   SEARCH HOME Math Central Quandaries & Queries  Question from Jose, a student: How would I solve the following problem? How would I write an equation? Mr. Appleberry brought in pumpkins for science class, one for each group. The fist task wa to weigh the pumpkins, but the scales they had wouldn't register any weight under 7 pounds, and it seemed that two or three pumpkins were under that weight. The class decided that they could weigh the pumpkins two at a time, so they weighed each pumpkin with each of other pumpkins. The weights registered on the scale were: 7 pounds 8 pounds 9 pounds 10 pounds 11 pounds 12 pounds How many pumpkins were there, and how much did each weigh? Hi Jose,

How many pumpkins were there?

Suppose there were 3 and you label them A, B and C. Weighing 2 at a time would be pairs AB, AC and BC. But there were 6 pairs so 3 pumpkins is not enough.

Suppose there were 4 pumpkins and you label them A, B, C and D. Weighing 2 at a time would be pairs AB, AC, AD, BC, BD and CD. This is 6 pairs so there must be 4 pumpkins.

Something else I notice is that each of the pairs gave different weights so no 2 pumpkins weigh the same. Suppose the weights are a, b, c and d then I can assume a < b < c < d.

Look at the pairs above. A was weighed 3 times, B was weighed 3 times, and the same for C and D. Thus the weights 7, 8, 9, 10, 11 and 12 come from weighing each pumpkin 3 times. Hence 7 + 8 + 9 + 10 + 11 + 12 = 57 pounds results from weighing each pumpkin 3 times. Thus 3 × (a + b + c + d) = 57 and a + b + c + d = 57/3 = 19 pounds.

Can you finish the problem now?

Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.