The average marks of 45 students was found to be 66. If the marks of two students were incorrectly entered as 28 and 64 instead of 82 and 46respectively, then what is the correct average?

Let the marks of the students are $$X_1,X_2, X_3......X_{45}$$

It is given that the marks of the two student which entered incorrectly that is $$28$$ and $$64$$

Let$$X_1=28$$ and $$X_2=64$$

Average marks of student when wrongly entered $$=\dfrac{28+64+ X_3+......+X_{45}}{45}=66$$
$$28+64+ X_3+......+X_{45}=66\times 45$$
$$X_3+......+X_{45}=66\times 45-28-64 ----------------------(i)$$

It is given that correct marks are $$82$$ and $$46$$

After entering correct marks, new average will be $$=\dfrac{82+46+ X_3+......+X_{45}}{45}  --------------(ii)$$

From equation (i) and (ii)

$$=\dfrac{82+46+66\times 45-28-64}{45}$$
$$=\dfrac{3006}{45}=66.8$$