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

Step 1: Set up the cumulative frequency table

Total students = 230

| Marks | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 10–20 | 12 | 12 |
| 20–30 | 30 | 42 |
| 30–40 | x | 42 + x |
| 40–50 | 65 | 107 + x |
| 50–60 | y | 107 + x + y |
| 60–70 | 25 | 132 + x + y |
| 70–80 | 18 | 150 + x + y |

Step 2: Form equations

Since total = 230:
$$150 + x + y = 230 \implies x + y = 80 \quad \cdots (1)$$

Step 3: Find median class

$\dfrac{n}{2} = \dfrac{230}{2} = 115$

The cf just greater than 115 is (107 + x), which falls in class 40–50.

So median class = 40–50; $l = 40$, $f = 65$, $cf = 42 + x$, $h = 10$

Step 4: Apply median formula

$$\text{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}\right) \times h$$

$$46 = 40 + \left(\frac{115 - (42 + x)}{65}\right) \times 10$$

$$6 = \frac{(73 - x) \times 10}{65}$$

$$6 \times 65 = 10(73 - x)$$

$$390 = 730 - 10x$$

$$10x = 340 \implies \boxed{x = 34}$$

From (1): $y = 80 - 34 = \boxed{46}$

∴ x = 34 and y = 46

Source: Chapter 13, Section 13.4 Median of Grouped Data

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• Two conditions are always used: (i) sum of all frequencies = total, giving one equation; (ii) median formula, giving the second equation.
• Identify the median class first by finding $n/2 = 115$ and locating which class's cumulative frequency first exceeds 115. Since cf up to 30–40 is $42+x$ and up to 40–50 is $107+x$, for the median class to be 40–50 we need $42+x < 115 \leq 107+x$, which is satisfied for $x = 34$.
• Show all substitution steps clearly — CBSE awards marks for each step.