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

Exercise 3.1

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

Ans:

We will find the range of heights by getting the difference between the highest height and lowest height.

Range= Highest height – Lowest Height

 $ =5.3-4.2\dfrac{z}{3} $ 

$ =1.1 $ 

Hence, the range of height of ten students of the class is 1.1 feet.

2.Organize the following marks in a class statement, 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 highest?

(ii) Which number is lowest?

(iii) What is the range of the lowest?

Ans: 

1. The highest number among the given data is 9.

2. The lowest number among the given data is 1.

3. We will find the range of marks by getting the difference between the highest value and lowest value.

Range= Highest value – Lowest value

$ =9-1 \\ $

$ =8 \\ $

Hence, the range of marks is 8.

4. We know that, arithmetic mean is the sum of all observations divided by the total number of observations.

$ \text{Mean=}\dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\ $

$ \text{=}\dfrac{4+6+7+5+3+5+4+5+2+6+2+5+1+9+6+5+8+4+6+7}{20} \\ $

$ \text{=}\dfrac{100}{20} \\ $

$ \text{=5} \\ $

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

Ans: The first five whole numbers are 0, 1, 2, 3, 4.

Arithmetic mean is the sum of all observations divided by total number of observations.

$\text{Mean=}\dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\ $

$ \text{=}\dfrac{0+1+2+3+4}{5} \\ $

$ \text{=}\dfrac{10}{5} \\ $

$ \text{=2} \\ $

Hence, the mean of the first five natural numbers is 2.

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

Ans: Arithmetic mean is sum of all observations divided by total number of observations.

Number of Observations= 8

$ \text{Mean=}\dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\ $

$ \text{=}\dfrac{58+76+40+35+46+45+0+100}{8} \\ $

$ \text{=}\dfrac{400}{8} \\ $

$ \text{=50} \\ $

Hence, the mean score of the cricketer is 50.

5. Following table shows the points 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 47? Why? 

(iii) B played in all the four games. How would you find the meaning? 

(iv) Who is the best performer?

Ans:

(i) The average of player A can be determined by dividing the sum of all scores of A by the number of games he played. 

 $ \text{Mean of player A=}\dfrac{\text{Sum of scored by A}}{\text{Number of games played by A}} \\ $

 $ \text{=}\dfrac{14+16+10+10}{4} \\ $

$ \text{ =}\dfrac{50}{4} \\ $

$ \text{ =12}\text{.5} \\ $

Hence, the average number of points scored by A is 12.5.

(ii)For finding average points scored, we divide the total scores by the total number of games played. In this case, C has played 3 games and therefore we will divide the total points by 3 and not by 4.

(iii)B played all 4 games. So, we will divide the total points scored by 4.

$ \text{Mean of player B=}\dfrac{\text{Sum of scored by B}}{\text{Number of games played by B}} \\ $

$ \text{=}\dfrac{0+8+6+4}{4} \\ $

$ \text{=}\dfrac{18}{4} \\ $

$ \text{=4}\text{.5} \\ $

(iv)Best performer is the one who has the highest average score among all the other players.

$ \text{Mean of player A=}\dfrac{\text{Sum of scored by A}}{\text{Number of games played by A}} \\ $

$ \text{=}\dfrac{14+16+10+10}{4} \\$ 

$ \text{=}\dfrac{50}{4} \\ $

$ \text{=12}\text{.5} \\ $

$ \text{Mean of player B=}\dfrac{\text{Sum of scored by B}}{\text{Number of games played by B}} \\ $

$ \text{=}\dfrac{0+8+6+4}{4} \\ $

$ \text{=}\dfrac{18}{4} \\ $

$ \text{=4}\text{.5} \\ $

$  \text{Mean of player C=}\dfrac{\text{Sum of scored by C}}{\text{Number of games played by C}} \\ $

$ \text{=}\dfrac{8+11+13}{3} \\ $

$ \text{=}\dfrac{32}{3} \\ $

$ \text{=10}\text{.67} \\ $

On comparing means of all players, we can see that the average score of player A is the highest, i.e., 12.5.

Hence, player 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) The highest and the lowest marks obtained by the students. 

(ii) Range of the marks obtained. 

(iii) Mean marks obtained by the group.

Ans:

(i) Highest marks obtained by the student = 95

Lowest marks obtained by the student = 39

(ii) We will find the range of marks by getting the difference between the highest value and lowest value.

Range= Highest value – Lowest value

$ =95-39 \\ $

$ =56 \\ $

Hence, the range of marks obtained is 56.

