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?
(In reply to
by Matt)
Aside from any requirement that all the weights be integers, your answer of 7.5, 8.5, 10.5, 11.5 and 15.5 adds up to 53.5. In the set of 10 weighings, each pumpkin was weighed 4 times (once in conjunction with each of the others), so if the total weight of the pumpkins were 53.5, the total of all 10 weighings would be 4 times that, or 214. But they actually add up to 216 pounds.
I think there is sufficient algebra available to actually eliminate the need for the extra proviso that all the individual weights are integers. Look at the other solutions.
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Posted by Charlie
on 2003-05-03 04:17:06 |