Gurpreet is very fond of doing research on plants. She collected some leaves from different plants and measured their lengths in mm. The data obtained is represented in the following table: Length (in mm) 70-80 80-90 90-100 100-110 110-120 120-130 130-140 --- --- --- --- --- --- --- --- Number of leaves 3 5 9 12 5 4 2 Based on the above information, answer the following questions: (i) Write the median class of the data. (ii) How many leaves are of length equal to or more than 10 cm ? (iii) (a) Find median of the data. (iii) (b) Write the modal class and find the mode of the data.

MathsCBSE Class 10 · Basic · Board exam 2024, set 430/1/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2024: question 37

Set in the CBSE Class 10 board exam of 2024, paper 430/1/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedStatistics

Gurpreet is very fond of doing research on plants. She collected some leaves from different plants and measured their lengths in mm.

The data obtained is represented in the following table:

Length (in mm)	70-80	80-90	90-100	100-110	110-120	120-130	130-140
Number of leaves	3	5	9	12	5	4	2

Based on the above information, answer the following questions:

(i)
Write the median class of the data.
1 mark
(ii)
How many leaves are of length equal to or more than 10 cm ?
1 mark
(iii)
(a) Find median of the data.
2 marks
OR
(iii)
(b) Write the modal class and find the mode of the data.
2 marks

Figure in the paper: Photograph of many leaves of different shapes and sizes spread out on a surface, collected from various plants. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/1/1.

Hints

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

Hint 1

Make a cumulative frequency column and find N/2.

Hint 2

Convert 10 cm to mm before counting the leaves.

Hint 3

Use Median $= l + \frac{N /2 - c f}{f} h$ and Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} h$ .

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 cumulative frequencies are 3, 8, 17, 29, 34, 38, 40, so $N = 40$ and $\frac{N}{2} = 20$ . The cf first exceeds 20 in the class 100–110, so the median class is 100–110.
1 mark
(ii) 10 cm = 100 mm. Leaves of length ≥ 100 mm $= 12 + 5 + 4 + 2 = 23$ .
1 mark
(iii)(a) Cumulative frequency table: 70–80: f = 3, cf = 3; 80–90: f = 5, cf = 8; 90–100: f = 9, cf = 17; 100–110: f = 12, cf = 29; 110–120: f = 5, cf = 34; 120–130: f = 4, cf = 38; 130–140: f = 2, cf = 40.
½ mark
(iii)(a) For the median class, $l = 100$ , $c f = 17$ , $f = 12$ and $h = 10$ . Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 100 + \frac{20 - 17}{12} \times 10 = 100 + 2.5 = 102.5$ mm.
1½ marks
(iii)(b) The highest frequency is 12, so the modal class is 100–110.
½ mark
(iii)(b) Here $l = 100$ , $f_{1} = 12$ , $f_{0} = 9$ , $f_{2} = 5$ and $h = 10$ . Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 100 + \frac{12 - 9}{24 - 9 - 5} \times 10 = 100 + 3 = 103$ mm.
1½ marks

Answer: (i) 100–110; (ii) 23; (iii)(a) median = 102.5 mm; (iii)(b) modal class 100–110, mode = 103 mm

Where the marks are

Identifies the median class 100–110 (1 mark)
Counts 23 leaves of length ≥ 100 mm (1 mark)
(iii)(a) Correct cumulative frequency table (½ mark)
(iii)(a) Substitutes in the median formula and gets 102.5 (1½ marks)
(iii)(b) Modal class 100–110 (½ mark)
(iii)(b) Substitutes in the mode formula and gets 103 (1½ marks)

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 paper2024 board paper, set 430/1/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 37

Hints

Step-by-step solution

Revise the chapter

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