Problem Solving

ok, so you should NEVER use more than one variable for these sorts of problems. trust me, you won't need more than one, pretty much no matter what.

here's the deal:
(1) make a chart that reads R x T = D across the top
(2) fill in the definitive numerical values first
(3) notice all the RELATIONSHIPS in the problem
(4) use ONE of the relationships to define ONE VARIABLE
(5) use R x T = D (NOT the other relationship(s)) to fill in the rest of the chart
(6) use leftover relationship(s) to define EQUATION(s) that you can solve

this is a lot of steps, but the process WILL successfully solve ANY rtd problem that the test can throw at you. i mean this.
these six steps will solve ANY rtd problem, with an effort level that's roughly equal for each problem.
the only difference in effort level results from variations in the creativity of the "relationships" - but the relationships aren't usually that creative.

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NOTE
this forum won't allow horizontal formatting, and i don't feel like making handwritten charts and uploading them right now, so use your imagination - or get out a sheet of paper and transcribe these charts onto it.

steps 1 and 2:
R x T = D
gear P: 1/6 rev/sec; blank; blank
gear Q: 2/3 rev/sec; blank; blank
(notice that i've already converted the units, so that the seconds don't pose a problem)

step 3:
TIME RELATIONSHIP = times for the 2 gears are the SAME (because they're turning simultaneously)
DISTANCE RELATIONSHIP = gear q turns 6 more revolutions than does gear p

step 4: **
i'm going to pick the TIME RELATIONSHIP to define a variable, because it's easy: just call both of them "t".
chart:
R x T = D
gear p: 1/6, t, blank
gear q: 2/3, t, blank

step 5:
since rt = d, just multiply across the rows to generate expressions for "d". DO NOT DEFINE ANOTHER VARIABLE to represent distance. DO NOT.
chart:
R x T = D
gear p: 1/6, t, t/6
gear q: 2/3, t, 2t/3

step 6:
the "leftover relationship" is the fact that gear q travels 6 more revolutions than does gear p.
therefore, t/6 + 6 = 2t/3
make common denominator: t/6 + 6 = 4t/6
6 = 3t/6
12 = t
done


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**notice that, at step 4, you could have decided to use the "distance" relationship instead, so that the numbers of revolutions would be x and x + 6. this wouldn't work out as nicely, though, because (a) time is the desired quantity at the end of the problem anyway, and, more noticeably, (b) you'd have to do division, creating fractions, to fill in the missing quantities in the time column.

vanesslondon wrote:Hello, the above methods are great but I did this in a slightly different way (this won't work for all qs).

The first thing I do w/ RDT questions is check that all the units align. here I changed everything to seconds.

So cog P can do 1 rotation in 6 seconds.

Cog Q can do 4 rotations in 6 seconds.

The difference here is insufficient.

So I went a step further:

Cog P can do 2 rotations in 12 seconds.

Cog Q can do 8 rotations in 12 seconds.

There is our answer, 8-2=6

Someone better equipped than me feel free to correct me if you don't recommend this method though.

no, that's fine, but it will only work on problems that are decently similar to this one. if you take a problem in which the rotations don't begin at the same time, or a distance problem involving the average rate of a combined 2 or more separate steps, then i don't see how you could generalize this method to that sort of situation.
the above method is a lot of artillery for a problem like this, but it will get the job done on any rtd problem that the gmat decides to toss into your path.