If the average (arithmetic mean) of three positive integers is 35, how

From the question stem, we know that the average of three positive integers is 35.

A few key inferences
a. all 3 numbers are integers
b. all numbers are greater than 0. None of the numbers is zero or negative.

Let the three numbers to be x, y, and z. Then \(\frac{(x + y + z)}{3}\) = 35
Therefore, x + y + z = 105

Data is sufficient if we are able to uniquely determine how many of these 3 numbers are greater than 10.
Data is not sufficient if we are not able to determine how many of these 3 numbers are greater than 10 or if we get more than one possible answer.

STATEMENT-1: The sum of two of the numbers is 75.

If x + y = 75, then z = 105 - (x + y) = 105 -75 =30
z is clearly greater than 10.

But we cannot conclusively determine whether one out of x, y is greater than 10 or both are greater than 10.

Example: x = 1, y = 74 and z = 30 satisfies the given criteria. Here 2 out of 3 are greater than 10.

Counter-Example: x = 40, y = 35, and z = 30 satisfies the given criteria.
Here all 3 numbers are greater than 10.

So, we are left with the answer that either 2 or all 3 numbers are greater than 10. We do not have a unique answer to our question.

Using statement 1 alone we cannot determine how many of x, y, and z is greater than 10.
Eliminate choices A and D.

STATEMENT-2: None of the numbers is greater than 40.
Let us pick the maximum possible value for two of the numbers. Say, x = 40 and y = 40, then the value of z will be 25.

Even when two of the numbers take the maximum possible value (which will minimize the third number), the third number comes out as greater than 10.
If the value of x or y decreases, value of z will increase concomitantly.

So, if none of the numbers is greater than 40, every number has to be greater than 10 for their average to be 35.
So all 3 numbers are greater than 10.

We are able to answer the question with the help of statement 2 alone
Choice-B is the answer