The average of 42 numbers is 65. The average of the first 21 numbers is 56 and that of the next 18 numbers is 70. The 41st number is 3 more than the 40th number and 6 less than the 42nd number. What is the average of the 40th and 42nd numbers?

Correct option is D

Given :

Total numbers = 42

Average of 42 numbers = 65

Average of first 21 numbers = 56

Average of next 18 numbers = 70

41st number is 3 more than 40th number

41st number is 6 less than 42nd number

Formula Used :
​$$\text{Sum} = \text{Average} \times \text{Number of terms}$$​​
Solution :

Total sum of 42 numbers

42$$\times 65$$​ = 2730
Sum of first 21 numbers

​$$21 \times 56 =$$​ 1176

Sum of next 18 numbers
​$$18 \times 70$$​ = 1260

Sum of first 39 numbers
1176 + 1260 = 2436

Sum of last 3 numbers (40th, 41st, 42nd)
2730 - 2436 = 294

Let 40th number = x
Then,
41st number = (x + 3)
42nd number = (x + 9)

Sum of 40th, 41st and 42nd numbers:
x + (x+3) + (x+9) = 3x + 12 = 294

3x = 282

x = 94

Therefore,
40th number = 94
42nd number = 94 + 9 = 103

Required average
​$$\frac{94 + 103}{2} = \frac{197}{2} = 98.5$$​​