Several students have taken an exam. There was an error in the answer key which affected the marks of 48 students, and their average marks reduced from 78 to 66. The average of remaining students increased by 3.5 marks. This resulted the reduction of the average of all students by 4.5 marks. The number of students that attended the exam is:

Let the total number of students = T

Number of students whose average is reduced = 48

Number of students whose average is increased = T-48

Using Alligation and Mixture rule,

Ratio of 48 and T-48 = 16 : 15

$$=$$>  $$\frac{48}{T-48}=\frac{16}{15}$$

$$=$$>  T-48 = 45

$$=$$>  T = 93

$$\therefore\ $$Total number of students = 93

Hence, the correct answer is Option D

Another method

Let the total number of students = T

Without error

Average of 48 students = 78

Sum of 48 students = 78 x 48

Let average of remaining T-48 students = p

Sum of T-48 students = p(T-48)

Total marks = (78 x 48) + p(T-48)

Let the average of total students = q

$$\Rightarrow$$ (78 x 48) + p(T-48) = Tq.........(1)

With error

Average of 48 students = 66

Sum of 48 students = 66 x 48

Average of T-48 students = (p+3.5)

Sum of T-48 students = (p+3.5)(T-48)

Total marks = (66 x 48)+(p+3.5)(T-48)

Average of total students = (q-4.5)

$$\Rightarrow$$ (66 x 48)+(p+3.5)(T-48) = T(q-4.5)

$$\Rightarrow$$ (66 x 48)+(p+3.5)(T-48) = Tq - 4.5T

$$\Rightarrow$$ (66 x 48)+(p+3.5)(T-48) = (78 x 48) + p(T-48) - 4.5T [From (1)]

$$\Rightarrow$$ (p+3.5)(T-48) - p(T-48) + 4.5T = (78 x 48) - (66 x 48)

$$\Rightarrow$$ (T-48)(p+3.5-p) + 4.5T = 48(78 - 66)

$$\Rightarrow$$ (T-48)3.5 + 4.5T = 48 x 12

$$\Rightarrow$$ 3.5T - 48 x 3.5 + 4.5T = 48 x 12

$$\Rightarrow$$ 8T = 48 x 15.5

$$\Rightarrow$$ T = 93

Total number of students = 93

Hence, the correct answer is Option D