 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Pumpkins (Posted on 2003-05-01) Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers.

How much does each pumpkin weigh?

 See The Solution Submitted by Ravi Raja Rating: 3.0000 (7 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Puzzle Solution | Comment 15 of 16 | Let the weights of the 5 pumpkins in decreasing order of magnitude be a,b,c,d,e.

Summing over all the weights of the 10 pairs of pumpkins, we have:

4(a+b+c+d+e) = 216, giving:
a+b+c+d+e = 54

Now, the largest pair sum must correspond to a+b, while the smallest pair sum is given by d+e
Thus, (a+b, c+d) = (27, 16), so that:
c = 54-27-16 = 11

Second largest pair sum is 26, and so: a+c = 26, giving a = 26-11= 15, so that: b = 27-15 = 12

Second smallest apir sum is 18, so that: c+e = 18, giving: e = 18-11 = 7, so that: d = 16-7 = 9

Consequently, the required weights in decreasing order of magnitude are 15, 12, 11, 9 and 7.

 Posted by K Sengupta on 2007-12-01 05:14:37 Please log in:

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