Mathematics — CBSE Class 10 board question

Modal Age:

The highest frequency is 33, corresponding to class 35–40.
So, modal class = 35–40, $l = 35$, $h = 5$, $f_1 = 33$, $f_0 = 21$, $f_2 = 11$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 35 + \frac{33-21}{66-21-11} \times 5 = 35 + \frac{12}{34} \times 5 = 35 + 1.76 = \textbf{36.76 years}$$

Median Age:

Cumulative frequency table:

| Age | f | cf |
|---|---|---|
| 15–20 | 2 | 2 |
| 20–25 | 4 | 6 |
| 25–30 | 18 | 24 |
| 30–35 | 21 | 45 |
| 35–40 | 33 | 78 |
| 40–45 | 11 | 89 |
| 45–50 | 3 | 92 |
| 50–55 | 6 | 98 |
| 55–60 | 2 | 100 |

$n = 100$, so $\frac{n}{2} = 50$.
cf just below 50 is 45 (class 30–35); hence median class = 35–40.

$l = 35$, $cf = 45$, $f = 33$, $h = 5$

$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h = 35 + \frac{50 - 45}{33} \times 5 = 35 + \frac{25}{33} = 35 + 0.76 = \textbf{35.76 years}$$

Source: Chapter 13, Sections 13.3 (Mode) and 13.4 (Median of Grouped Data)

---

• Modal class = class with highest frequency (33 → 35–40). Apply the standard mode formula.
• Median class = class where cumulative frequency first crosses $n/2 = 50$; cf reaches 45 after 30–35, so 35–40 is the median class.
• Examiners award marks for: correct identification of modal/median class, correct substitution in formula, and correct final values. Always show the cf table for median.