(iii) For finding average marks scored, we divide the total scores by the total number of observations.

 $ \text{Mean of obtained marks=}\dfrac{\text{Sum of marks}}{\text{Total number of marks}} \\ $

$ \text{=}\dfrac{85+76+90+85+39+48+56+95+81+75}{10} \\ $

$ \text{=}\dfrac{730}{10} \\$ 

$ \text{=73} \\ $

Hence, the mean marks obtained by the group of students is 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.

Ans: To find the mean enrolment, we will divide the sum of numbers of enrolment by the number of years of enrolment period.

$ \text{Mean enrolment=}\dfrac{\text{Sum of number of enrolment}}{\text{Total number of years}} \\$ 

$ \text{=}\dfrac{1555+1670+1751+2013+2540+2820}{6} \\ $

$ \text{=}\dfrac{12348}{6} \\ $

$ \text{=2058} \\ $

Hence, the mean enrolment of school over 6 years is 2058.

9. The height 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 data? 

(iv) What is the mean height of the girls? 

(v) How many girls have heights more than the mean height?

Ans: 

(i)As seen from the above given data, the height of the tallest girl is 151cm.

(ii)As seen from the above given data, the height of the shortest girl is 128cm.

(iii)We will find the range of heights by getting the difference between the highest value and lowest value.

$ \text{Range= Highest Value}-\text{Lowest Value} \\ $

$ \text{=151-128} \\ $

$ \text{=23} \\ $

Hence, the range of heights is 23cm.

(iv)To find the mean height, we will divide the sum of heights of all girls by the number of girls.

$ \text{Mean of height=}\dfrac{\text{Sum of heights of all girls}}{\text{Total number of girls}} \\ $

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

$ \text{=}\dfrac{1414}{10} \\ $

$ \text{=141}\text{.4} \\ $

Hence, the mean height of girls is 141.4cm.

(v) Mean height is 141.4cm. From the above data, we can see that heights 150, 151, 146, 149 and 143 are more than the mean value. 

Hence, five girls have heights more than the mean height.

Exercise 3.2

1. The scores in mathematics test (out of 25) of students are as follows: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 Find the mode and median of this data. Are they the same?

Ans; First, we shall arrange the given data in ascending order,

5, 9, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25

For Mode,

Mode is the observation occurred the highest number of times.

Mode= observation with highest frequency

=20

For Median,

Median is the middle observation.

Number of observations= 14 observations

$ \text{Median= Middle Observation} \\ $

$ \text{=}{{\left( \dfrac{14}{2} \right)}^{th}}\text{observation} \\ $

$ ={{7}^{th}}\text{observation} \\ $

$ \text{=20} \\ $

Yes, Mode and Median are the same for the given observations.

2. The runs scored in a cricket match by 11 players is as follows: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15 Find the mean, mode and median of this data. Are the three the same?

Ans: Arranging the given data in ascending order,

6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120

For Mean,

$ \text{Mean=}\dfrac{\text{Sum of observations}}{\text{Number of observations }} \\ $

$ \text{=}\dfrac{\text{6+8+10+10+15+15+15+80+100+120}}{11} \\ $

$ \text{=}\dfrac{429}{11} \\ $

$ \text{=39} \\ $

For Median, 

Total number of observations = n = 11 observations

$ \text{Median= Middle Observation} \\ $

$ \text{=}{{\left( \dfrac{n+1}{2} \right)}^{th}}\text{Observation} \\ $

$ \text{=}{{\left( \dfrac{11+1}{2} \right)}^{th}}\text{observation} \\$ 

$ \text{=}{{\left( \dfrac{12}{2} \right)}^{th}}\text{observation} \\ $

$ ={{6}^{th}}\text{observation} \\ $

$ \text{=15} \\ $

For Mode,

 Mode= observation with highest frequency

=15

We can see from answers above that mean, median and mode are not the same.

3. The weight (in kg) of 15 students of a class are: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47 

(i) Find the mode and median of this data. 

(ii) Is there more than one mode? 

Ans; Arranging the given data in ascending order,

32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50

(i) For Mode,

Mode is the observation occurred the highest number of times.

Mode= observation with highest frequency

= 38kg and 43kg

For Median,

Median is the middle observation.

