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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.
(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.
13 Jan 2026, 10:33
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Official Solution:
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 \(= \frac{21c + 7b}{3c + b} = 7\).
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 \(= \frac{56k}{8k} = 7\).
Answer: D
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 \(= \frac{21c + 7b}{3c + b} = 7\).
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 \(= \frac{56k}{8k} = 7\).
Answer: D
