# 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:

131 comments Page 1 of 14.
AnkitJ said:
4 years ago

First when average can be Zero:

0(Sum of observation)/20 Number of Observations =0 is answer (Average).

Than means we must get 0 sum on the top of /20 that will make average Zero.

Even single nagative number can make 0 on top of /20. If sum of 19 numbers is +1 one nagative number -1 can make it zero you can assume sum of 19 numbers as positive and put same nagative number to make it Zero.

Now coming to answer, they are asking how many number can be above 0 i.e. in positive (numbers below zero are nagative). 19 Max number can be positive to make one nagative number 0 so that we get average zero.

Hope it's clear now.

0(Sum of observation)/20 Number of Observations =0 is answer (Average).

Than means we must get 0 sum on the top of /20 that will make average Zero.

Even single nagative number can make 0 on top of /20. If sum of 19 numbers is +1 one nagative number -1 can make it zero you can assume sum of 19 numbers as positive and put same nagative number to make it Zero.

Now coming to answer, they are asking how many number can be above 0 i.e. in positive (numbers below zero are nagative). 19 Max number can be positive to make one nagative number 0 so that we get average zero.

Hope it's clear now.

Akash said:
8 years ago

Why don't all you guys understand that, the question is being asked on the basis of "at the most". I too do agree that, there may be chances of more than one number of appropriate positive and negative number i.e. ten positive and ten negative, fifteen positive and five negative etc. But for at the most, the answer will be 19.

Because there will be the one negative number of all 19 positive number, so that the sum become zero. For the very beginning I also used to think like what you guys thinking now. Hope you will understand.

Because there will be the one negative number of all 19 positive number, so that the sum become zero. For the very beginning I also used to think like what you guys thinking now. Hope you will understand.

Ashwani said:
8 years ago

The question is rhetoric, at the first glance it looks like well calculated but actually it requires analytical skill. Let's see.

Our average should be 0 and the total number is 20. Hence, we should understand that dividing by 20, if we have 20 then it will be 1.

If we have more than 20 then it will be more than 1 and the most crucial is that it will never be 0.

Then what can be more than 1, and it can be less than 20 so that we can avoid 1.

Therefore 19/20 =. ___ <1.

Our average should be 0 and the total number is 20. Hence, we should understand that dividing by 20, if we have 20 then it will be 1.

If we have more than 20 then it will be more than 1 and the most crucial is that it will never be 0.

Then what can be more than 1, and it can be less than 20 so that we can avoid 1.

Therefore 19/20 =. ___ <1.

Ankit Dere said:
6 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)

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.

(1)

PRABHAT KR.SINGH said:
8 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.

Shudhu said:
8 years ago

I suggest the answer may be 10.

Eg. If Avg of 20 numbers can be denoted as:

a + b + c + d + e + f + g + h + I + j + k + l + m + n + o + p + q + r + s + t.

Between j and k the zero will be present because we know that middle number will be the average.

So when zero is present after j so from k the number is greater than zero.

So the answer will be assumed as 10.

Eg. If Avg of 20 numbers can be denoted as:

a + b + c + d + e + f + g + h + I + j + k + l + m + n + o + p + q + r + s + t.

Between j and k the zero will be present because we know that middle number will be the average.

So when zero is present after j so from k the number is greater than zero.

So the answer will be assumed as 10.

MRP said:
3 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.

(4)

Rahul kumar said:
9 years ago

I think the solution is wrong as its just a assumption that out of 20 only one is 20. But it may also occurs that out of 20 10 may be positive and rest be negative.

Average of 20 numbers = (Sum of 20)/20.

Sum of 20 numbers = (20*21)/2.

So A/Q === (20*21)/(2*20) = 0.

=> 21/2 = 0.

=> 21 = 0.

Not possible :-).

Average of 20 numbers = (Sum of 20)/20.

Sum of 20 numbers = (20*21)/2.

So A/Q === (20*21)/(2*20) = 0.

=> 21/2 = 0.

=> 21 = 0.

Not possible :-).

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.

(18)

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