Speed and Distance Practice Questions

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.<br><br>

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

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.<br><br>

1) \(xb = \frac{d}{2}\)<br><br>

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

Solving equation 1) for \(b\) gives us:<br><br>

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

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

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

Multiply both sides by \(2x\):<br><br>

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

Distribute the \(2xy\)<br><br>

6) \(2xyt - dy = dx\)<br><br>

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

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

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

The correct answer is C.

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.<br><br>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.<br><br>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.<br><br>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.<br><br>The correct answer is D.

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.<br><br>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.<br><br>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.<br>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.<br><br>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.<br><br>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.<br><br>The correct answer is A.