These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.1

Question 1.

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

Answer:

Do it yourself, one simple solution may be to consider the heights (in cm) of ten students as follows:

111, 118, 110, 120, 114, 113, 118, 117, 115, 119

Average height

= 115.5 cm

Question 2.

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.

Answer:

(i) The highest number = 9

(ii) The lowest number = 1

(iii) The range = Highest number – Lowest number = 9 – 1 = 8

(iv) Arithmetic Mean

= \(\frac{\text { Sum of the marks }}{\text { Number of students }}\)

= \(\frac{100}{20}\) = 5

Thus, the mean marks = 5.

Question 3.

Find the mean of the first five whole numbers.

Answer:

First five whole numbers are 0, 1, 2, 3 and 4

Sum 0 + 1 + 2 + 3 + 4 = 10

Mean = \(\frac{\text { Sum of the number }}{\text { Number of whole numbers }}\)

= \(\frac{10}{5}\) = 2

Question 4.

A cricketer scores the following runs in eight innings.

58, 76, 40, 35, 46, 45, 0, 100

Find the mean score.

Answer:

Mean score = \(\frac{\text { Sum of the scores }}{\text { Number of innings }}\)

= \(\frac{58+76+40+35+46+45+0+100}{8}\)

= \(\frac{400}{8}\) = 50

Mean score = 50

Question 5.

Following table shows the points of each player scored in four games:

Now answer the following questions:

(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 the four games. How would you find the mean?

(iv) Who is the best performer?

Answer:

(i) Mean = \(\frac{\text { Sum of the observations }}{\text { Number of observations }}\)

= \(\frac{14+16+10+10}{4}\) = \(\frac{50}{4}\) = 12.5

A’s average score per game is 12.5

(ii) Since C played only 3 games (He did not play the 3rd game)

Total will be divided by 3

(iii) Mean score of B

= \(\frac{\text { Sum of the all observations }}{\text { Number of observations }}\)

= \(\frac{0+8+6+4}{4}\) = \(\frac{18}{4}\) = 4.5

Thus, th average’tnumber of points scored by is B is 4.5

(iv) Mean score of C

= \(\frac{\text { Sum of the all observations }}{\text { Number of observations }}\)

= \(\frac{8+11+13}{4}\) = \(\frac{32}{3}\) = 10.67

Thus, mean number of points scored by C is 10.67

A’s average score is = 12.5

B’s average score is = 4.5

C’s average score is = 10.67

The best performer is A.

Question 6.

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) Highest and the lowest marks obtained by the students.

(ii) Range of the marks obtained.

(iii) Mean marks obtained by the group.

Answer:

Write the given marks in ascending order, we get 39, 48, 56, 75, 76, 81, 85, 85, 90, 95

(i) Highest marks = 95, Lowest marks = 39

(ii) Range = Highest marks – Lowest marks = 95 – 39 = 56

(iii) Mean = \(\frac{\text { Sum of all the marks }}{\text { Number of students }}\)

= \(\frac{730}{10}\) = 73

Mean marks obtained by the group is 73.

Question 7.

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.

Answer:

∴ The mean enrolment is 2058 per year.

Question 8.

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

Day | Rainfall (in mm) |

Monday | 0.0 |

Tuesday | 12.2 |

Wednesday | 2.1 |

Thursday | 0.0 |

Friday | 20.5 |

Saturday | 5.5 |

Sunday | 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.

Answer:

(i) Range = Highest rainfall – Lowest rainfall = 20.5 – 0.0 = 20.5 mm.

= \(\frac{41.3}{7}\) = 5.9mm

Mean rainfall = 5.9 mm

(iii) The rainfall was less than the mean rainfall on 5 days.

Question 9.

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.

Answer:

Arrange the heights of girls in ascending order, we get

128, 132, 135, 139, 141, 143, 146, 149, 150, 151

(i) Height of the tallest girl = 151cm

(ii) Height of the shortest girl = 128cm

(iii) Range= Highest height – Lowest height = 151cm – 128cm = 23cm

The mean height of the girls is 141.4cm

(v) The heights of 5 girls are more than the mean height.