NCERT solutions

Chapter 3 Data Handling Math Class 7 Solutions to NCERT Chapter 3 Exercise 3.1

Chapter 3 Data Handling Math Class 7 Solutions to NCERT Chapter 3 Exercise 3.1 Are the questions from all the exercises (Exercise 3.1.

Solutions to NCERT Chapter 3 Exercise 3.1

Exercise 3.1

Find the range of heights of any ten students in your class.

Solution:
Since actual heights are not provided, assume a sample set of heights (in cm) for ten students: 150, 155, 148, 160, 152, 157, 149, 153, 158, 145.

Arrange in ascending order: 145, 148, 149, 150, 152, 153, 155, 157, 158, 160.
Highest height = 160 cm, Lowest height = 145 cm.
Range = Highest – Lowest = 160 – 145 = 15 cm.
Answer: Range = 15 cm (assuming sample data).

Organise the following marks in a class assessment, in a tabular form: 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.

Solution:

(i) Highest number: 9 (from the table).
(ii) Lowest number: 1 (from the table).
(iii) Range: Highest – Lowest = 9 – 1 = 8.
(iv) Arithmetic mean:
Sum of marks = 1×1 + 2×2 + 3×1 + 4×3 + 5×4 + 6×4 + 7×2 + 8×1 + 9×1
= 1 + 4 + 3 + 12 + 20 + 24 + 14 + 8 + 9 = 95.
Mean = Sum / Number of observations = 95 / 20 = 4.75.
Answer: (i) 9, (ii) 1, (iii) 8, (iv) 4.75.

Find the mean of the first five whole numbers.

Solution:
First five whole numbers: 0, 1, 2, 3, 4.
Sum = 0 + 1 + 2 + 3 + 4 = 10.
Number of observations = 5.
Mean = Sum / Number of observations = 10 / 5 = 2.
Answer: Mean = 2.

A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.

Solution:
Sum of runs = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400.
Number of innings = 8.
Mean = Sum / Number of innings = 400 / 8 = 50.
Answer: Mean score = 50 runs.

(i) A’s mean:
Sum of points = 14 + 16 + 10 + 10 = 50.
Number of games = 4.
Mean = 50 / 4 = 12.5.
Answer: A’s average = 12.5 points per game.
(ii) C’s mean:
C played 3 games (Game 1, Game 2, Game 4).
Sum of points = 8 + 11 + 13 = 32.
Mean = 32 / 3 ≈ 10.67.
Divide by 3 because C did not play in Game 3.
Answer: Divide by 3, as C played only 3 games.
(iii) B’s mean:
Sum of points = 0 + 8 + 6 + 4 = 18.
Number of games = 4.
Mean = 18 / 4 = 4.5.
Answer: Sum B’s points and divide by 4.
(iv) Best performer:
A’s mean = 12.5, B’s mean = 4.5, C’s mean ≈ 10.67.
A has the highest mean.
Answer: A is the best performer.

The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81, and 75.

Find the:
(i) The highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.

Solution:
Marks: 85, 76, 90, 85, 39, 48, 56, 95, 81, 75.

(i) Arrange in ascending order: 39, 48, 56, 75, 76, 81, 85, 85, 90, 95.
Highest = 95, Lowest = 39.
Answer: Highest = 95, Lowest = 39.
(ii) Range = Highest – Lowest = 95 – 39 = 56.
Answer: Range = 56.
(iii) Sum = 85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75 = 730.
Number of students = 10.
Mean = 730 / 10 = 73.
Answer: Mean = 73 marks.

The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2540, 2820. Find the mean enrolment of the school for this period.

Solution:
Sum = 1555 + 1670 + 1750 + 2013 + 2540 + 2820 = 11,348.
Number of years = 6.
Mean = 11,348 / 6 ≈ 1891.33.
Answer: Mean enrolment ≈ : 1891.33 students.

(i) Highest = 20.5 mm, Lowest = 0.0 mm.
Range = 20.5 – 0.0 = 20.5 mm.
Answer: Range = 20.5 mm.
(ii) Sum = 0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0 = 41.3.
Number of days = 7.
Mean = 41.3 / 7 ≈ 5.9 mm.
Answer: Mean rainfall ≈ 5.9 mm.
(iii) Mean ≈ 5.9 mm. Compare:
Mon (0.0), Wed (2.1), Thurs (0.0), Sat (5.5), Sun (1.0) are less than 5.9.
Number of days = 5.
Answer: 5 days.

The heights of 10 girls were measured in cm, and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height?

Solution:
Heights: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141.

(i) Arrange in ascending order: 128, 132, 135, 139, 141, 143, 146, 149, 150, 151.
Tallest = 151 cm.
Answer: 151 cm.
(ii) Shortest = 128 cm.
Answer: 128 cm.
(iii) Range = 151 – 128 = 23 cm.
Answer: Range = 23 cm.
(iv) Sum = 135 + 150 + 139 + 128 + 151 + 132 + 146 + 149 + 143 + 141 = 1414.
Number of girls = 10.
Mean = 1414 / 10 = 141.4 cm.
Answer: Mean height = 141.4 cm.
(v) Mean = 141.4 cm. Heights > 141.4: 143, 146, 149, 150, 151 (5 heights).
Answer: 5 girls.

The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day	Mon	Tue	Wed	Thurs	Fri	Sat	Sun
Rainfall (in mm)	0.0	12.2	2.1	0.0	20.5	5.5	1.0

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?

Solution:
Rainfall: 0.0, 12.2, 2.1, 0.0, 20.5, 5.5, 1.0.

The following table shows the points each player scored in four games:

Player	Game 1	Game 2	Game 3	Game 4
A	14	16	10	10
B	0	8	6	4
C	8	11	Did not Play	13

(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all four games. How would you find the mean?
(iv) Who is the best performer?

Solution:

Tabular form: