• Try this problem by trial and error.
• Try to solve this problem in a non-trial and error way.
• Is there more than one solution?

Bad way

Could the masses of the three pumpkins be \(4\) kg, \(6\) kg, and \(10\) kg? No — we don't get any of the three totals.

Could the masses of the three pumpkins be \(1\) kg, \(7\) kg, and \(14\) kg? No — we only get one of the three totals.

Better way

Lightest pumpkin: \(2\) kg

• Second pumpkin: \(10\) kg
• Third pumpkin: \(11\) kg
•  \(10\) kg \(+\) \(11\) kg  $\ne$ \(15\) kg 

Lightest pumpkin: \(4\) kg

• Second pumpkin: \(8\) kg
• Third pumpkin: \(9\) kg 
•  \(8\) kg \(+\) \(9\) kg \(\neq\) \(15\) kg

Lightest pumkin: \(5\) kg

• Second pumpkin: \(7\) kg
• Third pumpkin:  \(8\) kg
• \(7\) kg \(+\) \(8\) kg = \(15\) kg.

Therefore, the mass of the lightest pumpkin could be \(5\) kg.

In kilograms, let the masses of the three pumpkins in increasing order be \(A\), \(B\), and \(C\).

The lightest combined mass, \(12\) kg, must come from weighing the two lightest pumpkins.

That is, \(A+B=12\).

The heaviest combined mass, \(15\) kg, must come from weighing the two heaviest pumpkins.

That is, \(B+C=15\).

Then the third given mass, \(13\) kg, is the combined mass of the lightest and heaviest pumpkins.

That is, \(A+C=13\).

Since \(A+B=12\) and \(A+C=13\), then \(C\) is one more than \(B\).

Since \(B+C=15\) and \(C\) is one more than \(B\), then \(B=7\) and \(C=8\).

Since \(A+B=12\), then \(A+7=12\), or \(A=5\).

When \(A=5\), \(B=7\), and \(C=8\), the masses of the pairs of pumpkins are \(12\) kg, \(13\) kg, and \(15\) kg as we were given.

Therefore, the mass of the lightest pumpkin is \(5\) kg.