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Sunday Times Teaser 2822 – Losing Weight

by Michael Fletcher

Published: 23 October 2016 (link)

The members of our local sports club are split into two groups. Originally the groups had the same number of members, and the first group contained member Waites who weighed 100kg (and indeed that was the average weight of the members of the first group). Then the club trainer decided that the groups were overweight and that, for each of the next ten weeks, for each group the average weight of the members should be reduced by one kilogram.

This was achieved by simply transferring a member from the first group to the second group each week, and Waites was the tenth member to be transferred.

How many members are there in the club?

2 Comments Leave one →
  1. Brian Gladman permalink

    Here is manual solution.

    Let the club have \(2n\) members with the average weight of the people in the two equal size sub-groups being \(u\) and \(v\) kilograms respectively. If \(w_k\) is the weight of the member moved from the first to the second sub-group in week \(k\) (\(0 <= k <= 9\)), we have: \[(u-k)(n-k)-w_k = (u-k-1)(n-k-1)\] Hence: \[w_k=u+n-(2k+1)\] Since we are told that \(u\) is 100 kilograms and that the weight of Waites (\(w_9\)) is also 100 kilograms, this shows that \(n\) is 19 and the number of club members is 38.

    For the second sub-group, we have: \[(v-k)(n+k) + w_k=(v-k-1)(n+k+1)\] which simplifies to \[w_k=v-n-(2k+1)\]
    Eliminating \(w_k\) now shows that \(v=u+2n\), giving the initial average weight of the members in sub-group two as 138 kilograms.

  2. Not a very satisfying puzzle this week.As it quite quickly boils down to solving:

    \[ 90(n – 10) = 91(n – 9) – 100 \];

    Nevertheless I wrote a program to investigate solutions for increasing sized groups, until it finds the group described by the puzzle:

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