If the mode of the following distribution is 55, then find the value of x. Hence, find the mean. Class Interval 0 − 15 15 − 30 30 − 45 45 − 60 60 − 75 75 − 90 --- --- --- --- --- --- --- Frequency 10 7 x 15 10 12

MathsCBSE Class 10 · Standard · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2025-26: question 35(A)

From the official CBSE sample question paper for 2025-26, section D, question 35(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(A) · 5 marks · Long answerStatistics

If the mode of the following distribution is 55, then find the value of $x$ . Hence, find the mean.

Class Interval	0 − 15	15 − 30	30 − 45	45 − 60	60 − 75	75 − 90
Frequency	10	7	$x$	15	10	12

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

The modal class is the class that contains the mode, 55.

Hint 2

In the mode formula, the frequency before the modal class, $f_{0}$ , is the unknown $x$ .

Hint 3

After finding $x$ , use the class marks and $\overset{x}{ˉ} = \sum f_{i} x_{i} / \sum f_{i}$ .

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.

The mode is 55, which lies in the class 45–60. So the modal class is 45–60.
$l = 45$ , $h = 15$ , $f_{1} = 15$ , $f_{0} = x$ , $f_{2} = 10$
½ mark
Mode formula: $Mode = l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h$
Substituting: $55 = 45 + \frac{15 - x}{30 - x - 10} \times 15$
1 mark
$10 = \frac{15 ( 15 - x )}{20 - x}$
$\Rightarrow 200 - 10 x = 225 - 15 x$
$\Rightarrow 5 x = 25$
$\Rightarrow x = 5$
1 mark
Table for the mean (using $x = 5$ ):
C.I. $f_{i}$ $x_{i}$ $f_{i} x_{i}$
0–15 10 7.5 75
15–30 7 22.5 157.5
30–45 5 37.5 187.5
45–60 15 52.5 787.5
60–75 10 67.5 675
75–90 12 82.5 990
Total 59 2872.5
1½ marks
Mean $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2872.5}{59} \approx 48.69$
1 mark

C.I.	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0–15	10	7.5	75
15–30	7	22.5	157.5
30–45	5	37.5	187.5
45–60	15	52.5	787.5
60–75	10	67.5	675
75–90	12	82.5	990
Total	59		2872.5

Answer: $x = 5$ ; mean ≈ 48.69 (48.68 if truncated)

Where the marks are

Identifies modal class 45–60 with $l = 45$ , $h = 15$ (½ mark)
Substitutes correctly in the mode formula (1 mark)
Solves to get $x = 5$ (1 mark)
Correct table with $\sum f_{i} = 59$ and $\sum f_{i} x_{i} = 2872.5$ (1½ marks)
Mean ≈ 48.69 (1 mark)

Revise the chapter

Chapter: Statistics (notes, formulas, where marks are lost and every past question)
Topic: Relation Between Mean, Median and Mode

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 2025-26 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2025-26: question 35(A)

Hints

Step-by-step solution

Revise the chapter

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