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In the country of Mulligan, 90% of the golfing population 26 Nov 2003, 11:18
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In the country of Mulligan, 90% of the golfing population have handicaps equal or greater than 90 and are considered "Average", while 8% of the golfing population, who are considered "Good" golfers, have handicaps less than 90 but greater than or equal to 10. In addition, "Exceptional" golfers have who have handicaps less than 10 are only 2% of the golfing population. Furthermore, 30% of the population of Mulligan belong to a country club. If 60% of "Exceptional" golfers belong to a country club and 30% of "Good" golfers do not belong to a country club what percent of the golfing population of Mulligan are average golfers that do not belong to a country club?



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In the country of Mulligan, 90% of the golfing population 26 Nov 2003, 13:31
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60% of 2% is 1,2% of exceptional golfers who belong to the club.
30% do not belong , so 70% of the good golfers belong to the club or 70%x8%=5,6% since there are total 30% club members who belong to the club so 30-6,8=23,2% who are average club members and belong to the club, or 90%-23,2%=76,8% who are average and do not belong to the club
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In the country of Mulligan, 90% of the golfing population 26 Nov 2003, 13:36
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90% - 23.2% = 66.8%. Nice work BG. :wink:
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In the country of Mulligan, 90% of the golfing population 26 Nov 2003, 13:44
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in the club:
exp = 2 * 60% = 1.2%
good = 8 * 70% = 5.6%
makes 6.8% + average% in the club = 30%
average% = 23.2%

out of 90% = 90%-23.2% = 66.8%
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In the country of Mulligan, 90% of the golfing population 26 Nov 2003, 14:33
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Was this problem too easy for you guys? Seemed like a cake walk. I'll have to come up with more "difficult" questions. Good work!



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