Set A contains 50 different integers, the arithmetic mean of

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,

Oh ..missed the fact that the numbers are all "different"...
Neat one..but isn't this too time consuming a question? Could there be some other elegant solution

Cicerone is right - would have taken me 45 minutes on the exam to muscle through this one

(1) If ten of the integers are above 85, the sum of these ten must be at least 86+87+...+95 =60(10) +26+27+...+35

Suppose that the other 40 were not less than 35, then the sum of these 40 would be at least 35+36+...74=60(40) -25-24-...+13+14= 60(40)-25-24-23-...-15

Thus the sum of all 50 is at least 60(50)+(26+...+35)-(15+16+...25) which is clearly greater than 60(50). Thus the average of the 50 is greater than 60, which contradicts what we are told. Thus there must be at least one integer below 35. SUFF

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.

Hey Kevin what's wrong with it.
By the way it is not time consuming .............
Since i want everyone to understand this.................

You should suppose that the remaning numbers are greater than 34, not 35, though this small error does not detract from your excellent work on this question. However, a lot of calculations are involved in your approach.

I am grateful for your presence on this forum, Cicerone! You are helping everyone and I hope we can return the favour.