The average of 41 numbers is 58. The average of the first 16 numbers is 46 and the average of the last 26 numbers is 66. If the 16th number from the beginning is excluded, then what is the average of the remaining numbers?

Correct option is D

Given:

Total numbers = 41

Average of all 41 numbers = 58

Average of the first 16 numbers = 46

Average of the last 26 numbers = 66

The 16th number is excluded.

Formula Used:

Sum of observations = Average × Number of observations

Sum of remaining numbers = Total sum - 16th number

Solution:

Total sum of 41 numbers:

Total sum = 58 × 41 = 2378

Sum of the first 16 numbers:

Sum of first 16 = 46 × 16 = 736

Sum of the last 26 numbers:

Sum of last 26 = 66 × 26 = 1716

The 16th number is common to both the first 16 and last 26 numbers.

Let the 16th number be x

Total sum = Sum of first 16 + Sum of last 26 - x

2378 = 736 + 1716 - x

2378 = 2452 - x

x = 2452 - 2378 = 74

Sum of the remaining numbers after excluding the 16th number:

Sum of remaining numbers = 2378 - 74 = 2304

Number of remaining numbers = 40

New Average = $$\frac{2304}{40} = 57.6$$