Mathematics — CBSE Class 10 board question

The class with the highest frequency is 40–50 (frequency = 17), so the modal class = 40–50.

Here, $l = 40,\ f_1 = 17,\ f_0 = 12,\ f_2 = 4,\ h = 10$

$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 40 + \left(\frac{17-12}{34-12-4}\right) \times 10$$

$$= 40 + \frac{5}{18} \times 10 = 40 + 2.78 = \mathbf{42.78}$$

Source: Chapter 13, Section 13.3

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• Examiners award 1 mark for correctly identifying the modal class and writing the values of $l, f_1, f_0, f_2, h$, and 1 mark for correct substitution and final answer.
• The modal class is always the one with the highest frequency (here 40–50, f = 17), not the first class.
• $f_0$ is the frequency of the class before the modal class (30–40 = 12) and $f_2$ is the class after (50–60 = 4). Getting these mixed up is a common error.