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D
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Difficulty:
95%
(hard)
Question Stats:
29% (01:41) correct
71%
(01:20)
wrong
based on 201
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Each of the boxes in a shipment weighs either 5 pounds, 7 pounds, or 11 pounds. What is the average (arithmetic mean) weight per box in the shipment?
(1) The number of boxes that weigh 5 pounds is twice as many as the number that weigh 11 pounds.
(2) The ratio of the number of boxes weighing 5 pounds, 7 pounds, and 11 pounds is 4 to 2 to 2.
I31-07
(1) The number of boxes that weigh 5 pounds is twice as many as the number that weigh 11 pounds.
(2) The ratio of the number of boxes weighing 5 pounds, 7 pounds, and 11 pounds is 4 to 2 to 2.
I31-07
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GMAT Club Official Explanation:
Each of the boxes in a shipment weighs either 5 pounds, 7 pounds, or 11 pounds. What is the average (arithmetic mean) weight per box in the shipment?
(1) The number of boxes that weigh 5 pounds is twice as many as the number that weigh 11 pounds.
Let a, b, c be the numbers of 5 pound, 7 pound, and 11 pound boxes. We are told a = 2c.
Total weight = 5a + 7b + 11c = 5(2c) + 7b + 11c = 21c + 7b.
Total boxes = a + b + c = 2c + b + c = 3c + b.
Average = (21c + 7b)/(3c + b) = 7.
Sufficient.
Alternative way:
Assume there are some number of 7-pound boxes. There are twice as many 5-pound boxes as 11-pound boxes. So for every one 11-pound box that pulls the average up by 4 pounds from 7, there are two 5-pound boxes that together pull it down by an equal amount. These pulls cancel out, so the average stays at 7 pounds.
(2) The ratio of the number of boxes weighing 5 pounds, 7 pounds, and 11 pounds is 4 to 2 to 2.
Let the counts be 4k, 2k, 2k.
Total weight = 5(4k) + 7(2k) + 11(2k) = 56k.
Total boxes = 8k.
Average = 56k/(8k) = 7.
Sufficient.
Answer: D.
General Discussion
18 Dec 2025, 09:25
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I. Let the number of 11-pound boxes be x. Then 5-pound boxes are 2x, but the number of 7-pound boxes is unknown.
The average cannot be uniquely determined.
Not sufficient
II. The ratio of boxes weighing 5,7, and 11 pounds is 4:2:2.
Total boxes =8
Total weight = (4*5) + (2*7) + (2*11)
=> 20+14+22
=> 56
Average weight 56/8 = 7
Unique value
Sufficient
Answer: B
The average cannot be uniquely determined.
Not sufficient
II. The ratio of boxes weighing 5,7, and 11 pounds is 4:2:2.
Total boxes =8
Total weight = (4*5) + (2*7) + (2*11)
=> 20+14+22
=> 56
Average weight 56/8 = 7
Unique value
Sufficient
Answer: B
