Chapter 2 Data Handling

Solution:

The given data is: 9, 8, 4, 3, 2, 1, 6, 4, 8, 10, 12, 15, 4, 3.

The range of a data set is the difference between the maximum and minimum values in the set.

Maximum value in the data = $15$

Minimum value in the data = $1$

Range = Maximum value - Minimum value

Range = $15 - 1 = 14$

Thus, the range of the given data is $14$.

The correct option is (b).

Solution:

The given data is: 2, 5, 15, 25, 20, 12, 8, 7, 6, 16, 21, 17, 30, 32, 23, 40, 51, 15, 2, 9, 57, 19, 25.

The class intervals are given as 0–5, 5–10, 10–15, etc. This indicates that the upper limit of each class is exclusive.

We need to find the frequency of the class 20–25. This means we need to count how many data points are greater than or equal to 20 but strictly less than 25.

Let's examine the data points and see which ones fall into the interval $[20, 25)$:

2, 5, 15, 25 (Excluded), 20, 12, 8, 7, 6, 16, 21, 17, 30, 32, 23, 40, 51, 15, 2, 9, 57, 19, 25 (Excluded).

The data points in the interval $[20, 25)$ are 20, 21, and 23.

The number of such data points is 3.

Therefore, the frequency of the class 20–25 is 3.

The correct option is (c).

Solution:

The pie chart shows the distribution of viewers watching different types of channels on TV. Each sector represents a different type of programme, and the size of the sector corresponds to the proportion of viewers for that type.

To find which type of programmes are viewed the most, we need to identify the sector with the largest angle (or area) in the pie chart.

Upon examining the pie chart, the sector representing Entertainment is clearly the largest among all the sectors.

This indicates that the Entertainment type of programmes is viewed by the largest number of people.

Therefore, the type of programmes viewed the most is Entertainment.

The correct option is (c).

Solution:

The histogram represents the distribution of heights of girls. The horizontal axis shows the height in cm (class intervals), and the vertical axis shows the number of girls (frequency).

We are asked to find the number of girls having height 145 cm and above. This includes girls whose height falls in the class intervals starting from 145 cm.

Looking at the histogram, the class intervals that represent heights of 145 cm and above are:

145–150 cm

150–155 cm

155–160 cm

Now, we read the frequency (number of girls) for each of these intervals from the height of the corresponding bars:

Number of girls in the 145–150 cm interval = $10$

Number of girls in the 150–155 cm interval = $7$

Number of girls in the 155–160 cm interval = $2$

To find the total number of girls with height 145 cm and above, we add the frequencies of these intervals:

Total number of girls $\ge 145$ cm = (Number of girls in 145–150 cm) + (Number of girls in 150–155 cm) + (Number of girls in 155–160 cm)

Total number of girls $\ge 145$ cm = $10 + 7 + 2 = 19$

The number of girls having height 145 cm and above is $19$.

The correct option is (d).

Solution:

When a single dice is thrown, the possible outcomes are the numbers on its faces, which are $1, 2, 3, 4, 5, 6$. There are $6$ possible outcomes for a single throw.

When a dice is thrown two times, the outcome of the two throws can be represented as an ordered pair $(a, b)$, where $a$ is the result of the first throw and $b$ is the result of the second throw.

For the first throw, there are $6$ possible outcomes ($1$ to $6$).

For the second throw, there are also $6$ possible outcomes ($1$ to $6$).

Since the two throws are independent, the total number of possible ordered pairs $(a, b)$ is the product of the number of outcomes for each throw.

Total number of outcomes = (Number of outcomes for 1st throw) $\times$ (Number of outcomes for 2nd throw)

Total number of outcomes = $6 \times 6 = 36$

These outcomes are the pairs:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Although the sum is noted, the fundamental sample space consists of these $36$ equally likely pairs.

The number of possible outcomes is $36$.

The correct option is (d).

Solution:

When a dice is rolled, the possible outcomes are the numbers on its faces: $1, 2, 3, 4, 5, 6$.

The total number of possible outcomes is $6$.

We are interested in the event of getting a multiple of 2. The multiples of 2 among the possible outcomes are $2, 4, 6$.

The number of favourable outcomes (getting a multiple of 2) is $3$.

The probability of an event is calculated as:

Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

Probability of getting a multiple of 2 = $\frac{\text{Number of multiples of 2}}{\text{Total number of outcomes}}$

Probability = $\frac{3}{6}$

Simplifying the fraction:

Probability = $\frac{\cancel{3}^{1}}{\cancel{6}_{2}} = \frac{1}{2}$

The probability of getting a multiple of 2 when a dice is rolled is $\frac{1}{2}$.

The correct option is (c).

Solution:

The first class interval is 10–15.

The second class interval is 15–20.

From these two intervals, we can observe the pattern:

The lower limit of the second interval (15) is the upper limit of the first interval (15).

The class width is the difference between the upper and lower limits of an interval: $15 - 10 = 5$, or $20 - 15 = 5$.

So, the class width is $5$.

The class intervals are formed by adding the class width to the previous limits.

First class: 10–15

Second class: 15–20 ($15 = 10 + 5$, $20 = 15 + 5$)

Third class: The lower limit will be the upper limit of the second class, which is 20. The upper limit will be $20 + 5 = 25$. So, the third class is 20–25.

Fourth class: The lower limit will be the upper limit of the third class, which is 25. The upper limit will be $25 + 5 = 30$. So, the fourth class is 25–30.

The fourth class interval is 25–30.

The complete statement is: The fourth class interval for a grouped data whose first and second class intervals are 10 –15 and 15 –20 respectively is 25–30.

Solution:

In a class interval, the smaller number is called the lower limit, and the larger number is called the upper limit.

The given class interval is 250–275.

Here, 250 is the smaller number, and 275 is the larger number.

Therefore, 250 is the lower limit of the class interval.

The complete statement is: In the class interval 250 – 275, 250 is known as the lower limit.

Solution:

In statistics, when we collect a set of data, we often count how many times each distinct value or category appears.

This count of the occurrence of a particular observation is known as its frequency.

The complete statement is: The number of times a particular observation occurs in the given data is called its frequency.

Solution:

A pie chart represents the proportion of different categories within a whole. The whole circle represents the sum of all categories, and its central angle is $360^\circ$.

Each sector in a pie chart represents a category, and its central angle is proportional to the value of that category relative to the total value of all categories.

The formula for the central angle of a sector is:

Central angle = $\frac{\text{Value of the category}}{\text{Total value}} \times 360^\circ$

Here, $\frac{\text{Value of the category}}{\text{Total value}}$ is a fraction representing the proportion of the category out of the total.

Since this fraction is multiplied by $360^\circ$, the central angle of each sector will be a fraction of $360^\circ$.

The statement is true.

Solution:

When a standard dice is thrown, the possible outcomes are $1, 2, 3, 4, 5, 6$.

The total number of possible outcomes is $6$.

We are interested in the event of getting an odd number. The odd numbers among the possible outcomes are $1, 3, 5$.

The number of favourable outcomes (getting an odd number) is $3$.

The probability of an event is given by:

Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

Probability of getting an odd number = $\frac{\text{Number of odd outcomes}}{\text{Total number of outcomes}}$

Probability = $\frac{3}{6}$

Simplifying the fraction:

Probability = $\frac{\cancel{3}^{1}}{\cancel{6}_{2}} = \frac{1}{2}$

The calculated probability of getting an odd number is $\frac{1}{2}$, which matches the statement.

The statement is true.

Solution:

A pie chart is a circular graph divided into sectors, where each sector's area is proportional to the quantity it represents. It is used to show parts of a whole.

A pictograph is a graph that uses pictures or symbols to represent data. Each picture or symbol typically represents a certain number of items.

These two types of data representation are fundamentally different in how they display information.

Therefore, a pie chart is not the same as a pictograph.

The statement is false.

Solution:

We are given the weekly wages of 30 workers and asked to create a frequency distribution table using class intervals 800–810, 810–820, and so on. The upper limit of each class interval is considered exclusive.

We will list the class intervals, use tally marks to count the number of workers whose wages fall into each interval, and then write down the frequency.

The frequency distribution table shows the number of workers in each wage group.

Given:

Total marks obtained by the student = 540.

The distribution of marks among different subjects is shown in the pie chart. The marks in each subject are proportional to the central angle of the corresponding sector.

Central angles for the subjects are:

• Hindi: $70^\circ$
• English: $55^\circ$
• Social Science: $65^\circ$
• Science: $80^\circ$
• Mathematics: $90^\circ$

The sum of the central angles is $70^\circ + 55^\circ + 65^\circ + 80^\circ + 90^\circ = 360^\circ$, which represents the total marks.

To Find:

(i) The subject in which the student scored 105 marks.

(ii) The difference in marks between Mathematics and Hindi.

Solution:

The total central angle of the pie chart ($360^\circ$) corresponds to the total marks obtained (540).

We can find the value of marks per degree of the central angle:

Marks per degree $= \frac{\text{Total Marks}}{\text{Total Angle}} = \frac{540}{360}$

Marks per degree $= \frac{\cancel{540}^{3}}{\cancel{360}_{2}} = \frac{3}{2} = 1.5$ marks per degree.

Now, we calculate the marks obtained in each subject by multiplying the central angle by 1.5:

Marks in Hindi $= 70 \times 1.5 = 105$ marks.

Marks in English $= 55 \times 1.5 = 82.5$ marks.

Marks in Social Science $= 65 \times 1.5 = 97.5$ marks.

Marks in Science $= 80 \times 1.5 = 120$ marks.

Marks in Mathematics $= 90 \times 1.5 = 135$ marks.

Answer to part (i):

We are looking for the subject where the student scored 105 marks.

From our calculations, the marks obtained in Hindi are 105.

Therefore, the student scored 105 marks in Hindi.

Answer to part (ii):

We need to find the difference between the marks in Mathematics and Hindi.

Marks in Mathematics = 135

Marks in Hindi = 105

Difference = Marks in Mathematics - Marks in Hindi

Difference $= 135 - 105 = 30$ marks.

The student obtained 30 more marks in Mathematics than in Hindi.

Final Answers:

(i) The student scored 105 marks in Hindi.

(ii) The student obtained 30 more marks in Mathematics than in Hindi.

Solution:

To draw a pie chart, we first need to calculate the central angle for each sector. The total number of people surveyed is required to find the proportion for each category.

