Someone in this forum previously asked this question but didn't need an answer. I've also submitted this for the live session with Ron but I'm not sure yet if I'll be able to attend or that he'll cover it.

For the the Rate questions types such as "The Kiss" or "The Chase." The answers for 5,14,15 indicate using the table to set up the equations and then either adding the equations or setting them equal to each other.

The original templates say to add/subtract the rates etc. but not if one starts earlier than the other. Is there a way to solve these problems using just the tables? I've gotten comfortable calculating work/travel rates using the table format and while I can use the equations, I'm worried that will trip me up on test day.

Please let me know if you need clarification on the question.

Thank you!


**I wasn't sure if this required a new thread since the previous question wasn't answered.

I would also like to see some tables for these problems, esp. 5 and 14 both of which I did not get right. I have a knack for doing the time relationship in reverse from what the solution has. The chapter covers the kiss and the chase. It also covers time relations but doesn't have an example where both are involved. Any help would be awesome. Thank you.

You'll never be able to use JUST the tables; the tables are only a tool to get you to the final equation you need, whether adding two things, setting them equal, or expressing some other relationship. In the case where two people start at different times you'll need to express the relationship either in the chart somewhere or in the final equation.

Consider problem 5. We'll set up a chart with R,T,D as our columns and Adrienne, James, Total as the rows (always include a total row even if you don't know whether you'll need it). The problem tells us what their rates are, and we know that their times differ by 1. Give Adrienne a variable t for time and call James's time t-1.

.....................Rate.............Time............Distance

Adrienne...........3..................t....................?

James...............4.................t-1..................?

Total.................?...................?...................?

Now multiplying through their distances are 3t and 4(t-1). We know the distances are equal, so set 3t=4(t-1) and we can get t, which means we can get the distance as well.

14 is very similar. You are given both rates, and you know by how much the times differ. Create variable expressions for the times (remember to avoid the pitfall of confusing hours and minutes), then you can get expressions for each train's distance. Add these to equal 300 and you can solve for t.

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Tim Sanders
Manhattan GMAT Instructor

For this type of problem when one starts ahead or after the other, how do you distinguish when to use t-1 vs. t+1?

If A leaves 1 hour earlier than B, is A's rate t+1 because they are travelling an hour more than B, or is B t-1 because they are travelling one hour less than A?

either way; it depends on which one you want to call t. if A's time is t, B's time will be t-1. if B's time is t, A's time will be t+1..

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Tim Sanders
Manhattan GMAT Instructor

I was just looking through this topic and I wanted to know if you could help me determine when to set the distances equal to each other and when to add the two times up and have them equal a distance.

For instance in Q14 the trains times are added and set equal to the distance of 300 miles. However, in 15 the two distances are set equal to each other.

How do I determine which method to pick for these problems?

Sure. If the two trains/people/whatever are working together to accomplish something, add the the two distances and have them equal the total distance. For example, two trains are moving toward each other and covering 300 miles. Where will they meet? When they meet, they will have together covered 300 miles.

The same situation occurs when two people start back to back and move away from each other. How long does it take before they cover 300 miles? Again, they are working together to get 300 miles away from each other. Maybe one is fast and covers 200 miles, and the other is slow and covers 100 miles, but together they have covered 300 miles.

A "chase" situation mandates that we set the distances equal (assuming they both started at the same place). For example, Bob starts walking at 3mph. One hour later Jim starts walking at 4mph. How long will it take Jim to catch Bob? Well, when Jim does catch Bob they will have both gone the same distance. Thus, Bob's rate * (t+1) = Jim's rate * (t). Bob gets "t+1) because he has walked one more hour than Jim. Therefore, 3(t+1)=4t.

Hopes this helps!

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Jamie Nelson
ManhattanGMAT Instructor

Referring to question 5, while I was able to solve using the footsteps template, I was completely lost on the alternate explanation given to the answer.

I have the following few questions related to the ALTERNATE explanation only:

Q1. Why does the explanations say that Adrienne has a 3-mile headstart?

Q2 How do we get the time as 3hrs for James as a travel time.

On the templates show for different RT=W questions, the Chase templates are confusing.

For example it says if the "Time" is same then its "t", but if the times are different, than how do we use the Chase templates.

I) James didn’t start until after Adrienne had walked 3 miles that first hour. This is explained in the setup to the problem..

II) James has 3 miles to catch up (the head start). He catches up 1 mile per hour, so it will take 3 hours (3 miles / 1 mile/hr)..

Your best bet for using a chase setup is to figure out how far apart the parties are when they both start moving. At that point you will be able to use the same time for both..

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Tim Sanders
Manhattan GMAT Instructor