Mr. T has a wrong weighing pan. One arm is lengthier than the other. 1…

A defective ("false") balance with unequal arms behaves as if one pan always carries a fixed extra weight compared to the other, no matter what is placed on it. Any single reading where the pans appear level is really one linear equation in TWO unknowns -- the true unit weight AND this fixed bias -- so one reading alone cannot pin down the unit weight. Two independent readings give two such equations, and subtracting them cancels the unknown bias and leaves the true weight directly.

Let m be the weight of one melon (in kg), and let x be the fixed weight-equivalent bias by which the left pan reads heavier than the right pan, so an apparent balance really means (true weight on the left) + x = (true weight on the right).

1. Reading 1 -- 1 kg on the left pan balances 8 melons on the right pan: 1 + x = 8m ...(i)

2. Reading 2 -- 1 kg on the right pan balances 2 melons on the left pan: 2m + x = 1 ...(ii)

3. Subtract (ii) from (i) to cancel x: (1 + x) - (2m + x) = 8m - 1, so 1 - 2m = 8m - 1, giving 2 = 10m.

4. So m = 0.2 kg = 200 g.

Substituting m = 0.2 back into (ii) gives x = 1 - 2(0.2) = 0.6 kg; checking (i): 1 + 0.6 = 1.6, and 8(0.2) = 1.6 -- the same bias value emerges from both readings, confirming 200 g is the unique weight consistent with both observations.