Find the missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode. Daily Allowance Number of Children --- --- 11 – 13 7 13 – 15 6 15 – 17 9 17 – 19 13 19 – 21 f 21 – 23 5 23 – 25 4

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

CBSE Class 10 Mathematics Standard 2025: question 35

Set in the CBSE Class 10 board exam of 2025, paper 30/1/1, 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 missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode.

Daily Allowance	Number of Children
11 – 13	7
13 – 15	6
15 – 17	9
17 – 19	13
19 – 21	f
21 – 23	5
23 – 25	4

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

Hints

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

Hint 1

Find the class marks and the products $f_{i} x_{i}$ , keeping f as an unknown.

Hint 2

Put $\frac{\sum f _{i} x _{i}}{\sum f _{i}} = 18$ and solve for f.

Hint 3

Once you know f, find the modal class. Each class is 2 wide, so use $h = 2$ in the mode formula.

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.

Make the table using class marks $x_{i}$ :
Class $x_{i}$ $f_{i}$ $f_{i} x_{i}$
11–13 12 7 84
13–15 14 6 84
15–17 16 9 144
17–19 18 13 234
19–21 20 f 20f
21–23 22 5 110
23–25 24 4 96
Total 44 + f 752 + 20f
1½ marks
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , so
$18 = \frac{752 + 20 f}{44 + f}$
1 mark
Cross-multiply: $792 + 18 f = 752 + 20 f$
So $2 f = 40$ , which gives $f = 20$ .
½ mark
The highest frequency is now 20, so the modal class is 19–21.
Here $l = 19$ , $f_{1} = 20$ , $f_{0} = 13$ , $f_{2} = 5$ , and the class width is $h = 2$ .
½ mark
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$
$= 19 + \frac{20 - 13}{40 - 13 - 5} \times 2 = 19 + \frac{14}{22}$
1 mark
$= 19 + 0.636 \approx 19.64$
So the mode is $19 \frac{7}{11} \approx 19.64$ .
½ mark

Class	$x_{i}$	$f_{i}$	$f_{i} x_{i}$
11–13	12	7	84
13–15	14	6	84
15–17	16	9	144
17–19	18	13	234
19–21	20	f	20f
21–23	22	5	110
23–25	24	4	96
Total		44 + f	752 + 20f

Answer: f = 20; Mode ≈ 19.64 (= 19 7/11)

Where the marks are

Writes a correct table of $x_{i}$ and $f_{i} x_{i}$ with totals $44 + f$ and $752 + 20 f$ (1½ marks)
Sets up $\frac{752 + 20 f}{44 + f} = 18$ (1 mark)
Solves to get $f = 20$ (½ mark)
Identifies the modal class 19–21 (½ mark)
Substitutes correctly into the mode formula with $h = 2$ (1 mark)
Gets mode ≈ 19.64 (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 30/1/1 (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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