In a previous post, we took a look at two different types of “averages” you might see on the ACT Math Test, one of which was a fairly rare challenging concept known as “average speed. Often found in complex word problems, this type of question is one many students are less familiar with so don’t get nervous if you don’t know how to approach it yet! Let’s try a couple practice questions to give ourselves an ACT Math challenge!

But first, remember that the formula for Average Speed = Total Distance / Total Time. It might help to remember it as **AS = D / T**, or come up with your own fun mnemonic device!

*Suzanne drove 40 miles to see her aunt and was going 20 mph. It took her 2 hours to get to her aunt’s house. Then, she left and drove another 30 miles to the pet store, but this time only drove at 10 mph. If it took Suzanne 3 hours to arrive at the pet store, what was her average speed in miles per hour for the entire car ride from her home to the pet store?** *

*(A) 10*

*(B) 11*

*(C) 12*

*(D) 14*

*(E) 15*

Average Speed = Total Distance / Total Time.

40 miles + 30 miles so the Total Distance was 70 miles. Suzanne drove for 2 hours + 3 hours so the Total Time was 5 hours. 70/5 = 14.

The Average Speed for the whole trip was 14 mph. The correct answer is (D).

Notice how the test-maker has made this problem tricky! The average speed in this problem is 14 mph, which is *different* from simply taking the mean of the two speeds. If we had just averaged the two speeds (10mph and 20mph) we would have gotten 15mph. Average Speed is a weighted average. Since Suzanne spent more time in the problem going 10mph than 20mph, it makes sense that the Average Speed would be closer to 10mph.

Let’s try another word problem!

*Students at Thomas Jefferson High School boarded the bus for a field trip that went 15mph through a 30 mile section of the city. The bus then stopped for lunch in a suburb before continuing on a 3 hour tour of countryside at a constant speed of 10mph. Finally, the bus drove 40 miles straight back to the high school. If the trip back to Thomas Jefferson High school took two hours, approximately what was the average speed for the entire field trip?*

*(A) 11*

*(B) 12*

*(C) 13*

*(D) 14*

*(E) 15*

To find the average speed of the bus, we know we will need to find the Total Distance and the Total Time, so we can start by using another formula (**Distance = Rate x Time**) to help us find the pieces we’re missing for each part of the trip.

For the first part of the field trip: 30 miles = 15mph x T, so we know that T = 2 hours. For the middle part of the trip, we know that D = 10mph x 3 hours, so we know that D = 30 miles. For the last part of the trip, we know that 40 miles = R x 2 hours, so we know that R = 20mph. Now we can find the Total Distance and the Total Time:

Total Distance = 30 miles + 30 miles + 40miles = 100 miles.

Total Time = 2 hours + 3 hours + 2 hours = 7 hours.

The Average Speed = 100 miles/ 7 hours = 14.28mph. Since the question used the word “approximately,” we can round to the nearest integer: 14. The correct answer is (D).

On your 2nd problem, I’m struggling to understand the wording of the last sentence: “students arrived back … three hours after they arrived. Is this a misprint? Or is there some calculation that I’m missing. I noticed in your answer you had the 40 miles taking 2 hours instead of 3.

Hi there,

You’re right, there’s something funky about this question. I’m not sure what the original intent was, but I’m going to send this to our blog team so that they can fix it. Thanks for pointing it out to us 🙂