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?

The correct option is (b) - 8.

[as per provisional answerkey]

Solution

We need to measure exactly 20 kg using weights of 1 kg, 2 kg, 5 kg, and 10 kg. The constraint is that the number of 1 kg weights (let's call this 'n') must be 8, 9, 10, or 11. We need to find the total number of combinations for the remaining weight ($20-n$) using 2 kg, 5 kg, and 10 kg weights.

Case 1: n = 8 (Remaining weight = 12 kg)
Possible combinations of (10kg, 5kg, 2kg):
1. (1, 0, 1) -> $10+0+2=12$
2. (0, 2, 1) -> $0+10+2=12$
3. (0, 1, x) -> $5+(2\times x)=12$ (No integer solution for x)
4. (0, 0, 6) -> $0+0+12=12$
Total for Case 1: 3 ways

Case 2: n = 9 (Remaining weight = 11 kg)
Possible combinations of (10kg, 5kg, 2kg):
1. (1, 0, x) -> $10+(2\times x)=11$ (No integer solution)
2. (0, 2, x) -> $10+(2\times x)=11$ (No integer solution)
3. (0, 1, 3) -> $0+5+6=11$
Total for Case 2: 1 way

Case 3: n = 10 (Remaining weight = 10 kg)
Possible combinations of (10kg, 5kg, 2kg):
1. (1, 0, 0) -> $10+0+0=10$
2. (0, 2, 0) -> $0+10+0=10$
3. (0, 1, x) -> $5+(2\times x)=10$ (No integer solution)
4. (0, 0, 5) -> $0+0+10=10$
Total for Case 3: 3 ways

Case 4: n = 11 (Remaining weight = 9 kg)
Possible combinations of (10kg, 5kg, 2kg):
1. (0, 1, 2) -> $0+5+4=9$
2. (0, 0, x) -> $0+0+(2\times x)=9$ (No integer solution)
Total for Case 4: 1 way

Total number of ways = $3+1+3+1=8$.

Why the other options are incorrect

• Option (a) - 7: This count misses one valid combination, likely failing to account for the case where 10 kg is made solely of 2 kg weights or the 10 kg single weight.
• Option (c) - 9: This count overestimates the possibilities, likely by including a combination that results in an odd remainder (like 11 or 9) being solved with only 2 kg weights, which is mathematically impossible.
• Option (d) - 10: This is a significant overcount, possibly ignoring the constraint that the number of 1 kg weights must be between 8 and 11 inclusive.

Key Concept

This is a Partition Problem with constraints, where the total sum is fixed and the parity (even/odd nature) of the remaining weight determines the feasibility of using 2 kg weights.