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16 Dec 2025, 23:40
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We need to find No. of packages>420gms.
let's look at the statements
1. Given Mean= lightest package
And we know Min<=Mean<=Max
As per given info Min=lightest is only possible when there is only package or all package has same weight but then there will not be lightest or heaviest. Also, we dont know what is the lightest weight. So insufficient
2. Given, Median=410 Gms
we dont know how many package are there, neither we know what is the range or heaviest package, so Insufficient again
combine 1&2
Since Mean=Lightest and median=410, so either all the packages has same weight as median or there is one item, Then we clearly have no package more than 420, because if we had package more than 420, then mean can't be equal to the lightest package, it would have been more than 410. SO option C is correct
let's look at the statements
1. Given Mean= lightest package
And we know Min<=Mean<=Max
As per given info Min=lightest is only possible when there is only package or all package has same weight but then there will not be lightest or heaviest. Also, we dont know what is the lightest weight. So insufficient
2. Given, Median=410 Gms
we dont know how many package are there, neither we know what is the range or heaviest package, so Insufficient again
combine 1&2
Since Mean=Lightest and median=410, so either all the packages has same weight as median or there is one item, Then we clearly have no package more than 420, because if we had package more than 420, then mean can't be equal to the lightest package, it would have been more than 410. SO option C is correct
Bunuel
msignatius
16 Dec 2025, 23:47
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No matter how you slice it, if the arithmetic mean equals to the weight of the lightest package - the weight of all packages is the same.
For instance, let's see the weight of the lightest package is 400 grams. There are 3 packages (assumed). As 400 is the lightest, the other two have to be 400 (and thus, also the lightest) or more, like 401. But anything more than 400 will also shift the arithmetic mean ahead of 400. There's no other way to look at this.
Now, coming back to the stem - we need to find how many packages weigh more than 420 grams. Note the 'how many'. We need an absolute number. Zero is also an absolute number. This will come in handy later.
Statement I: This is what I was highlighting above. We know that each package has the same weight, basis my logic. However, we don't know if they're all 430 grams (or more than 420 grams) or 400 grams (or less than 420 grams) or even the same as 420 grams. We eliminate A and D.
Statement II: Oh, nice. Another trick. The median of all packages is 410 grams. Now, lets say there are two more packages. One is 409 grams. The only, 2100 grams. That means one package will be more 420 grams. But what if the larger package is 411 grams; then no package will be more than 420 grams. This won't be sufficient either.
Combining both Statements: Okay, now, with Statement I we can tell all packages weigh the same. With II, we can tell all packages weigh 410 grams - as that is the Median and all packages weigh the same. Great!
And then the KEY logic - if all packages are 410 grams, exactly ZERO are above 420 grams. That's an absolute number. I and II together are enough. C is the right answer.
For instance, let's see the weight of the lightest package is 400 grams. There are 3 packages (assumed). As 400 is the lightest, the other two have to be 400 (and thus, also the lightest) or more, like 401. But anything more than 400 will also shift the arithmetic mean ahead of 400. There's no other way to look at this.
Now, coming back to the stem - we need to find how many packages weigh more than 420 grams. Note the 'how many'. We need an absolute number. Zero is also an absolute number. This will come in handy later.
Statement I: This is what I was highlighting above. We know that each package has the same weight, basis my logic. However, we don't know if they're all 430 grams (or more than 420 grams) or 400 grams (or less than 420 grams) or even the same as 420 grams. We eliminate A and D.
Statement II: Oh, nice. Another trick. The median of all packages is 410 grams. Now, lets say there are two more packages. One is 409 grams. The only, 2100 grams. That means one package will be more 420 grams. But what if the larger package is 411 grams; then no package will be more than 420 grams. This won't be sufficient either.
Combining both Statements: Okay, now, with Statement I we can tell all packages weigh the same. With II, we can tell all packages weigh 410 grams - as that is the Median and all packages weigh the same. Great!
And then the KEY logic - if all packages are 410 grams, exactly ZERO are above 420 grams. That's an absolute number. I and II together are enough. C is the right answer.
Bunuel
17 Dec 2025, 00:35
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Bunuel
Let weight of each package = W
Let number of packages with W > 420 = N
We need, N = ?
(1) Mean = Lightest weight
=> This is only possible when all the weights are equal
We do not know anything about the weight of the lightest package, so we cannot get N
Statement (1) is NOT sufficient
(2) Median = 410
=> This information is not enough to get the value of N, since we do not know the total number of packages.
Statement (2) is NOT sufficient
(1) + (2)
All packages have equal weight and Median is 410
=> All the packages weigh 410 grams
So, the N = 0 (number of packages with weight greater than 420 is 0)
(1) + (2) is SUFFICIENT
Therefore, answer C) Both statements together are sufficient
