The average of n numbers is 42. If 75% of the numbers are increased by 4 each and the remaining numbers are decreased by 8 each, then what is the average of the numbers, so obtained?

Total numbers = n 

Average of n numbers = 42 

A sum of n numbers = 42 n

Now 75% of numbers = $$ \dfrac{75}{100}\times n = \dfrac {3}{4}n $$

sum of $$\dfrac{3}{4} n numbers increased = \dfrac{3}{4}\times n \times 42 \times 4 \times \dfrac{3}{4}n $$

$$\Rightarrow \dfrac {63n}{2} + 3n = \dfrac{69}{2}n $$ 

sum of left 25% numbers = $$\dfrac{25}{100}\times n \times 42 -8\dfrac{1}{4}n $$

$$\Rightarrow \dfrac{21n}{2} - 2 n $$

$$\Rightarrow \dfrac{17 n} {2}$$

Now total sum = $$\dfrac{69 n}{2} + \dfrac {17 n}{2} $$

$$\Rightarrow\dfrac {69n + 17n}{2}$$

$$\Rightarrow \dfrac{86n}{2}$$ = 43n

Now Average = $$\dfrac{43n}{n}$$ = 43 Ans