If the average (arithmetic mean) of three positive integers is 35, how many of the numbers are greater than 10 ?

[BTGmoderatorDC]

If the average (arithmetic mean) of three positive integers is 35, how many of the numbers are greater than 10 ?

(1) The sum of two of the numbers is 75.
(2) None of the numbers is greater than 40.


OA B

Source: Princeton Review

[swerve]

If the average (arithmetic mean) of three positive integers is 35, how many of the numbers are greater than 10 ?

(1) The sum of two of the numbers is 75.
(2) None of the numbers is greater than 40.


OA B

Source: Princeton Review

From the given information we have, \(a+b+c = 105\)

Statement 1:
\(a+b = 75,\) this means that \(c = 30> 10\) so, we don't know whether \(a,b >10\) or one of them is less than \(10 \Longrightarrow\) Not sufficient \(\Large{\color{red}\chi}\)

Statement 2:
\(a,b,c < 40\)
If we assume that \(a<10\), then \(b+c > 95\), this assumption is rejected because the sum of any two must be \(< 80\), this forces that all \(a,b,c\) must be \(>10 \Longrightarrow\) Sufficient \(\Large{\color{green}\checkmark}\)

Therefore, B

[deloitte247]

Let the positive integers = a, b and c
$$\frac{a+b+c}{3}=35;\ and\ a+b+c=105$$
Target question=> How many of the numbers are greater than 10?

Statement 1=> The sum of two of the numbers is 75.
a+b = 75; so 75+c = 105, and c = 30
If a+b = 5+70; then 2 of the numbers are >10, but if a+b=25+50, the all the numbers are > 10.
Since the is not definite, statement 1 is NOT SUFFICIENT.

Statement 2=> None of the numbers is greater than 40.
If a<10, then b+c>95. For a+b+c=105
But since none of the numbers is > 40, the maximum possible sum of two numbers = 40+40 = 80.
So, the sum of any 2 numbers must be < 80. Hence, the third number will be > 10.
In view of this, all the 3 numbers are > 10.

Therefore, statement 2 is SUFFICIENT. Hence, option B is the correct answer.