Fahim2 wrote:
In the football league of 2010, ManU won 40% of their games. In 2011 they won 60% of their Games. If there were thrice as many games played in the second as in the first, What percentage of the games did ManU win the 2 Years?
A) 115%
B) 60%
C) 57.5%
D) 55%
E) 40%
Don't know the Original source.
The original source was probably an assistant coach at Liverpool.
Anyway:
I. Assign Values:Pick simple numbers
Let the # of 1st year games = 10
Let the # of 2nd year games = 30 (= "thrice" 10)
# of games won, 1st year:
40% = (.4 * 10) = 4
# of games won, 2nd yr:
60% = (.6 * 30) = 18
Total games won: (18+4) = 22
Total games played: (10+30) = 40
Overall percent won?
\((\frac{GamesWon}{Total Games}*100)=(\frac{22}{40}*100)=\)
\(.55*100=55\) %
Answer D
II. Weighted average"Weight" the win percents from the ratio of number of games, which is 1 : 3
40% = .40 weighs 1
60% = .60 weighs 3
Total weight= (1 + 3) = 4
(You can use variables for weights but they're not necessary. x and 3x would get canceled out in the division by the total of the ratio parts, 4x)
\(\frac{.40(1) + .60(3)}{4}=\frac{.4+1.8}{4}=\frac{2.2}{4}=.55=\)
55 percent
Answer D
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