Total number of people = $30 + 40 + 25 + 25 = 120$

The sum of the central angles in a pie chart is $360^\circ$. The central angle for each category is given by:

Central Angle = $\frac{\text{Number of people for the category}}{\text{Total number of people}} \times 360^\circ$

Calculating the central angle for each food type:

For North Indian food: $\frac{30}{120} \times 360^\circ = \frac{\cancel{30}^{1}}{\cancel{120}_{4}} \times 360^\circ = \frac{1}{4} \times 360^\circ = 90^\circ$

For South Indian food: $\frac{40}{120} \times 360^\circ = \frac{\cancel{40}^{1}}{\cancel{120}_{3}} \times 360^\circ = \frac{1}{3} \times 360^\circ = 120^\circ$

For Chinese food: $\frac{25}{120} \times 360^\circ = \frac{\cancel{25}^{5}}{\cancel{120}_{24}} \times 360^\circ = \frac{5}{24} \times 360^\circ = 5 \times \frac{\cancel{360}^{15}}{\cancel{24}_{1}}^\circ = 5 \times 15^\circ = 75^\circ$

For Others: $\frac{25}{120} \times 360^\circ = \frac{\cancel{25}^{5}}{\cancel{120}_{24}} \times 360^\circ = \frac{5}{24} \times 360^\circ = 5 \times \frac{\cancel{360}^{15}}{\cancel{24}_{1}}^\circ = 5 \times 15^\circ = 75^\circ$

Let's verify the sum of the central angles: $90^\circ + 120^\circ + 75^\circ + 75^\circ = 360^\circ$.

Now, we can create a table summarizing the data and angles:

To draw the pie chart:

1. Draw a circle of any convenient radius.

2. Draw a radius from the centre to the circle.

3. Using a protractor, draw the first sector with a central angle of $90^\circ$ (for North Indian food) starting from the radius.

4. From the new radius (boundary of the first sector), draw the next sector with a central angle of $120^\circ$ (for South Indian food).

5. Continue this process for the remaining sectors: $75^\circ$ (for Chinese) and $75^\circ$ (for Others).

6. Label each sector with the name of the food type and/or its central angle or the number of people.

Solution:

The frequency distribution table from Example 13 is:

To draw a histogram, we represent the class intervals on the horizontal axis and the frequency on the vertical axis. Since the class intervals are continuous (800-810, 810-820, etc.), we will draw adjacent bars with heights corresponding to the frequency of each class interval.

On the horizontal axis, mark the points 800, 810, 820, ..., 900. On the vertical axis, choose a suitable scale to represent the frequency (number of workers).

Draw a bar above each class interval with height equal to its frequency. For example, above the interval 800–810, draw a bar of height 3; above 810–820, draw a bar of height 2, and so on.

Based on the frequency distribution table (and the histogram if drawn):

(i) Which class interval has the maximum number of workers?

We look for the highest frequency in the table. The maximum frequency is 9, which corresponds to the class interval 830–840.

The class interval with the maximum number of workers is 830–840.

(ii) How many workers earn Rs. 850 and more?

Workers earning Rs. 850 and more fall into the class intervals 850–860, 860–870, 870–880, 880–890, and 890–900.

Number of workers = (Frequency of 850–860) + (Frequency of 860–870) + (Frequency of 870–880) + (Frequency of 880–890) + (Frequency of 890–900)

Number of workers = $1 + 3 + 1 + 1 + 4 = 10$

10 workers earn Rs. 850 and more.

(iii) How many workers earn less than Rs. 850?

Workers earning less than Rs. 850 fall into the class intervals 800–810, 810–820, 820–830, 830–840, and 840–850.

Number of workers = (Frequency of 800–810) + (Frequency of 810–820) + (Frequency of 820–830) + (Frequency of 830–840) + (Frequency of 840–850)

Number of workers = $3 + 2 + 1 + 9 + 5 = 20$

Alternatively, total workers minus workers earning 850 or more: $30 - 10 = 20$.

20 workers earn less than Rs. 850.

(iv) How many workers earn Rs. 820 or more but less than Rs. 880?

Workers earning Rs. 820 or more but less than Rs. 880 fall into the class intervals 820–830, 830–840, 840–850, 850–860, and 860–870. (Note: the interval 870–880 includes wages up to less than 880).

Number of workers = (Frequency of 820–830) + (Frequency of 830–840) + (Frequency of 840–850) + (Frequency of 850–860) + (Frequency of 860–870) + (Frequency of 870–880)

Number of workers = $1 + 9 + 5 + 1 + 3 + 1 = 20$

20 workers earn Rs. 820 or more but less than Rs. 880.

Solution:

We are given a frequency distribution table and asked to answer questions based on it.

(i) Class interval which has the lowest frequency.

Looking at the 'Frequency' column, the frequencies are 1, 5, 5, 4, 0, 8, 2. The lowest frequency is 0.

The class interval corresponding to the frequency 0 is 65–75.

(ii) Class interval which has the highest frequency.

Looking at the 'Frequency' column, the frequencies are 1, 5, 5, 4, 0, 8, 2. The highest frequency is 8.

The class interval corresponding to the frequency 8 is 75–85.

(iii) What is the class size of the intervals?

The class size is the difference between the upper limit and the lower limit of a class interval.

For the interval 35–45, class size = $45 - 35 = 10$.

For the interval 45–55, class size = $55 - 45 = 10$.

And so on for the following intervals (55-65, 65-75, 75-85, 85-95), the class size is consistently 10.

Note: The first class interval '35-35' is unusual and has a 'size' of $35-35=0$, which is inconsistent with the other intervals. Assuming the question refers to the standard class size used in the majority of intervals.

The class size of the intervals is 10.

(iv) What is the upper limit of the fifth class?

Let's list the class intervals in order:

1st class: 35–35

2nd class: 35–45

3rd class: 45–55

4th class: 55–65

5th class: 65–75

The upper limit of the fifth class (65–75) is 75.

(v) What is the lower limit of the last class?

Let's list the class intervals in order:

1st class: 35–35

2nd class: 35–45

3rd class: 45–55

4th class: 55–65

5th class: 65–75

6th class: 75–85

7th class: 85–95

The last class is the 7th class (85–95). The lower limit of this class is 85.

Solution:

We are given a pie chart showing the reasons for lower back injury among surveyed people. The total number of people surveyed is 600.

We need to find the number of people who injured their back while either bending and lifting. This means we need to consider the categories "Bending" and "Lifting".

From the pie chart, we can read the percentages for each category:

Percentage for Bending = $30\%$

Percentage for Lifting = $10\%$

To find the number of people for each category, we use the formula:

Number of people = $\frac{\text{Percentage}}{100} \times \text{Total number of people}$

Number of people who injured their back while Bending = $\frac{30}{100} \times 600$

Number of people for Bending = $0.30 \times 600 = 180$

Number of people who injured their back while Lifting = $\frac{10}{100} \times 600$

Number of people for Lifting = $0.10 \times 600 = 60$

The total number of people who injured their back while either bending and lifting is the sum of the numbers for these two categories.

Total number of people (Bending and Lifting) = Number for Bending + Number for Lifting

Total number = $180 + 60 = 240$

Alternatively, we can add the percentages first:

Total percentage for Bending and Lifting = Percentage for Bending + Percentage for Lifting

Total percentage = $30\% + 10\% = 40\%$

Total number of people (Bending and Lifting) = $\frac{\text{Total percentage}}{100} \times \text{Total number of people}$

Total number = $\frac{40}{100} \times 600 = 0.40 \times 600 = 240$

The number of people who injured their back while either bending and lifting is 240.

Solution:

In a histogram, the horizontal axis represents the class intervals, and the vertical axis represents the frequency.

The width of each rectangle corresponds to the width of the class interval.

The height of each rectangle in a histogram represents the number of observations (frequency) that fall within that specific class interval.

Thus, the height of a rectangle in a histogram shows the frequency of the class.

The correct option is (d).

Solution:

A pie chart is a circular graph that is divided into sectors, where each sector represents a proportion of the whole. The entire circle represents the whole, and the size of each sector shows the relationship of the part (category) to the whole.

A histogram is used for continuous data and shows the distribution of frequencies over intervals. A bar graph is used to compare quantities among different categories. A pictograph uses pictures to represent data quantities.

The geometric representation that shows the relationship between a whole and its parts is a pie chart.

The correct option is (a).

Solution:

A pie chart is a representation of data using a circle. The entire circle represents the sum of all the parts or categories.

The angle around the centre of a circle is always $360^\circ$.

In a pie chart, the sum of the central angles of all sectors corresponds to the total value represented by the circle.

Therefore, the total angle at the centre of the circle in a pie chart is $360^\circ$.

The correct option is (b).

Solution:

The given data set is: 30, 61, 55, 56, 60, 20, 26, 46, 28, 56.

The range of a data set is the difference between the maximum (highest) value and the minimum (lowest) value in the set.

Let's find the maximum value in the data:

The values are 30, 61, 55, 56, 60, 20, 26, 46, 28, 56.

The maximum value is 61.

Let's find the minimum value in the data:

The values are 30, 61, 55, 56, 60, 20, 26, 46, 28, 56.

The minimum value is 20.

Range = Maximum value - Minimum value

Range = $61 - 20 = 41$

The range of the data is 41.

The correct option is (c).

Solution:

A random experiment is an experiment in which the outcome cannot be predicted with certainty, but the set of all possible outcomes is known.

Let's examine each option:

(a) Tossing a coin: The outcomes can be Head or Tail. We know the possible outcomes, but we cannot predict which one will occur. This is a random experiment.

(b) Rolling a dice: The outcomes can be 1, 2, 3, 4, 5, or 6. We know the possible outcomes, but we cannot predict which one will occur. This is a random experiment.

(c) Choosing a card from a deck of 52 cards: The outcome is one of the 52 cards. We know the set of all possible outcomes, but we cannot predict which specific card will be chosen. This is a random experiment.

(d) Throwing a stone from a roof of a building: Assuming gravity is present, the stone will fall downwards. While factors like wind might affect the exact landing spot, the general outcome (falling down) is predictable based on physics principles. The motion is deterministic under standard conditions, not random in the probabilistic sense like the other options.

Throwing a stone from a roof of a building is not a random experiment.

The correct option is (d).

Solution:

The English alphabet consists of 26 letters.

The vowels in the English alphabet are A, E, I, O, U.

Total number of possible outcomes (choosing any letter from the alphabet) = 26.

Number of favourable outcomes (choosing a vowel) = 5.

The probability of an event is given by:

Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

Probability of choosing a vowel = $\frac{\text{Number of vowels}}{\text{Total number of alphabets}}$

Probability = $\frac{5}{26}$

The probability of choosing a vowel from the alphabets is $\frac{5}{26}$.

The correct option is (b).

Solution:

The problem states that for every 5 students, 3 can participate in a competition.

Total number of students considered = $5$

Number of students who can participate = $3$

Number of students who do not make it to the competition = Total students - Students who can participate

Number of students who do not make it = $5 - 3 = 2$

The probability of an event is given by:

Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

In this case, the favourable outcome is that a student does not make it to the competition.

Probability (student does not make it) = $\frac{\text{Number of students who do not make it}}{\text{Total number of students}}$

Probability = $\frac{2}{5}$

To express this probability as a decimal, we divide 2 by 5:

Probability = $2 \div 5 = 0.4$

The probability of the students who do not make it to the competition is $\frac{2}{5}$ or $0.4$.

The correct option is (b).

Solution:

In a pie chart, the fraction of the total that a category represents is equal to the fraction that its central angle is of the total central angle ($360^\circ$).

