**
Q.1 Use the bar graph (Fig 3.3) to answer the following questions.
(a) Which is the most popular pet?
(b) How many students have dog as a pet?
**

Open in new tab *link*

(a) From bar graph, the most popular pet is Cat.

(b) From the bar graph, 8 students have dog as a pet out of 12 students.

**
Q.2 Read the bar graph (Fig 3.4) which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:
(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?
**

Open in new tab *link*

(i) By observing bar graph,

175 books were sold in the year 1989.

475 books were sold in the year 1990.

225 books were sold in the year 1992.

(ii) By observing bar graph,

475 books were sold in the year 1990.

225 books were sold in the year 1992.

(iii) By observing bar graph,

In the years 1989 and 1992, the number of books sold were less than 250.

(iv) By observing bar graph

From the graph, it can be concluded that the number of books sold in the year 1989 is about 1 and 3/4th part of 1 cm.
We know that the scale is taken as 1 cm = 100 books.
100 + 3/4× 100 = 100 + 75 = 175
Therefore, about 175 books were sold in the year 1989.

**
Q.3 Number of children in six different classes are given below. Represent the data on a bar graph.
(a) How would you choose a scale?
(b) 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 eight.
**

Open in new tab *link*

(i) We will take the scale as 1 unit = 10 children.

(ii) (a) Fifth class has the maximum number of children i.e., 135.

(b) Tenth class has the minimum number of children i.e., 80.

(c) Number of children in class eight = 100
Therefore Ratio of class sixth to the students of class eighth

120/100= 6/5= 6:5

**
Q.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?
**

Open in new tab *link*

(i) In Math, the performance of the students improved the most.

(ii) In social science, the performance of the students improved the least.

(iii) Yes, in Hindi the performance of the students has gone down.

**
Q.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?
**

Open in new tab *link*

(i) The above bar graph depicts the number of people who are watching and who are participating in sports.

(ii) Cricket

(iii) watching

**
Q.6 Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this chapter. Plot a double bar graph using the data and answer the following:
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and maximum temperature.
**

Open in new tab *link*

(i) Jammu

(ii) Hottest city is Jammu with 41°C temperature and coldest city is Bengaluru with 21°C temperature.

(iii) Bengaluru having its maximum temperature 28°C is less than the minimum temperature 29°C in Ahmedabad.

Bengaluru having its maximum tem-perature 28°C is less than the maximum temperature 29°C in Jaipur.

(iv) Mumbai has the least difference between its minimum and maximum temperatures i.e. 32°C – 27°C = 5°C

**
Q.1 Tell whether the following situations are certain to happen, impossible to happen, can happen but not certain.
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A dice 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.
**

Open in new tab *link*

(i) certain to happen

(ii)Can happen but not certain.

(iii) impossible

(iv) can happen but not certain.

(v) can happen but not certain

**
Q.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?
**

Open in new tab *link*

(i) Total number of marbles marked with the number from 1 to 6 = 6 Therefore n(S) = 6 Number of marble marked with 2=1 Therefore n(E) = 1 Therefore Required probability = \(\frac{n(E)}{n(S)}=\frac{1}{6} \) (ii) Number of marble marked with 5 = 1 Therefore n(E) = 1 Therefore Required probability = \(\frac{n(E)}{n(S)}=\frac{1}{6} \)

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

Open in new tab *link*

Coin has 2 faces—Head (H) and Tail (T)

Therefore Sample space S(n) = 2

Number of successful event n(E) = 1

Therefore Required probability = \(\frac{n(E)}{n(S)}=\frac{1}{2} \)

**
Q.4 A box contains pairs of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that it will make a pair?
**

Open in new tab *link*

It can be observed that while closing the eyes, one can draw either a black sock or a white sock. Therefore, there are two possible cases.

\(\frac{n(e)}{n(s)} = \frac{1}{2} \)

Let us have the heights of 10 students are as follows:

139 cm, 140.5 cm, 138 cm, 135.3 cm, 162 cm,
138 cm, 140.5 cm, 135.5 cm, 160 cm, 158 cm

Here, minimum height = 135.3 cm
Maximum height =162 cm
Therefore Range = Maximum height – Minimum height
= 162 cm — 135.3 cm = 26.7 cm
Hence, the required range = 26.7 cm.

