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The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
A. 0 B. 1 C. 10 D. 19
7 Réponses
+ 8
@ChaoticDawg is correct.
+ 5
D 19
if -19 was 1 number and the rest of the numbers were 1
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + -19 = 0
0/20 = 0
+ 4
Your answer is right but your logic is not right Dude !
+ 2
Can you tell me Which Logic are you applying ???
+ 1
20+20+20+20+20-20-20-20-20-20 = 0
avg=0/20;
avg=0
+ 1
The Exact Explanation:
Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a)
0
ans is 19
take any 19 numbers and add them. let the sum of those numbers be x. let the 20th number be -x. so when you find the AVG of 20 Numbers it will be 0