**NCERT Solutions for Class 7 Maths**

Chapter 3 Data Handling

Exercise 3.1

Chapter 3 Data Handling

Exercise 3.1

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

**Answer**

S No. |
Name of Student |
Height (in feet) |

1 | Rahul Khanna | 5.2 |

2 | Arvind Kishore | 5.1 |

3 | Arundhati Kumari | 4.8 |

4 | Pragya Kumari | 4.4 |

5 | Rohit Sharma | 5.3 |

6 | Virat Kohli | 5.4 |

7 | Sachin Sharma | 5.2 |

8 | Rita Arora | 4.1 |

9 | Dolly Devi | 4.4 |

10 | Dimple Chauhan | 4.2 |

Range = Highest height – Lowest height

= 5.4 – 4.1 = 1.3 feet

**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**

Marks x_{i} |
Tally Marks |
Frequency (No. of Students) f_{i} |
f_{i}x_{i} |

1 | | | 1 | 1 |

2 | || | 2 | 4 |

3 | | | 1 | 3 |

4 | ||| | 3 | 12 |

5 | 5 | 25 | |

6 | |||| | 4 | 24 |

7 | || | 2 | 14 |

8 | | | 1 | 8 |

9 | | | 1 | 9 |

20 | 100 |

(i) The highest number is 9

(ii) The lowest number is 1

(iii) The range of data is 9-1 = 8

(iv) Arithmetic mean = $\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{100}{20}=5 $

**3. Find the mean of the first five whole numbers.**

**Answer**

First 5 whole numbers are 0, 1, 2, 3, 4

∴ Mean = $ =\frac{0+1+2+3+4}{5}=\frac{10}{5}=2$

Hence, the required mean = 2

**4. A cricketer scores the following runs in eight innings:**

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

**Find the mean score.**

**Answer**

Following are the scores of the runs in eight innings:

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

Mean $ =\frac{\text { Sum of all runs }}{\text { Number of innings }} $

$\begin{aligned} &=\frac{58+76+40+35+46+45+0+100}{8} \\ &=\frac{400}{8}=50 \end{aligned}$

**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?**

**Answer**

(i) Mean of player A

$=\frac {\text {Sum of Scores By A}}{\text {No. of games played by A}}$

$=\frac{14+16+10+10}{4}$

$=\frac{50}{4}=12.5$

(ii) Since, C did not play Game 3, he played only 3 games. So, the total 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 $=\frac{0+8+6+4}{4}=\frac{18}{4}=4.5$

(iv) Mean score of C

$=\frac{8+11+13}{3}=\frac{32}{3}=10.67$

Mean score of C = 10.67

While mean score of A = 12.5

Clearly, A is the best performer.

**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**

Marks obtained are:

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

(i) Highest marks = 95

Lowest marks = 39

(ii) Range of the marks

= Highest marks – Lowest marks

= 95 – 39 = 56

(iii) Mean marks

$=\frac{\text {Sum of all marks scored}} {\text {No. of students}}$

$=\frac{85+76+90+85+39+48+56+95+81+75}{10}=\frac{730}{10}=73$

**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**

Mean enrolment

$=\frac {\text {Sum of the enrollments of all years}}{\text{No. of years}}$

$=\frac{1555+1670+1750+2013+2540+2820}{6}$

$=\frac{12348}{6}=2058$

Thus, the required mean = 2058

**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.**

**Answer**

(i) Maximum rainfall = 20.5 mm

Minimum rainfall = 0.0 mm

∴ Range = Maximum rainfall – Minimum rainfall

= 20.5 mm – 0.0 mm = 20.5 mm

(ii) Mean rainfall

$=\frac{\text {Sum of rainfalls (in mm)}}{\text {No. of days}}$

$=\frac{0.0+12.2++2.1+0.0+20.5+5.5+1.0}{7}$

$=\frac{41.3mm}{7}=5.9 mm$

(iii) Number of days on which the rainfall was less than the mean rainfall = Monday, Wednesday, Thursday, Saturday, Sunday = 5 days.

**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**

(i) Height of the tallest girl = 151 cm.

(ii) Height of the shortest girl = 128 cm.

(iii) Range = Height of tallest girl – Height of the shortest girl

= 151 cm – 128 cm = 23 cm.

$\text {Mean height}=\frac {\text {Sum of all heights}}{\text {No. of girls}}$

$=\frac{135+150+139+128+151+132+146+149+143+141}{10}$

$=\frac{1410}{10}=141.4 cm$

(v) Number of girls having more height than the mean height

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

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