The table given below shows the daily expenditure on food of 25 households in a locality : Daily expenditure (₹) 100–150 150–200 200–250 250–300 300–350 --- --- --- --- --- --- Number of household 4 5 12 2 2 (i) Find the mean daily expenditure on food. (ii) Find the mode of the data.

MathsCBSE Class 10 · Basic · Board exam 2023, set 430/1/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2023: question 34

Set in the CBSE Class 10 board exam of 2023, paper 430/1/1, section D, question 34, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34 · 5 marks · Long answerCompetency-basedStatistics

The table given below shows the daily expenditure on food of 25 households in a locality :

Daily expenditure (₹)	100–150	150–200	200–250	250–300	300–350
Number of household	4	5	12	2	2

(i)
Find the mean daily expenditure on food.
3 marks
(ii)
Find the mode of the data.
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/1/1.

Hints

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

Hint 1

For the mean, find the class mark (mid-point) of each class and compute $\sum f_{i} x_{i}$ .

Hint 2

For the mode, pick the class with the highest frequency and use $l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ .

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.

(i) Table with class marks $x_{i}$ and $f_{i} x_{i}$ : 100–150: $f = 4$ , $x = 125$ , $f x = 500$ ; 150–200: $5$ , $175$ , $875$ ; 200–250: $12$ , $225$ , $2700$ ; 250–300: $2$ , $275$ , $550$ ; 300–350: $2$ , $325$ , $650$ . $\sum f_{i} = 25$ , $\sum f_{i} x_{i} = 5275$ .
1½ marks
(i) Mean $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{5275}{25} = 211$ . Mean daily expenditure = ₹211.
1½ marks
(ii) Highest frequency is 12, so modal class = 200–250; $l = 200$ , $f_{1} = 12$ , $f_{0} = 5$ , $f_{2} = 2$ , $h = 50$ .
½ mark
(ii) Mode $= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) h = 200 + (\frac{12 - 5}{24 - 5 - 2}) \times 50 = 200 + \frac{350}{17}$
1 mark
(ii) Mode $= \frac{3750}{17} \approx ₹220.59$
½ mark

Answer: (i) Mean = ₹211; (ii) Mode = 3750/17 ≈ ₹220.59

Where the marks are

Correct table of $x_{i}$ and $f_{i} x_{i}$ with totals 25 and 5275 (1½ marks)
Mean $= \frac{5275}{25} = 211$ (1½ marks)
Identifies modal class 200–250 (½ mark)
Substitutes correctly in mode formula (1 mark)
Mode $\approx 220.59$ (½ mark)

Revise the chapter

Chapter: Statistics (notes, formulas, where marks are lost and every past question)
Topic: Mode 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 paper2023 board paper, set 430/1/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2023: question 34

Hints

Step-by-step solution

Revise the chapter

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