There are four types of weights, namely 1 kg, 2 kg, 5 kg and 10 kg. What is the maximum number of different ways one can measure 20 kg, if at least eight but not more than eleven weights of 1 kg are to be used while measuring?

Correct Answer: (B) 8

Let $w_1$ be the number of 1 kg weights used, with $8$$\le$ w_1 $\le$ 11. The remaining weight $20 - w_1$ must be formed by weights of 2, 5, and 10 kg ($2a + 5b + 10c = 20 - w_1$). Testing each possible value of $w_1$: For $w_1 = 8$ (remaining 12 kg), there are 3 non-negative integer solutions: $(1, 0, 1)$, $(6, 0, 0)$, and $(1, 2, 0)$. For $w_1 = 9$ (remaining 11 kg), parity requires $b = 1$, yielding 1 solution: $(3, 1, 0)$. For $w_1 = 10$ (remaining 10 kg), there are 3 solutions: $(0, 0, 1)$, $(5, 0, 0)$, and $(0, 2, 0)$. For $w_1 = 11$ (remaining 9 kg), there is 1 solution: $(2, 1, 0)$. Summing all cases: $3 + 1 + 3 + 1 = 8$ ways.