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

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

19

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $19$
Let the $20$ numbers be $a_{1}, a_{2}, . . ., a_{20}$

Given that average of 20 numbers is zero.

Therefore, $\frac{a_{1} + a_{2} + . . . + a_{20}}{20} = 0$

$⟹ a_{1} + a_{2} + . . . + a_{20} = 0$

$⟹ a_{1} + a_{2} + . . . + a_{19} = - a_{20}$

Therefore, at the most $19$ numbers can be greater than zero.

Suggest Corrections

Similar questions

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

(a) 0
(b) 1
(c) 10
(d) 19

The average of 20 numbers is zero. If one of them is -19, then find the average of the remaining 19 numbers?

Q. The average of 'n' terms is zero. How many maximum numbers of terms can be negative?

Let $S$ be the set of prime numbers greater than or equal to $2$ and less than $100$ . Multiply all elements of $S$ . With how many consecutive zeros will the product end?

Q. How many numbers greater than 1000, but not greater than 4000 can be formed with the digits 0, 1, 2, 3, 4, repetition of digits being allowed?

Join BYJU'S Learning Program

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

The correct option is D 19Let the 20 numbers be a1,a2,...,a20Given that average of 20 numbers is zero.Therefore, a1+a2+...+a2020=0⟹a1+a2+...+a20=0⟹a1+a2+...+a19=−a20Therefore, at the most 19 numbers can be greater than zero.

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