For distance word problems, it is important to remember the formula for speed:

**Definition: **\(Speed = \Large \frac{{Distance}}{{Time}}\)

We can use this definition to solve different types of problems. Let’s jump straight to an example:

**Example:** Huong drove to the ferry office and back. The trip there took three hours and the trip back took four hours. She averaged 10 mph faster on the trip there than on the return trip. What was Huong’s average speed on the outbound trip?

**Solution:** First make sure you understand which of the two trips was the outbound trip. The outbound trip was the *first* trip Houng made, when she was travelling toward the ferry office. This trip took three hours, and the speed of this trip is the speed we are concerned with in this problem. We will denote Huong’s speed on the outbound trip as \(S\). Now we can set up an equation to model the outbound trip.

\(S = \Large \frac{{Distance Travelled}}{{3 hours}}\)

On the return trip, Houng averaged 10 mph *less* than on the outbound trip. But the distance travelled remained the same. So we can write, for the return trip:

\(S - 10 = \Large \frac{{Distance Travelled}}{{4 hours}}\)

Now we have two equations, but it *appears* that we have two variables. But let’s simplify both of these equations by multiplying them by values that will eliminate the denominators and clean things up. We have

\(\left( {3 hours} \right)\left( S \right) = Distance Travelled\)

and

\(\left( {4 hours} \right)\left( {S - 10} \right) = Distance Travelled\)

Do you see where this is going? We have two equations which represent the distance travelled. But since the distance travelled is the same in both cases, we can conclude that the left sides of the equations equal each other as well! That is,

\(\left( {3 hours} \right)\left( S \right) = (4 hours)(S - 10)\)

Now we have an equation with only one unknown value, the outbound speed! This is what we want to find anyway. Let’s solve for \(S\).

\({\text{}}3S = 4S - 40\)

\(S = 40\) *miles per hour*

**Example:** An Air Force plane left Singapore and flew toward the maintenance facility at an average speed of 150 mph. A cargo plane left some time later flying in the same direction at an average speed of 180 mph. After flying for five hours the cargo plane caught up with the Air Force plane. How long did the Air Force plane fly before the cargo plane caught up?

**Solution:** For the Air Force Plane, we have

\(Speed = \Large \frac{{Distance}}{{Time}}\)

so that

\(Distance = Speed \cdot Time\)

\(Distance = 150 mph \cdot x hours\)

For the cargo plane, we have

\(Distance = Speed \cdot Time\)

\(Distance = 180 mph \cdot 5 hrs\)

Since at the moment they caught up, their distance travelled would be equal, we can conclude that

\(150x = 180 \cdot 5\)

\(150x = 900\)

\(x = 6 hours\)

That is, the Air Force Plane flew for six hours before the cargo plane caught up.

Below you can **download** some** free** math worksheets and practice.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**