## IM Commentary

For sixth grade, this is presented as a series of problems leading up to the last one. This last question is appropriate without scaffolding for 7th grade; see "7.RP.3 Friends Meeting on Bikes."

Most students should be able to answer the first two questions without too much difficulty. The decimal numbers may cause some students trouble, but if they make a drawing of the road that the girls are riding on, and their positions at the different times, it may help.

The third question has a bit of a challenge in that students won't land on the exact meeting time by making a table with distance values every hour.

The fourth question addresses a useful concept for problems involving objects moving at different speeds which may be new to sixth grade students. This question provides one way to answer the next one, but not the only way. This question also addresses Standard for Mathematical Practice 3.

The story context is intended to make the problem more interesting to
students, but it can also serve several mathematical purposes. A
student who doesn't know where to start on a problem like this can
guess or estimate using the story as a guide, and students who have
found an answer can check it to see if it makes sense in the story.
For example, Anya's speed in the second ride should be greater than in
the first. Also, in comparing the two bike rides in July and in
September, students can recognize that Taylor's speed doesn't change
but Anya's does, and this changes the meeting time.

## Solutions

Solution:
Problem a, first method.

They started 63 miles apart, so $63 - 5.5 - 12.5 = 45$ miles.

Solution:
Problem a, second method.

In the first hour, Anya went 5.5 miles and Taylor went 12.5 miles, so
the two of them together went 18 miles, and $63 - 18 = 45$ miles.

Solution:
Problem b, first method.

The table should show 63 miles at 0 hours, 45 miles at
1 hour, 27 miles at 2 hours, and 9 miles at 3 hours. Some of the
students will probably recognize that the distance is decreasing by 18 miles
each time, and this fact can be brought up in class discussion.

Solution:
Problem b, second method.

Here are a couple of ways to offer extra scaffolding to students who might need it.

Draw a line representing the route that Taylor and Anya will take, with their starting points labeled as being 63 miles apart. Then mark Taylor's position at 0, 1, 2 and 3 hours, and likewise for Tanya's.

Make a table with four columns instead of just two. The first column would be time in hours, then how far Anya has traveled, then how far Taylor has traveled, and finally the distance between them.

Solution:
Problem c, first method.

First, students should recognize from the table that the friends are 9 miles
apart after three hours, and therefore they will meet in less than an hour.
With a little more thought and discussion, they should recognize that 9 is
half of 18, and so they will meet in 3.5 hours, or at 11:30 am.

If this 3.5 hour solution isn't clear, then one approach would be to re-do
the table from Problem 2, but this time use half-hour increments instead of
hour increments.

Solution:
Problem c, second method.

Since the friends are moving toward each other at 18 miles per hour,
we can write $63 / 18 = 3.5$ hours, so they will meet at 11:30.

Solution:
Problem d

What Taylor really means is that the distance between them is decreasing by 18 miles every hour, so the amount of time it will take them to meet is the same as if one person stays put and the other rides at 18 miles per hour. However, the place they meet will not be the same.

Solution:
Problem e

There are two shorter solutions listed for this problem
in the 7th grade version. A student who doesn't see how to do the
shorter versions might go back and make a table, as in Problem b.
At 0 hours the friends are 63 miles apart, and at 3 hours they are
0 miles apart. Using this, a student could figure out the missing
entries, and then reason that the friends are getting closer at
21 miles per hour. Since Taylor is riding 12.5 miles per hour, Anya
must be riding 8.5 miles per hour. Adding the two extra columns
described in the second solution of Problem b might also help.

## Comments

Log in to comment## Robert Kaplinsky says:

over 6 yearsI think that the solution for Problem D is actually the solution for Problem E. For Problem D, I would hope a discussion brings out that what is "the same" is that the two girls would meet at the same time but not at "the same" location.

## Kristin says:

over 6 yearsThanks for catching that--I'm not sure what happened to the solution for that question. I've added a solution in.