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

(i) Modal Class:
The class with the highest frequency is 600 – 800 (frequency = 7).
∴ Modal class = 600 – 800

(ii) Mean of the given data:

| Class | $x_i$ (mid-value) | $f_i$ | $f_i x_i$ |
|-------|-------------------|--------|------------|
| 200–400 | 300 | 3 | 900 |
| 400–600 | 500 | 4 | 2000 |
| 600–800 | 700 | 7 | 4900 |
| 800–1000 | 900 | 4 | 3600 |
| 1000–1200 | 1100 | 3 | 3300 |
| 1200–1400 | 1300 | 3 | 3900 |
| | | $\Sigma f_i = 24$ | $\Sigma f_i x_i = 18600$ |

$$\text{Mean} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{18600}{24} = \textbf{775 mm}$$

(iii) Sub-divisions with at least 800 mm rainfall = 4 + 3 + 3 = 10 sub-divisions

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• (i) Modal class is simply the class with the highest frequency — just identify it, no calculation needed.
• (ii) Use Direct Method for mean: calculate mid-values, multiply by frequency, sum up, then divide by total frequency. Median would require cumulative frequency and locating the median class; mean is simpler here.
• (iii) "At least 800 mm" means classes 800–1000, 1000–1200, and 1200–1400, so add their frequencies: 4+3+3 = 10. Examiners look for correct addition of relevant classes.