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

Step 1: Form equations using total frequency = 90

| Class | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 20–30 | p | p |
| 30–40 | 15 | p + 15 |
| 40–50 | 25 | p + 40 |
| 50–60 | 20 | p + 60 |
| 60–70 | q | p + 60 + q |
| 70–80 | 8 | p + 68 + q |
| 80–90 | 10 | p + 78 + q |

Sum of frequencies: $p + 15 + 25 + 20 + q + 8 + 10 = 90$

$$p + q + 78 = 90 \implies p + q = 12 \quad \cdots (1)$$

Step 2: Identify median class

$\dfrac{n}{2} = \dfrac{90}{2} = 45$

Median = 50, so the median class is 40–50 (cf just before it is $p + 15$, and cf after it is $p + 40$, which must be ≥ 45).

Here: $l = 40,\ f = 25,\ cf = p + 15,\ h = 10$

Step 3: Apply median formula

$$50 = 40 + \left(\frac{45 - (p+15)}{25}\right) \times 10$$

$$10 = \frac{(30 - p)}{25} \times 10$$

$$25 = 30 - p \implies p = 5$$

Step 4: Find q

From (1): $5 + q = 12 \implies q = 7$

$$\boxed{p = 5, \quad q = 7}$$

Source: Statistics, Section 13.4 Median of Grouped Data, Chapter 13

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• Two equations are needed for two unknowns: one from total frequency, one from the median formula.
• The median class is the class containing the 45th observation. Since cf up to 40–50 is $p+40$ and the median lies in this interval, apply the formula with $cf = p+15$ (cf of the preceding class).
• Examiners award marks for: setting up the cf table correctly (1 mark), forming equation (1) from total frequency (1 mark), identifying median class (1 mark), applying formula and solving for p (1 mark), finding q (1 mark).
• Always write the final boxed values clearly.