Aptitude - Average - Discussion

@Hema.

Can you explain me your method once again?

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 :-).

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.

1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1-19.

0>1

Here 1 is 19
Negative number -19

SO answer is 19.

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.

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.

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.

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1+ 1 + 1 + 1 + 1 + 1 - 19 = 0.

+ve numbers are 19, So, D is the answer.

What about the answer if the question would be, "The average of 20 numbers is zero. Of them, at least, how many may be greater than zero"?

Hi,

In this question, they are asking about the number which is at most (it means maximum numbers which are greater than 0).

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.