MGMAT CAT 1 "Flat Tire "

[yannickdago]

Hi MGMAT Team,

I tried to solve the below RTD problem with "the 3rd grade method" explained by Ron on "Thurday with Ron" study hall but I cant find the right answer.
Please see below both the question stem and my calculation details. Can you please let me know what is wrong with my approach ? (John is supposed to catch up Jacob by the time they ride the same distance but this never happens on my chart)

John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. It takes John one hour to fix the flat tire and Jacob continues to ride during this time. Once John has resumed his ride, at a rate of 15 miles per hour, how many hours will it take him to catch up with Jacob? (consider John's deceleration/acceleration before/after the flat tire to be negligible)

(A) 3 (B) 3 1/3 (C) 3 1/2 (D) 4 (E) 4 1/2


My answer with the 3rd grade method :

Distance achieved (in miles)
Time passed John Jacob
20 min 5 4
40 min 10 8
1h 10 12
1h20 10 16
1h40 10 20
2hours 15 24
2h20 20 28
2h40 25 32
3h00 30 36
3h20 35 40
3h40 40 44
4h00 45 48
4h20 50 52
4h40 55 56

Thanks for your feedback

Yannick

[RonPurewal]

Weirdly, you stopped making the table right before the golden moment. If you had made one more row, it would say:
5h00 60 60

Hey!
There it is.

Perhaps you didn't read the question carefully enough:

Once John has resumed his ride, at a rate of 15 miles per hour, how many hours will it take him to catch up with Jacob?

The purple thing means that you're measuring time from "1h40", not from time zero. So that's an additional 3 hours and 20 minutes, so (b).

[RonPurewal]

Although we should definitely include "5 hours" as a wrong answer choice to this problem.

[yannickdago]

Ok I get it now...

Thanks for your feedback !

Yannick

[tim]

:)

[BrianW172]

Hi guys,

I am trying to tackle this problem using the traditional RT=D table. I used two separate tables. The first table provided me with the information I needed up to the point where Jon got a flat tire. Table 1 is below:

Rate x time = distance
Jon 15 m/h 2/3hr = 10 miles
Jac 12 m/h 2/3hr = 8 miles

I made a second table for everything after Jon's flat tire. For table 2 I have the following:

Rate x time = distance
Jon 15 m/h xhr = 15x
Jac 12 m/h 1 +xhr = 12 + 2 + 12x

Since Jon took 1hr to fix his bike and Jacob was biking that entire time I have (1+x) for Jacob's time. Furthermore, since Jacob needed to travel an additional 2 more miles than Jon (see distances in table 1) I have (12 + 2 + 12x) for Jacob's distance. I believe I made my mistake in this table but I am not sure where?

I know the total distances for Jon and Jacob are equal so:

15x = 14 + 12x
3x + 14
x = 4.6 which is not a possible answer choice.

Any help would be much appreciated. Thanks!!

[RonPurewal]

you're adding the 2 miles to the wrong side.

to make this more clear, you should draw a diagram of the situation, in which your (12 + 12x) and (15x) represent the distances traveled by each person from the moment of the flat tire.

AT the moment of the flat tire, jacob is BEHIND john.
at the END of everything, jacob and john are IN THE SAME PLACE.
this means that jacob travels a LONGER distance than john does, over the time that you are mapping with the variable x.

if you draw a diagram, this will be quite clear. you'll see, in your diagram, that jacob's distance is exactly 2 miles longer than john's.

so, you'll have
12 + 12x = 15x + 2

if you solve that, you'll get x = 3 1/3 hours, exactly in keeping with the correct answer.