Mathematics — CBSE Class 10 board question

Step 1: Cumulative Frequency Table

| Class | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 65–85 | 4 | 4 |
| 85–105 | 5 | 9 |
| 105–125 | x | 9 + x |
| 125–145 | 20 | 29 + x |
| 145–165 | 14 | 43 + x |
| 165–185 | y | 43 + x + y |
| 185–205 | 4 | 47 + x + y |

Step 2: Using total frequency = 68

$$47 + x + y = 68 \implies x + y = 21 \quad \text{...(1)}$$

Step 3: Finding median class

$\dfrac{n}{2} = \dfrac{68}{2} = 34$

cf just before 125–145 is $9 + x$, and cf of 125–145 is $29 + x$.
Since median = 137 lies in class 125–145, this is the median class.

Here: $l = 125,\ f = 20,\ cf = 9 + x,\ h = 20$

Step 4: Applying median formula

$$137 = 125 + \left(\frac{34 - (9+x)}{20}\right) \times 20$$

$$12 = 25 - x$$

$$x = 13$$

Step 5: From (1):

$$y = 21 - 13 = 8$$

$$\boxed{x = 13, \quad y = 8}$$

Source: Chapter 13, Section 13.4 Median of Grouped Data

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• First form the equation from total frequency (sum = 68) to get x + y = 21.
• Identify the median class by locating where n/2 = 34 falls in cumulative frequencies — it must be 125–145 since median 137 lies there.
• Apply the median formula with correct l, cf, f, h values to get x = 13.
• Substitute back into equation (1) to get y = 8.
• Examiner checks: correct cf column, right median class, correct substitution, and both values stated clearly.