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:
135 comments Page 2 of 14.
Bires said:
3 years ago
{(1+2+3+....+19)-(x)}/20 =0.
Adding the series 1 - 19,
Sum=n/2 (a+L) {n-no. Of term, a-1st term, L-last term},
S=20/2 (1+19).
= 200.
So, (200 - x)/20 = 0.
x = -200.
Therefore, 19 nos. Of term can be possible to be greater than 0.
The 20th is negative which is less than zero.
Adding the series 1 - 19,
Sum=n/2 (a+L) {n-no. Of term, a-1st term, L-last term},
S=20/2 (1+19).
= 200.
So, (200 - x)/20 = 0.
x = -200.
Therefore, 19 nos. Of term can be possible to be greater than 0.
The 20th is negative which is less than zero.
(6)
Ankit Dere said:
7 years ago
It is asked in the question.
Avg of 20 numbers is Zero,
Given avg=0.
Numbers=20.
Explanation:
Avg=sum÷number i.e.
0=sum÷20
Now, cross multiply.
Sum=20x0=0.
This means the sum of 20 numbers is equal to 0.
{1+2+...+20}= is 0.
{1+2+...+19}= Will not be 0.
But the sum of rest 19 number will not be 0 it will be greater than 0(as the 2nd part of the question says how many may be greater than 0?)
Avg of 20 numbers is Zero,
Given avg=0.
Numbers=20.
Explanation:
Avg=sum÷number i.e.
0=sum÷20
Now, cross multiply.
Sum=20x0=0.
This means the sum of 20 numbers is equal to 0.
{1+2+...+20}= is 0.
{1+2+...+19}= Will not be 0.
But the sum of rest 19 number will not be 0 it will be greater than 0(as the 2nd part of the question says how many may be greater than 0?)
(5)
Tarun said:
1 decade ago
Dear read the question carefully.
The question ask "at the most".
So you have to give maximum numbers that can be positive,.
And that is 19.
For example:
take any 19 positive numbers (11, 10, 9, 8, 5, 12, 18, 16, 1, 3, 5, 6, 3, 7, 8, 22, 3, 4, 1).
Avg = (152+x) /20.
Now put x = -152.
So, Avg= (152-152) /20=0.
The question ask "at the most".
So you have to give maximum numbers that can be positive,.
And that is 19.
For example:
take any 19 positive numbers (11, 10, 9, 8, 5, 12, 18, 16, 1, 3, 5, 6, 3, 7, 8, 22, 3, 4, 1).
Avg = (152+x) /20.
Now put x = -152.
So, Avg= (152-152) /20=0.
(4)
N A said:
5 years ago
Here, it is asked that at the most, how many may be greater than zero.
So let see.
(1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1-19)/20 = 0.
Now see it's 19 number which is greater 0.
So let see.
(1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1-19)/20 = 0.
Now see it's 19 number which is greater 0.
(3)
Saurabh said:
4 years ago
I don't understand this, please explain shortly.
(3)
Praveen said:
1 decade ago
@ hema ...
see here the sum is zero ...
sum of 20 numbers has to be 0
ex: 1+2+3+4+5+6...+20 = 0
let 1+2+3+4+5...19=a and the 20th number should be '-a' in order to get the sum as zero .
(a(1st 19numbers)-a(last 20th munber)=0 , hence a-a=0)
hence all 1st 19 numbers are positive means greater than zero and the last 20th is negative .
hence answer is 19 positive i.e greater than zero numbers.
see here the sum is zero ...
sum of 20 numbers has to be 0
ex: 1+2+3+4+5+6...+20 = 0
let 1+2+3+4+5...19=a and the 20th number should be '-a' in order to get the sum as zero .
(a(1st 19numbers)-a(last 20th munber)=0 , hence a-a=0)
hence all 1st 19 numbers are positive means greater than zero and the last 20th is negative .
hence answer is 19 positive i.e greater than zero numbers.
(2)
Pradnyesh said:
4 years ago
@Tarun.
Good explanation.
Good explanation.
(2)
Navin said:
4 years ago
Answer is 10.
(2)
Shalini said:
4 years ago
But how? Explain please.
(2)
Harika said:
2 decades ago
Hi Hema,
Here average of 20 numbers= 0.
So we have avg = sum/20.
That implies sum = 0.
Now for the sum of 20 numbers to be equal to zero, there may be all 19 numbers of them >0 and only one number which is -ve of all the 19 numbers sum.
So the answer is 19.
Hope you understood now.
All the best.
Here average of 20 numbers= 0.
So we have avg = sum/20.
That implies sum = 0.
Now for the sum of 20 numbers to be equal to zero, there may be all 19 numbers of them >0 and only one number which is -ve of all the 19 numbers sum.
So the answer is 19.
Hope you understood now.
All the best.
(1)
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