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Each • in the mileage table above represents an entry indica

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Each • in the mileage table above represents an entry indica [#permalink] New post 02 Jun 2010, 09:32
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A
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65% (02:19) correct 35% (01:50) wrong based on 386 sessions
Attachment:
Table.png
Table.png [ 23.23 KiB | Viewed 6227 times ]
Each • 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 between all pairs 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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 12 Dec 2012, 09:09, edited 3 times in total.
Renamed the topic and edited the question.
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Re: PS Q: OG-12 #116 [#permalink] New post 02 Jun 2010, 10:42
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snkrhed wrote:
So there's a chart that looks a lot like this:

-E D C B A
A * * * *
B * * *
C * *
D *
E

Each * in the mileage table above represents an entry indicating the distance between all pairs 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


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
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Re: PS Q: OG-12 #116 [#permalink] New post 02 Jun 2010, 10:54
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snkrhed wrote:
Attachment:
img.jpg

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


We are there 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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Re: PS Q: OG-12 #116 [#permalink] New post 30 Jun 2010, 01:16
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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?
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Re: PS Q: OG-12 #116 [#permalink] New post 30 Jun 2010, 06:45
This can be done in n * (n - 1) / 2 ways.

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

Correct answer choice is B. Thank You.

Thanks,
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Re: PS Q: OG-12 #116 [#permalink] New post 03 Jan 2013, 11:54
Bunuel wrote:
snkrhed wrote:
Attachment:
img.jpg

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


We are there 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.



Hello, Bunuel
Which formula did you use here ?

Thanks,
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Re: Each • in the mileage table above represents an entry indica [#permalink] New post 03 Jan 2013, 12:37
Combinations formula.

30 C 2. Answer 435.
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Re: Each • in the mileage table above represents an entry indica [#permalink] New post 10 Nov 2013, 22:42
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.
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Re: Each • in the mileage table above represents an entry indica [#permalink] New post 16 Feb 2014, 01:47
stevennu wrote:
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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Re: PS Q: OG-12 #116 [#permalink] New post 13 Apr 2014, 11:30
Bunuel wrote:
snkrhed wrote:
Attachment:
img.jpg

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


We are there 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.


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)!}"
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Re: PS Q: OG-12 #116 [#permalink] New post 14 Apr 2014, 00:04
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russ9 wrote:
Bunuel wrote:
snkrhed wrote:
Attachment:
img.jpg

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


We are there 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.


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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Re: PS Q: OG-12 #116 [#permalink] New post 09 May 2014, 14:22
Bunuel wrote:

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.


Hi Bunuel,

Unfortunately, still not clear. Why are we choosing 2 out of 30?
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Re: PS Q: OG-12 #116 [#permalink] New post 10 May 2014, 04:06
Expert's post
russ9 wrote:
Bunuel wrote:

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.


Hi Bunuel,

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


Consider the table given in the original post:
Image
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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: PS Q: OG-12 #116 [#permalink] New post 15 May 2014, 15:02
Bunuel wrote:
russ9 wrote:
Bunuel wrote:

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.


Hi Bunuel,

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


Consider the table given in the original post:
Image
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?


Yes, now it does. Thanks!
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Re: Each • in the mileage table above represents an entry indica [#permalink] New post 13 Oct 2014, 02:08
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.
Re: Each • in the mileage table above represents an entry indica   [#permalink] 13 Oct 2014, 02:08
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