We are looking for the colour that received $\frac{1}{5}$ of the votes. This means the central angle for that colour's sector should be $\frac{1}{5}$ of $360^\circ$.

Central angle for $\frac{1}{5}$ of votes = $\frac{1}{5} \times 360^\circ$

Central angle = $\cancel{360}^{72}/5 = 72^\circ$

Now, we look at the pie chart to see which colour corresponds to a central angle of $72^\circ$.

From the pie chart:

Red has an angle of $90^\circ$

Blue has an angle of $120^\circ$

Green has an angle of $60^\circ$

Yellow has an angle of $72^\circ$

Orange has an angle of $18^\circ$

The colour Yellow has a central angle of $72^\circ$, which is $\frac{1}{5}$ of the total angle.

Therefore, Yellow received $\frac{1}{5}$ of the votes.

The correct option is (d).

Solution:

The total number of students who voted is 400.

We need to find the number of students who voted for 'Others' colour.

From the pie chart, the central angle for 'Others' is $18^\circ$.

The number of students for a category is proportional to the central angle of its sector compared to the total central angle ($360^\circ$).

Number of students for 'Others' = $\frac{\text{Central angle for Others}}{\text{Total central angle}} \times \text{Total number of students}$

Number of students for 'Others' = $\frac{18^\circ}{360^\circ} \times 400$

Simplify the fraction of the angle:

$\frac{18}{360} = \frac{\cancel{18}^{1}}{\cancel{360}_{20}} = \frac{1}{20}$

Number of students for 'Others' = $\frac{1}{20} \times 400$

Number of students for 'Others' = $\frac{\cancel{400}^{20}}{\cancel{20}_{1}} = 20$

20 students voted for 'Others' colour as their favourite.

The correct option is (b).

Solution:

Let's analyse each option based on the central angles given in the pie chart:

• Red: $90^\circ$
• Blue: $120^\circ$
• Green: $60^\circ$
• Yellow: $72^\circ$
• Orange: $18^\circ$

The total angle is $360^\circ$. The number of students for each colour is proportional to its central angle.

Let's evaluate each option:

(a) $\frac{1}{20}$ th student voted for blue colour.

The fraction of votes for Blue is $\frac{120^\circ}{360^\circ} = \frac{1}{3}$.

Since $\frac{1}{3} \neq \frac{1}{20}$, this statement is false.

(b) Green is the least popular colour.

The popularity is determined by the frequency (or central angle). The angles are $90^\circ, 120^\circ, 60^\circ, 72^\circ, 18^\circ$.

The smallest angle is $18^\circ$, which corresponds to Orange. So, Orange is the least popular colour.

This statement is false.

(c) The number of students who voted for red colour is two times the number of students who voted for yellow colour.

The ratio of votes for Red to Yellow is the ratio of their central angles: $\frac{90^\circ}{72^\circ} = \frac{90}{72} = \frac{5}{4} = 1.25$.

The number of students for Red is 1.25 times the number of students for Yellow, not two times.

This statement is false.

(d) Number of students liking together yellow and green colour is approximately the same as those for red colour.

Sum of angles for Yellow and Green = Angle for Yellow + Angle for Green = $72^\circ + 60^\circ = 132^\circ$.

Angle for Red = $90^\circ$.

We compare $132^\circ$ with $90^\circ$. The difference is $|132 - 90| = 42^\circ$. While $132^\circ$ and $90^\circ$ are not very close, compared to the other options which are definitively false based on the exact data, this is the most reasonable conclusion, considering the term "approximately the same".

Based on the elimination of other options and the use of "approximately", option (d) is the most reasonable conclusion.

The correct option is (d).

Solution:

The given temperature data in °C is: –6, –8, 0, 3, 2, 0, 1, 5, 4, 4.

The range of the data is the difference between the maximum (highest) temperature and the minimum (lowest) temperature.

Let's find the maximum temperature in the data:

The values are –6, –8, 0, 3, 2, 0, 1, 5, 4, 4.

The maximum value is 5°C.

Let's find the minimum temperature in the data:

The values are –6, –8, 0, 3, 2, 0, 1, 5, 4, 4.

The minimum value is –8°C.

Range = Maximum value - Minimum value

Range = $5^\circ\text{C} - (-8^\circ\text{C})$

Range = $5^\circ\text{C} + 8^\circ\text{C} = 13^\circ\text{C}$

The range of the temperature data is 13°C.

The correct option is (b).

Given:

Number of blue buttons = $4$

Number of red buttons = $7$

Number of black buttons = $3$

Number of white buttons = $6$

To Find:

The probability that the button on the floor is blue.

Solution:

Total number of buttons on the table = Number of blue buttons + Number of red buttons + Number of black buttons + Number of white buttons

Total number of buttons = $4 + 7 + 3 + 6$

Total number of buttons = $20$

The number of favorable outcomes (getting a blue button) is the number of blue buttons, which is $4$.

The total number of possible outcomes is the total number of buttons, which is $20$.

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability (getting a blue button) = $\frac{\text{Number of blue buttons}}{\text{Total number of buttons}}$

$P(\text{Blue}) = \frac{4}{20}$

Simplify the fraction:

$P(\text{Blue}) = \frac{\cancel{4}^{1}}{\cancel{20}_{5}}$

$P(\text{Blue}) = \frac{1}{5}$

Thus, the probability that the button on the floor is blue is $\frac{1}{5}$.

Comparing this with the given options, we find that option (c) is correct.

The correct answer is (c) $\frac{1}{5}$.

Given:

Rahul wins if the spinner lands on red.

Varun wins if the spinner lands on blue.

Yash wins if the spinner lands on green.

To Find:

Which spinner should be used to make the game fair.

Solution:

A game is considered fair if each player has an equal probability of winning.

In this game, for it to be fair, the probability of the spinner landing on red, blue, and green must be equal for Rahul, Varun, and Yash to have an equal chance of winning.

This means the sections corresponding to red, blue, and green on the spinner must occupy equal areas or contain an equal number of outcomes if the spinner is divided into discrete sectors.

Let's examine the spinners:

Spinner (i) shows three sections labeled Red, Blue, and Green. Visually, these three sections appear to be of equal size, each representing approximately one-third of the circle's area.

Spinner (ii) shows sections for Red, Blue, and Green, but their sizes are clearly unequal. The red section appears smaller than the blue and green sections.

Spinner (iii) shows sections for Red, Blue, Green, and likely other colors (like Yellow). The sizes of the sections are unequal.

Spinner (iv) shows sections for six different colors (Red, Blue, Green, Yellow, Orange, Purple). The sizes of these sections are unequal.

For Spinner (i), if the sections are indeed of equal size, the probability of landing on each color is:

$P(\text{Red}) = \frac{\text{Area of Red section}}{\text{Total Area of circle}} = \frac{1}{3}$

$P(\text{Blue}) = \frac{\text{Area of Blue section}}{\text{Total Area of circle}} = \frac{1}{3}$

$P(\text{Green}) = \frac{\text{Area of Green section}}{\text{Total Area of circle}} = \frac{1}{3}$

Since $P(\text{Red}) = P(\text{Blue}) = P(\text{Green})$, the probability of each player winning is equal when using Spinner (i).

For the other spinners, the areas occupied by the red, blue, and green sections are not equal to each other, and in some cases, the total area is divided among more than three colors. Therefore, the probabilities of landing on red, blue, and green would not be equal, making the game unfair.

Thus, the spinner that should be used to make the game fair is Spinner (i).

The correct answer is (a) (i).

Given:

Frequency distribution classes are 0–10, 10–20, etc.

Size of the class intervals = 10.

To Find:

The lower limit of the fourth class.

Solution:

The classes are given in a continuous form, where the upper limit of one class is the lower limit of the next class.

The first class is 0 – 10.

The lower limit of the first class is $0$. The upper limit is $10$.

The class size is Upper limit - Lower limit = $10 - 0 = 10$. This matches the given information.

Let's list the first few classes based on the pattern and class size 10:

First class: $0$ – $10$ (Lower limit = $0$)

Second class: $10$ – $20$ (Lower limit = $10$)

Third class: $20$ – $30$ (Lower limit = $20$)

Fourth class: The lower limit of the fourth class will be the upper limit of the third class, which is $30$. The upper limit of the fourth class will be its lower limit plus the class size ($30 + 10 = 40$). So, the fourth class is $30$ – $40$.

The lower limit of the fourth class is $30$.

Comparing this with the given options, we find that option (d) is correct.

The correct answer is (d) 30.

Given:

Total number of times the coin is tossed = $200$

Number of times head appeared = $120$

To Find:

The probability of getting a head in this experiment.

Solution:

The experimental probability of an event is given by:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this case, the event is getting a head.

Number of times the event (getting a head) occurred = $120$

Total number of trials (total tosses) = $200$

Probability of getting a head = $\frac{\text{Number of heads}}{\text{Total number of tosses}}$

$P(\text{Head}) = \frac{120}{200}$

Simplify the fraction:

$P(\text{Head}) = \frac{\cancel{120}^{12}}{\cancel{200}_{20}}$

$P(\text{Head}) = \frac{12}{20}$

Further simplification:

$P(\text{Head}) = \frac{\cancel{12}^{3}}{\cancel{20}_{5}}$

$P(\text{Head}) = \frac{3}{5}$

The probability of getting a head in this experiment is $\frac{3}{5}$.

Comparing this with the given options, we find that option (b) is correct.

The correct answer is (b) $\frac{3}{5}$.

Given:

Percentage of buyers interested in a particular brand of toothpaste = $40\%$

To Find:

The central angle of the sector of the pie chart representing this information.

Solution:

A pie chart represents the whole as a circle, which has a total angle of $360^\circ$ at the center.

The proportion of each category is represented by the area of a sector, and the central angle of this sector is proportional to the value the category represents.

To find the central angle for a given percentage, we calculate that percentage of the total angle $360^\circ$.

Central angle for the sector = Percentage $\times 360^\circ$

Central angle = $40\%$ of $360^\circ$

Central angle = $\frac{40}{100} \times 360^\circ$

Calculate the value:

Central angle = $\frac{40}{100} \times 360^\circ$

Central angle = $\frac{4}{10} \times 360^\circ$

Central angle = $4 \times 36^\circ$

Central angle = $144^\circ$

The central angle of the sector representing $40\%$ of the buyers is $144^\circ$.

Comparing this with the given options, we find that option (c) is correct.

The correct answer is (c) $144^\circ$.

Given:

Monthly salary of a person = $\textsf{₹}15000$

Central angle of the sector representing expenses on food and house rent = $60^\circ$

To Find:

The amount spent on food and house rent.

Solution:

In a pie chart, the central angle of a sector is proportional to the value it represents compared to the total value.

The total angle of a circle in a pie chart is $360^\circ$, which represents the total monthly salary of $\textsf{₹}15000$.

The fraction of the salary spent on food and house rent is represented by the ratio of the central angle of the sector to the total angle of the circle.

