The median of the following data is 50. Find the values of 'p' and 'q', if the sum of all frequencies is 90. Also find the mode of the data. Marks obtained Number of students --- --- 20 – 30 p 30 – 40 15 40 – 50 25 50 – 60 20 60 – 70 q 70 – 80 8 80 – 90 10

MathsCBSE Class 10 · Standard · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2023-24: question 35

From the official CBSE sample question paper for 2023-24, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerCompetency-basedStatistics

The median of the following data is 50. Find the values of 'p' and 'q', if the sum of all frequencies is 90. Also find the mode of the data.

Marks obtained	Number of students
20 – 30	p
30 – 40	15
40 – 50	25
50 – 60	20
60 – 70	q
70 – 80	8
80 – 90	10

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

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

Hint 1

Write the cumulative frequencies in terms of p and q, and use the total of 90.

Hint 2

A median of 50 tells you which class is the median class. Its lower limit is 50.

Hint 3

Once p and q are known, the class with the highest frequency gives the mode.

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:
Marks f cf
20–30 p p
30–40 15 p+15
40–50 25 p+40
50–60 20 p+60
60–70 q p+q+60
70–80 8 p+q+68
80–90 10 p+q+78
1 mark
Sum of frequencies $= 90$ :
$p + q + 78 = 90 \Rightarrow p + q = 12$ ...(1)
½ mark
The median is 50, so the median class is 50–60, with $\frac{n}{2} = 45$ .
Here $l = 50$ , $c f = p + 40$ , $f = 20$ , $h = 10$ .
½ mark
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h$
$\Rightarrow 50 = 50 + \frac{45 - ( p + 40 )}{20} \times 10$
½ mark
$\Rightarrow 45 - (p + 40) = 0 \Rightarrow p = 5$
½ mark
From (1): $q = 12 - 5 = 7$
½ mark
The highest frequency is 25, so the modal class is 40–50.
Here $l = 40$ , $f_{1} = 25$ , $f_{0} = 15$ , $f_{2} = 20$ , $h = 10$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$
½ mark
Mode $= 40 + \frac{25 - 15}{50 - 15 - 20} \times 10 = 40 + \frac{100}{15} = 40 + 6.67 = 46.67$ marks
1 mark

Marks	f	cf
20–30	p	p
30–40	15	p+15
40–50	25	p+40
50–60	20	p+60
60–70	q	p+q+60
70–80	8	p+q+68
80–90	10	p+q+78

Answer: p = 5, q = 7; mode ≈ 46.67

Where the marks are

Cumulative frequency column (1 mark)
p + q = 12 (½ mark)
Identifies the median class 50–60 with n/2 = 45 (½ mark)
Substitutes correctly into the median formula (½ mark)
p = 5 (½ mark)
q = 7 (½ mark)
Modal class 40–50 and the mode formula (½ mark)
Mode = 46.67 (1 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 2023-24 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2023-24: question 35

Hints

Step-by-step solution

Revise the chapter

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