Find mean and mode of the following data: Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80 --- --- --- --- --- --- --- --- Frequency 5 4 10 13 12 10 6

MathsCBSE Class 10 · Basic · Board exam 2026, set 430/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2026: question 35

Set in the CBSE Class 10 board exam of 2026, paper 430/5/1, 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 mean and mode of the following data:

Class	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	4	10	13	12	10	6

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/5/1.

Hints

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

Hint 1

For the mean, find class marks and use the step-deviation (or assumed mean) method.

Hint 2

For the mode, pick the class with the highest frequency and 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.

Take assumed mean $a = 45$ , $h = 10$ , $u_{i} = \frac{x _{i} - 45}{10}$ :
Class $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$
10-20 5 15 −3 −15
20-30 4 25 −2 −8
30-40 10 35 −1 −10
40-50 13 45 0 0
50-60 12 55 1 12
60-70 10 65 2 20
70-80 6 75 3 18
Total 60 17
1½ marks
Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 45 + \frac{17}{60} \times 10$
1 mark
$= 45 + 2.83 = 47.83$ (approx.)
½ mark
Highest frequency is 13, so modal class is 40-50: $l = 40$ , $f_{1} = 13$ , $f_{0} = 10$ , $f_{2} = 12$ , $h = 10$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{13 - 10}{26 - 10 - 12} \times 10$
1½ marks
$= 40 + \frac{3}{4} \times 10 = 47.5$
½ mark

Class	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$
10-20	5	15	−3	−15
20-30	4	25	−2	−8
30-40	10	35	−1	−10
40-50	13	45	0	0
50-60	12	55	1	12
60-70	10	65	2	20
70-80	6	75	3	18
Total	60			17

Answer: Mean ≈ 47.83; Mode = 47.5

Where the marks are

Correct table with $\sum f_{i} = 60$ , $\sum f_{i} u_{i} = 17$ (1½ marks)
Correct mean formula with substitution (1 mark)
Mean ≈ 47.83 (½ mark)
Modal class 40-50 and correct mode formula substitution (1½ marks)
Mode = 47.5 (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 paper2026 board paper, set 430/5/1 (Basic): 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 Basic 2026: question 35

Hints

Step-by-step solution

Revise the chapter

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