pumpkin riddle
If i brought 5 pumpkins to a weight contest,and instead of weighing them 1 at a time i weight them 2 at a time.I got the following weights 110,112,113,114,115,117,118,120 and 121pounds.How much did each pumpkin weighed?
2007-03-25 12:46:13 · 4 answers · asked by ccrider251 1 in Entertainment & Music ➔ Jokes & Riddles
49
50
55.5
53
60
2007-03-25 13:03:57 · answer #1 · answered by godschild 5 · 0⤊ 0⤋
Hi,
Label the pumpkins A- E, with A being lightest and E being heaviest.
Because the sums are so close, we know there is very little difference among the weights. No two weights are exactly the same or we wouldn't have 10 different sums.
The two lightest A + B = 110 (both odd or both even)
The two heaviest D + E = 121 (one odd and one even)
The lightest and the middle one A + C = 112
Since A + B = 110, we know that C is 2 lbs more than B.
Likewise, the heaviest and the middle one C + E = 120
Since D + E = 121, we know that D is 1 lb more than C.
A and B must be at least 2 apart because their sum is even. I'm betting on 2 apart and not 4 because the sums of the lightest 2 and the heaviest 2 are only 11 apart.
Try 54 and 56 for A and B, then C = 58, D = 59 and E = 62
Check this out and see if it works. It may take a little more juggling of the weights to make it work completely.
2007-03-25 12:52:39 · answer #2 · answered by Anonymous · 0⤊ 0⤋
119
2007-03-25 12:48:47 · answer #3 · answered by Katie S 3 · 0⤊ 0⤋
117...idk...just a lucky guess!!!
2007-03-25 12:55:57 · answer #4 · answered by ♪-~`*guitar♫chick*`~-♪ 3 · 0⤊ 0⤋