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.
Guna said:
3 years ago
Average means Sum of numbers divided by total;
Ex: average of first 5 numbers is;
(1+2+3+4+5)/5.
Similarly for the average of 20 numbers "any 20 numbers" is;
Let's say (1+1+1+1+1...+1-19(20th number)= 0
Because 19*1 = 19 and adding it to -19 so the average is 0/20=0.
So there 19 numbers that are greater than 0.
Ex: average of first 5 numbers is;
(1+2+3+4+5)/5.
Similarly for the average of 20 numbers "any 20 numbers" is;
Let's say (1+1+1+1+1...+1-19(20th number)= 0
Because 19*1 = 19 and adding it to -19 so the average is 0/20=0.
So there 19 numbers that are greater than 0.
(27)
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)
King praba said:
1 decade ago
Hi,
We know the formula.average=sum of observations/number of observations.
sum=0(given)..no.of observation=20..sub in above formula 0/20=0..sum of 20 numbers is
0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19.........
among above numbers 0<remaining numbers.
so the remaining numbers are 19(ans)
We know the formula.average=sum of observations/number of observations.
sum=0(given)..no.of observation=20..sub in above formula 0/20=0..sum of 20 numbers is
0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19.........
among above numbers 0<remaining numbers.
so the remaining numbers are 19(ans)
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)
Swathy said:
1 decade ago
@Tarun your answer s perfect but i have one doubt that if 10 nos are positive n 10 nos are negative means also the average will be zero? i mean for example (11,12,14,15,16,4,6,8,7,9,-11,-12,-14,-15,-16,-4,-6,-8,-7,-9)it mean also the average will be zero, so y the answer may be 10 for this problem?
Vipin said:
9 years 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.
Ashline said:
1 decade ago
@Jeevanandham.
your logic was nice but the answer will be 10 in your case which is True but they asked for" At the most" So you need to tell maximum possible numbers that can be greater than zero in order to maintain the sum zero :).
Hope that helps :).
your logic was nice but the answer will be 10 in your case which is True but they asked for" At the most" So you need to tell maximum possible numbers that can be greater than zero in order to maintain the sum zero :).
Hope that helps :).
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)
Seethu said:
6 years ago
@Akshay.
It is also a possible case but in the question it is mentioned as "atmost positive numbers".
Suppose if we consider.
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19.
As 19 positive numbers and the 20th number as -189.
Then we will get maximum positive numbers.
It is also a possible case but in the question it is mentioned as "atmost positive numbers".
Suppose if we consider.
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19.
As 19 positive numbers and the 20th number as -189.
Then we will get maximum positive numbers.
Ganesh said:
1 decade ago
I agree with @Swathy, there might be 10 +ve nos. and 10 -ve nos.
And questions says at most nos. greater than 0 i.e., at most +ve nos. i.e., maximum +ve nos. then it can be 10 too, why 19 then.
Whats at most means actually, according to me its maximum, please tell me.
And questions says at most nos. greater than 0 i.e., at most +ve nos. i.e., maximum +ve nos. then it can be 10 too, why 19 then.
Whats at most means actually, according to me its maximum, please tell me.
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