cicerone wrote:
Folks, I think the answer must be D (either statement alone)
Now look at this.
Set A contains 50 different +ve integers and their mean 60.
So sum of the numbers must be 3000.
Statement 1: 10 of the digits are above 85.
For the remaining 40 values to be max these 10 must be minimum
i.e 86,87,88,.............,95.
So the sum of these numbers is 905.
So the sum of the remaining 40 numbers must be 3000-905= 2095.
Now if I suppose the remaining 40 numbers greater than 35 and take the min numbers let's see what happens?
So the remaining 40 numbers are
36,37,..................,75.
The sum of these numbers is 2220 which is > than 2095.
So definitely the remaining 40 numbers contain numbers less than 35.
Hence statement 1 alone is sufficient.
Let's take statement 2:
8 of the digits are above 90.
Proceeding along the same lines, let the 9 numbers be
91,92,...........,98.
The sum of these numbers is 756.
So the sum of the remaining 40 numbers must be 3000-756=2244.
Now if I suppose the remaining 42 numbers greater than 35 and take the min numbers let's see what happens?
So the remaining 42 numbers are
36,37,.................,77.
The sum of the above 42 numbers is 2373 which is > 2244.
So the remaining 42 numbers definitely contain numbers less than 35.
Hence statement 2 alone is sufficient.
Ha, I think every thing is clear.
So the answer must be D.
Regards,
Watch these small mistakes, though in this question your mistake didn't have an impact on the answer. In addition, this method is a bit too time-consuming.