Lots of people sent in correct solutions for this one. There are many ways to work through this problem, so the best solutions included a description of the thinking that was used.
The answer is:
Amil - 7 eggs
Jack - 8 eggs
Peter - 9 eggs
Melanie - 14 eggs (Lucky girl!)
Abbie (St Peters School, St Albans UK) first used logic to work out which child had the most eggs, then used this as a starting point for using the numbers. Abbie said:
I worked it out first by doing this ...
Jack < Peter (has less than)
Peter < Melanie
Amil < Melanie
Amil < Peter
Then you can see that the people with most eggs are Melanie and Peter, but Peter has five less than Melanie, so Melanie has most eggs. Then you can do this - Melanie has most, Peter has five less than her (Melanie -5), Jack has 1 less than Peter (Melanie -6), and Amil has 2 less than Peter (Melanie -7). As we know that Amil has half as many eggs as Melanie then Melanie must have 14 eggs, and then you can work the rest out! I checked by adding all the eggs together and they make 38.
Timothy used the first letter of each child's name, together with the numbers given in the clues, to make some equations. Timothy said:
To work out the puzzle I used Algebra.
J + 1 = P
P + 5 = M
M/2 = A
A + 2 = P
I knew that the two 'P's (standing for Peter) must equal the same so I guessed this,
6 = J, so J + 1 = P so P = 7 so
7 + 5 = 12 so M = 12 so
12/2 = 6 so A = 6 so
6 + 2 = 8 so P = 8
The guess was wrong as Peter is 7 and 8.
8 = J so J + 1 = P = 9
P + 5 = 14 = M
M/2 = 7 = A
A + 2 = 9 = P
As P are now the same, all the numbers were added up to show
8 + 9 + 14 + 7 = 38 so
Peter had 9 eggs
Melanie had 14 eggs
Jack had 8 eggs
Amil had 7 eggs
Correct solutions were also sent in by:
Susannah (Headington Junior School)
Laura (All Saints School, Whiteparish)
William, Byron & Kerry (Downing Primary, Ipswich)
Harry (St Peters College, Adelaide, Australia)
Mark (St Johns, Walsall Wood)