Each • in the mileage table above represents an entry indica

City B, the second city has 1 point
City C the third city has 2 points
City D, the fourth city has 3 points

What's the pattern?

Number of cities minus 1 so the 30th city is going to have 29 points

Then it becomes a matter of adding the consecutive integers from 1 to 29
The sum is the average * number of terms
average = 15
number of terms = 29
29*15 = 435

We are told that there should be one entry for each pair. How many entries would the table then have? Or how many different pairs can 30 cities give?

\(C^2_{30}=435\)

Answer: B.
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Because the GMAT is multiple choice.

I first found the pattern mentioned each additional city adds (total cities - 1) to the table.

So I figure with 30 cities the last 3 cities added 29+28+27 to our total number of entries so (A) is ridiculous

And I knew the whole table would be 30x30 with 900 entries. Since I know I will not have entries for each box I ruled out (E) 900

Looking at the remaining choices I thought

(B) 435 = less than half the table is filled
(C) 450 = exactly half the table is filled
(D) 465 = more than half the table is filled

The table given shows a 5x5 table with only 10 entries. 10< .5(25)

So I chose (B).

This method works for these answer choices, but if the choices were 430, 435, 440 I would be screwed right?

APPROACH #1:
Each entry in the mileage table denotes a distinct pair of cities.
We can determine the total number of distinct pairs of cities by using combinations.
When there are 5 cities, the total number of distinct pairs of cities = 5C2 = 10
Noticed that in the given table there are 10 entries. Perfect!

Likewise, if we have a mileage table consisting of 30 cities, the total number of distinct pairs of cities = 30C2 [to learn how to mentally calculate combinations like 30C2, watch the video below]
= (30)(29)/(2)(1)
= 435
Answer: B



APPROACH #2:
Notice that, when there are 5 cities in the mileage table, the number of entries = 1 + 2 + 3 + 4
Likewise, if we have a mileage table consisting of 30 cities, the number of entries = 1 + 2 + 3 + . . . . + 28 + 29

One way to calculate this is to apply the following formula:
The sum of the integers from 1 to n inclusive = (n)(n+1)/2

So, 1+2+...........+28+29 = (29)(29+1)/2
= (29)(30)/2
= (29)(15)
= 435
Answer: B

Cheers,
Brent

RELATED VIDEO

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This can be done in n * (n - 1) / 2 ways.

Hence -> 30 * 29 / 2 = 435 ways.

Correct answer choice is B. Thank You.

Thanks,
Akhil M.Parekh

How would this question be solved using a consecutive integer format? Can you find the average on the consecutive integers and then multiply by the number of terms? I ask because this question is listed as a consecutive integer question in the MGAMT quant guide.

If second entry =1
third entry = 2
30th entry = 29 etc

Thus S(n)=n/2(2a+(n-1)d) where a=1, d=1, n=29
plug in and you get the answer.
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Hi Bunuel,

Can you please elaborate on how this formula works?

Thanks!

EDIT: I did it via the table method but i've seen your formula pop up quite often and I'm failing miserably at it. That might explain the horrible score in NP.

I understand what formula to use but i'm having a hard time connecting the formula to the problem "\(C^n_k = \frac{n!}{k!(n-k)!}\)"

\(C^2_{30}\) is choosing 2 out of 30. There are 30 cities and each pair of cities need an entry, hence 30 cites need \(C^2_{30}\) entries.

Hope it's clear.
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Hi Bunuel,

Unfortunately, still not clear. Why are we choosing 2 out of 30?

Consider the table given in the original post:

A and B have 1 entry;
A and C have 1 entry;
A and D have 1 entry;
A and E have 1 entry;
B and C have 1 entry;
B and D have 1 entry;
B and E have 1 entry;
C and D have 1 entry;
C and E have 1 entry;
D and E have 1 entry.

So, each pair of letters from {A, B, C, D, E} has 1 entry, total of 10 entries. How many pairs can we have? \(C^2_5=10\).