Total number of observations = n = 15 observations

 $ \text{Median= Middle Observation} \\ $

 $ \text{=}{{\left( \dfrac{n+1}{2} \right)}^{th}}\text{Observation} \\ $

 $ \text{=}{{\left( \dfrac{15+1}{2} \right)}^{th}}\text{observation} \\ $

 $ \text{=}{{\left( \dfrac{16}{2} \right)}^{th}}\text{observation} \\ $

 $ ={{8}^{th}}\text{observation} \\ $

 $ \text{=40} \\ $

(ii)Yes, there are two modes, i.e., 38 and 43. This is because both 38 and 43 occur for the same highest number of times, 3.

4. Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14 

Ans:  Arranging the given data in ascending order,

12, 12, 13, 13, 14, 14, 14, 16, 19

For Mode,

Mode is the observation occurred the highest number of times.

Mode= Observation with highest frequency

= 14

For Median,

Median is the middle observation.

Total number of observations = n = 9 observations

  $  \text{Median= Middle Observation} \\ $

 $ \text{=}{{\left( \dfrac{n+1}{2} \right)}^{th}}\text{Observation} \\ $

 $ \text{=}{{\left( \dfrac{9+1}{2} \right)}^{th}}\text{observation} \\ $

 $ \text{=}{{\left( \dfrac{10}{2} \right)}^{th}}\text{observation} \\ $

 $ ={{5}^{th}}\text{observation} \\ $

 $ \text{=14} \\ $

5. Tell whether the statement is true or false: 

(i) The mode is always one of the numbers in a data. 

(ii) The mean is one of the numbers in the data. 

(iii) The median is always one of the numbers in a data. 

(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 have mean 9.

Ans: (i) True, it is the number that occurs most frequently in the data

(ii) False, mean is the ratio of sum of all observations to the number of observations. Hence, it can or cannot be one of the observations.

(iii) True, median is the central value of data after the observations are placed in an ascending or descending order.

(iv) False, mean is the ratio of sum of all observations to the number of observations.

$ \text{Mean=}\dfrac{\text{Sum of observations}}{\text{Number of observations }} \\ $

$ \text{=}\dfrac{6+4+3+8+9+12+13+9}{8} \\ $

$ \text{=}\dfrac{64}{8} \\ $

$ \text{=8} \\ $

The mean of given observations is 8.

Hence, the given statement is false.

Exercise 3.3

1. Use the bar graph to answer the following questions: 

(image will be uploaded soon)

(a) Which is the most popular pet? 

(b) How many students have dogs as pets? 

Ans: 

1. As seen from the graph, the bar of the cat is the tallest.

Hence, cats are the most popular pet.

2. As seen from the graph, the bar of the dog is till 8.

 Hence, 8 students have a dog as their pet.

2. Read the bar graph which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:

(image will be uploaded soon)

(i) About how many books were sold in 1989? 1990? 1992? 

(ii) In which year were about 475 books sold? About 225 books sold? 

(iii) In which years were fewer than 250 books sold? 

(iv) Can you explain how you would estimate the number of books sold in 1989?

Ans: (i) From the graph, we can see that

Number of books sold in 1989 = 180

Number of books sold in 1990 = 475

Number of books sold in 1992 = 225

(ii) On observing the graph, we can find that 475 books were sold in 1990 and 225 books were sold in 1992. 

(iii) From the graph, we find that in the years 1989 and 1992, books less than 250 were sold.

(iv) From the graph, we can calculate that 180 books were sold in 1989.

3. Number of children in six different classes are given below. Represent the data on a bar graph.

1. How would you choose a scale? 

2. Answer the following questions: 

(i)Which class has the maximum number of children? And the minimum? 

(ii)Find the ratio of students of class sixth to the students of class eighth.

Ans:

(image will  be uploaded soon)

1. Scale = Difference between upper value and lower value of 1 unit on y-axis 

= 25-0

=25 children

2. (i) We can see from the graph that fifth class has the tallest bar and hence it has the maximum number of children.

We can see from the graph that the smallest bar corresponds to the tenth class and hence it has a minimum number of children.

(ii)Ratio of sixth class to eighth class

$ \text{Ratio=}\dfrac{\text{Number of students in class sixth}}{\text{Number of students in class eighth}} \\ $

$ \text{= }\dfrac{120}{100} \\ $

$ \text{=}\dfrac{6}{5} \\ $

$ \text{=6:5} \\$ 

4. The performance of a student in 1st term and 2nd term is given. Draw a double bar graph choosing appropriate scale and answer the following:

(i) In which subject has the child improved his performance the most? 

(ii) In which subject is the improvement the least? 

(iii) Has the performance gone down in any subject? 

