Find the Mean and Mode of the following frequency distribution : Class Frequency --- --- 0 – 10 8 10 – 20 7 20 – 30 15 30 – 40 20 40 – 50 12 50 – 60 8 60 – 70 10

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/2/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2025: question 35

Set in the CBSE Class 10 board exam of 2025, paper 30/2/1, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerStatistics

Find the Mean and Mode of the following frequency distribution :

Class	Frequency
0 – 10	8
10 – 20	7
20 – 30	15
30 – 40	20
40 – 50	12
50 – 60	8
60 – 70	10

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/2/1.

Hints

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

Hint 1

For the mean, find the class marks and choose a central one, such as 35, as the assumed mean.

Hint 2

Use $\overset{x}{ˉ} = a + \frac{Σ f _{i} u _{i}}{Σ f _{i}} \times h$ with $u_{i} = \frac{x _{i} - a}{h}$ .

Hint 3

For the mode, the class with frequency 20 is modal. Use $l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ .

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.

Mean (step-deviation method) with assumed mean $a = 35$ and class size $h = 10$ , where $u_{i} = \frac{x _{i} - 35}{10}$ :
Class $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$
0–10 8 5 −3 −24
10–20 7 15 −2 −14
20–30 15 25 −1 −15
30–40 20 35 0 0
40–50 12 45 1 12
50–60 8 55 2 16
60–70 10 65 3 30
Total 80 5
So $Σ f_{i} = 80$ and $Σ f_{i} u_{i} = 5$ .
1½ marks
Mean $= a + \frac{Σ f _{i} u _{i}}{Σ f _{i}} \times h = 35 + \frac{5}{80} \times 10$
1 mark
$= 35 + 0.625 = 35.625$
½ mark
Mode: The highest frequency is 20, so the modal class is 30–40.
½ mark
Here $l = 30$ , $f_{1} = 20$ , $f_{0} = 15$ , $f_{2} = 12$ and $h = 10$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 30 + \frac{20 - 15}{40 - 15 - 12} \times 10$
1 mark
$= 30 + \frac{50}{13} = \frac{440}{13} \approx 33.85$
½ mark

Class	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$
0–10	8	5	−3	−24
10–20	7	15	−2	−14
20–30	15	25	−1	−15
30–40	20	35	0	0
40–50	12	45	1	12
50–60	8	55	2	16
60–70	10	65	3	30
Total	80			5

Answer: Mean = 35.625; Mode = 440/13 ≈ 33.85

Where the marks are

Correct table of class marks, $u_{i}$ and $f_{i} u_{i}$ , with $Σ f_{i} = 80$ and $Σ f_{i} u_{i} = 5$ (1½ marks)
Correct mean formula with values substituted (1 mark)
Mean $= 35.625$ (½ mark)
Identifies the modal class 30–40 (½ mark)
Correct mode formula with values substituted (1 mark)
Mode $= \frac{440}{13} \approx 33.85$ (½ mark)

Revise the chapter

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

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 30/2/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2025: question 35

Hints

Step-by-step solution

Revise the chapter

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