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Three in, one out (Posted on 2013-04-29) Difficulty: 2 of 5
The average age of our group of n people was 37 yrs at the end of the previous month.

This month 3 new members enrolled (aged 33, 38, and 40 yrs) and one member (aged m yrs) left.

Now the average age of our group became lower by 18 months.

Given that both m and n are square numbers, find their values.

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.5000 (2 votes)

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Solution possible solutions? | Comment 2 of 5 |

After having difficulty finding a solution where the calculation of averages took into consideration ages by month, I thought maybe ages were calculated as of the last birthday, and that the 18-month difference in averages was really a 1.5-year difference in averages of integral years.

I assumed that in the "group of n people" n was the number of people last month so that n+2 is the number this month.

DEFDBL A-Z
FOR srm = 2 TO 11
  m = srm * srm
  FOR srn = 2 TO 11
    n = srn * srn
    lastmo = n * 37
    FOR bd = 0 TO n
      thismo = lastmo + bd + 38 + 33 + 40 - m
      'IF thismo / (n + 2) = lastmo / (n) - 1.5 THEN PRINT m, n
      IF thismo * (n) = lastmo * (n + 2) - 1.5 * n * (n + 2) THEN PRINT m; "years", n; "people", bd; "birthdays"
    NEXT bd
  NEXT
NEXT

finds

49 years      4 people      3 birthdays
64 years      16 people     0 birthdays
100 years     36 people     6 birthdays
121 years     36 people     27 birthdays

for m, n and the number of birthdays taking place in the intervening month.

For example, with 4 people originally the total ages were 4*37=148; when 33+38+40-49 are added in, the total becomes 210, but with 3 birthdays having passed, it's 213. With the group now at 6, the average age is 35.5, or 18 months younger.

If you want a solution with no birthdays having passed, m would be 64 and n = 16.


  Posted by Charlie on 2013-04-29 14:41:50
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