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 13 of 14.
MRP said:
4 years ago
Given avg of 20 number is 0.i.e 20 * 0 = 0.
And also they given almost i. e less than 20.so 19 because to get the sum as zero if we take 19 +ve numbers and 20th is -ve number which is equivalent to the sum of 19no.s(152+ve means -152ve). Then we get sum as 0.
Mainly in atmost they didn't mention almost 12 like that so it's better to take up to 19.
And also they given almost i. e less than 20.so 19 because to get the sum as zero if we take 19 +ve numbers and 20th is -ve number which is equivalent to the sum of 19no.s(152+ve means -152ve). Then we get sum as 0.
Mainly in atmost they didn't mention almost 12 like that so it's better to take up to 19.
(6)
Pradnyesh said:
4 years ago
@Tarun.
Good explanation.
Good explanation.
(2)
Saurabh said:
4 years ago
I don't understand this, please explain shortly.
(3)
Navin said:
4 years ago
Answer is 10.
(2)
Shalini said:
4 years ago
But how? Explain please.
(2)
Geetanjali said:
4 years ago
I don't understand this, Anyone, please explain shortly.
(10)
Vrush said:
4 years ago
You are right. @Swathy.
But they are asking for at most means maximum how many numbers. You have taken 10 but till 19 numbers you can take then the sum will get cancelled by the 20th negative number. And the average becomes zero.
But they are asking for at most means maximum how many numbers. You have taken 10 but till 19 numbers you can take then the sum will get cancelled by the 20th negative number. And the average becomes zero.
(10)
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)
Teja Reddy said:
3 years ago
@All.
Why can't we take the same 10 positives and 10 negative numbers?
Their sum will be 0, the answer may be 10.
Why can't we take the same 10 positives and 10 negative numbers?
Their sum will be 0, the answer may be 10.
(57)
Guna said:
2 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)
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