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 answer by K Sengupta)

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