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?

• Let $n_1, n_2, n_5, n_{10}$ be the number of 1 kg, 2 kg, 5 kg, and 10 kg weights used respectively.

• The total weight equation is $1n_1 + 2n_2 + 5n_5 + 10n_{10} = 20$.

• The constraint on 1 kg weights is $8 \le n_1 \le 11$.

• Since $n_1 \ge 8$, the remaining weight to be measured by other weights is $20 - n_1$. Let $W_{rem} = 20 - n_1$.

• The equation for the remaining weights is $2n_2 + 5n_5 + 10n_{10} = W_{rem}$.

• We analyze each valid value of $n_1$:

• If $n_1 = 8$, $W_{rem} = 12$. Solutions for $2n_2 + 5n_5 + 10n_{10} = 12$:

• $n_{10}=0 \Rightarrow 2n_2 + 5n_5 = 12$. $n_5=0 \Rightarrow n_2=6$ (1 way). $n_5=2 \Rightarrow 2n_2=2 \Rightarrow n_2=1$ (1 way). Total 2 ways.

• $n_{10}=1 \Rightarrow 2n_2 + 5n_5 = 2$. $n_5=0 \Rightarrow n_2=1$ (1 way). Total 1 way.

• Total for $n_1=8$ is $2+1=3$ ways.

• If $n_1 = 9$, $W_{rem} = 11$. Solutions for $2n_2 + 5n_5 + 10n_{10} = 11$:

• $n_{10}=0 \Rightarrow 2n_2 + 5n_5 = 11$. $n_5=1 \Rightarrow 2n_2=6 \Rightarrow n_2=3$ (1 way). $n_5=3 \Rightarrow 2n_2=-4$ (invalid). Total 1 way.

• $n_{10}=1 \Rightarrow 2n_2 + 5n_5 = 1$. No integer solution for non-negative $n_2, n_5$. Total 0 ways.

• Total for $n_1=9$ is 1 way.

• If $n_1 = 10$, $W_{rem} = 10$. Solutions for $2n_2 + 5n_5 + 10n_{10} = 10$:

• $n_{10}=0 \Rightarrow 2n_2 + 5n_5 = 10$. $n_5=0 \Rightarrow n_2=5$ (1 way). $n_5=2 \Rightarrow 2n_2=0 \Rightarrow n_2=0$ (1 way). Total 2 ways.

• $n_{10}=1 \Rightarrow 2n_2 + 5n_5 = 0 \Rightarrow n_2=0, n_5=0$ (1 way).

• Total for $n_1=10$ is $2+1=3$ ways.

• If $n_1 = 11$, $W_{rem} = 9$. Solutions for $2n_2 + 5n_5 + 10n_{10} = 9$:

• $n_{10}=0 \Rightarrow 2n_2 + 5n_5 = 9$. $n_5=1 \Rightarrow 2n_2=4 \Rightarrow n_2=2$ (1 way). Total 1 way.

• $n_{10} \ge 1$ is impossible as $10 > 9$.

• Total for $n_1=11$ is 1 way.

• Summing the ways: $3 + 1 + 3 + 1 = 8$.

• Therefore, the maximum number of different ways is 8, which corresponds to option (b).