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Difficulty:
95%
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Question Stats:
39% (02:26) correct
61%
(02:28)
wrong
based on 64
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A bakery packed 150 dessert boxes in two batches. If each box contained between 2 and 5 eclairs, inclusive, was the average (arithmetic mean) number of eclairs per box greater than 3.5?
(1) The median number of eclairs per box was 4.5.
(2) The second batch contained 110 boxes and had an average (arithmetic mean) of 4.2 eclairs per box.
(1) The median number of eclairs per box was 4.5.
(2) The second batch contained 110 boxes and had an average (arithmetic mean) of 4.2 eclairs per box.
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16 Apr 2026, 02:45
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GMAT Club Official Solution:
A bakery packed 150 dessert boxes in two batches. If each box contained between 2 and 5 eclairs, inclusive, was the average (arithmetic mean) number of eclairs per box greater than 3.5?
We need to know whether the average number of eclairs per box was greater than 3.5. Since there were 150 boxes, this is equivalent to asking whether the total number of eclairs was greater than 150 * 3.5 = 525.
(1) The median number of eclairs per box was 4.5.
Since there are 150 boxes, the median is the average of the 75th and 76th values after the boxes are arranged in increasing order.
Because each box contains an integer number of eclairs, and each box contained between 2 and 5 eclairs, a median of 4.5 means:
75th value = 4
76th value = 5
Let's check whether the total number of eclairs could have been 525 or less. To make the total as small as possible:
• the first 74 boxes must each contain 2 eclairs,
• the 75th box contains 4,
• and boxes 76 through 150 must each contain 5.
So the minimum possible total is 74 * 2 + 4 + 75 * 5 = 527
Since even the minimum possible number of eclairs, 527, is greater than 525, the answer to the question is YES. Sufficient.
(2) The second batch contained 110 boxes and had an average (arithmetic mean) of 4.2 eclairs per box.
So the second batch contained 110 * 4.2 = 462 eclairs.
That leaves 150 - 110 = 40 boxes in the first batch. Since each box had at least 2 eclairs, the first batch had at least 40 * 2 = 80 eclairs.
So the total number of eclairs was at least 462 + 80 = 542.
Since 542 > 525, the answer to the question is YES. Sufficient.
Answer: D.
A bakery packed 150 dessert boxes in two batches. If each box contained between 2 and 5 eclairs, inclusive, was the average (arithmetic mean) number of eclairs per box greater than 3.5?
We need to know whether the average number of eclairs per box was greater than 3.5. Since there were 150 boxes, this is equivalent to asking whether the total number of eclairs was greater than 150 * 3.5 = 525.
(1) The median number of eclairs per box was 4.5.
Since there are 150 boxes, the median is the average of the 75th and 76th values after the boxes are arranged in increasing order.
Because each box contains an integer number of eclairs, and each box contained between 2 and 5 eclairs, a median of 4.5 means:
75th value = 4
76th value = 5
Let's check whether the total number of eclairs could have been 525 or less. To make the total as small as possible:
• the first 74 boxes must each contain 2 eclairs,
• the 75th box contains 4,
• and boxes 76 through 150 must each contain 5.
So the minimum possible total is 74 * 2 + 4 + 75 * 5 = 527
Since even the minimum possible number of eclairs, 527, is greater than 525, the answer to the question is YES. Sufficient.
(2) The second batch contained 110 boxes and had an average (arithmetic mean) of 4.2 eclairs per box.
So the second batch contained 110 * 4.2 = 462 eclairs.
That leaves 150 - 110 = 40 boxes in the first batch. Since each box had at least 2 eclairs, the first batch had at least 40 * 2 = 80 eclairs.
So the total number of eclairs was at least 462 + 80 = 542.
Since 542 > 525, the answer to the question is YES. Sufficient.
Answer: D.
General Discussion
09 Apr 2026, 09:22
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Statement 1:
When median = 4.5:
75th = 4 and 76th = 5.
Take the absolute lowest now possible:
2*74 = 148
5*74 = 370
Since both of them are equal in number the mean would be in between = 3.5
Now we also have a 4 and 5 that would take mean to be >3.5
SUFFICIENT.
Statement 2:
Second batch = 110*4.2 = 462
First batch = 40*2 = 80 [Minimum]
462+80 = 542/150 > 3.5.
SUFFICIENT.
Answer: Option D
When median = 4.5:
75th = 4 and 76th = 5.
Take the absolute lowest now possible:
2*74 = 148
5*74 = 370
Since both of them are equal in number the mean would be in between = 3.5
Now we also have a 4 and 5 that would take mean to be >3.5
SUFFICIENT.
Statement 2:
Second batch = 110*4.2 = 462
First batch = 40*2 = 80 [Minimum]
462+80 = 542/150 > 3.5.
SUFFICIENT.
Answer: Option D
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