Case Study 3. The India Meteorological Department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results. The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023. Rainfall (mm) No. of Sub-divisions --- --- 200 – 400 3 400 – 600 4 600 – 800 7 800 – 1000 4 1000 – 1200 3 1200 – 1400 3 Based on the information given above, answer the following questions : (i) Write the modal class. (ii) (a) Find the median of the given data. (ii) (b) Find the mean rainfall in the season. (iii) If a sub-division having at least 800 mm rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall ?

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/3/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2025: question 38

Set in the CBSE Class 10 board exam of 2025, paper 30/3/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedStatistics

Case Study 3. The India Meteorological Department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results.

The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023.

Rainfall (mm)	No. of Sub-divisions
200 – 400	3
400 – 600	4
600 – 800	7
800 – 1000	4
1000 – 1200	3
1200 – 1400	3

Based on the information given above, answer the following questions :

(i)
Write the modal class.
1 mark
(ii)
(a) Find the median of the given data.
2 marks
OR
(ii)
(b) Find the mean rainfall in the season.
2 marks
(iii)
If a sub-division having at least 800 mm rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall ?
1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/3/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

The modal class is the class with the highest frequency.

Hint 2

For the median, build cumulative frequencies and find the class containing the $N /2$ -th value. Then use $l + \frac{N /2 - c f}{f} \times h$ .

Hint 3

For the mean, use class marks (midpoints) and $\frac{\sum f _{i} x _{i}}{\sum f _{i}}$ . For (iii), add the frequencies of the classes from 800 mm upwards.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) The highest frequency is 7, which belongs to the class 600–800. So the modal class is 600–800.
1 mark
(ii)(a) Cumulative frequencies: 200–400: 3; 400–600: 7; 600–800: 14; 800–1000: 18; 1000–1200: 21; 1200–1400: 24. So $N = 24$ .
½ mark
(ii)(a) $\frac{N}{2} = 12$ . The first cumulative frequency of at least 12 is 14, so the median class is 600–800. Here $l = 600$ , $c f = 7$ , $f = 7$ , $h = 200$ .
½ mark
(ii)(a) Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 600 + \frac{12 - 7}{7} \times 200$
½ mark
(ii)(a) $= 600 + \frac{1000}{7} = \frac{5200}{7} \approx 742.86$ mm
½ mark
(ii)(b) Class marks $x_{i}$ : 300, 500, 700, 900, 1100, 1300. Products $f_{i} x_{i}$ : 900, 2000, 4900, 3600, 3300, 3900. So $\sum f_{i} = 24$ and $\sum f_{i} x_{i} = 18600$ .
1 mark
(ii)(b) Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{18600}{24} = 775$ mm
1 mark
(iii) The classes with at least 800 mm are 800–1000, 1000–1200 and 1200–1400. Number of sub-divisions $= 4 + 3 + 3 = 10$ .
1 mark

Answer: (i) 600–800; (ii)(a) Median = 5200/7 ≈ 742.86 mm; (ii)(b) Mean = 775 mm; (iii) 10

Where the marks are

(i) States the modal class as 600–800 (1 mark)
(ii)(a) Draws up the correct cumulative frequency table (½ mark)
(ii)(a) Identifies the median class 600–800 (½ mark)
(ii)(a) Substitutes correctly into the median formula (½ mark)
(ii)(a) Gets the median as 5200/7 ≈ 742.86 mm (½ mark)
(ii)(b) Draws up the correct $x_{i}$ , $f_{i} x_{i}$ table with $\sum f_{i} x_{i} = 18600$ (1 mark)
(ii)(b) Gets the mean as 775 mm (1 mark)
(iii) Counts 10 sub-divisions (1 mark)

Revise the chapter

Chapter: Statistics (notes, formulas, where marks are lost and every past question)
Topic: Median of Grouped Data

Practise this chapterBoard-style questions from Statistics, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 30/3/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2025: question 38

Hints

Step-by-step solution

Revise the chapter

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