Find median and mode of the following distribution: Class Interval 0–15 15–30 30–45 45–60 60–75 75–90 90–105 --- --- --- --- --- --- --- --- Frequency 15 10 12 9 8 10 6

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

CBSE Class 10 Mathematics Standard 2026: question 34(b)

Set in the CBSE Class 10 board exam of 2026, paper 30/5/1, section D, question 34(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(b) · 5 marks · Long answerStatistics

Find median and mode of the following distribution:

Class Interval	0–15	15–30	30–45	45–60	60–75	75–90	90–105
Frequency	15	10	12	9	8	10	6

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/5/1.

Hints

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

Hint 1

For the median, first make a cumulative frequency column and find $n /2$ .

Hint 2

The median class is the first class whose cumulative frequency exceeds $n /2$ ; the modal class is the one with the highest frequency.

Hint 3

For the mode, the modal class here is the first class, so the frequency of the class before it, $f_{0}$ , is 0.

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.

Cumulative frequency table:
Class $f$ $c f$
0–15 15 15
15–30 10 25
30–45 12 37
45–60 9 46
60–75 8 54
75–90 10 64
90–105 6 70
$n = \sum f = 70$
1 mark
$\frac{n}{2} = \frac{70}{2} = 35$ . The first $c f$ greater than 35 is 37, so the median class is 30–45.
Here $l = 30$ , $c f = 25$ , $f = 12$ , $h = 15$ .
½ mark
Median $= l + (\frac{\frac{n}{2} - c f}{f}) \times h = 30 + \frac{35 - 25}{12} \times 15$
1 mark
$= 30 + 12.5 = 42.5$
½ mark
The highest frequency is 15, so the modal class is 0–15.
Here $l = 0$ , $f_{1} = 15$ , $f_{0} = 0$ (no class before it), $f_{2} = 10$ , $h = 15$ .
½ mark
Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h = 0 + \frac{15 - 0}{30 - 0 - 10} \times 15$
1 mark
$= \frac{15}{20} \times 15 = \frac{45}{4} = 11.25$
Hence, median = 42.5 and mode = 11.25.
½ mark

Class	$f$	$c f$
0–15	15	15
15–30	10	25
30–45	12	37
45–60	9	46
60–75	8	54
75–90	10	64
90–105	6	70

Answer: Median = 42.5; Mode = 11.25

Where the marks are

Correct cumulative frequency table (1 mark)
Finds $n /2 = 35$ and identifies median class 30–45 (½ mark)
Substitutes correctly in the median formula (1 mark)
Median = 42.5 (½ mark)
Identifies modal class 0–15 (½ mark)
Substitutes correctly in the mode formula with $f_{0} = 0$ (1 mark)
Mode = 11.25 (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 paper2026 board paper, set 30/5/1 (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 34(b)

Hints

Step-by-step solution

Revise the chapter

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