Fraction of salary spent = $\frac{\text{Central angle}}{\text{Total angle of circle}}$

Fraction of salary spent = $\frac{60^\circ}{360^\circ}$

Simplify the fraction:

Fraction of salary spent = $\frac{\cancel{60}^{1}}{\cancel{360}_{6}}$

Fraction of salary spent = $\frac{1}{6}$

Now, calculate the amount spent on food and house rent by multiplying the total salary by this fraction.

Amount spent = Fraction of salary spent $\times$ Total salary

Amount spent = $\frac{1}{6} \times \textsf{₹}15000$

Amount spent = $\frac{15000}{6} \textsf{₹}$

Amount spent = $\textsf{₹}2500$

The amount the person spends on food and house rent is $\textsf{₹}2500$.

Comparing this with the given options, we find that option (b) is correct.

The correct answer is (b) $\textsf{₹}2500$.

Given:

The pie chart shows the distribution of constituents in the human body with the following percentages:

Water = $60\%$

Protein = $16\%$

Carbohydrates = $14\%$

Other constituents = $10\%$

To Find:

The central angle of the sector showing the distribution of protein and other constituents.

Solution:

First, find the combined percentage of protein and other constituents.

Combined percentage = Percentage of Protein + Percentage of Other constituents

Combined percentage = $16\% + 10\%$

Combined percentage = $26\%$

Note: Rechecking the pie chart image provided, the percentages are: Water 60%, Protein 16%, Fats 15%, Minerals 5%, Carbohydrates 3%, Other constituents 1%. Let's use these values from the image instead of the text description in the previous turn.

From the pie chart image:

Percentage of Protein = $16\%$

Percentage of Other constituents = $14\%$

(It seems the question text or my previous interpretation had a different value for 'Other constituents'. Let's trust the image percentages for calculation as per standard practice for such questions. The image shows 'Other' as 14% combined with 'Carbohydrates' 3%, 'Fats' 15%, 'Minerals' 5% etc., making up the rest after Water 60% and Protein 16%. Let's assume "other constituents" in the question refers to the segment labelled "Other 14%" in the chart.)

Combined percentage of protein and other constituents = Percentage of Protein + Percentage of 'Other' from chart

Combined percentage = $16\% + 14\%$

Combined percentage = $30\%$

The total angle in a pie chart is $360^\circ$, representing $100\%$ of the data.

To find the central angle for the combined percentage, we calculate $30\%$ of $360^\circ$.

Central angle = Percentage (as a decimal) $\times 360^\circ$

Central angle = $\frac{30}{100} \times 360^\circ$

Calculate the value:

Central angle = $\frac{30}{100} \times 360^\circ$

Central angle = $\frac{3}{10} \times 360^\circ$

Central angle = $3 \times 36^\circ$

Central angle = $108^\circ$

The central angle of the sector showing the distribution of protein and other constituents is $108^\circ$.

Comparing this with the given options, we find that option (a) is correct.

The correct answer is (a) $108^\circ$.

Given:

There are 5 cards in total.

The cards are shown in the figure. Assuming the cards are numbered 1, 2, 3, 4, 5 based on the options provided.

To Find:

The probability of Rohan picking a card that has the number 2 on it.

Solution:

The total number of possible outcomes is the total number of cards, which is $5$.

Total number of outcomes = $5$

The number of favorable outcomes is the number of cards that have the number 2 on them.

Assuming the cards are numbered 1, 2, 3, 4, 5, there is only one card with the number 2.

Number of favorable outcomes (card with 2) = $1$

The probability of an event is calculated as:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability of picking a card with the number 2 = $\frac{\text{Number of cards with 2}}{\text{Total number of cards}}$

$P(\text{Card with 2}) = \frac{1}{5}$

Thus, the probability of Rohan picking a card that has the number 2 on it is $\frac{1}{5}$.

Comparing this with the given options, we find that option (b) is correct.

The correct answer is (b) $\frac{1}{5}$.

Given:

From the pie chart, the distribution of proteins in different parts of the human body is given by percentages:

Percentage of protein in Muscles = $30\%$

Percentage of protein in Bones = $20\%$

To Find:

The ratio of distribution of proteins in the muscles to that of proteins in the bones.

Solution:

The ratio of the distribution of proteins in the muscles to that of proteins in the bones is determined by the ratio of their respective percentages from the pie chart.

Ratio = (Percentage in Muscles) : (Percentage in Bones)

Ratio = $30\% : 20\%$

To find the simplified ratio, we can write the percentages as a fraction and reduce it:

Ratio = $\frac{30\%}{20\%}$

Ratio = $\frac{30}{20}$

Simplifying the fraction by dividing the numerator and the denominator by their greatest common divisor, which is 10:

Ratio = $\frac{\cancel{30}^{3}}{\cancel{20}_{2}}$

Ratio = $\frac{3}{2}$

Thus, the ratio of distribution of proteins in the muscles to that of proteins in the bones, based on the provided percentages, is $3:2$.

Comparing this with the given options, the correct answer is (d) 2 : 1.

Given:

From the pie chart provided in Question 20, the distribution of proteins in different parts of the human body is given by percentages:

Percentage of protein in Skin = $10\%$

Percentage of protein in Bones = $20\%$

To Find:

The central angle of the sector representing skin and bones together.

Solution:

First, find the combined percentage of protein in skin and bones.

Combined percentage = Percentage of Skin + Percentage of Bones

Combined percentage = $10\% + 20\%$

Combined percentage = $30\%$

The total angle in a pie chart is $360^\circ$, representing $100\%$ of the data.

To find the central angle for the combined percentage, we calculate $30\%$ of $360^\circ$.

Central angle = Percentage (as a decimal) $\times 360^\circ$

Central angle = $\frac{30}{100} \times 360^\circ$

Calculate the value:

Central angle = $\frac{30}{100} \times 360^\circ$

Central angle = $\frac{3}{10} \times 360^\circ$

Central angle = $3 \times 36^\circ$

Central angle = $108^\circ$

Based on the percentages shown in the pie chart, the central angle for skin and bones together is $108^\circ$. However, this value is not among the given options.

There seems to be an inconsistency between the pie chart percentages and the provided options. Assuming that one of the options is correct, let's consider the possibility of a typo in the percentages presented in the chart such that the combined percentage for Skin and Bones leads to one of the given central angles.

Let's look at the angles corresponding to the options:

(a) $36^\circ \implies \frac{36}{360} \times 100 = 10\%$ (This is the percentage for Skin only).

(b) $60^\circ \implies \frac{60}{360} \times 100 = \frac{100}{6} \approx 16.67\%$

(c) $90^\circ \implies \frac{90}{360} \times 100 = 25\%$

(d) $96^\circ \implies \frac{96}{360} \times 100 = \frac{80}{3} \approx 26.67\%$

If the combined percentage for Skin and Bones was $25\%$, the central angle would be $90^\circ$ (Option c). Given that Skin is shown as $10\%$, this would imply Bones should be $15\%$ (instead of $20\%$) for their sum to be $25\%$. This is a plausible scenario involving a single number change.

Assuming the intended combined percentage for Skin and Bones is $25\%$, the central angle is:

Central angle = $25\%$ of $360^\circ$

Central angle = $\frac{25}{100} \times 360^\circ$

Central angle = $\frac{1}{4} \times 360^\circ$

Central angle = $90^\circ$

Based on the most likely assumed intended value leading to one of the options, option (c) seems to be the intended answer, despite the inconsistency with the percentages shown in the pie chart.

The correct answer is (c) $90^\circ$ (assuming an intended combined percentage of $25\%$ for Skin and Bones).

Given:

From the pie chart provided in Question 20, the distribution of proteins in different parts of the human body is given by percentages:

Percentage of protein in Blood = $6\%$

Percentage of protein in Liver = $20\%$

Percentage of protein in Other proteins (Enzymes, Hormones, etc.) = $14\%$

Percentage of protein in Muscles = $30\%$

Percentage of protein in Bones = $20\%$

Percentage of protein in Skin = $10\%$

To Find:

The central angle of the sector representing hormones enzymes and other proteins.

Solution:

The phrase "hormones enzymes and other proteins" likely refers to the combined category of functional proteins in the body, as opposed to structural proteins (like those in muscles, bones, skin) or proteins in specific organs (like the liver and blood, which contain functional as well as structural proteins). However, given the options and the specific categories in the chart, the most plausible interpretation that leads to one of the answers is that this refers to the sum of the categories Blood, Liver, and Other proteins (which are typically rich in various enzymes, hormones, and other functional proteins).

Let's calculate the combined percentage of protein in Blood, Liver, and Other proteins:

Combined percentage = Percentage in Blood + Percentage in Liver + Percentage in Other proteins

Combined percentage = $6\% + 20\% + 14\%$

Combined percentage = $40\%$

The total angle in a pie chart is $360^\circ$, representing $100\%$ of the data.

To find the central angle for the combined percentage, we calculate $40\%$ of $360^\circ$.

Central angle = Percentage (as a decimal) $\times 360^\circ$

Central angle = $\frac{40}{100} \times 360^\circ$

Calculate the value:

Central angle = $\frac{40}{100} \times 360^\circ$

Central angle = $\frac{4}{10} \times 360^\circ$

Central angle = $4 \times 36^\circ$

Central angle = $144^\circ$

The central angle of the sector representing these combined proteins is $144^\circ$.

Comparing this with the given options, we find that option (b) is correct.

The correct answer is (b) $144^\circ$.

Given:

Total number of times the coin is tossed = $12$

The sequence of outcomes is T, H, H, T, H, T, H, H, H, T, H, T.

To Find:

The chance (probability) of occurrence of Head in this experiment.

Solution:

The total number of trials in this experiment is the total number of times the coin was tossed, which is $12$.

Total number of outcomes = $12$

Now, count the number of times Head appeared in the given sequence of outcomes.

Sequence: T, H, H, T, H, T, H, H, H, T, H, T

Number of Heads = Count the 'H's in the sequence = $7$

Number of favorable outcomes (getting a Head) = $7$

The experimental probability of an event is calculated as:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Probability of getting a Head = $\frac{\text{Number of Heads}}{\text{Total number of tosses}}$

$P(\text{Head}) = \frac{7}{12}$

The probability of occurrence of Head in this experiment is $\frac{7}{12}$.

Comparing this with the given options, we find that option (c) is correct.

The correct answer is (c) $\frac{7}{12}$.

Given:

Number of red balls = $3$

Number of black balls = $5$

Number of blue balls = $4$

To Find:

Total number of outcomes when a ball is drawn from the bag.

Solution:

When a single ball is drawn from the bag, each ball in the bag represents a possible outcome.

The total number of possible outcomes is the total number of balls in the bag.

Total number of outcomes = Number of red balls + Number of black balls + Number of blue balls

Total number of outcomes = $3 + 5 + 4$

Total number of outcomes = $12$

Thus, the total number of outcomes when a ball is drawn from the bag is $12$.

