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 Three in, one out (Posted on 2013-04-29)
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.

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 I've done something wrong. | Comment 1 of 5
At the end of the previous month the total age of the group is 37n.
This month (before the change in memberships) the average age is 37+1/12 and the total age is (445/12)n since everyone got a month older.
After adding and subtracting members the total age increases by 111-m and there are n+2 members so the new average is
[(445/12)n+111-m]/(n+2)

Since they are 18 months lower (than which?  I've assumed than last month)
[(445/12)n+111-m]/(n+2) = 37-1.5 = 71/2
409.5n+40=m
Which doesn't seem to have any values where both are squares. n=55^2, m=1113^2 is close but that person who left is over a million years old.

 Posted by Jer on 2013-04-29 14:04:46

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