Thanks a lot rich...I took 30 terms instead of 29...

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.

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

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

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

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.