**
Q.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.
**

Open in new tab *link*

(i) 9 is the highest marks.

(ii) 1 is the lowest marks.

(iii) Range = Max. marks – Min. marks
= 9 – 1 = 8

(iv) Arithmetic mean= \(\frac{sum of (f)(x)}{sum of (x)}= \frac{100}{25}=5 \)

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

Therefore , mean \(\frac{0+1+2+3+4}{5}= \frac{10}{5}= 2 \)

Hence, the required mean = 2.

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

Open in new tab *link*

Following are the scores of the runs in eight innings:

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

\(Mean=\frac{sum of all runs}{number of innings}=\frac{58+76+40+35+46+45+0+100}{8}=\frac{400}{8} =50

Hence, the required mean = 50.

**
Q.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?
**

Open in new tab *link*

(i) Number of points scored by A in all games are
Game 1 = 14, Game 2 = 16, Game 3 = 10, Game 4 = 10

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

(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

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

(iv) Mean score of C \(=\frac{8+11+14}{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.

**
Q.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.
**

Open in new tab *link*

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 Mark's \(=\frac{sum of all obtained mark}{number of students} \)

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

**
Q.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.
**

Open in new tab *link*

Mean enrolment = $$\frac{sum of all enrolment}{number of enrolment} $$ $$= \frac{1555+1670+1750+2013+2540+2820}{6}=\frac{12348}{6}=2058 $$ Thus the required mean = 2058

**
Q.8 The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
(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?
**

Open in new tab *link*

(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
$$\frac{sum of all observations}{number of observation}$$
$$=\frac{0.0+12.2+2.1+0.0+20.0+5.5+1.0}{7}=\frac{41.3}{7}=5.9mm $$
iii) Number of days on which the rainfall was less than the mean rainfall = Monday, Wednesday, Thursday, Saturday, Sunday = 5 days.

**
Q.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?
**

Open in new tab *link*

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

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

$$= 151 cm – 128 cm = 23 cm.$$
(iv) mean height $$= \frac{135+150+139+128+151+132, 146+149+143+141}{10}$$
$$=\frac{1414}{10}=141.4 cm. $$
(v) Number of girls having more height

than the mean height
= 150, 151, 146, 149 and 143 = 5 girls

**
Q.1 The scores in mathematics test (out of 25) of 15 students is 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 same?
**

Open in new tab *link*

Arranging the given scores in an ascending order, we get

5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25

Since 20 occurs 4 times (highest)

Therefore Mode = 20

n = 15 (odd)

Therefore Median \(= \frac{n+1}{2} \)th term\( = \frac{15+1}{2} = 8th \) term =20

Thus, median = 20 and mode = 20

Therefore Mode and median are same.

**
Q.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 same?
**

Open in new tab *link*

Given data:

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

Mean = sum of total observations / number of observations =

6+15+120+50+100+80+10+15+8+10+15/11 =429/11=39

Arranging the given data in increasing order, we get

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

Here, 15 occurs 3 times (highest)

Therefore Mode = 15

n = 11 (odd)

Therefore Median \(=\frac{11+1}{2} \)^{th} term = 6^{th} term = 15

Thus mean = 39, mode = 15 and median = 15

$$ No, they are not same. $$

**
Q.3 The weights (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?
**

Open in new tab *link*

Arranging in increasing order, we get

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

(i) Here, 38 and 42 occur 3 times (highest)

Thus mode = 38 and 43

n = 15(odd)

Median = (n+1)/2th term =(15+1)/2th term

= 8th term = 40

Thus mode 38 and 43 and median = 40

(ii) Yes, the given data has two modes i.e. 38 and 43.

Arranging the given data in ascending order we get:

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

Here, 14 occur 3 times (highest)

Thus, mode = 14

n = 9(odd)

Therefore Median = (n+1)/2 th term =(9+1)/2 th term
= 5th term = 14
Hence, mode = 14 and median = 14.

**
Q.5 Tell whether the statement is true or false.
(i) The mode is always one of the number in a data.
(ii) The mean is one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9.
**

Open in new tab *link*

(i) True

(ii) False

(iii) True

(iv) False