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Speed and Distance Practice Questions

Practice Speed and Distance questions with worked explanations and timing guidance for Quantitative Reasoning.

Five-question preview. Answer 3 now without an account.
Question 1 of 3 free Medium

Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home?

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Correct answer: C

Let \(b\) be the number of hours Bob spends biking. Then \((t - b)\) is the number of hours he spends walking. Let \(d\) be the distance in miles from his home to school.



Since he had the flat tire halfway to school, he biked \(\frac{d}{2}\) miles and he walked \(\frac{d}{2}\) miles.



Now we can set up the equations using the formula rate × time = distance. Remember that we want to solve for \(d\), the total distance from Bob's home to school.



1) \(xb = \frac{d}{2}\)



2) \(y(t - b) = \frac{d}{2}\)



Solving equation 1) for \(b\) gives us:



3) \(b = \frac{d}{2x}\)



Substituting this value of \(b\) into equation 2 gives:



4) \(y\left(t - \frac{d}{2x}\right) = \frac{d}{2}\)



Multiply both sides by \(2x\):



5) \(2xy\left(t - \frac{d}{2x}\right) = dx\)



Distribute the \(2xy\)



6) \(2xyt - dy = dx\)



7) \(2xyt = dx + dy\) Add \(dy\) to both sides to collect the \(d\)'s on one side.



8) \(2xyt = d(x + y)\) Factor out the \(d\)



9) \(\frac{2xyt}{(x + y)} = d\) Divide both sides by \((x + y)\) to solve for \(d\)



The correct answer is C.
Question 2 of 3 free Medium

Lexy walks 5 miles from point A to point B in one hour, then bicycles back to point A along the same route at 15 miles per hour. Ben makes the same round trip, but does so at half of Lexy’s average speed. How many minutes does Ben spend on his round trip?

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Correct answer: D

We begin by figuring out Lexy's average speed. On her way from A to B, she travels 5 miles in one hour, so her speed is 5 miles per hour. On her way back from B to A, she travels the same 5 miles at 15 miles per hour. Her average speed for the round trip is NOT simply the average of these two speeds. Rather, her average speed must be computed using the formula RT = D, where R is rate, T is time and D is distance. Her average speed for the whole trip is the total distance of her trip divided by the total time of her trip.

We already know that she spends 1 hour going from A to B. When she returns from B to A, Lexy travels 5 miles at a rate of 15 miles per hour, so our formula tells us that 15T = 5, or T = 1/3. In other words, it only takes Lexy 1/3 of an hour, or 20 minutes, to return from B to A. Her total distance traveled for the round trip is 5+5=10 miles and her total time is 1+1/3=4/3 of an hour, or 80 minutes.

We have to give our final answer in minutes, so it makes sense to find Lexy's average rate in miles per minute, rather than miles per hour. 10 miles / 80 minutes = 1/8 miles per minute. This is Lexy's average rate.

We are told that Ben's rate is half of Lexy's, so he must be traveling at 1/16 miles per minute. He also travels a total of 10 miles, so (1/16)T = 10, or T = 160. Ben's round trip takes 160 minutes.

The correct answer is D.
Question 3 of 3 free Medium

Triathlete Dan runs along a 2-mile stretch of river and then swims back along the same route. If Dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour, what is his average rate for the entire trip in miles per minute?

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Correct answer: A

There is an important key to answering this question correctly: this is not a simple average problem but a weighted average problem. A weighted average is one in which the different parts to be averaged are not equally balanced. One is "worth more" than the other and skews the "simple" average in one direction. In addition, we must note a unit change in this problem: we are given rates in miles per hour but asked to solve for rates in miles per minute.

Average rate uses the same D = RT formula we use for rate problems but we have to figure out the different lengths of time it takes Dan to run and swim along the total 4-mile route. Then we have to take the 4 miles and divide by that total time.

First, Dan runs 2 miles at the rate of 10 miles per hour. 10 miles per hour is equivalent to 1 mile every 6 minutes, so Dan takes 12 minutes to run the 2 miles.
Next, Dan swims 2 miles at the rate of 6 miles per hour. 6 miles per hour is equivalent to 1 mile every 10 minutes, so Dan takes 20 minutes to swim the two miles.

Dan's total time is 12 + 20 = 32 minutes. Dan's total distance is 4 miles. Distance/time = 4 miles / 32 minutes = 1/8 miles per minute.

Note that if you do not weight the averages but merely take a simple average, you will get 2/15, which corresponds to incorrect answer choice B. 6 mph and 10 mph average to 8mph. (8mph)(1h/60min) = 8/60 miles/minute or 2/15 miles per minute.

The correct answer is A.
Question 4 of 5 preview Medium

Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover half of the distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

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Question 5 of 5 preview Medium

It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have travelled than the regular train when the two trains pass each other?

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