In a company of Clergymen, Doctors and Lawyers it is found that the sum of the ages of all present is 2160; their average age is 36; the average age of the Clergymen and the Doctors is 39; the average age of the Doctors and the Lawyers is (32 + 8/11); the average age of the Clergymen and the Lawyers is (36 + 2/3).
If each Clergymen had been 2 years older, each Lawyer 5 years older, and each Doctor 7 years older, then the average age of everyone present would have been greater by 5 years.
Determine the number of each profession present and their average ages.
Clearly there are 60 people all together (2160 / 36 i.e. sum of all "ages" divided by the "average age"). To get the "average age" we must apparently assume that we are averaging the "whole-year" age of each, which is not a true average age. If this were my birthday then I would be "age X"; but I would also be "age X" 364 days hence, or anywhere in between. The "average age" of a group is not the average of the years+fractions ages of its members. I guess, if I were aged "30 and a day" or "30 and 300 days" I would still be averaged as though age 30.000. Or, perhaps not?? We can't have fractional people (I hope).
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Since there are 60 people, and their average age increases by 5 years, there will be a total of 300 more years in the sum of their ages. There are 11 ways in which integer values for the number of clerics,docs,and legals, given respectively 2, 7, and 5 more years come to a total of 300 years (e.g. 2 C, 3 D, 55 L ... 22 C, 33 D, 5 L). From that, the next step would be to see which of these eleven (presumably only one set) allows the results given for the three groups of two occupations.
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The eleven sets of possible values for C, D, and L are
2 3 55
4 6 50
6 9 45
8 12 40
10 15 35
12 18 30
14 21 25
16 24 20 **
18 27 15
20 30 10
22 33 5
The long weekend is now beginning, so I'll look back in next Monday. On partial grounds I would select 16 - 24 - 20 above since that is the only group for which the sum of doctors and lawyers (i.e. 44) is divisible by 11, but one would still need to figure a compatible set of average ages. Duty calls.
Edited on September 16, 2010, 7:57 pm
Edited on September 16, 2010, 8:42 pm
a hexy problem