Comparing this with the given options, we find that option (d) is correct.

The correct answer is (d) 12.

Let's analyze the options to determine which type of graph is used to show two sets of data simultaneously:

(a) A Pictograph uses symbols to represent data, often for a single set of categories or data points.

(b) A Histogram is used for grouped continuous data and shows the frequency distribution of a single variable over intervals.

(c) A Pie chart represents parts of a whole as sectors of a circle, showing the proportion of different categories within a single dataset.

(d) A Double bar graph uses two bars side-by-side for each category to compare two different sets of data related to the same categories.

Based on the definitions, a Double bar graph is specifically designed to display and compare two sets of data simultaneously for the same categories.

The correct answer is (d) Double bar graph.

Given:

The class interval is $150$ – $175$.

To Find:

The size of the class $150$ – $175$.

Solution:

In a class interval represented as Lower Limit – Upper Limit, the size or width of the class interval is the difference between the upper limit and the lower limit.

For the class interval $150$ – $175$:

Lower Limit = $150$

Upper Limit = $175$

Size of the class = Upper Limit $-$ Lower Limit

Size of the class = $175 - 150$

Size of the class = $25$

The size of the class interval $150$ – $175$ is $25$.

Comparing this with the given options, we find that option (c) is correct.

The correct answer is (c) 25.

Given:

A standard dice is thrown.

To Find:

The probability of getting the number 7.

Solution:

When a standard dice is thrown, the possible outcomes are the numbers on its faces.

The faces of a standard dice are numbered 1, 2, 3, 4, 5, and 6.

The set of all possible outcomes is $\{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is the count of elements in this set.

Total number of possible outcomes = $6$

The event we are interested in is getting the number 7.

Check if the number 7 is present in the set of possible outcomes $\{1, 2, 3, 4, 5, 6\}$.

The number 7 is not present on a standard dice.

The number of favorable outcomes (getting the number 7) is $0$.

The probability of an event is calculated as:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability of getting the number 7 = $\frac{\text{Number of outcomes where 7 appears}}{\text{Total number of possible outcomes}}$

$P(7) = \frac{0}{6}$

$P(7) = 0$

The probability of getting the number 7 in a throw of a standard dice is 0.

Comparing this with the given options, we find that option (d) is correct.

The correct answer is (d) 0.

Let's consider the different types of data representation listed:

(a) A Bar graph uses rectangular bars to compare different categories. It does not use circles for the main representation.

(b) A Histogram is similar to a bar graph but is used for continuous data, where the bars are adjacent. It also uses rectangles, not circles.

(c) A Pictograph uses pictures or symbols to represent data, not circles as the primary structure.

(d) A Pie chart represents data as sectors of a circle, where the entire circle represents the whole dataset. This method uses a circle as its fundamental shape for data representation.

Therefore, data represented using circles is known as a Pie chart.

The correct answer is (d) Pie chart.

Given:

The question asks what tally marks are used to find.

The options are Class intervals, Range, Frequency, and Upper limit.

Solution:

Let's consider the purpose of tally marks in data handling:

Tally marks are a quick way to count items in groups of five. They are typically used when collecting raw data or when organizing data into a frequency distribution table.

Let's examine how tally marks relate to each option:

(a) Class intervals: Class intervals are ranges used to group data in frequency distributions or histograms (e.g., 0-10, 10-20). Tally marks are used *within* these intervals to count how many data points fall into each range, but the tally marks themselves do not determine or "find" the class intervals.

(b) Range: The range of a data set is the difference between the highest and lowest values. Tally marks are used for counting individual data points, not for calculating the overall range.

(c) Frequency: Frequency is the number of times a particular data point or value occurs, or the number of data points that fall within a specific class interval. Tally marks are the standard method used to record these counts as data is collected or sorted. The sum of the tally marks for a specific category or interval gives its frequency.

(d) Upper limit: The upper limit is the highest value in a class interval. Tally marks do not help in determining the upper limit of a class.

Therefore, tally marks are primarily used as a tool for counting the number of occurrences of each observation or within each category/class interval, which is the definition of Frequency.

The correct answer is (c) Frequency.

Given:

The class interval is $75$ – $85$.

To Find:

The upper limit of the class interval $75$ – $85$.

Solution:

A class interval is typically represented in the form:

Lower Limit – Upper Limit

In the given class interval $75$ – $85$:

The first number, $75$, is the lower limit.

The second number, $85$, is the upper limit.

Therefore, the upper limit of the class interval $75$ – $85$ is $85$.

Comparing this with the given options, we find that option (d) is correct.

The correct answer is (d) 85.

Given:

Numbers written on separate slips are 1, 2, 3, 4, 5.

Total number of slips = $5$.

To Find:

The probability that the slip picked bears an odd number.

Solution:

The total number of possible outcomes is the total number of slips in the box, which is $5$.

Total number of outcomes = $5$

The event is picking a slip that bears an odd number.

From the numbers 1, 2, 3, 4, 5, the odd numbers are 1, 3, and 5.

The favorable outcomes are the slips with numbers 1, 3, or 5.

Number of favorable outcomes (getting an odd number) = $3$

The probability of an event is calculated as:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability of picking an odd number = $\frac{\text{Number of odd numbers}}{\text{Total number of slips}}$

$P(\text{Odd number}) = \frac{3}{5}$

The probability that the slip bears an odd number is $\frac{3}{5}$.

Comparing this with the given options, we find that option (c) is correct.

The correct answer is (c) $\frac{3}{5}$.

Given:

Number of red marbles = $6$

Number of green marbles = $5$

Number of blue marbles = $4$

Number of yellow marbles = $5$

To Find:

The probability that the chosen marble is of red colour.

Solution:

First, find the total number of marbles in the jar.

Total number of marbles = Number of red marbles + Number of green marbles + Number of blue marbles + Number of yellow marbles

Total number of marbles = $6 + 5 + 4 + 5$

Total number of marbles = $20$

The total number of possible outcomes when Hari takes out a marble is the total number of marbles in the jar, which is $20$.

Total number of possible outcomes = $20$

The event we are interested in is choosing a red marble.

The number of favorable outcomes is the number of red marbles, which is $6$.

Number of favorable outcomes (choosing a red marble) = $6$

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability of choosing a red marble = $\frac{\text{Number of red marbles}}{\text{Total number of marbles}}$

$P(\text{Red}) = \frac{6}{20}$

Simplify the fraction:

$P(\text{Red}) = \frac{\cancel{6}^{3}}{\cancel{20}_{10}}$

$P(\text{Red}) = \frac{3}{10}$

The probability that the chosen marble is of red colour is $\frac{3}{10}$.

Comparing this with the given options, we find that option (b) is correct.

The correct answer is (b) $\frac{3}{10}$.

Given:

A coin is tossed two times.

To Find:

The number of possible outcomes when a coin is tossed two times.

Solution:

When a coin is tossed once, there are two possible outcomes: Head (H) or Tail (T).

The set of possible outcomes for a single toss is $\{H, T\}$.

When the coin is tossed two times, we consider the outcome of the first toss and the outcome of the second toss together.

Let the outcome of the first toss be represented first, and the outcome of the second toss be represented second.

The possible outcomes are:

1. Head on the first toss and Head on the second toss (HH)

2. Head on the first toss and Tail on the second toss (HT)

3. Tail on the first toss and Head on the second toss (TH)

4. Tail on the first toss and Tail on the second toss (TT)

The set of all possible outcomes for tossing a coin two times is $\{HH, HT, TH, TT\}$.

The total number of distinct outcomes in this set is $4$.

Alternatively, using the multiplication principle:

Number of outcomes for the first toss = $2$

Number of outcomes for the second toss = $2$

Total number of outcomes for two tosses = (Outcomes for 1st toss) $\times$ (Outcomes for 2nd toss)

Total number of outcomes = $2 \times 2 = 4$

The number of possible outcomes when a coin is tossed two times is $4$.

Comparing this with the given options, we find that option (d) is correct.

The correct answer is (d) 4.

Given:

A coin is tossed three times.

To Find:

The number of possible outcomes when a coin is tossed three times.

Solution:

When a coin is tossed once, there are two possible outcomes: Head (H) or Tail (T).

The set of possible outcomes for a single toss is $\{H, T\}$.

When the coin is tossed three times, we consider the sequence of outcomes for each toss.

The possible outcomes for the three tosses are:

1. Head on the 1st toss, Head on the 2nd toss, Head on the 3rd toss (HHH)

2. Head on the 1st toss, Head on the 2nd toss, Tail on the 3rd toss (HHT)

3. Head on the 1st toss, Tail on the 2nd toss, Head on the 3rd toss (HTH)

4. Head on the 1st toss, Tail on the 2nd toss, Tail on the 3rd toss (HTT)

5. Tail on the 1st toss, Head on the 2nd toss, Head on the 3rd toss (THH)

6. Tail on the 1st toss, Head on the 2nd toss, Tail on the 3rd toss (THT)

7. Tail on the 1st toss, Tail on the 2nd toss, Head on the 3rd toss (TTH)

8. Tail on the 1st toss, Tail on the 2nd toss, Tail on the 3rd toss (TTT)

The set of all possible outcomes for tossing a coin three times is $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.

The total number of distinct outcomes in this set is $8$.

Alternatively, using the multiplication principle:

Number of outcomes for the first toss = $2$

Number of outcomes for the second toss = $2$

Number of outcomes for the third toss = $2$

Total number of outcomes for three tosses = (Outcomes 1st toss) $\times$ (Outcomes 2nd toss) $\times$ (Outcomes 3rd toss)

Total number of outcomes = $2 \times 2 \times 2 = 2^3 = 8$

The number of possible outcomes when a coin is tossed three times is $8$.

Comparing this with the given options, we find that option (d) is correct.

The correct answer is (d) 8.

Given:

A standard die is tossed two times.

To Find:

The number of possible outcomes when a die is tossed two times.

Solution:

When a standard six-sided die is tossed once, there are 6 possible outcomes: the numbers 1, 2, 3, 4, 5, or 6.

The set of possible outcomes for a single toss is $\{1, 2, 3, 4, 5, 6\}$.

Number of outcomes for a single toss = $6$

When the die is tossed two times, the outcome is a pair of results, where the first element is the result of the first toss and the second element is the result of the second toss.

Using the multiplication principle for independent events:

Number of possible outcomes for two tosses = (Number of outcomes for 1st toss) $\times$ (Number of outcomes for 2nd toss)

Number of possible outcomes = $6 \times 6$

Number of possible outcomes = $36$

The set of possible outcomes can be listed as ordered pairs $(a, b)$, where $a$ is the result of the first toss and $b$ is the result of the second toss:

$\{(1,1), (1,2), ..., (1,6), (2,1), (2,2), ..., (2,6), ..., (6,1), (6,2), ..., (6,6)\}$

There are $6$ possibilities for the first element and $6$ possibilities for the second element, making a total of $6 \times 6 = 36$ unique ordered pairs.

