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21 Feb 2019, 00:10
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Difficulty:
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Question Stats:
65% (00:52) correct
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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
E. 18
A. 0
B. 1
C. 10
D. 19
E. 18
Archit3110
Major Poster
Updated on: 21 Feb 2019, 12:51
Originally posted by Archit3110 on 21 Feb 2019, 08:41.
Last edited by Archit3110 on 21 Feb 2019, 12:51, edited 1 time in total.
Last edited by Archit3110 on 21 Feb 2019, 12:51, edited 1 time in total.
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[quote="Bunuel"]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
E. 18[/quote]
so total 20 no.s the avg = 0 means sum = 0
and the equal -ve and +ve integers
which would be possible -10 to 10 range only
so max +10 is possible in case of consective no .
But here it can be sum of 1-19 integers and -ve value of its sum so avg 0
IMO D
A. 0
B. 1
C. 10
D. 19
E. 18[/quote]
so total 20 no.s the avg = 0 means sum = 0
and the equal -ve and +ve integers
which would be possible -10 to 10 range only
so max +10 is possible in case of consective no .
But here it can be sum of 1-19 integers and -ve value of its sum so avg 0
IMO D
Archit3110
Major Poster
21 Feb 2019, 09:07
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mohitranjan05
mohitranjan05
what will be range according to you?
its given that total 20 nos are there and for sum to be = 0 ; it would be only possible when we have equal integers of both signs isnt it?
