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Profession and Average Age (Posted on 2010-09-16) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
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a hexy problem | Comment 1 of 2

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).


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.


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
  Posted by ed bottemiller on 2010-09-16 18:44:30

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