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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
Joined
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Posted:
20 messages

OG 13th Q124

Post Tue Aug 05, 2014 9:25 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post Tue Aug 05, 2014 9:32 pm
    Hi anurag_7,

    Your calculation involves a minor math mistake:

    1+2+3+….29 does NOT total 450.

    I'm going to explain why below, but you should retry your calculation first to see if you can find the mistake:


    1 to 29, inclusive is 29 terms…
    The average of those terms is 15….
    29 x 15 = 435


    Also, you'll notice that the first column in that chart is completely empty, so…
    FEWER than HALF of the 900 squares would have a dot in them

    Final Answer: B

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    anurag_7 Junior | Next Rank: 30 Posts Default Avatar
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    Posted:
    20 messages
    Post Tue Aug 05, 2014 9:40 pm
    Thanks a lot rich...I took 30 terms instead of 29...

    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.

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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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    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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    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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    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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    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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