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16 Dec 2025, 10:43
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Statement 1 says that the AM is equal to the lowest number. We can check for multiple cases to understand that this is only possible if all the values in the set are equal to each other. But we still do not know the repeating number in the set. This statement is not sufficient.
Statement 2 says that the median weight is 410 gms. As such, it doesn’t say anything about the number of packages weighing more than 420 gms. Not sufficient.
Combing both the statements, we can say that the median of 410 gets repeated in the set. Therefore, we can answer the question as none of the packages weighing more than 420 gms.
Therefore, Option C
Statement 2 says that the median weight is 410 gms. As such, it doesn’t say anything about the number of packages weighing more than 420 gms. Not sufficient.
Combing both the statements, we can say that the median of 410 gets repeated in the set. Therefore, we can answer the question as none of the packages weighing more than 420 gms.
Therefore, Option C
16 Dec 2025, 10:43
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Think i got it:
Each package has a weight in grams how many packages over 420 in weight:
1) for the average to be equal to the smallest quanity every entry in the set must be the same, as any number, even fractionally higher would shift the mean away smallest number eg
(1,1,1,1,1) is valid but (1,1,1,1,1.1) is invalid since now the mean is no longer equal to the smallest number.
this gives us the distribution but we dont actually know anything about the numbers itself or even how many items are there, the list could be
(500,500,500) or (100,100,100,100)
Hence not sufficeint
2) This give us the meadian but tells us nothing about the number of items in the distribution, since the number of items could drastically change the answer
hence not sufficent
1+2:
we know that all the entries in the series are the same, and now we know one term that is 410 which is less than 420.
This implies there is no item that weighs more than 420, we get our answer, 0
Hence (c)
Each package has a weight in grams how many packages over 420 in weight:
1) for the average to be equal to the smallest quanity every entry in the set must be the same, as any number, even fractionally higher would shift the mean away smallest number eg
(1,1,1,1,1) is valid but (1,1,1,1,1.1) is invalid since now the mean is no longer equal to the smallest number.
this gives us the distribution but we dont actually know anything about the numbers itself or even how many items are there, the list could be
(500,500,500) or (100,100,100,100)
Hence not sufficeint
2) This give us the meadian but tells us nothing about the number of items in the distribution, since the number of items could drastically change the answer
hence not sufficent
1+2:
we know that all the entries in the series are the same, and now we know one term that is 410 which is less than 420.
This implies there is no item that weighs more than 420, we get our answer, 0
Hence (c)
16 Dec 2025, 10:45
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Bunuel
AD ---> If average weight is equal to lightest package that means all package must be of same weight. But we do not whether all the packages weighs less or more than 420 grams. --> Insufficient
B ---> No information about the packages more or less than 410 grams ---> Insufficient
C ---> Upon combining, we can conclude that all packages weight 410 grams -- Hence, there are no packages or 0 packages that weigh more than 0 --> Answer is C
