MathsCBSE Class 10 · Standard · Sample paper, session 2026-27 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard sample paper 2026-27: question 38

From the official CBSE sample question paper for 2026-27, 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

A school conducted a weekly test for Class X students, before the commencement of the pre-board examination and recorded their scores (out of 50). To analyse performance patterns, the academic coordinator grouped the marks into intervals. The grouped frequency distribution is as below:

Marks Obtained	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Number of students	3	6	12	15	14

The academic coordinator computed the central tendencies to judge overall learning level, the most common performance range and to understand consistency across the batch for planning the remedial sessions.
Based on the above information, answer the following questions:

(i)
Identify the class with the most common performance range.
1 mark
(ii)
Find the class interval containing the median.
1 mark
(iii)
(A) What is the average performance of the students?
2 marks
OR
(iii)
(B) Find the mode of the data.
2 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

The 'most common' range is the class with the largest frequency.

Hint 2

For the median class, find N/2 and look at the cumulative frequencies.

Hint 3

Mean = Σfᵢxᵢ / Σfᵢ using class marks; mode = $l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ .

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.

(i) The most common performance range is the class with the highest frequency (15).
So the modal class is 30 – 40.
1 mark
(ii) Cumulative frequencies are 3, 9, 21, 36, 50, so $N = 50$ and $\frac{N}{2} = 25$ .
The first cf greater than 25 is 36, which belongs to class 30 – 40. So the median class is 30 – 40.
1 mark
(iii)(a) Table using class marks $x_{i}$ :
Class $f_{i}$ $x_{i}$ $f_{i} x_{i}$
0 – 10 3 5 15
10 – 20 6 15 90
20 – 30 12 25 300
30 – 40 15 35 525
40 – 50 14 45 630
Total 50 1560
1 mark
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{1560}{50} = 31.2$ marks
1 mark
(iii)(b) Modal class is 30 – 40, with:
- $l = 30$ , $h = 10$
- $f_{1} = 15$ , $f_{0} = 12$ , $f_{2} = 14$
½ mark
Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 30 + (\frac{15 - 12}{30 - 12 - 14}) \times 10$
1 mark
$= 30 + \frac{3}{4} \times 10 = 37.5$ marks
½ mark

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0 – 10	3	5	15
10 – 20	6	15	90
20 – 30	12	25	300
30 – 40	15	35	525
40 – 50	14	45	630
Total	50		1560

Answer: (i) 30 – 40; (ii) 30 – 40; (iii)(a) 31.2; (iii)(b) 37.5

Where the marks are

(i) Identifies modal class 30 – 40 (1 mark)
(ii) Uses cumulative frequency and N/2 = 25 to get median class 30 – 40 (1 mark)
(iii)(a) Correct table with class marks and fᵢxᵢ, total 1560 (1 mark)
(iii)(a) Mean = 31.2 (1 mark)
(iii)(b) Correct l, f₀, f₁, f₂ (½ mark)
(iii)(b) Correct substitution in the mode formula (1 mark)
(iii)(b) Mode = 37.5 (½ 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 paperSample paper 2026-27 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2026-27: question 38

Hints

Step-by-step solution

Revise the chapter

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