RTD problem - any Manhattan GMAT users??

[bada_buddy]

This is a relatively easy OG work rate problem. But I am trying to follow the RTD table technique suggested by Manhattan GMAT. I am not able to reach a solution using that technique for this question.

The question is #223 in OG 11th edition Problem Solving section (page 182)

Car A is 20 miles behind car B, which is traveling in the same direction along the same route as car A. Car A is traveling at a constant speed of 58 miles per hour and car B is traveling at a constant speed of 50 miles per hour. How many hours will it take for car A to overtake and drive 8 miles ahead of car B?

The easy solution is to find the relative speed and the extra distance traveled by car A to figure out the time taken to catch up with Car B.

But I am trying to learn the manhattan GMAT RTD table technique and I want to make sure this technique works for this problem

A B
---------------------------------------
Rate 58 50

Time ( T + 28/58 ) T

Distance ( X + 28 ) X

I basically come up with two equations here. The first one for car B :
X = 50T

The second equation for car A : 58*( T + 28/58 ) = X + 28. I substitute for X and solve for T in the second equation.

But this method does not yield the right answer. Why? Anybody familiar with this technique? Your help is much appreciated.

Thanks in advance.

[jazzcat4u]

A B
R 58 50
T T T
D D D-28 *(28 bc 20 miles ahead + 8 miles to overtake)

RELATIONSHIPS TO LOOK FOR
A: 58T = D
B: 50T = D-28

SUBSTITUTE TO SOLVE FOR T
50T = 58T-28
28 = 58T - 50T
28 = 8T
3.5 = T

familiar with it, took me some time to get use to spotting certain details - like the whole "D-28" part..good luck

[bada_buddy]

Thanks a bunch jazzcat. I realized what I was doing wrong. I guess it does not make a difference if I used d+28 for A or d-28 for B. Both state the same fact.

But the mistake I made was in assigning values for T.

Thanks again.

[kalyan.gmat]

Cann't we do it in this way,

Coz both Cars are going in the same direction, their relative speed is Diff b/n their speed and distance Car A shud travel is 20 + 8(with the same speed).
So we have Distance and Speed ...T = Dist/Speed
T = 28/8 = 3.5

Cheers,
Kalyan.

[Morgoth]

Cann't we do it in this way,

Coz both Cars are going in the same direction, their relative speed is Diff b/n their speed and distance Car A shud travel is 20 + 8(with the same speed).
So we have Distance and Speed ...T = Dist/Speed
T = 28/8 = 3.5

Cheers,
Kalyan.

Absolutely, in fact this strategy is fastest on the more complicated distance and speed problems.

[bada_buddy]

I totally agree - but like i said in the first post, i wanted to figure how to solve it specifically thru the RTD method.

[superrrom]

Cann't we do it in this way,

Coz both Cars are going in the same direction, their relative speed is Diff b/n their speed and distance Car A shud travel is 20 + 8(with the same speed).
So we have Distance and Speed ...T = Dist/Speed
T = 28/8 = 3.5

Cheers,
Kalyan.

can somebody explain this problem? How do you get 20+8

[Morgoth]

Cann't we do it in this way,

Coz both Cars are going in the same direction, their relative speed is Diff b/n their speed and distance Car A shud travel is 20 + 8(with the same speed).
So we have Distance and Speed ...T = Dist/Speed
T = 28/8 = 3.5

Cheers,
Kalyan.

can somebody explain this problem? How do you get 20+8

A-B = 20 miles

A's speed = 58mph
B's speed = 50mph

increasing or decreasing speed = 58-50 = 8,[since they are traveling in the same direction]

T = 20/8 = 2.5hrs [at this point A and B meet each other]

A has to get ahead of B by 8 miles

T = 8/8 = 1hr

Total Time = 2.5 + 1 = 3.5 hrs.

or the shorter way, which helps you save one step

20 miles + 8 miles = 28 miles

Total time = 28/8 = 3.5 hrs.

Hope its clear.

[nadib002]

A B
R 58 50
T T T
D D D-28 *(28 bc 20 miles ahead + 8 miles to overtake)


I was trying to solve the above mentioned RTD problem using the RTD chart. My chart looks a bit different from the one described above--

The problem states that "Car A is 20 miles behind Car B". If we assume that Car B traveled "d" miles, shouldn't the distance traveled by Car A be "d-20" miles( i am only trying to set up the distance to overtake, not the overtake + 8). In the chart above, B's distance is taken as d-28 (20 + 8) instead of A's distance. Is my assumption correct or am I missing something? For now i am trying to get my fundamentals clear with these kind of problems.

Please help...

Thank you

[nadib002]

Can someone please shed some light on the RTD chart question above...


Thank you

[Brent@GMATPrepNow]

Car A is 20 miles behind car B, which is traveling in the same direction along the same route as car A. Car A is traveling at a constant speed of 58 miles per hour and car B is traveling at a constant speed of 50 miles per hour. How many hours will it take for car A to overtake and drive 8 miles ahead of car B?
A. 1.5
B. 2.0
C. 2.5
D. 3.0
E. 3.5

APPROACH #1: Start with a word equation
The question tells us that Car A's total travel distance is 28 miles greater than Car B's total travel distance.
So we can write: (Car A's travel distance) - (Car B's travel distance) = 28 miles

Let t = Car A's travel time, which means t = Car B's travel also.
Distance = (rate)(time)

Plug our values into the word equation to get: 58t - 50t = 28
Subtract 50t from both sides: 8t = 28
Solve: t = 28/8 = 7/2 = 3.5
Answer: E

APPROACH #2: Answer the question in two parts
58 mph - 50 mph = 8 mph.
So, the original 20 mile gap between the two cars shrinks at a rate of 8 mph.
Time to reduce the gap to zero = distance/rate = 20/8 = 2.5 hours.
Time to increase the gap from 0 to 8 miles = distance/rate = 8/8 = 1 hour.
So, the total time for car A go from being 20 miles behind to being 8 miles ahead = 2.5 hours + 1 hour = 3.5 hours
Answer: E