The Question: Did more than half of the fragile boxes contain electronics?
Let:
F = number of fragile boxes
NF = number of non-fragile boxes
E = number of boxes containing electronics
NE = number of boxes not containing electronics
F_E = number of fragile boxes containing electronics
NF_E = number of non-fragile boxes containing electronics
Total boxes = F + NF
We want to determine if F_E > 0.5 * F.
Statement (1): Two-thirds of the boxes in the shipment are fragile.
This means F / (F + NF) = 2/3.
So, F = (2/3) * Total boxes.
And NF = (1/3) * Total boxes.
This statement tells us the proportion of fragile boxes, but nothing about electronics.
Therefore,
Statement (1) alone is not sufficient.Statement (2): Four-fifths of the boxes in the shipment contain electronics.
This means E / (F + NF) = 4/5.
So, E = (4/5) * Total boxes.
We know that E = F_E + NF_E.
This statement tells us the total proportion of boxes with electronics, but doesn't distinguish between fragile and non-fragile.
Therefore,
Statement (2) alone is not sufficient.Combining Statements (1) and (2):
From (1), F = (2/3) * Total boxes.
From (2), E = (4/5) * Total boxes.
We know that F_E + NF_E = E.
And we also know that F_E <= F (the number of fragile boxes with electronics cannot exceed the total number of fragile boxes).
And NF_E <= NF (the number of non-fragile boxes with electronics cannot exceed the total number of non-fragile boxes).
We want to know if F_E > 0.5 * F.
Let's assume the total number of boxes is 15 (a common multiple of 3 and 5 to make calculations easier).
From (1):
F = (2/3) * 15 = 10 fragile boxes.
NF = (1/3) * 15 = 5 non-fragile boxes.
From (2):
E = (4/5) * 15 = 12 boxes contain electronics.
Now, we need to see how these 12 electronic boxes are distributed between the 10 fragile and 5 non-fragile boxes.
We know F_E + NF_E = 12.
Consider the minimum possible value for F_E.
The maximum number of electronics that can be in non-fragile boxes is 5 (NF_E <= NF).
If NF_E = 5, then F_E = 12 - 5 = 7.
In this case, F_E = 7, and F = 10. Is 7 > 0.5 * 10 (which is 5)? Yes, 7 > 5.
Consider the maximum possible value for F_E.
The maximum number of electronics that can be in fragile boxes is 10 (F_E <= F).
If F_E = 10, then NF_E = 12 - 10 = 2. This is possible since NF is 5, and 2 <= 5.
In this case, F_E = 10, and F = 10. Is 10 > 0.5 * 10 (which is 5)? Yes, 10 > 5.
Let's try to find a scenario where F_E is NOT greater than 0.5 * F.
We need F_E <= 5 (since 0.5 * F = 0.5 * 10 = 5).
If F_E = 5 (or less), then NF_E = 12 - 5 = 7.
But NF (non-fragile boxes) is only 5. So, NF_E cannot be 7. This scenario is impossible.
This implies that F_E must be greater than 5.
Since F_E + NF_E = 12, and NF_E cannot exceed NF (which is 5), the minimum value for F_E is 12 - 5 = 7.
Since F_E (minimum 7) is always greater than 0.5 * F (which is 5), the answer to the question is always "Yes".
Therefore, both statements combined are sufficient.The final answer is C