiplied by 3 hours equals 333 miles:

\[
333 = 3(2x + 113)
\]

This simplifies correctly as:

\[
333 = 6x + 339
\]

Reevaluating the problem step by step confirms that there seems to be an error that needs review.

Would you like to discuss this further or explore a different approach?

Discover how to solve a classic distance-speed-time problem where two buses leave a station at the same time, travel in opposite directions, and are 333 miles apart after 3 hours. This problem involves key algebraic concepts and calculations of relative speed. Suitable for students in Grades 7-10.

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