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 [ 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.)