Does this make sense?
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Combination formula is no doubt easiest and fastest. But other method is

Imagine it was an excel spreadsheet. Remove Cells A1, B2, C3, D4 etc, basically a diagonal across. Total will be 30 such cells.
So now we have 900 - 30 = 870.

On both sides of the diagonal distance (between cities) is shown twice.

Therefore divide 870 into half.

Answer 435.

How if as:
the pattern is 5*5= 25-5=20/2=10
Thus, 30*30= 900-30= 870/2 = 435
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Hi Mates,

How are we getting this 29 for 30 cities?

Hi kirtivardhan,

The prompt gives us a 5x5 table, then tells us that it will be EXTENDED to a 30x30 table. From the table, you can see the 'pattern' involving the dots - as you go 'to the right', each column has one more dot in it than the column before. This means the final column will have 29 dots in it.

There are actually several different ways to answer this question. Noticing that LESS than HALF of the squares have dots in them and using the "spread" of the answer choices can help you to avoid most of the "math" involved in this question.

GMAT assassins aren't born, they're made,
Rich
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I used one of the Kaplan formulas:

Sum of Sequence = Mean of terms * number of terms

Mean of terms (when evenly spaced) = (Largest Number - Smallest Number)/2
Number of terms = (Largest Number - Smallest Number) + 1

Mean = (1 + 29)/2 = 15
Terms = (29-1)+1 = 29

Sum of Sequence = 15*29 = 435

I used 1 and 29, because there are 30 rows, but the first is blank so R1 has 0, R2 has 1, R3 has 2..... R29 has 28, and R30 has 29.

Is this method correct to use?

Hi stephyw,

Your approach absolutely works. If you read the other posts in this thread, you'll see that there are several ways to approach this prompt (and most of the questions that you'll see on Test Day will also be approachable in multiple ways). This goes to show that just because you got a question correct doesn't necessarily mean that you can't improve on your process. There might be a faster, more strategic way to approach the prompt; there might be a way that requires less 'work', etc. As you continue to study, you should commit some of your time to improving your overall tactical knowledge - those other methods can help you to increase your scores, improve your pacing and get "unstuck" when dealing with a question in which "your" approach doesn't seem to get you to the solution.

GMAT assassins aren't born, they're made,
Rich
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Solution:

This problem can best be solved using combinations.

This problem is similar to one in which 30 sports teams are playing in a tournament where every team plays every other team exactly once. No team plays itself, obviously, and the order of each pairing doesn't matter. [For example, if Team A plays Team B, the pairing of (Team A vs. Team B) is identical to (Team B vs. Team A)]. We would calculate 30C2, or the number of combinations of 30 items taken 2 at a time.

We also can solve this problem in the same way:

30C2 = 30! / [2! x (30 – 2)!]

(30 x 29 x 28!) / [2! x 28!]

(30 x 29)/2!

(30 x 29) / 2

15 x 29 = 435

Answer B
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Solution:

This problem can best be solved using combinations.

This problem is similar to one in which 30 sports teams are playing in a tournament in which every team plays every other team exactly once. No team plays itself, obviously, and the order of each pairing doesn't matter. [For example, if Team A plays Team B, the pairing of (Team A vs. Team B) is identical to (Team B vs. Team A)]. We would calculate 30C2, or the number of combinations of 30 items taken 2 at a time.

We can solve this problem in the same way:

30C2 = (30 x 29) / 2! = 15 x 29 = 435

Alternate Solution:

Notice that when there are 5 cities, there are 1 + 2 + 3 + 4 entries.

If there were 6 cities, the last column would have contained 5 entries; therefore there would have been 1 + 2 + 3 + 4 + 5 entries. Notice that the last column always contains one less entry than the number of cities.

Thus, if there were 30 cities, there would have been 1 + 2 + 3 + … + 29 = (29 * 30)/2 = 29 * 15 = 435 entries.

Answer: B
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