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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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

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

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,
Rich
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Thanks ! My bad

Will edit the mistake

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

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

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

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?

(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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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.

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

First thought: 30*30 grid means 900 divided by 2 = 450. But that's too easy, why?
This would be the case if the graph was exactly half covered in dots. So the top right would be covered, the bottom left would be empty and the 'middle line' where each city represents itself would be half filled with dots.
We want to remove that half filled portion. Half of the 30 cities is 15, so 450 - 15 = 435. Now the middle line is completely empty.
Drawing a picture with only a few lines is useful to visualize this.

Alternatively, we have a 30*30 for a total of 900. Since we can't choose the same city to represent itself we have Choose 1(30)*Choose 2(29) /2 (to get rid of arrangements) = 435

Number of entries would be equal to number of selection of 2 cities out of 30 = 30C2 = 435

IMO B

Posted from my mobile device
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When it is 5 pairs (total 25 entry):
Here diagonal carries 5 entries. So, this could be 25-5 (removing diagonal part)=20. So, each side of diagonal carries 20/2=10 entry (10 entry in filled . part and 10 entry in non-filled part).

When it is 30 pairs (total 30*30=900 entry):
Total diagonals=30 entry.
Removing diagonal=900-30==>870 entry. So, entries of table=870/2=435. The correct choice is B.
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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 in which 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 x 29) / 2! = 15 x 29 = 435

Alternate Solution:

Notice that when there are 5 cities, there are 1 + 2 + 3 + 4 entries.

If there were 6 cities, the last column would have contained 5 entries; therefore there would have been 1 + 2 + 3 + 4 + 5 entries. Notice that the last column always contains one less entry than the number of cities.

Thus, if there were 30 cities, there would have been 1 + 2 + 3 + … + 29 = (29 * 30)/2 = 29 * 15 = 435 entries.

Answer: B
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Progression method for me.

\(a1=0 \)(there's no * on the first row/column)
progression looks like this: 0,1,2,3,4,5.....
\(an=a1+ (n-1)d\)
common difference = 1
\(n=30\)
\(an=29\)

Sum when the set is evenly spaced= mean* number of elements.
\(Mean= a1+an/2\)
Number of elements= 30

\(0+29/2*30\) \(= 435\)