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…

The correct option is (b) - 8.

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$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$10+0+2=12
2. (0, 2, 1) -> $0 + 10 + 2 = 12$0+10+2=12
3. (0, 1, x) -> $5 + (2 \times x) = 12$5+(2×x)=12 (No integer solution for x)
4. (0, 0, 6) -> $0 + 0 + 12 = 12$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$10+(2×x)=11 (No integer solution)
2. (0, 2, x) -> $10 + (2 \times x) = 11$10+(2×x)=11 (No integer solution)
3. (0, 1, 3) -> $0 + 5 + 6 = 11$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$10+0+0=10
2. (0, 2, 0) -> $0 + 10 + 0 = 10$0+10+0=10
3. (0, 1, x) -> $5 + (2 \times x) = 10$5+(2×x)=10 (No integer solution)
4. (0, 0, 5) -> $0 + 0 + 10 = 10$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$0+5+4=9
2. (0, 0, x) -> $0 + 0 + (2 \times x) = 9$0+0+(2×x)=9 (No integer solution)
Total for Case 4: 1 way

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