Ans:

(image will be uploaded soon)

Difference of marks of 1st term and 2nd term

English = 70 – 67 = 3 

Hindi = 65 – 72 = -7 

Maths = 95 – 88 = 7

Science = 85 – 81 = 4 

S. Science = 75 – 73 = 2

(i)We can see that Maths has a maximum difference.

Hence, the child has improved the most in maths.

(ii)We can see that S. Science has a minimum difference.

Hence, the child has improved the least in maths.

(ii)We can see that Hindi has a decrease in the size of bars.

Hence, children's performance has gone down in Hindi.

5. Consider this data collected from a survey of a colony.

(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph? 

(ii) Which sport is most popular? 

(iii) Which is more preferred, watching or participating in sports? 

Ans; 

(i)

 (image will be uploaded soon)

The double bar graph represents the number of people who like watching and participating in various sports.

(ii) We can observe from the graph that cricket has the tallest bar.

Hence, cricket is the most popular sport.

(iii) From the double bar graph graph, we can see that the watching (blue) bars are taller in comparison to participating (orange) bars.

Hence, watching sports is more preferred.

6. Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this Chapter. Plot a double bat graph using the data and answer the following:

Temperature of cities as on 20.6.2006

(i) Which city has the largest difference in the minimum and maximum temperature on the given data? 

(ii) Which is the hottest city and which is the coldest city? 

(iii) Name two cities where the maximum temperature of one was less than the minimum temperature of the order. 

(iv) Name the city which has the least difference between its minimum and the maximum temperature.

Ans;

(image will be uploaded soon) 

(i)Jammu has the largest difference in temperature i.e., 

Maximum temperature = 41℃ and 

Minimum temperature = 26℃.

Difference = 41℃ – 26℃ = 15℃

(ii)Jammu is the hottest city because the maximum temperature bar is the tallest and Bangalore is the coldest city because maximum temperature is shortest.

(iii)Maximum temperature of Bangalore - 28℃

Minimum temperature of two cities whose minimum temperature is higher than the maximum temperature of Bangalore are Ahmedabad and Jaipur = 29℃

(iv)Mumbai has the least difference in temperature i.e., 

Maximum temperature = 32℃ and

Minimum temperature = 27℃

Difference = 32℃ – 27℃ = 5℃

Exercise 3.4

1. Tell whether the following is certain to happen, impossible can happen but not certain. 

(i) You are older today than yesterday. 

(ii) A tossed coin will land heads up. 

(iii) A die when tossed shall land up with 8 on top. 

(iv) The next traffic light seen will be green. 

(v) Tomorrow will be a cloudy day. 

Ans: Let us first define the terms- 

Certain to happen means that the situation will definitely happen.

Impossible means that the situation can never happen.

Can happen but not certain means that the happening of the situation is not definite, i.e., it may or may not happen.

(i)We grow older every day. Hence, the statement is certain to happen.

(ii)A coin has two faces, heads and tails. The tossed coin can land on either heads or tails. Hence, the situation in the statement can happen but is not certain.

(iii)A die has six faces marked 1, 2, 3, 4, 5 and 6. The tossed die can land on either of the faces. The die does not have 8 on any of the faces. Hence, the situation in the statement is impossible.

(iv)A traffic light turns green after it has turned yellow. But we don’t know if the traffic light has turned yellow or red. Hence, the situation in the statement can happen but is not certain.

(v)We can never give the surety of the weather of an upcoming day. Hence, the situation in the statement is impossible.

2. There are 6 marbles in a box with numbers from 1 to 6 marked on each of them. 

(i) What is the probability of drawing a marble with number 2? 

(ii) What is the probability of drawing a marble with number 5? 

Ans: Total marbles from 1 to 6 marked in a box = 6

(i) $\text{Probability of drawing marble with number 2=}\dfrac{\text{Number of favourable outcome}}{\text{Number of possible outcomes}}$

Hence, $\text{P(appearance of 2)=}\dfrac{1}{6}$

(ii) $\text{Probability of drawing marble with number 5=}\dfrac{\text{Number of favourable outcome}}{\text{Number of possible outcomes}}$

Hence, $\text{P(appearance of 5)=}\dfrac{1}{6}$

3. A coin is flipped to decide which team starts the game. What is the probability that your team will start?

Ans: A coin has two possible outcomes: Head and Tail.

Probability of getting Head or Tail is equal.

$\text{Probability =}\dfrac{\text{Number of favourable outcome}}{\text{Number of possible outcomes}}$

Hence, $\text{P(starting game)=}\dfrac{1}{2}$

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