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OG12 #116 - A Different Approach?

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by Jim@StratusPrep » Wed Jun 13, 2012 8:34 am
Your first solution works, but you can make it more manageable:

Column 1 -> 0 dots
Column 2 -> 1 dots
Column 3 -> 2 dots
Column 4 -> 3 dots
...
Column 30 -> 29 dots

Essentially you will add the consecutive integers from 1 to 29 inclusive, which is 15 x 29 = 435

Hope this helps
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by Bill@VeritasPrep » Wed Jun 13, 2012 8:34 am
I looked at it differently. If we have 30 cities, how many routes are there between them?

30 (cities for the starting point) * 29 (cities for the ending point; we can't end in the starting city)

However, 30*29 gives us a lot of repeats. It will include both City A to City B and City B to City A. Since we want each distance represented only once, we need to divide by 2:

(30*29)/2 = 15*29 = 435.
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by GMATGuruNY » Wed Jun 13, 2012 12:44 pm
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Each dot in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900
Every PAIR of cities must be represented by a dot.
To determine the number of dots that are needed, we simply need to count how many PAIRS can be formed from the 30 cities.
The number of combinations of 2 that can be formed from 30 options = 30C2 = (30*29)/(2*1) = 435.

The correct answer is B.
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by lunarpower » Wed Jun 13, 2012 1:36 pm
Hi,

Of all the solutions I see here, I like the mentality behind this one most:
Jim@StratusPrep wrote:Your first solution works, but you can make it more manageable:

Column 1 -> 0 dots
Column 2 -> 1 dots
Column 3 -> 2 dots
Column 4 -> 3 dots
...
Column 30 -> 29 dots

Essentially you will add the consecutive integers from 1 to 29 inclusive
this is the point: you should be able to LOOK FOR PATTERNS, without needing a memorized "template" or "problem type" for everything on the test.
there are other ways to do the problem, including the combinatorial solutions that some other instructors have offered up on this thread, but pattern recognition is really the name of the game here.

the idea of the book is to contain patterns that are *more common* than others -- but you definitely need to be able to recognize patterns WITHOUT the benefit of "categories" or whatever else.
in fact, you should first look to see whether you can make any intuitive sense of the problem yourself -- i.e., find the patterns without having to bother with classifying the problem. you should really only be classifying the problems if you don't have the requisite intuition.

this is clearly not something we can print in a book -- i.e., we can't write a book that says "the primary point is to get it, but here are some templates/categories if you don't get it when you first look at it". we have to fill the book with objective techniques that people can execute.
but, that's the point. in fact, the point of the entire GMAT, really.
Ron has been teaching various standardized tests for 20 years.

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by gmattesttaker2 » Fri Jun 15, 2012 11:10 pm
Jim@StratusPrep wrote:Your first solution works, but you can make it more manageable:

Column 1 -> 0 dots
Column 2 -> 1 dots
Column 3 -> 2 dots
Column 4 -> 3 dots
...
Column 30 -> 29 dots

Essentially you will add the consecutive integers from 1 to 29 inclusive, which is 15 x 29 = 435

Hope this helps

Hello Jim,

I was just wondering why we are taking 15 here. Is it because 15 = 30/2? I just wasn't very clear about the reasoning. Thanks for your help.

Regards,
Sri
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by lunarpower » Sat Jun 16, 2012 4:13 am
gmattesttaker2 wrote:I was just wondering why we are taking 15 here. Is it because 15 = 30/2? I just wasn't very clear about the reasoning. Thanks for your help.

Regards,
Sri
i'm not jim, but he is just using the universal formula for the sum of any set of numbers when the average is known:
Sum = (Average) x (How many numbers)
if you have the numbers from 1 to 29, then there are 29 numbers, whose average is 15.
Ron has been teaching various standardized tests for 20 years.

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by gmattesttaker2 » Sat Jun 16, 2012 8:55 pm
lunarpower wrote:
gmattesttaker2 wrote:I was just wondering why we are taking 15 here. Is it because 15 = 30/2? I just wasn't very clear about the reasoning. Thanks for your help.

Regards,
Sri
i'm not jim, but he is just using the universal formula for the sum of any set of numbers when the average is known:
Sum = (Average) x (How many numbers)
if you have the numbers from 1 to 29, then there are 29 numbers, whose average is 15.
Hello Ron,

Thank you very much for the explanation. It is clear now.

Best Regards,
Sri
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