A survey regarding heights (in cm) of 51 girls of class X of a school was conducted and the following data was obtained: Heights (in cm) Number of girls --- --- less than 140 04 less than 145 11 less than 150 29 less than 155 40 less than 160 46 less than 165 51 Find the median height of girls. If mode of the above distribution is 148.05, find the mean using empirical formula.

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

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

From the official CBSE sample question paper for 2025-26, 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

A survey regarding heights (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:

Heights (in cm)	Number of girls
less than 140	04
less than 145	11
less than 150	29
less than 155	40
less than 160	46
less than 165	51

Find the median height of girls. If mode of the above distribution is 148.05, find the mean using empirical formula.

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

Subtract consecutive 'less than' counts to get the frequency of each 5 cm class.

Hint 2

Find $N /2$ , then the class whose cumulative frequency first passes it.

Hint 3

For the mean, rearrange $3 Median = Mode + 2 Mean$ .

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.

Convert the cumulative ('less than') data into class frequencies:
C.I. $f$ c.f.
135–140 4 4
140–145 7 11
145–150 18 29
150–155 11 40
155–160 6 46
160–165 5 51
1 mark
$N = 51$ , so $\frac{N}{2} = 25.5$ .
The first c.f. greater than 25.5 is 29, so the median class is 145–150.
$l = 145$ , $c f = 11$ , $f = 18$ , $h = 5$
Median formula: $Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h$
1 mark
$Median = 145 + \frac{25.5 - 11}{18} \times 5 = 145 + \frac{72.5}{18} \approx 145 + 4.03 = 149.03$
1 mark
Median height ≈ 149.03 cm
1 mark
Empirical formula: $3 Median = Mode + 2 Mean$
$3 (149.03) = 148.05 + 2 Mean$
$\Rightarrow 2 Mean = 447.09 - 148.05 = 299.04$
$\Rightarrow Mean = 149.52$ cm
1 mark

C.I.	$f$	c.f.
135–140	4	4
140–145	7	11
145–150	18	29
150–155	11	40
155–160	6	46
160–165	5	51

Answer: Median ≈ 149.03 cm; Mean ≈ 149.52 cm

Where the marks are

Converts the cumulative data into class intervals with frequencies (1 mark)
Identifies the median class 145–150 and writes the median formula (1 mark)
Substitutes correctly (1 mark)
Median ≈ 149.03 cm (1 mark)
Uses $3 Median = Mode + 2 Mean$ to get Mean ≈ 149.52 cm (1 mark)

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Chapter: Statistics (notes, formulas, where marks are lost and every past question)

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(B)

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Step-by-step solution

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