The Peytonville Peacocks and the Adenville Aardvarks have each played the same number of games so far this season.
The Peacocks have a .664 average, and the Aardvarks have won 70 games. Which team is ahead?
Note that each team plays 162 games in a season, and that the team's average is the number of games won divided by the number of games played rounded off to three decimal places.
By the problem, both the teams each played between 70 games and 162 games inclusively. Let the total number of games won and played by the Peacocks be W and L respectively. Then, in terms of provisions of the problem:
(i) 70<=W<L<=162 ; (ii) W/L = 0.664.
Now, 70/106 = 0.6603773 < 0.6666666 = 70/105;
or, 70/106 ~ 0.660 < 0.667~ 70/105.
Accordingly, W cannot be 70. We also observe in terms of trial and error that the minimum solution satisfying (i) and (ii) ocurs at W=71 and L=107 since 71/107 = 0.6635514 ~ 0.664.
Consequently, W must always be greater than 70. sine, it is given in terms of provisions of the problem that the Aardvarks won precisely 70 games, it follows that the total number of games won by the Peacocks will always be greater than that of the Aardvarks.
Having regard to the fact that equal number of games were played by both the teams, it follows that the PEACOCKS WILL ALWAYS BE AHEAD OF THE AARDVARKS IN TERMS OF MATCH WINNING AVERAGE.l
Edited on March 29, 2024, 12:57 pm
FULL SOLUTION
