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Each • in the mileage table above represents an entry indica

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

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New post 04 May 2016, 08:16
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snkrhed wrote:
Attachment:
Table.png
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


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

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New post 11 May 2016, 21:00
I solved it using patterns -
5 cities
4 , 3, 2, 1 for each city per say.
extending to 30 cities, would mean 1,2,3,4, ..............,29
thats the series of consecutive integers - now calculating the sum -
no of terms = 29-1 +1 29
Average = 1+29 / 2 = 15
sum = 29 * 15 = 435
choice B :-)
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Re: Each • in the mileage table above represents an entry indica  [#permalink]

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New post Updated on: 16 Jun 2016, 18:06
kirtivardhan

If you look at the table City A(1st city) has 0 points under it , City B(2nd City) has 1 , City C(3rd City) has 2 and so on . Therefore the 30th city will have 29 points under it.

Sum of these points is given by [ 2a + (n-1)d] * (n/2)

If you take first term a= 0 then n=30 and d=1;

Sum = [2(0) +29] * 30/2

=435

Or if you wish to take first term a=1 ( i.e. starting from city b) n= 29 d=1

Sum = [ 2 +28] *29/2

=435
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Originally posted by rishi02 on 15 Jun 2016, 21:39.
Last edited by rishi02 on 16 Jun 2016, 18:06, edited 1 time in total.
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Re: Each • in the mileage table above represents an entry indica  [#permalink]

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New post 16 Jun 2016, 09:19
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rishi02 wrote:
kirtivardhan

If you look at the table City A(1st city) has 0 points under it , City B(2nd City) has 1 , City C(3rd City) has 2 and so on . Therefore the 30th city will have 29 points under it.

Sum of these points is given by [ 2a + (n-1)d] * (n/2)

If you take first term a= 0 then n=30 and d=1;

Sum = [2(0) +29] * 30/2

=485

Or if you wish to take first term a=1 ( i.e. starting from city b) n= 29 d=1

Sum = [ 2 +28] *29/2

=485


Hi rishi02,

Your set-up is correct, but you should double-check your math (485 is NOT the result of either of those calculations).

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

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New post 16 Jun 2016, 18:05
EMPOWERgmatRichC wrote:
rishi02 wrote:
kirtivardhan

If you look at the table City A(1st city) has 0 points under it , City B(2nd City) has 1 , City C(3rd City) has 2 and so on . Therefore the 30th city will have 29 points under it.

Sum of these points is given by [ 2a + (n-1)d] * (n/2)

If you take first term a= 0 then n=30 and d=1;

Sum = [2(0) +29] * 30/2

=485

Or if you wish to take first term a=1 ( i.e. starting from city b) n= 29 d=1

Sum = [ 2 +28] *29/2

=485


Hi rishi02,

Your set-up is correct, but you should double-check your math (485 is NOT the result of either of those calculations).

GMAT assassins aren't born, they're made,
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Thanks ! My bad

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

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New post 25 Mar 2017, 11:04
we can solve it in 2 approaches:
1) find the sum of consecutive integers: (1+29/2)*29=435
or
2) how many matches are for 30 items: 2C30=29*30/2=435
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Re: Each • in the mileage table above represents an entry indica  [#permalink]

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New post 02 May 2017, 19:48
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

My 2 cents.

We see that the number of dots in city E (which is the 5th term) is 4.
From this, we can conclude that 30 cities will have 29 dots.
Now, the issue is how are we going to add 1+2+3.....29.
Well, we have a short cut in adding.
When you look at the below number, we see that if we add the first term and the last term, we get 30.

