Averages

Question

The average of $20$ numbers is zero.Of them, at the most, how many may be greater than zero?

• A. $10$
• B. $19$
• C. $17$
• D. $9$

Comments

b) 19 ??

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Yes

Please explain the answer

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Sum of any 19 positive numbers is equal to 20th number in negative then average is 0.
Let x1, x2,.....x20 are 20 postive numbers.
Average is zero means
A= (x1 + x2 + ......x20)/20 = 0
x1 + x2 +.......x20 = 0
Maximum 19 numbers can be positive.
For 5 numbers ex. is 1,2,3,4, -10 (max 4 are positive)

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Thanks I got it

Answer 1

The average of 20 numbers is zero:

So,we can have :      [1+1+1+.............. +1 (19 times) -19 ] / 20

                                = [19 -19 ] / 20

                                = 0

b) 19 numbers are greater than 0.

Answer 2

Sum of 20 numbers = $20 * 0$ = $0$

∴ We've to find out atmost how many numbers from the 20 numbers are greater than $0$.

If the question asks us to find out at least then it could be 1. As we've to find out atmost, the answer will be $19$.

Because then only the sum of all 19 numbers would be something (assume $x$), and the 20th number will be $-x$

and hence, the sum of all possible numbers will be $0$ [∵ $x + (-x)$ = $0$]

Answer 3

Its 19 because 19 numbers can be any value of +ve integer and the 20th no. will be -ve of sum of all the 19 numbers