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 the exact distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

Hi, I took the MGMAT CAT and this is a problem in which I'm not sure how the second part of the problem was calculated.

So, there are two parts to this question.

1. Time for Tom to cover Linda's distance, which I was able to calculate - 30 minutes
2. Time for Tom to cover twice Linda's distance, which I am confused about.

In the explanation, the equation shows up as : 6T=2(2(T+1). I don't understand why the doubling the distance that Linda travels, you also double the time....because ultimately this equation becomes 6T = 4T+4. Doesn't this actually double count the time that Linda spent walking?

The correct answer is 90 minutes [120-30 minutes].

Greatly appreciate any help!

Hi,

The key point to understand is that, the instant Tom starts moving, the distance between Tom and Linda is 2 miles.

First Part:

Now say in time "t" after Tom starts , Linda moves 2t distance and Tom moves 6t distance.

How much has Linda moved in total?

2t +2

From (1)
6t = 2t +2


t= 30 min, which you have rightly calculated.


Similarly

Say in time "T" after Tom starts , Linda moves 2T distance and Tom moves 6T distance.

How much has Linda moved in total?

2T +2

From (2)
6T = 2 x (2T +2)

2T= 4

T = 2 Hours.

T (-) t = 1.5 Hours = 90 Mon.


Aditya

Two points:

1) Wendy, you're not doubling Linda's time when you write 2(2(T+1). 2 is Linda's rate, T+1 is Linda's time, and 2(T+1) is Linda's distance, rate*time. The other T makes this twice Linda's distance, precisely what you're solving for here.

2) Though I'm not sure that I follow every step of Aditya's work, the basic approach is sound. One method of solving problems that involve a time head-start is to rewrite them so that they instead include a distance head-start. This works more intuitively when one person is running the other down, but it'll work here too.

Can someone please show me how to do this question using the "relative rates" method? The assumption here would be that since they are walking in a straight line, Tom can be viewed as "catching up" to Linda after 1 hour, so in 1 hour, Tom would be catching up to
R (linda's rate) X T = D
2 x 1 = 2 miles

From there, I tried to use the relative rate method

Difference in rate x T = D

(6-2) x T= 2 miles
T=1/2 hours

To cover twice of Linda's distance, Tom would need to cover 2 x 2 = 4 miles ,

using the relative rate method again

(6-2)x T = 4
T= 1

So the difference would be 1 - 1/2 = 1/2 hours = 30 minutes.

looks like you're doing things fine here, you just need to remember that when we talk about Tom covering twice Linda's distance that means we are measuring the moment when Tom's distance so far is twice Linda's distance so far. You are comparing Tom's distance at one point in time with Linda's distance at a DIFFERENT point in time..

_________________
Tim Sanders
Manhattan GMAT Instructor

hi - I just have a basic question here? Can I pick a nos for Linda's time here and equate it will tom

If so Please correct me where I'm going wrong please.

Linda's speed = 2 MPH
Lets assume she has traveled 4 miles
So 4/2 = 2 hrs Linda has traveled. Now only half of which Tom has traveled as per the question

so
6 * 1 hr = 6 miles
Now the difference
amount of time it takes Tom to cover half of the distance that Linda has covered

Linda has traveled 4 miles so

6 * t1 = 2
ti = 2/6 * 60 = 20 mins

the amount of time it takes Tom to cover twice the distance that Linda has covered

Linda has traveled 4 miles so

6 * t2 = 8
t2 = 8/6 * 60 = 80 mins

So the + Diff 80-20= 60 mins

I think its safe to assume the Linda's distance because when we are calculating Tom's time we are anyways equating to Linda's distance.. hence should work right? Please help!

Cheers
Jp

No. If it is ever possible to compute something directly in a problem, you are not allowed to just make up whatever number you want. All the values here are directly calculable, so picking numbers is not going to work..

_________________
Tim Sanders
Manhattan GMAT Instructor