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OG 13th Q124

This topic has 5 expert replies and 4 member replies
anurag_7 Junior | Next Rank: 30 Posts Default Avatar
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OG 13th Q124

Post Tue Aug 05, 2014 9:25 pm
OG 13th Q124- 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
(0 450
(D) 465
(E) 900

What is wrong in my method????
2 cities can be shown via one entry. For 3 cities we need two entries and so on. So for 30 cities we need 29 entries. Thus 1+2+...........+28+29 = 450. But the answer is wrong...

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GMATinsight Legendary Member
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Post Wed Aug 06, 2014 10:52 am
ALTERNATE

For a Table with 2x2 Matrix, number of entries = 1 = 2x1/2
For a Table with 3x3 Matrix, number of entries = 3 = 3x2/2
For a Table with 4x4 Matrix, number of entries = 6 = 4x3/2
For a Table with 5x5 Matrix, number of entries = 10 = 5x4/2
.....
.....
For a Table with 30x30 Matrix, number of entries = 30x29/2 = 435

Answer: Option B

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melguy Master | Next Rank: 500 Posts
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Post Fri Oct 28, 2016 8:53 pm
Hello All

I am still a bit lost with this Q even after reading all the explanations. If I may ask

- What is this question even asking?
- Which topic of GMAT is this question covering (P&C?)
- How did we get to '30 cities'. We can pair A -> B , A -> C, A -> D, A -> E and the same for rest 4 cities so there should be a total of 4 x 4 = 16 cities?


Please help.

Thanks
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GMATinsight Legendary Member
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Post Wed Aug 06, 2014 10:52 am
ALTERNATE

For a Table with 2x2 Matrix, number of entries = 1 = 2x1/2
For a Table with 3x3 Matrix, number of entries = 3 = 3x2/2
For a Table with 4x4 Matrix, number of entries = 6 = 4x3/2
For a Table with 5x5 Matrix, number of entries = 10 = 5x4/2
.....
.....
For a Table with 30x30 Matrix, number of entries = 30x29/2 = 435

Answer: Option B

_________________
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Testimonials e-mail: info@GMATinsight.com I Mobile: +91-9999687183 / +91-9891333772
To register for One-on-One FREE ONLINE DEMO Class Call/e-mail
One-On-One Private tutoring fee - US$40 per hour & for FULL COURSE (38 LIVE Sessions)-US$1000

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melguy Master | Next Rank: 500 Posts
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Post Fri Oct 28, 2016 8:53 pm
Hello All

I am still a bit lost with this Q even after reading all the explanations. If I may ask

- What is this question even asking?
- Which topic of GMAT is this question covering (P&C?)
- How did we get to '30 cities'. We can pair A -> B , A -> C, A -> D, A -> E and the same for rest 4 cities so there should be a total of 4 x 4 = 16 cities?


Please help.

Thanks
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GMAT/MBA Expert

Post Tue Nov 01, 2016 5:17 am
anurag_7 wrote:
OG 13th Q124- 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
(0 450
(D) 465
(E) 900
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 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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Post Sat Oct 29, 2016 2:44 am
melguy wrote:
Hello All

I am still a bit lost with this Q even after reading all the explanations. If I may ask

- What is this question even asking?
- Which topic of GMAT is this question covering (P&C?)
- How did we get to '30 cities'. We can pair A -> B , A -> C, A -> D, A -> E and the same for rest 4 cities so there should be a total of 4 x 4 = 16 cities?

Thanks
Prompt:
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?


The portion in blue indicates that there are 30 cities in total.

In the mileage chart, every distance between two cities must be represented by a dot:
The distance between A AND B --> dot.
The distance between A AND C --> dot.
The distance between A AND D --> dot.
And so on.

Since every pair of cities requires a dot, the total number of dots is equal to the total number of PAIRS that can be formed from the 30 cities.
From 30 cities, the number of combinations of 2 that can be formed = 30C2 = (30*29)/(2*1) = 435.
Thus, the mileage chart requires a total of 435 dots.

This problem asks us to count COMBINATIONS: the total number of pairs that can be formed from 30 options.

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GMATinsight Legendary Member
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Post Wed Aug 06, 2014 10:46 am
anurag_7 wrote:
OG 13th Q124- 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
(0 450
(D) 465
(E) 900

What is wrong in my method????
2 cities can be shown via one entry. For 3 cities we need two entries and so on. So for 30 cities we need 29 entries. Thus 1+2+...........+28+29 = 450. But the answer is wrong...
To make an entry you need one Start point City and one destination City and two Cities out of 30 Cities can be chosen in 30C2 ways
30C2 = 30!/(2!x28!) = 435

Answer: Option B

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Testimonials e-mail: info@GMATinsight.com I Mobile: +91-9999687183 / +91-9891333772
To register for One-on-One FREE ONLINE DEMO Class Call/e-mail
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Post Wed Aug 06, 2014 8:11 am
anurag_7 wrote:
OG 13th Q124- 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
(0 450
(D) 465
(E) 900

What is wrong in my method????
2 cities can be shown via one entry. For 3 cities we need two entries and so on. So for 30 cities we need 29 entries. Thus 1+2+...........+28+29 = 450. But the answer is wrong...
Another way to find the sum 1+2+...........+28+29 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

Cheers,
Brent

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Post Wed Aug 06, 2014 2:42 am
anurag_7 wrote:
OG 13th Q124- 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
Each PAIR OF CITIES requires a dot in the table.
From 30 cities, the number of pairs that can be formed = 30C2 = (30*29)/(2*1) = 435.

The correct answer is B.

_________________
Mitch Hunt
GMAT Private Tutor
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.

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