**Data Handling Class 7 Ex. 3.1;**

**New Ncert Solution for Class 7 Maths Chapter 3 Data Handling Free Solution;**

**Exercise 3.1**

**Question 1 :- Find the range, of heights of any ten-students of your class.**

**Solution 1:-**

Let’s take the heights of 10 students are as follows:

Student 1 = 140 cm,

Student 2 = 141.5 cm,

Student 3 = 138 cm,

Student 4 = 150 cm,

Student 5 = 161 cm,

Student 6 = 138 cm,

Student 7 = 140.5 cm,

Student 8 = 135.5 cm,

Student 9 = 160 cm,

Student 10 = 158 cm

Now, the minimum height of student = 135.5 cm

The maximum height of student =161 cm

So, **Range = Maximum height – Minimum height**

= 161 cm – 135.5 cm = 25.5 cm

Hence, the required range = 25.5 cm.

**Question-2:- Organize, 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 2:- **

Let’s make a frequency distribution table:

(i) Marks 9, is the highest marks.

(ii) Marks 1, is the lowest marks.

(iii) Range = Max. marks – Min. marks

= 9 – 1 = 8

(iv) Arithmetic mean = Σ(fixi)/Σ(fi)=100/20 = 5.

**Question-3:- Find the mean of first five whole numbers.**

**Solution-3:-**

The first 5-whole numbers are = 0, 1, 2, 3, 4

So, the **mean of first five numbers = Sum of number/ total number**

= (0+1+2+3+4)/5 = 10/5 = 2

Hence, the required mean = 2.

**Question 4 :- A cricketer-scores, the following runs in eight innings are :**

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

**Find the mean-score.**

**Solution 4:-**

Total number of inning = 8

Following are the scores of the runs in 8 innings:

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

So, Mean = Sum of all scores/Number of innings

= (58+76+40+35+46+45+0+100)/8 = 400/8 = 50

Hence, the required mean = 50.

**Question 5 :- Following-table shows, the points of 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 |

**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?**

**Solution 5:- **

(i)

(ii) Since, player C did not play Game 3, he played only 3 games. So, the sum of all digit will be divided by 3.

(iii) Number of points scored by B, (in all the games) are

Game 1 = 0, Game 2 = 8, Game 3 = 6, Game 4 = 4

Average score = (0+8+6+4)/4 = 18/4 = 4.5

(iv) Mean score of player A = 12.5

Mean score of player B = 4.5

Mean score of player C = (8+11+13)/3 = 32/3 = 10.67

Clearly, A is the best performer.

**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.**

**Solution 6:-**

Marks obtained by the students are:

85, 76, 90, 85, 39, 48, 56, 95, 81 and 75

(i) Highest marks obtain by student = 95

Lowest marks obtain by the student = 39

(ii) Range of the marks = ( Highest-marks – Lowest-marks )

= 95 – 39 = 56

(iii)

So, the mean-marks, obtain by the group = 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.**

**Solution 7:-**

Thus, the required-mean = 2058.

**Question 8 :- The rainfall (in mm) in a city on 7 days 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?**

**Solution 8:-**

(i) Since, maximum rainfall = 20.5 mm

Minimum rainfall = 0.0 mm

So, Range = Maximum rainfall – Minimum rainfall

= 20.5 mm – 0.0 mm = 20.5 mm

(ii) Mean rainfall = (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0)/7 = (41.3) / 7 = 5.9 mm.

(iii) Total number of days on which the rainfall was less than the mean rainfall =

Monday, Wednesday, Thursday, Saturday, Sunday = 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?**

**(Hi) 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 9:-**

(i) The height of the tallest-girl = 151 cm.

(ii) The height of the shortest-girl = 128 cm.

(iii) The range = Height of tallest-girl – Height of the shortest-girl

= 151 cm – 128 cm = 23 cm.

(iv) Mean of the height = (135+150+139+128+151+132+146+149+143+141)/10 = 1414/10 = 141.4 cm.

(v) Total number of girls having more height than the mean height

= 150, 151, 146, 149 and 143 = 5 girls.