In a class, the number of students who scored 61 or more marks is 30 and there are atmost 39 students who have scored below 61 marks. If the average marks of the class is 48, then determine that what is the largest possible average marks of students who have scored below 61 marks?

38

The correct answer is Option (4) → 38

Given:

• Students scoring ≥ 61 marks = 30
• Students scoring < 61 marks ≤ 39
• Average marks of entire class = 48

Let:

• $x$ = number of students scoring < 61 marks
• $S_1$ = total marks of students scoring < 61
• $S_2$ = total marks of students scoring ≥ 61

Total students = $x + 30$

Total marks = $S_1 + S_2 = 48(x + 30)$

Let each of the 30 students scoring ≥ 61 score exactly 61 (to maximize average of students scoring < 61):

Then $S_2 = 30 \cdot 61 = 1830$

So, total marks: $S_1 + 1830 = 48(x + 30)$

⇒ $S_1 = 48(x + 30) - 1830$

Average of students scoring < 61 = $\frac{S_1}{x} = \frac{48(x + 30) - 1830}{x}$

To maximize this expression, minimize $x$ (since numerator increases with $x$, but denominator increases faster)

Minimum value of $x = 1$

Maximum value of $x = 39$

Check $x = 30$:

$\frac{48 \cdot 60 - 1830}{30} = \frac{2880 - 1830}{30} = \frac{1050}{30} = 35$

Check $x = 35$:

$\frac{48 \cdot 65 - 1830}{35} = \frac{3120 - 1830}{35} = \frac{1290}{35} = 36.857$

Check $x = 39$:

$\frac{48 \cdot 69 - 1830}{39} = \frac{3312 - 1830}{39} = \frac{1482}{39} = 38$