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Each • in the mileage table above represents an entry indica [#permalink]
02 Jun 2010, 09:32

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

67% (02:11) correct
32% (01:50) wrong based on 155 sessions

Attachment:

Table.png [ 23.23 KiB | Viewed 3517 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?

Re: PS Q: OG-12 #116 [#permalink]
02 Jun 2010, 10:42

4

This post received KUDOS

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

Re: PS Q: OG-12 #116 [#permalink]
02 Jun 2010, 10:54

2

This post received KUDOS

Expert's post

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?

Re: PS Q: OG-12 #116 [#permalink]
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?

Re: Each • in the mileage table above represents an entry indica [#permalink]
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.

Re: Each • in the mileage table above represents an entry indica [#permalink]
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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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)!}"

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.