The number of possible outcomes when a die is tossed two times is $36$.

Comparing this with the given options, we find that option (c) is correct.

The correct answer is (c) 36.

Data available in an unorganised form is called raw data.

Raw data refers to data that has been collected directly from a source and has not undergone any processing, sorting, or organization.

It is in its initial state, exactly as it was gathered, without any classification, grouping, or calculation performed on it.

Before raw data can be effectively used for analysis, interpretation, or presentation (e.g., in tables, charts, or graphs), it needs to be organized and processed.

In the class interval 20 – 30, the lower class limit is 20.

A class interval is represented in the form of [Lower Limit] – [Upper Limit].

In the interval $20$ – $30$, the number $20$ is the starting or minimum value included in this class (or the boundary before this class in an exclusive interval), and it is called the lower limit.

The number $30$ is the ending or maximum value included (or the boundary after this class in an inclusive interval), and it is called the upper limit.

In the class interval 26 – 33, 33 is known as Upper limit.

A class interval is written in the format [Lower Limit] – [Upper Limit].

In the interval $26$ – $33$, the number $26$ is the lower limit, and the number $33$ is the upper limit.

Given Data:

The data set is 6, 8, 16, 22, 8, 20, 7, 25.

To Find:

The range of the given data.

Solution:

The range of a data set is the difference between the highest value and the lowest value in the set.

First, identify the lowest value in the data set:

Lowest value = $6$

Next, identify the highest value in the data set:

Highest value = $25$

Calculate the range:

Range = Highest value $-$ Lowest value

Range = $25 - 6$

Range = $19$

The range of the data 6, 8, 16, 22, 8, 20, 7, 25 is 19.

A pie chart is used to compare parts to a whole.

In a pie chart, the entire circle represents the total value (the whole), and it is divided into sectors, where each sector represents a part or category of the data.

The size of each sector (both in area and central angle) is proportional to the value it represents, allowing for easy visual comparison of the size of each part relative to the whole and to other parts.

In the experiment of tossing a coin one time, the outcome is either Head or Tail.

A standard coin has two distinct sides, commonly known as Head (usually showing a face or emblem) and Tail (usually showing the reverse side).

When a coin is tossed, assuming it doesn't land on its edge, there are only two possible results: either it lands with the Head facing up, or it lands with the Tail facing up.

These two results, Head and Tail, constitute the complete set of possible outcomes for a single coin toss.

When a dice is rolled, the six possible outcomes are 1, 2, 3, 4, 5, 6.

A standard die is a cube with six faces, and each face is marked with a different number from 1 to 6.

When the die is rolled, the outcome is the number shown on the face that lands facing upwards.

Therefore, the set of all possible results from rolling a single die is $\{1, 2, 3, 4, 5, 6\}$.

Each outcome or a collection of outcomes in an experiment makes an event.

In probability and statistics, an experiment is a process that results in an outcome.

An outcome is a single result of an experiment.

An event is a specific outcome or a set of specific outcomes that we are interested in.

For example, when rolling a die:

The experiment is rolling the die.

The possible outcomes are 1, 2, 3, 4, 5, 6.

An event could be "getting a 4" (a single outcome) or "getting an even number" (a collection of outcomes: 2, 4, 6).

An experiment whose outcomes cannot be predicted exactly in advance is called a random experiment.

A random experiment (also known as a probabilistic experiment or a chance experiment) is an experiment or a process for which the outcome cannot be determined in advance, but all possible outcomes are known.

Examples include tossing a coin (outcome is Head or Tail, but cannot be predicted beforehand) or rolling a die (outcome is one of the numbers 1 to 6, but the specific number is random).

The difference between the upper and lower limit of a class interval is called the size (or width) of the class interval.

For a class interval given as [Lower Limit] – [Upper Limit], the size of the class is calculated as:

Size = Upper Limit $-$ Lower Limit

For example, in the class interval 10 – 20, the size is $20 - 10 = 10$.

Given:

The first two class intervals for a grouped data are 10 – 15 and 15 – 20.

To Find:

The sixth class interval.

Solution:

Observe the given class intervals:

1st class: 10 – 15

2nd class: 15 – 20

This is a continuous frequency distribution, where the upper limit of one class is the lower limit of the next class.

Calculate the class size (width) by finding the difference between the upper and lower limits of a class interval:

Class size = $15 - 10 = 5$

Class size = $20 - 15 = 5$

The class size is 5.

Now, determine the subsequent class intervals by adding the class size to the upper limit of the previous class to find the upper limit of the current class, and using the upper limit of the previous class as the lower limit of the current class.

1st class: 10 – 15

2nd class: 15 – 20

3rd class: 20 – ($20 + 5$) = 20 – 25

4th class: 25 – ($25 + 5$) = 25 – 30

5th class: 30 – ($30 + 5$) = 30 – 35

6th class: 35 – ($35 + 5$) = 35 – 40

The sixth class interval is 35 – 40.

The sixth class interval for a grouped data whose first two class intervals are 10 – 15 and 15 – 20 is 35 – 40.

To find the total number of people surveyed, we need to sum the frequencies (number of people) from each class interval shown in the histogram.

From the histogram:

Number of people owning 0-10 books (Frequency of 0-10) = 10

Number of people owning 10-20 books (Frequency of 10-20) = 20

Number of people owning 20-30 books (Frequency of 20-30) = 30

Number of people owning 30-40 books (Frequency of 30-40) = 25

Number of people owning 40-50 books (Frequency of 40-50) = 15

Number of people owning 50-60 books (Frequency of 50-60) = 5

Total number of people surveyed = Sum of all frequencies

Total surveyed = $10 + 20 + 30 + 25 + 15 + 5$

Total surveyed = $105$

The total number of people surveyed is 105.

To find the number of people owning more than 60 books, we need to look at the histogram for class intervals where the lower limit is 60 or greater.

The class intervals shown on the x-axis of the histogram are 0-10, 10-20, 20-30, 30-40, 40-50, and 50-60.

The histogram does not show any bar for class intervals starting from 60 or extending beyond 60.

This indicates that in the surveyed group, there were no people who owned more than 60 books as per the data represented by this histogram.

The number of people owning books more than 60 is 0.

To find the number of people owning books less than 40, we need to sum the frequencies of the class intervals where the number of books is less than 40.

The class intervals representing people owning less than 40 books are 0-10, 10-20, 20-30, and 30-40.

From the histogram:

Number of people owning 0-10 books = 10

Number of people owning 10-20 books = 20

Number of people owning 20-30 books = 30

Number of people owning 30-40 books = 25

Total number of people owning less than 40 books = Sum of frequencies for 0-10, 10-20, 20-30, and 30-40 intervals.

Total = $10 + 20 + 30 + 25$

Total = $95$

The number of people owning books less than 40 is 95.

To find the number of people having books more than 20 and less than 40, we need to sum the frequencies of the class intervals that fall within this range.

The class intervals from the histogram are 0-10, 10-20, 20-30, 30-40, 40-50, 50-60.

We are looking for people with books more than 20 and less than 40. This corresponds to the class intervals where the lower limit is 20 or more, and the upper limit is 40 or less.

The relevant class intervals are 20-30 and 30-40.

From the histogram:

Number of people owning 20-30 books (Frequency of 20-30) = 30

Number of people owning 30-40 books (Frequency of 30-40) = 25

Total number of people owning books more than 20 and less than 40 = Frequency (20-30) + Frequency (30-40)

Total = $30 + 25$

Total = $55$

The number of people having books more than 20 and less than 40 is 55.

The number of times a particular observation occurs in a given data is called its frequency.

Frequency is a measure of how often a specific value, observation, or event appears within a dataset.

For example, if the data set is {2, 3, 3, 4, 2, 3, 5}, the frequency of the number 2 is 2, the frequency of 3 is 3, the frequency of 4 is 1, and the frequency of 5 is 1.

When the number of observations is large, the observations are usually organised in groups of equal width called class intervals.

When dealing with a large amount of raw data, organizing each individual observation can be cumbersome.

To make the data more manageable and to visualize its distribution, observations are grouped into ranges or categories.

These ranges are called class intervals (or classes).

Grouping data into class intervals, often of equal width, is a common step in creating frequency distributions and drawing histograms.

The total number of outcomes when a coin is tossed is 2.

A standard coin has two sides, typically designated as Head (H) and Tail (T).

When a coin is tossed a single time, the result must be one of these two possibilities.

The set of all possible outcomes for one coin toss is $\{H, T\}$.

The total count of outcomes is 2.

Given:

The class interval is $80$ – $85$.

To Find:

The class size of the interval $80$ – $85$.

Solution:

The class size of a class interval is the difference between its upper limit and its lower limit.

In the interval $80$ – $85$:

Lower limit = $80$

Upper limit = $85$

Class size = Upper limit $-$ Lower limit

Class size = $85 - 80$

Class size = $5$

The class size of the interval 80 – 85 is 5.

In a histogram bars (or rectangles) are drawn with width equal to a class interval without leaving any gap in between.

A histogram is a graphical representation of the distribution of numerical data.

It is composed of adjacent vertical bars, where:

• The base (width) of each bar represents a class interval.
• The height of each bar represents the frequency of data points falling within that class interval.
• The bars are drawn adjacent to each other (without gaps) to indicate the continuous nature of the data across the class intervals.

When a dice is thrown, outcomes 1, 2, 3, 4, 5, 6 are equally likely (or probable).

In probability, when the chances of occurrence of all possible outcomes in an experiment are the same, these outcomes are said to be equally likely.

A standard, fair die is designed so that each of its six faces (numbered 1 through 6) has the same chance of landing facing up when rolled.

Therefore, the outcomes 1, 2, 3, 4, 5, and 6 are equally likely, each having a probability of $\frac{1}{6}$.

In a histogram, class intervals and frequencies are taken along horizontal axis and vertical axis, respectively.

When constructing a histogram:

• The horizontal axis (x-axis) represents the class intervals (e.g., ranges of values, age groups, etc.). This axis shows the different bins into which the data is grouped.
• The vertical axis (y-axis) represents the frequency or count of observations that fall within each corresponding class interval. The height of each bar corresponds to the frequency.

Therefore, class intervals are shown along the horizontal axis, and frequencies are shown along the vertical axis.

In frequency distributions with continuous class intervals like 10–20, 20–30, etc., a standard convention is followed to ensure that each data point falls into exactly one class interval.

The convention is that the lower limit of a class interval is included in that class, while the upper limit is excluded.

So, the class interval 10–20 includes values greater than or equal to 10 and strictly less than 20 ($10 \le x < 20$).

The class interval 20–30 includes values greater than or equal to 20 and strictly less than 30 ($20 \le x < 30$).

Based on this convention, the value 20 is included in the class interval where it is the lower limit.

Therefore, 20 lies in the class 20–30.

In the class intervals 10 –20, 20 –30, etc., respectively, 20 lies in the class 20–30.

True

Explanation:

A pie chart is a type of graph that visually represents the distribution of a dataset. It is called a "pie chart" because it resembles a pie that is sliced into different pieces.

