The marks of 10 students of a class in an examination arranged in ascending order is as follows:

13, 35, 43, 46, x, x + 4, 55, 61, 71, 80

If the median marks is 48, find the value of x. Hence, find the mode of the given data.

Here, n = 10, which is even.

By formula,

Median = $\dfrac{\dfrac{\text{n}}{2} \text{ th observation} + \Big(\dfrac{\text{n}}{2} + 1\Big) \text{ th observation}}{2}$

$= \dfrac{\dfrac{10}{2} \text{ th observation} + \Big(\dfrac{10}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{5 \text{ th observation} + (5 + 1) \text{ th observation}}{2} \\[1em] = \dfrac{5 \text{ th observation} + 6 \text{ th observation}}{2} \\[1em]$

Median = $\dfrac{x + (x + 4)}{2}$

$\Rightarrow 48 = \dfrac{\text{x + x + 4}}{2} \\[1em] \Rightarrow 48 = \dfrac{\text{2x + 4}}{2} \\[1em] \Rightarrow 48 \times 2 = \text{2x + 4}$

⇒ 96 = 2x + 4

⇒ 2x = 96 - 4

⇒ 2x = 92

⇒ x = $\dfrac{92}{2}$

⇒ x = 46

Set of observations : 13, 35, 43, 46, 46, 50, 55, 61, 71, 80

Here, 46 has the maximum frequency.

Hence, value of x = 46 and mode = 46.