The marks of 10 students of a class in an examination arranged in ascending order are 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 (no. of observations) = 10, which is even.

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

Given, median marks = 48.

∴ x + 2 = 48
⇒ x = 46.

Putting value of x in data we get,

13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data 46 is repeated more number of times than any other number.

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