In a pie chart, the entire circle represents the whole or $100\%$ of the data.

The circle is divided into several parts, and each part represents a category or a component of the data. These parts are called sectors.

A sector of a circle is the region bounded by two radii and the intercepted arc. Think of it as a slice of pie.

Each sector in a pie chart corresponds to a specific category. The size of each sector (specifically, the angle at the center of the circle) is proportional to the quantity it represents from the whole. The sum of the central angles of all the sectors in a pie chart is always $360^\circ$.

Therefore, the statement that in a pie chart a whole circle is divided into sectors is true because this is precisely how a pie chart is constructed to represent data proportions.

False

Explanation:

In a pie chart, the central angle of a sector is proportional to the value or frequency it represents in the dataset.

The entire circle represents the total value or $100\%$ of the data, and its central angle is $360^\circ$.

If a particular category represents more than $50\%$ of the total data, its corresponding sector will have a central angle greater than $180^\circ$.

For example, if a category represents $60\%$ of the data, the central angle for that sector would be calculated as:

Since $216^\circ$ is greater than $180^\circ$, a central angle in a pie chart can indeed be more than $180^\circ$.

True

Explanation:

A pie chart represents a dataset as a whole circle, which corresponds to the total quantity or $100\%$ of the data.

The circle is divided into different sectors, where each sector represents a part or category of the data.

The size of each sector's central angle is proportional to the value it represents within the total.

Since the sectors together make up the entire circle, the sum of the central angles of all the sectors must equal the total angle of the circle.

The total angle at the center of a circle is always $360^\circ$.

Therefore, the sum of all the central angles in any pie chart is always equal to $360^\circ$.

True

Explanation:

In a pie chart, the sum of all central angles must be $360^\circ$.

If a pie chart has only two sectors, and each sector has a central angle of $180^\circ$, then the sum of these angles is $180^\circ + 180^\circ = 360^\circ$.

This scenario is possible if the dataset is divided into two categories, and each category represents exactly half (or $50\%$) of the total.

For example, if you are representing the results of flipping a coin 100 times and got 50 heads and 50 tails, the pie chart could have two sectors: one for heads and one for tails. Each would represent $50\%$ of the total, corresponding to a central angle of $180^\circ$.

True

Explanation:

In a pie chart, the size of the central angle for each sector is directly proportional to the value or frequency that the sector represents in the dataset.

The formula to calculate the central angle for a sector is:

If two or more different categories in the dataset have the same value or frequency, then according to this formula, their corresponding central angles will also be equal.

For instance, if a survey shows that the number of people who prefer 'Apples' is the same as the number of people who prefer 'Bananas', then the central angle for the 'Apples' sector will be equal to the central angle for the 'Bananas' sector in the pie chart.

Therefore, it is possible and quite common for two or more central angles in a pie chart to be equal.

True

Explanation:

In probability, an experiment is a process that results in one of several possible outcomes. Throwing a die is an experiment.

The set of all possible outcomes of an experiment is called the sample space. When a standard six-sided die is thrown, the sample space is the set of possible numbers that can land face up, which is $\{1, 2, 3, 4, 5, 6\}$.

An event is defined as any subset of the sample space. It is a collection of one or more outcomes from the sample space.

The statement describes the outcome "Getting a prime number on throwing a die". We need to identify which outcomes in the sample space are prime numbers.

The prime numbers in the sample space $\{1, 2, 3, 4, 5, 6\}$ are those greater than 1 that have no positive divisors other than 1 and themselves. These are 2, 3, and 5.

So, the outcomes corresponding to "Getting a prime number" are $\{2, 3, 5\}$.

Since the set $\{2, 3, 5\}$ is a subset of the sample space $\{1, 2, 3, 4, 5, 6\}$, it fits the definition of an event.

Therefore, getting a prime number on throwing a die is indeed an event.

True

Explanation:

The provided table shows the marks obtained by students out of 10 and the frequency, which represents the number of students who obtained that particular mark.

The question asks about the number of students who got "full marks". Since the marks are out of 10, full marks means obtaining 10 marks.

Looking at the frequency table:

Locate the column where the mark is 10. The corresponding frequency in that column is 9.

This means that 9 students obtained 10 marks.

Since 10 marks is full marks, the statement "9 students got full marks" is correct.

False

Explanation:

The frequency table provided is:

We need to find the frequency of students who obtained less than 8 marks. This includes students who obtained 4, 5, or 7 marks.

From the table:

Frequency for 4 marks = 5

Frequency for 5 marks = 10

Frequency for 7 marks = 8

The total frequency for less than 8 marks is the sum of these frequencies:

The statement says the frequency of less than 8 marks is 29. Our calculation shows it is 23.

Therefore, the statement is false.

True

Explanation:

The provided frequency table shows the marks obtained by students and the number of students who achieved each mark.

We need to find the frequency of students who obtained more than 8 marks. This includes students who obtained marks of 9 or 10.

From the table:

Frequency for 9 marks = 12

Frequency for 10 marks = 9

The total frequency for more than 8 marks is the sum of these frequencies:

The statement says the frequency of more than 8 marks is 21. Our calculation confirms this.

Therefore, the statement is true.

False

Explanation:

The provided frequency table is:

We need to identify the highest frequency value from the "Frequency" row and see which mark it corresponds to.

The frequencies are 5, 10, 8, 6, 12, and 9.

Comparing these values, the highest frequency is 12.

Now, look at the "Marks" row to see which mark corresponds to the frequency 12.

The frequency 12 corresponds to the mark 9.

Therefore, 9 marks has the highest frequency (which is 12).

The statement says that 10 marks has the highest frequency. The frequency for 10 marks is 9, which is not the highest frequency.

Thus, the statement "10 marks the highest frequency" is false.

False

Explanation:

We are given the fourth and fifth class intervals:

Fourth class interval: 55 – 60

Fifth class interval: 60 – 65

From these intervals, we can observe the following:

1. The class intervals are continuous, meaning the upper limit of one interval is the lower limit of the next interval.

2. The class width is the difference between the upper and lower limits of an interval. For the fourth interval, the width is $60 - 55 = 5$. For the fifth interval, the width is $65 - 60 = 5$. The class width is constant and equal to 5.

To find the first class interval, we can work backward from the given intervals, subtracting the class width from the lower limit of each preceding interval.

Fifth class interval: 60 – 65

Fourth class interval: $60 - 5 = 55$. So it is 55 – 60.

Third class interval: $55 - 5 = 50$. So it is 50 – 55.

Second class interval: $50 - 5 = 45$. So it is 45 – 50.

First class interval: $45 - 5 = 40$. So it is 40 – 45.

Our calculation shows that the first class interval is 40 – 45.

The statement claims that the first class interval is 45 – 50.

Since 40 – 45 is not equal to 45 – 50, the statement is false.

False

Explanation:

The histogram shows the number of literate males in different age groups.

We need to find the total number of literate males above the age of 20.

From the histogram, the age groups above 20 are 20-30, 30-40, 40-50, 50-60, 60-70, and 70-80.

We read the frequency (number of literate males) for each of these age groups from the histogram bars:

Age group 20-30: Approximately 6000 literate males.

Age group 30-40: Approximately 3500 literate males.

Age group 40-50: Approximately 1500 literate males.

Age group 50-60: Approximately 1000 literate males.

Age group 60-70: Approximately 500 literate males.

Age group 70-80: Approximately 250 literate males.

To find the total number of literate males above the age of 20, we sum the frequencies for these age groups:

The statement claims that 1500 males above the age of 20 are literate.

Our calculation shows that the total number of literate males above the age of 20 is approximately 12750.

The number 1500 corresponds specifically to the age group 40-50, not the total number of males above 20.

Therefore, the statement is false.

True

Explanation:

The class size (also called class width or length) of a class interval is the difference between its upper limit and its lower limit.

For the given class interval 60 – 68:

Lower limit = 60

Upper limit = 68

Class size = Upper limit - Lower limit

The calculated class size is 8, which matches the value given in the statement.

Therefore, the statement is true.

False

Explanation:

When a single coin is tossed, there are two possible outcomes: Heads (H) or Tails (T).

When a pair of coins (which means two coins) is tossed simultaneously or one after the other, we need to consider the outcomes for both coins.

Let's list the possible outcomes:

1. The first coin shows Heads and the second coin shows Heads (HH).

2. The first coin shows Heads and the second coin shows Tails (HT).

3. The first coin shows Tails and the second coin shows Heads (TH).

4. The first coin shows Tails and the second coin shows Tails (TT).

The set of all possible outcomes, also known as the sample space, is $\{HH, HT, TH, TT\}$.

Counting these distinct outcomes, we find there are a total of 4 possible outcomes when a pair of coins is tossed.

The statement claims that the number of outcomes is 2, which is incorrect.

Therefore, the statement is false.

True

Explanation:

When a standard six-sided die is thrown once, the possible outcomes are the numbers on its faces.

The sample space (set of all possible outcomes) is $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of outcomes is the number of elements in the sample space, which is $n(S) = 6$.

We are interested in the event of getting an even number. Let this event be E.

The even numbers in the sample space are 2, 4, and 6.

So, the favourable outcomes for event E are $E = \{2, 4, 6\}$.

The number of favourable outcomes is the number of elements in E, which is $n(E) = 3$.

The probability of an event E is calculated as:

Substituting the values we found:

Simplifying the fraction:

The calculated probability of getting an even number is $\frac{1}{2}$, which matches the probability given in the statement.

Therefore, the statement is true.

False

Explanation:

When a standard six-sided die is thrown once, the set of all possible outcomes (sample space) is:

The total number of outcomes is $n(S) = 6$.

A composite number is a positive integer that has at least one divisor other than 1 and itself.

Let's identify the numbers in the sample space that are composite:

1 is neither prime nor composite.

2 is prime.

3 is prime.

4 is composite (divisors are 1, 2, 4).

5 is prime.

6 is composite (divisors are 1, 2, 3, 6).

The event of getting a composite number is E = $\{4, 6\}$.

The number of favourable outcomes is $n(E) = 2$.

The probability of event E is given by:

Substitute the values:

Simplify the fraction:

The probability of getting a composite number is $\frac{1}{3}$.

The statement claims the probability is $\frac{1}{2}$, which is incorrect.

Therefore, the statement is false.

False

Explanation:

The provided image is a pie chart showing the percentage distribution of Manganese production across five states: A, B, C, D, and E.

In a pie chart, the size of each sector is proportional to the quantity it represents. The state with the least production will be represented by the smallest sector, i.e., the sector with the smallest percentage.

Let's look at the percentages for each state from the pie chart:

State A: $25\%$

State B: $15\%$

State C: $30\%$

State D: $10\%$

State E: $20\%$

To find the state with the least production, we need to find the minimum percentage among all the states. Comparing the percentages ($25\%$, $15\%$, $30\%$, $10\%$, $20\%$), the minimum percentage is $10\%$.

