In the first half of the season a football team

Finally the team won 66.67% of the total matches, i.e. 2/3 of the total matches.
Let us assume total matches played = 3n

As number of matches in the 2nd half is double of 1st half, matches played in 1st half = n and matches played in the 2nd half = 2n

Number of matches lost = 60% of n + 2 = (0.6)*n + 2
This will be equal to 1/3 of total matches, i.e. n

So, (0.6)*n + 2 = n
So, (0.4)*n = 2
So, n = 5

Answer : B

guerrero wrote:In the first half of the season a football team participating in the local league won 40% of its matches. It then went on to win all its matches except two in the second half of the season. If it is known that for the entire season the percentage of matches won by the team was 66.67% and that the team played double the number of matches in the second half of the season as compared to the first half of the season then how many matches did the team play in the first half of the season?

(A) 2
(B) 5
(C) 10
(D) 15
(E) 20

OAB

An alternate approach is to plug in the answers, which represent the number of games played in the first half of the season.

Answer choice C: first half = 10 games
Since twice as many games are played in the second half of the season, second half = 20 games.
Total games = 10+20 = 30.

Since 2/3 of all of the games are won, total won = (2/3)30 = 20.
Since 2/5 of the games in the first half are won, the number won in the first half = (2/5)10 = 4.

Thus, the number won in the second half = total won - number won in the first half = 20-4 = 16.
Thus, the number LOST in the second half = total games in the second half - number won = 20-16 = 4.

Doesn't work: only 2 games should be lost in the second half.
Since the number lost in the second half is TWICE what is required, all of the values in the problem -- including the number of games played in the first half -- must decrease by 1/2.

The correct answer is B.