sgr21 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
We 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 2nd half, 2nd half = 20 games.
Whole season:
Total games = 10+20 = 30.
Since 2/3 are won, total won = (2/3) * 30 = 20.
1st half:
Since 2/5 are won, number won = (2/5) * 10 = 4.
2nd half:
Number won = total won - number won in the 1st half = 20-4 = 16.
Number lost = total games in the 2nd half - number won in the 2nd half = 20-16 = 4.
Doesn't work: only 2 games should be lost in the second half.
Since the result is TWICE what is required, all of the values in the problem -- including the value in the correct answer choice -- must decrease by 1/2.
The correct answer is
B.
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