1+2+3...27+28+29

Continuing this trend, we can go up to 14+16.
You should be careful in going up 15+15 because there are NO two 15s in this sequence, only 1.
So, we are almost done.
14 * 30 = 420 and we have to add 15
Hence 435
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Re: Each • in the mileage table above represents an entry indica  [#permalink]

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New post 11 Jun 2017, 05:03
1+2+3+4+5+6+7+8+9+10=55
11+12+...........20 =100+55
21+22+...........30 = 200+55
Total 465

Total - 30= 435
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Each • in the mileage table above represents an entry indica  [#permalink]

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New post 08 Apr 2018, 03:26
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snkrhed wrote:
Attachment:
Table.png
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



Hello generis,

I used this formula to get answer SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\)

i was almost there but something went wrong :) but i am stll happy :)

\(\frac{30}{2} (2 *1+(30-1)1)\)

where
\(30\) is number of terms
\(1\) is distance

so i get \(15(2*1+(30-1)*1)\) ---> \(15(2+29*1)\) ---> \(15*31 = 465\)


so i got 465 as an answer :) but correct one is 435 :?

hello generis, i think you missed my question :) so i decided to tag you one more time :-) have a great start of the week :)
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Each • in the mileage table above represents an entry indica  [#permalink]

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New post 15 Apr 2018, 10:16
1
dave13 wrote:
snkrhed wrote:
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

Hello generis,

I used this formula to get answer SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\)

i was almost there but something went wrong :) but i am stll happy :)

\(\frac{30}{2} (2 *1+(30-1)1)\)

where
\(30\) is number of terms
\(1\) is distance

so i get \(15(2*1+(30-1)*1)\) ---> \(15(2+29*1)\) ---> \(15*31 = 465\)

so i got 465 as an answer :) but correct one is 435 :?

hello generis, i think you missed my question :) so i decided to tag you one more time :-) have a great start of the week :)

dave13 - I did miss the tag.
Or my email has gone berserk. Sorry.

I'm glad you're happy.
You should be! You are very close!
(Plus, I do not think this formula
is intuitive at all. I use the "easier" version. :-) )
I uploaded the table, see below.

Notice . . .
how many dots are under City A? Zero (0)
Because we begin at City A, let's say City A = City 1
City 1? 0 dots
How many dots under City B = City 2? One (1) dot
City C = 3? 2 dots

There is one fewer dot than there are cities.
There is no distance between City A and City A :-)
Easy mistake.

You have to account for that disparity
when you think about the first term, \(a\), in your sum
A-to-A = 0 (as a first term in your sum of integers)

So your formula will work perfectly if
you alter one of two things.

\(\frac{n}{2} (2a+(n-1)d)\)

Either start with \(a\) equal to 0, i.e.
\(a = 0\), then
\(n = 30\)
\(d = 1\)

OR let \(a = 1\), in which case
\(n = 29\) (one fewer dot/distance between than number of cities)
\(d = 1\)

Try your approach with one of these two changes --
whichever seems most intuitive to you.
Does this explanation make sense?
Attachment:
TableCities.png
TableCities.png [ 23.23 KiB | Viewed 454 times ]

(Alternatively, although I think it is harder,
you could use your original answer and correct for over-counting.
You counted 30 city-city distances too many.
You have to subtract for A-A, B-B, C-C, etc.
So: (465 - 30) = 435.)
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Re: Each • in the mileage table above represents an entry indica  [#permalink]

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New post 20 Apr 2018, 01:27
Step 1: First of all - see the magic in this beautiful table. It’s amazing. We see that 1 city has no distance, two cities have ofcourse 1 distance. Let’s extend this logic to 30 cities. OMG - this means we must have 29 distances at 30 cities.
Step 2: ERMAHGERD - this is like a number set that goes till 29. Wow. Let’s use the formula that are dear to us to find the sum. Because the question is, how many entries would the table then have - aka - how many dots?
Step 3: Count the amount of integers using the magic formula of last-first/increment+1 = 29.
Step 4: Find the mean by using the magic lazy formula of Last + First / 2 = 30/2 = 15.
Step 5: Wow - magic - let’s multiply these numbers together. 29*15 = 435.
OMG - 435. Matches AC B.
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Re: Each • in the mileage table above represents an entry indica  [#permalink]

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New post 05 Jun 2018, 23:10
Approach - the question says - "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?"
Essentially it is asking for pairs to be represented by 1 dot, so we can use combinations method
30C2 --> 435
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Re: Each • in the mileage table above represents an entry indica &nbs [#permalink] 05 Jun 2018, 23:10

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