Gmat_mission wrote: ↑Thu May 21, 2020 1:11 am
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
[spoiler]OA=B[/spoiler]
Source: Official Guide
APPROACH #1:
Each entry in the mileage table denotes a distinct pair of cities.
We can determine the total number of distinct pairs of cities by using combinations.
When there are 5 cities, the total number of distinct pairs of cities = 5C2 = 10
Noticed that in the given table there are 10 entries. Perfect!
Likewise, if we have a mileage table consisting of 30 cities, the total number of distinct pairs of cities = 30C2
= (30)(29)/(2)(1)
= 435
Answer: B
APPROACH #2:
Notice that, when there are 5 cities in the mileage table, the number of entries = 1 + 2 + 3 + 4
Likewise, if we have a mileage table consisting of 30 cities, the number of entries =
1 + 2 + 3 + . . . . + 28 + 29
One way to calculate this 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
Answer: B
Cheers,
Brent
