# 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:

132 comments Page 1 of 14.
Jagadeesh said:
16 hours ago

In question, if they are consecutive. Definitely the answer will be 10, but they didn't mention what type of numbers are there, it's asking the possibility of max. Positive numbers satisfy the given condition, so in 20 numbers average is 0. So 20 positive numbers can't take place there is a possibility of 19 numbers positive and the remaining number is numerically equal to the sum of those 19 positive numbers but the sign is negative.

Nanthakumar said:
2 months ago

First 1-19 numbers = each number is -1.

Therefore Adding these numbers the answer is = -19.

Then the 20th number is +19.

Therefore -19+19 = 0

The average of 20 numbers is 0/20=0

Therefore Adding these numbers the answer is = -19.

Then the 20th number is +19.

Therefore -19+19 = 0

The average of 20 numbers is 0/20=0

(2)

Guna said:
1 year 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.

(19)

Teja Reddy said:
1 year 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.

(41)

Bires said:
2 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.

(3)

Vrush said:
2 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.

(9)

Geetanjali said:
2 years ago

I don't understand this, Anyone, please explain shortly.

(6)

Shalini said:
3 years ago

But how? Explain please.

(1)

Navin said:
3 years ago

Answer is 10.

(1)

Saurabh said:
3 years ago

I don't understand this, please explain shortly.

(1)

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