If the average (arithmetic mean) of five positive numbers is 30, how

Question

If the average (arithmetic mean) of five positive numbers is 30, how many of the numbers are greater than 30?
 
(1) None of the five numbers is equal to 30.
(2) Three of the numbers are equal to 20.

amanvermagmat · Answer

Lets say the numbers are a, b, c, d, e. Given that average = 30, so their sum = a+b+c+d+e = 30*5 = 150

(1) None of the five numbers is equal to 30. Its possible that  a=b=c=d=25 , and  e=50  (here one number is > 30). Its also possible that  a=b=c=20  and  d=e=45  (here two numbers are > 30). So we cant say how many numbers are > 30.  Insufficient .

(2) Lets say  a=b=c=20 , then  d+e=90 . Here both d & e could be 45 each (so both > 30) or we could have  d=30, e=60  (so one number > 30). So we cant say how many numbers are > 30.  Insufficient .

Combining the two statements, say  a=b=c=20 . Thus  d+e=90 , and no number could be 30. Here also we could have both d/e greater than 30 (say d=e=45), or we could have only one number > 30 (say d=25, e=65). Again we cant determine how many numbers are > 30. So  Insufficient .

Hence  E answer

vipulshahi · Answer

Bunuel ,

Can you please explain?

IanStewart · Answer

If five numbers average to 30, they sum to (5)(30) = 150.

Using both Statements, we know three numbers are 20, so three of the numbers sum to 60. The other two numbers must therefore sum to 90, to get the total sum of 150. But now it might be that both of those numbers are greater than 30 -- we might have this list of numbers:

20, 20, 20, 40, 50

for example, or it might be true that only one of the numbers is greater than 30 -- we might have this list of numbers:

20, 20, 20, 10, 80

for example. So using both Statements, while we know either one or two of the numbers is greater than 30, we can't be sure which, and the answer is E.

vipulshahi · Answer

IanStewart ,

Thank you ?

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If the average (arithmetic mean) of five positive numbers is 30, how

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