Aptitude - Average - Discussion
Discussion Forum : Average - General Questions (Q.No. 4)
4.
The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
Answer: Option
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).
Discussion:
136 comments Page 8 of 14.
Amin Ullah said:
10 years ago
While the actual answer is 20.
Suppose all 20 numbers are of value =1 then,
1+1+1....till 20. Thier sum will be 20.
Avg=sum of all number/total number.
Avg= 20/20.
= 0.
So, as for question is concern there is no option with 20. The nearest 19 will do.
Suppose all 20 numbers are of value =1 then,
1+1+1....till 20. Thier sum will be 20.
Avg=sum of all number/total number.
Avg= 20/20.
= 0.
So, as for question is concern there is no option with 20. The nearest 19 will do.
Kumar said:
10 years ago
Thanks for all of your explanations.
Divya said:
10 years ago
Is there any shortcut to find the answer for this question?
Shivu said:
10 years ago
@Prabhat.
Your explanation was good, I got it now. Thank you.
Your explanation was good, I got it now. Thank you.
Hemavathi said:
10 years ago
To get a value zero we can use,
(Sum of 19 positive numbers) - (-Value of sum)/(Sum of 18 positive numbers) - (Value of sum),
Here, the maximum probability of positive number is 19 (at most),
So the answer is 19.
(Sum of 19 positive numbers) - (-Value of sum)/(Sum of 18 positive numbers) - (Value of sum),
Here, the maximum probability of positive number is 19 (at most),
So the answer is 19.
PRABHAT KR.SINGH said:
10 years ago
19 will be appropriate answer.
Because in the question, it is mention that maximum number greater than zero.
If we take all positive number then sum will never be 0.
So, at least one number should be taken negative that should be equal to the sum of the 19 numbers.
So, (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1+ 1) + (-19) /20.
I hope you got it.
Because in the question, it is mention that maximum number greater than zero.
If we take all positive number then sum will never be 0.
So, at least one number should be taken negative that should be equal to the sum of the 19 numbers.
So, (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1+ 1) + (-19) /20.
I hope you got it.
Satyam said:
10 years ago
Hi,
In this question, they are asking about the number which is at most (it means maximum numbers which are greater than 0).
So, (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1+ 1) + (-19) /20.
I hope you got it.
In this question, they are asking about the number which is at most (it means maximum numbers which are greater than 0).
So, (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1+ 1) + (-19) /20.
I hope you got it.
Himanshu arora said:
10 years ago
What about the answer if the question would be, "The average of 20 numbers is zero. Of them, at least, how many may be greater than zero"?
Dsp said:
1 decade ago
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1+ 1 + 1 + 1 + 1 + 1 - 19 = 0.
+ve numbers are 19, So, D is the answer.
+ve numbers are 19, So, D is the answer.
Vipin said:
1 decade ago
The question is not very rigorous but the main adversity of this question is that answer is 19 because in the question only asking that the no. an average of only 20 numbers is zero not asking that 1 to 20 numbers is zero so the 20 = 0 and 0<19 value. i.e 1, 2, 3, 4, 5, 6, 6, 7, 8, 9. 19.
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