Difficulty: Medium
Correct Answer: 220
Explanation:
Introduction / Context:
This final problem in the set continues the pattern of subset-sum reasoning with four given box weights. Again, we must determine which listed total weight cannot be obtained by selecting some or all of the boxes, with each used at most once. The question tests careful enumeration and the ability to rule out impossible totals.
Given Data / Assumptions:
Concept / Approach:
As with previous questions of this type, the straightforward method is to list all possible sums of the four weights. Because there are only four boxes, the total number of non-empty subsets is limited and manageable. Once we have the full list, we compare it against the answer options and pick the value that does not appear.
Step-by-Step Solution:
Verification / Alternative check:
To verify, try constructing 220 kg directly. Three-box totals are 160, 170, 190 and 200 kg, none of which equals 220. Adding the remaining box to any three-box combination always yields either 240 kg (the four-box total) or a value greater than 220, which is not possible here. Therefore, 220 kg cannot be obtained by any allowed combination.
Why Other Options Are Wrong:
Option 240 kg is the sum of all four box weights: 40 + 50 + 70 + 80.
Option 160 kg is achieved as 40 + 50 + 70.
Option 200 kg is achieved as 50 + 70 + 80.
Common Pitfalls:
Examinees may overlook one of the triple combinations or may incorrectly assume that if a number lies between the smallest and largest possible totals it must be achievable. This is not true when selections are limited to specific discrete weights. Careful calculation of each combination prevents such errors.
Final Answer:
The total weight that cannot be formed is 220 kilograms.
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