The mean of 6 numbers is 16 with the removal of a number the mean of remaining numbers is 17. The number removed is :

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

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

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

Open in App

Solution

The correct option is C 11
Given that mean of $6$ numbers is $16$
Let the numbers be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$

Therefore the mean $= \frac{a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6}}{6} = 16$

$⟹ a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = 16 \times 6 = 96$ ....(1)

Let the removed number be $a_{6}$

Given that mean of the remaining numbers is $17$

Therefore the mean $= \frac{a_{1} + a_{2} + a_{3} + a_{4} + a_{5}}{5} = 17$

$⟹ a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 17 \times 5 = 85$ ....(2)

Substituting (2) in (1) we get

$85 + a_{6} = 96 ⟹ a_{6} = 96 - 85 = 11$

Suggest Corrections

Similar questions

Q. The mean of a set of

5

numbers is

90

. If one of the numbers is removed, the mean of the remaining numbers is

92

. Identify the removed number.

Q. Two consecutive numbers from

1, 2, 3, . . ., n

are removed. The Arithmetic mean of the remaining number is

\frac{105}{4}

The removed numbers are

Q. Two consecutive numbers from

1, 2, 3, . . . . . n

are removed. Arithmetic mean of the remaining numbers is

\frac{105}{4}

. Find

n

-(3 times of the sum of those removed numbers) ?

Arithmetic mean of 5 numbers in a series is 16. If 2 numbers (mean numbers) are removed from this series and replaced with 20 and 24, what will be the new mean?

Q. The average of a set of

10

numbers is

15

. If one number is removed the average of the remaining numbers is

16

. What is the value of the number removed?

Join BYJU'S Learning Program