RTD Chase problem

[THANU.KG]

Hi, I have a question on the #5 on page 45 in the MGMAT strategy guide book.

One hour after Adrienne started walking the 60 miles from x to y, James started walking from X to Y as well. Adrienne walks 3 miles per hour, and James walks 1 mile per hour faster than Adrienne. How far from X will James be when he catches up to Adrienne?
(A). 8 miles
(B). 9 miles
(C). 10 miles
(D). 11 miles
(E). 12 miles

Is there any reason why we would not set the total distance as 60 miles and do the following
3(t+1) + 4t = 60?
How do we use the given 60 miles in our problem or is it not a useful piece of information in this problem?

Please help.
Thanks

[atharshiraz]

Hi, I have a question on the #5 on page 45 in the MGMAT strategy guide book.

One hour after Adrienne started walking the 60 miles from x to y, James started walking from X to Y as well. Adrienne walks 3 miles per hour, and James walks 1 mile per hour faster than Adrienne. How far from X will James be when he catches up to Adrienne?
(A). 8 miles
(B). 9 miles
(C). 10 miles
(D). 11 miles
(E). 12 miles

Is there any reason why we would not set the total distance as 60 miles and do the following
3(t+1) + 4t = 60?
How do we use the given 60 miles in our problem or is it not a useful piece of information in this problem?

Please help.
Thanks

Your assumption is this : that "t" is the time that James has walked when he catches up with Adrienne.

So since distance = speed x time
3(t+1) = the distance that James and Adrienne are from X when James catches up with Adrienne.
4(t)= is also the distance that James and Adrienne are from X when James catches up with Adrienne.

Why are you assuming that 3(t+1)+4(t) is 60 miles?

In this case the total distance is of no consequence.

An approach to the solution is this:
3(t+1) = 4t (as we just noticed above)
t = 3 hrs

if "t" is the time that James has walked when he catches up with Adrienne then the distance traveled by James (his distance from x) is : 4 (m/hr) x 3 (hrs)= 12 miles

(E)

[jnelson0612]

atharshiraz is exactly right . . . we don't care at all about the 60 miles between x and y. The only thing of interest is how far both Adrienne and James will have gone when James catches up to her. If you draw a picture of this, you can see that when he catches up to her, they will have both gone the same distance.

As atharshiraz pointed out, Rate * Time = Distance.

James will walk 4 miles per hour (Rate) * T (the amount of time James walks) = Distance.

Adrienne will walk 3 miles per hour (Rate) * T+1 (the amount of time James walks PLUS the extra hour Adrienne walked) = Distance.

Thus, J walks a distance of 4T and Adrienne walks a distance of 3(T+1). They walk the same distance, or 4T=3(T+1). T=3. Thus, James will have walked 4mph * 3 hours or 12 miles when he catches up to her.

What you are proposing works well when you have two people walking toward each other to cover a distance and we are interested in where they meet on the continuum or how long each one walks. For example, if Adrienne and James start walking toward each other and are working together to cover 60 miles and to meet somewhere in the middle, then we would set up the equation just like you proposed. We would add her distance to his distance and set that equal to 60.

Please let us know if we can add further clarification.

[THANU.KG]

Thanks Jamie. I understand now.

[jlucero]

Awesome!