Find mean and mode of the following distribution : Class : 0 – 15 15 – 30 30 – 45 45 – 60 60 – 75 75 – 90 90 – 105 --- --- --- --- --- --- --- --- Frequency : 4 8 11 14 10 7 6

MathsCBSE Class 10 · Standard · Board exam 2026, set 30/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2026: question 35(b)

Set in the CBSE Class 10 board exam of 2026, paper 30/4/1, section D, question 35(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(b) · 5 marks · Long answerStatistics

Find mean and mode of the following distribution :

Class :	0 – 15	15 – 30	30 – 45	45 – 60	60 – 75	75 – 90	90 – 105
Frequency :	4	8	11	14	10	7	6

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/4/1.

Hints

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

Hint 1

Find the class marks first, then use the step-deviation method with a = 52.5 and h = 15.

Hint 2

The modal class is the class with the largest frequency.

Hint 3

Mode = l + ((f₁ − f₀)/(2f₁ − f₀ − f₂)) × 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 $a = 52.5$ and $h = 15$ , with $u_{i} = \frac{x _{i} - 52.5}{15}$ .
Class $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$
0–15 4 7.5 −3 −12
15–30 8 22.5 −2 −16
30–45 11 37.5 −1 −11
45–60 14 52.5 0 0
60–75 10 67.5 1 10
75–90 7 82.5 2 14
90–105 6 97.5 3 18
Total 60 3
2 marks
Mean $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 52.5 + 15 \times \frac{3}{60}$
½ mark
$= 52.5 + 0.75 = 53.25$
½ mark
The highest frequency is 14, so the modal class is 45–60.
Here $l = 45$ , $f_{1} = 14$ , $f_{0} = 11$ , $f_{2} = 10$ and $h = 15$ .
½ mark
Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 45 + \frac{14 - 11}{28 - 11 - 10} \times 15$
1 mark
$= 45 + \frac{45}{7} = 45 + 6.43 \approx 51.43$
Mean = 53.25 and Mode ≈ 51.43.
½ mark

Class	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$
0–15	4	7.5	−3	−12
15–30	8	22.5	−2	−16
30–45	11	37.5	−1	−11
45–60	14	52.5	0	0
60–75	10	67.5	1	10
75–90	7	82.5	2	14
90–105	6	97.5	3	18
Total	60			3

Answer: Mean = 53.25; Mode ≈ 51.43

Where the marks are

Correct table with class marks, u_i and f_iu_i, giving Σf = 60 and Σf_iu_i = 3 (2 marks)
Correct substitution in the step-deviation formula (½ mark)
Mean = 53.25 (½ mark)
Identifies modal class 45–60 (½ mark)
Correct substitution in the mode formula (1 mark)
Mode ≈ 51.4 (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 30/4/1 (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 2026: question 35(b)

Hints

Step-by-step solution

Revise the chapter

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