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,

$\text{Median} = \dfrac{\dfrac{n}{2}\text{th term} + \Big(\dfrac{n}{2} + 1\Big)\text{th term}}{2} \\[1em] = \dfrac{\dfrac{10}{2}\text{th term} + \Big(\dfrac{10}{2} + 1\Big)\text{th term}}{2} \\[1em] = \dfrac{\text{5th term + 6th term}}{2} \\[1em] = \dfrac{x + x + 4}{2} \\[1em] = \dfrac{2x + 4}{2} \\[1em] = x + 2.$

Given,

Median = 48

⇒ x + 2 = 48

⇒ x = 46.

Set of observations : 13, 35, 42, 46, 46, 50, 55, 61, 71, 80.

Here, 46 has the maximum frequency.

∴ Mode = 46.

Hence, x = 46 and mode = 46.