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A museum received 7 storage crates for an upcoming exhibit, including two crates labeled Atlas and Birch. If each of the 7 crates weighed at least 50 kilograms and the average (arithmetic mean) weight of the 7 crates was 60 kilograms, did Atlas weigh less than 90 kilograms?
(1) Atlas weighed 9 kilograms more than Birch.
(2) The median weight of the 7 crates was 65 kilograms.
(1) Atlas weighed 9 kilograms more than Birch.
(2) The median weight of the 7 crates was 65 kilograms.
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09 Apr 2026, 02:30
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GMAT Club Official Solution:
A museum received 7 storage crates for an upcoming exhibit, including two crates labeled Atlas and Birch. If each of the 7 crates weighed at least 50 kilograms and the average (arithmetic mean) weight of the 7 crates was 60 kilograms, did Atlas weigh less than 90 kilograms?
(1) Atlas weighed 9 kilograms more than Birch.
Since the average weight of the 7 crates was 60 kilograms, the total weight was
7 * 60 = 420
Also, each of the other 5 crates weighed at least 50 kilograms, so together they weighed at least
5 * 50 = 250
That means Atlas and Birch together weighed at most
420 - 250 = 170
Since Atlas weighed 9 kilograms more than Birch, if Atlas had weighed 90 kilograms or more, then Birch would have weighed 81 kilograms or more, and together they would have weighed at least 171 kilograms, which is impossible. So Atlas must have weighed less than 90 kilograms. Sufficient.
(2) The median weight of the 7 crates was 65 kilograms.
Arrange the 7 weights in increasing order. Then the fourth weight is 65.
Let's check whether Atlas could weigh more than 90 kilograms. To maximize Atlas weight, make the other 6 crates as light as possible while keeping the median equal to 65. So the 6 smallest possible weights are:
{50, 50, 50, 65, 65, 65}
Their total is:
50 + 50 + 50 + 65 + 65 + 65 = 345
Since the total for all 7 crates is 420, the maximum weight of Atlas is therefore:
420 - 345 = 75
So Atlas must weigh less than 90 kilograms. Sufficient.
Answer: D.
A museum received 7 storage crates for an upcoming exhibit, including two crates labeled Atlas and Birch. If each of the 7 crates weighed at least 50 kilograms and the average (arithmetic mean) weight of the 7 crates was 60 kilograms, did Atlas weigh less than 90 kilograms?
(1) Atlas weighed 9 kilograms more than Birch.
Since the average weight of the 7 crates was 60 kilograms, the total weight was
7 * 60 = 420
Also, each of the other 5 crates weighed at least 50 kilograms, so together they weighed at least
5 * 50 = 250
That means Atlas and Birch together weighed at most
420 - 250 = 170
Since Atlas weighed 9 kilograms more than Birch, if Atlas had weighed 90 kilograms or more, then Birch would have weighed 81 kilograms or more, and together they would have weighed at least 171 kilograms, which is impossible. So Atlas must have weighed less than 90 kilograms. Sufficient.
(2) The median weight of the 7 crates was 65 kilograms.
Arrange the 7 weights in increasing order. Then the fourth weight is 65.
Let's check whether Atlas could weigh more than 90 kilograms. To maximize Atlas weight, make the other 6 crates as light as possible while keeping the median equal to 65. So the 6 smallest possible weights are:
{50, 50, 50, 65, 65, 65}
Their total is:
50 + 50 + 50 + 65 + 65 + 65 = 345
Since the total for all 7 crates is 420, the maximum weight of Atlas is therefore:
420 - 345 = 75
So Atlas must weigh less than 90 kilograms. Sufficient.
Answer: D.
General Discussion
02 Apr 2026, 22:20
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Statement 1:
Take the extreme values:
81,90.
Take remaining as 50 since its minimum and check if its possible.
50,50,50,50,50,81,90
Sum = 421>420 [Original sum required 60*7 = 420]
Since at the very least we are having a 421 sum if we include 90, then all values are less than 90.
SUFFICIENT.
Statement 2:
Against minimize all values and check:
50,50,50,65,65,65,90.
Is this possible?
Sum = 150+195+90 = 435 > 420 again.
Thus 90 is not possible to be had as the weights for the 7 crates and should be much lesser to bring total sum down to 420.
SUFFICIENT.
Answer: Option D
Take the extreme values:
81,90.
Take remaining as 50 since its minimum and check if its possible.
50,50,50,50,50,81,90
Sum = 421>420 [Original sum required 60*7 = 420]
Since at the very least we are having a 421 sum if we include 90, then all values are less than 90.
SUFFICIENT.
Statement 2:
Against minimize all values and check:
50,50,50,65,65,65,90.
Is this possible?
Sum = 150+195+90 = 435 > 420 again.
Thus 90 is not possible to be had as the weights for the 7 crates and should be much lesser to bring total sum down to 420.
SUFFICIENT.
Answer: Option D
Bunuel
