Mathematics — Standard (041) — CBSE Class 10 board question

The highest frequency is 23, so the modal class = 45–55.

Here, $l = 45$, $f_1 = 23$, $f_0 = 22$, $f_2 = 14$, $h = 10$

$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$

$$= 45 + \left(\frac{23 - 22}{2(23) - 22 - 14}\right) \times 10 = 45 + \frac{1}{10} \times 10 = 45 + 1 = \mathbf{46}$$

Source: Chapter 13, Section 13.3

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• Modal class is the class with the highest frequency (23 → 45–55).
• Identify $f_0$ (preceding class frequency = 22) and $f_2$ (succeeding class frequency = 14) carefully — a common error is swapping these.
• Examiner awards 1 mark for correct identification of modal class/values and 1 mark for correct substitution and final answer.