+ 3

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

24th Jul 2017, 4:05 PM
~Sudo Bash
~Sudo Bash - avatar
7 Antworten
+ 8
@ChaoticDawg is correct.
24th Jul 2017, 4:27 PM
Hatsy Rei
Hatsy Rei - avatar
+ 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
24th Jul 2017, 4:26 PM
ChaoticDawg
ChaoticDawg - avatar
+ 4
Your answer is right but your logic is not right Dude !
24th Jul 2017, 4:28 PM
~Sudo Bash
~Sudo Bash - avatar
+ 2
Can you tell me Which Logic are you applying ???
24th Jul 2017, 4:13 PM
~Sudo Bash
~Sudo Bash - avatar
+ 1
20+20+20+20+20-20-20-20-20-20 = 0 avg=0/20; avg=0
24th Jul 2017, 4:37 PM
‎ ‏‏‎Anonymous Guy
+ 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)
27th Jul 2017, 5:54 AM
~Sudo Bash
~Sudo Bash - avatar
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
24th Jul 2017, 9:25 PM
Devbrath
Devbrath - avatar