250 apples of a box were weighed and the distribution of masses of the apples is given in the following table: Mass (in grams) 80–100 100–120 120–140 140–160 160–180 --- --- --- --- --- --- Number of apples 20 60 70 x 60 (i) Find the value of x and the mean mass of the apples. (ii) Find the modal mass of the apples.

MathsCBSE Class 10 · Standard · Board exam 2023, set 30/6/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2023: question 35

Set in the CBSE Class 10 board exam of 2023, paper 30/6/1, 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

250 apples of a box were weighed and the distribution of masses of the apples is given in the following table:

Mass (in grams)	80–100	100–120	120–140	140–160	160–180
Number of apples	20	60	70	$x$	60

(i)
Find the value of x and the mean mass of the apples.
3 marks
(ii)
Find the modal mass of the apples.
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/6/1.

Hints

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

Hint 1

The frequencies must add up to the total number of apples.

Hint 2

For the mean, find each class mark (midpoint of the class) and multiply it by its frequency.

Hint 3

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) The total number of apples is 250, so $20 + 60 + 70 + x + 60 = 250$ , which gives $210 + x = 250$ and $x = 40$ .
1 mark
(i) Table of class marks $x_{i}$ and products $f_{i} x_{i}$ :
Mass (g) $f_{i}$ $x_{i}$ $f_{i} x_{i}$
80–100 20 90 1800
100–120 60 110 6600
120–140 70 130 9100
140–160 40 150 6000
160–180 60 170 10200
Total 250 33700
1 mark
(i) Mean $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{33700}{250} = 134.8$ .
So the mean mass is $134.8$ g.
1 mark
(ii) The highest frequency is 70, so the modal class is 120–140.
Here $l = 120$ , $f_{1} = 70$ , $f_{0} = 60$ , $f_{2} = 40$ , $h = 20$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 120 + \frac{70 - 60}{140 - 60 - 40} \times 20$ .
1 mark
(ii) Mode $= 120 + \frac{10}{40} \times 20 = 120 + 5 = 125$ .
So the modal mass is $125$ g.
1 mark

Mass (g)	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
80–100	20	90	1800
100–120	60	110	6600
120–140	70	130	9100
140–160	40	150	6000
160–180	60	170	10200
Total	250		33700

Answer: (i) x = 40, mean mass = 134.8 g; (ii) modal mass = 125 g

Where the marks are

Uses the total of 250 to get $x = 40$ (1 mark)
Correct table of $x_{i}$ and $f_{i} x_{i}$ with $\sum f_{i} x_{i} = 33700$ (1 mark)
Mean $= 33700/250 = 134.8$ g (1 mark)
Identifies modal class 120–140 and substitutes correctly in the mode formula (1 mark)
Modal mass $= 125$ g (1 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 paper2023 board paper, set 30/6/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2023: question 35

Hints

Step-by-step solution

Revise the chapter

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