This minimum percentage ($10\%$) corresponds to State D.

Therefore, the production of Manganese is least in State D, not State B (which has $15\%$ production).

The statement claims that the production of Manganese is least in state B, which contradicts our finding that State D has the least production.

Thus, the statement is false.

True

Explanation:

In probability, an experiment is a process that has a well-defined set of possible results.

An outcome is a single result of an experiment.

The sample space is the set of all possible outcomes of an experiment.

An event is defined as any subset of the sample space. A subset can contain:

- Zero outcomes (this is the impossible event, represented by the empty set).

- Exactly one outcome (this is called an elementary or simple event).

- Two or more outcomes.

- All possible outcomes (this is the sure event, which is the sample space itself).

The statement says "One or more outcomes of an experiment make an event". This means that any collection of one outcome, or two outcomes, or more outcomes from the sample space constitutes an event.

This aligns with the definition of an event as a subset of the sample space, excluding only the impossible event (which has zero outcomes).

Therefore, the statement is true because a collection of one or more outcomes always forms a subset of the sample space and is, by definition, an event (specifically, a non-empty event).

False

Explanation:

When a standard six-sided die is thrown once, the set of all possible outcomes (sample space) is:

The total number of outcomes is $n(S) = 6$.

The probability of an event E is given by:

Let's evaluate the first part of the statement: probability of getting number 6.

The event of getting a 6 is E1 = $\{6\}$. The number of favourable outcomes is $n(E1) = 1$.

Using formula (1):

This matches the first part of the statement.

Now, let's evaluate the second part of the statement: probability of getting number 5.

The event of getting a 5 is E2 = $\{5\}$. The number of favourable outcomes is $n(E2) = 1$.

Using formula (1):

The second part of the statement claims the probability of getting a number 5 is $\frac{1}{5}$. Our calculation shows it is $\frac{1}{6}$.

Since the second part of the statement is false, the entire combined statement is false.

False

Explanation:

When a standard six-sided die is thrown once, the sample space (set of all possible outcomes) is the numbers on its faces:

The total number of possible outcomes is $n(S) = 6$.

The probability of an event E is given by:

First, let's find the probability of getting a prime number.

Prime numbers in the sample space $\{1, 2, 3, 4, 5, 6\}$ are numbers greater than 1 that have no positive divisors other than 1 and themselves. These are 2, 3, and 5.

Let E_prime be the event of getting a prime number. E_prime = $\{2, 3, 5\}$.

The number of favourable outcomes for E_prime is $n(E\_prime) = 3$.

Using formula (1), the probability of getting a prime number is:

Next, let's find the probability of getting a composite number.

Composite numbers are positive integers greater than 1 that are not prime. In the sample space $\{1, 2, 3, 4, 5, 6\}$, the composite numbers are 4 and 6.

Let E_composite be the event of getting a composite number. E_composite = $\{4, 6\}$.

The number of favourable outcomes for E_composite is $n(E\_composite) = 2$.

Using formula (1), the probability of getting a composite number is:

Comparing the two probabilities:

Since $\frac{1}{2} \neq \frac{1}{3}$, the probability of getting a prime number is not the same as that of getting a composite number when throwing a die once.

Therefore, the statement is false.

True

Explanation:

When a standard six-sided die is thrown once, the set of all possible outcomes (sample space) is:

The total number of possible outcomes is $n(S) = 6$.

The probability of an event E is given by:

First, let's find the probability of getting an even number.

The even numbers in the sample space $\{1, 2, 3, 4, 5, 6\}$ are 2, 4, and 6.

Let E_even be the event of getting an even number. E_even = $\{2, 4, 6\}$.

The number of favourable outcomes for E_even is $n(E\_even) = 3$.

Using formula (1), the probability of getting an even number is:

Next, let's find the probability of getting an odd number.

The odd numbers in the sample space $\{1, 2, 3, 4, 5, 6\}$ are 1, 3, and 5.

Let E_odd be the event of getting an odd number. E_odd = $\{1, 3, 5\}$.

The number of favourable outcomes for E_odd is $n(E\_odd) = 3$.

Using formula (1), the probability of getting an odd number is:

Comparing the two probabilities:

Since $\frac{1}{2} = \frac{1}{2}$, the probability of getting an even number is indeed the same as that of getting an odd number when throwing a die once.

Therefore, the statement is true.

False

Explanation:

A random experiment is an experiment or a process where the outcome is not known beforehand and cannot be predicted with certainty, even when the experiment is repeated under identical conditions. There must be multiple possible outcomes, and chance plays a role in determining which outcome occurs.

Examples of random experiments include:

- Tossing a coin (outcomes are Heads or Tails)

- Rolling a die (outcomes are numbers 1 through 6)

- Drawing a card from a shuffled deck (outcomes are different cards)

Verifying Pythagoras theorem involves checking if the relationship $a^2 + b^2 = c^2$ holds true for a right-angled triangle, where $a$ and $b$ are the lengths of the two shorter sides (legs) and $c$ is the length of the longest side (hypotenuse).

The process usually involves:

1. Having or constructing a right-angled triangle.

2. Measuring the lengths of its sides ($a$, $b$, and $c$).

3. Calculating $a^2$, $b^2$, and $c^2$.

4. Checking if $a^2 + b^2$ is equal to $c^2$.

Pythagoras theorem is a fundamental geometric property that is always true for any right-angled triangle. The outcome of verifying it for a given right triangle is always expected to be that the relationship $a^2 + b^2 = c^2$ holds (within the limits of measurement accuracy if done experimentally).

Since the outcome of whether the theorem is true for a right triangle is a known and predictable mathematical fact, the process of verifying it is not a random experiment. While there might be small variations in measurements if done physically, the core principle being verified is not subject to random chance in the same way as the outcome of a coin toss or die roll.

Therefore, verifying Pythagoras theorem is a deterministic process based on geometry, not a random experiment.

False

Explanation:

The given pictorial representation shows the number of students who like different sports (Cricket, Football, Tennis, Hockey).

This type of graph, which uses bars to represent the frequency of distinct categories, is called a bar graph or bar chart.

Key characteristics of a bar graph:

- Used for representing categorical or discrete data.

- There are typically gaps between the bars.

- The horizontal axis represents categories (sports in this case).

- The height of the bars represents the frequency or count for each category.

A histogram, on the other hand, is used to represent the frequency distribution of continuous data.

Key characteristics of a histogram:

- Used for representing continuous data grouped into class intervals.

- The bars are drawn adjacent to each other with no gaps (unless a class interval has zero frequency).

- The horizontal axis represents class intervals (ranges of numerical data).

- The area of each bar (which is proportional to its height if the class widths are equal) represents the frequency of the data in that interval.

Since the given graph represents categorical data (sports) with gaps between the bars, it is a bar graph, not a histogram.

Therefore, the statement is false.

(a) The frequency distribution table lists the class intervals and their corresponding frequencies. The class intervals are: 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60.

The second class interval is 20 - 30. The lower limit of this interval is the starting value.

The lower limit of the second class interval is 20.

(b) The class intervals are: 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60.

The last class interval listed is 50 - 60. The upper limit of this interval is the ending value.

The upper limit of the last class interval is 60.

(c) The third class interval in the table is 30 - 40.

We look at the frequency corresponding to the class interval 30 - 40 in the table.

The frequency of the third class (30 - 40) is 4.

(d) We need to find the class interval that has a frequency of 10.

Looking at the frequency column in the table, we find the value 10.

The class interval corresponding to the frequency 10 is 20 - 30.

The interval with a frequency of 10 is 20 – 30.

(e) We need to find the lowest frequency value in the frequency column and identify the corresponding class interval.

The frequencies listed are 5, 10, 4, 15, 12.

Comparing these values, the lowest frequency is 4.

The class interval corresponding to the frequency 4 is 30 - 40.

The interval with the lowest frequency is 30 – 40.

(f) The class size (or class width) is the difference between the upper limit and the lower limit of any class interval, assuming the class size is uniform throughout the distribution.

Let's take the first interval: 10 - 20. Class size = $20 - 10 = 10$.

Let's take the second interval: 20 - 30. Class size = $30 - 20 = 10$.

The class size is consistent for all intervals.

The class size is 10.

To draw a histogram for the given frequency distribution table, we follow these steps:

Steps to draw the histogram:

1. Draw the horizontal and vertical axes on a graph paper.

2. The horizontal axis represents the Class Intervals (Maximum Speed in km/h). Since the intervals are contiguous (the upper limit of one interval is the lower limit of the next), we mark the limits of the class intervals on the horizontal axis at equal distances (e.g., 10, 20, 30, 40, 50, 60, 70). Label this axis "Maximum Speed (km/h)".

3. The vertical axis represents the Frequency (Number of Animals). Choose a suitable scale for the vertical axis to accommodate the highest frequency (which is 10 in this table). Label this axis "Frequency".

4. Draw rectangles (bars) above each class interval on the horizontal axis.

5. The width of each rectangle corresponds to the width of the class interval (which is $20-10=10$ for all intervals). So, each bar will have a width spanning from the lower limit to the upper limit of the interval on the horizontal axis.

6. The height of each rectangle corresponds to the frequency of the class interval.

Description of the resulting Histogram:

Based on the table:

- For the class interval 10 - 20, draw a rectangle with a height of 5.

- For the class interval 20 - 30, draw a rectangle adjacent to the first one, with a height of 5.

- For the class interval 30 - 40, draw a rectangle adjacent to the previous one, with a height of 10.

- For the class interval 40 - 50, draw a rectangle adjacent to the previous one, with a height of 8.

- For the class interval 50 - 60, draw a rectangle adjacent to the previous one, with a height of 0. This will result in no bar or a bar with zero height for this interval.

- For the class interval 60 - 70, draw a rectangle adjacent to the previous space, with a height of 2.

The bars in the histogram will be touching each other (except where the frequency is zero), representing the continuous nature of the data (speed) grouped into intervals.

(Note: An actual image of the histogram cannot be generated here, but the description above explains how it would be constructed.)

The total number of children is 350. The pie chart shows the distribution of time spent in different games based on the central angle of each sector. To find the number of children in each category, we use the formula: Number of children = $\left(\frac{\text{Central Angle}}{360^\circ}\right) \times \text{Total number of children}$.

Calculating the number of children for each time category:

Less than 1 hour ($36^\circ$): $\frac{36}{360} \times 350 = \frac{1}{10} \times 350 = 35$ children.

1 hour ($72^\circ$): $\frac{72}{360} \times 350 = \frac{1}{5} \times 350 = 70$ children.

2 hours ($108^\circ$): $\frac{108}{360} \times 350 = \frac{3}{10} \times 350 = 105$ children.

3 hours ($90^\circ$): $\frac{90}{360} \times 350 = \frac{1}{4} \times 350 = 87.5$ children.

More than 3 hours ($54^\circ$): $\frac{54}{360} \times 350 = \frac{3}{20} \times 350 = 52.5$ children.

(a) How many children spend at least one hour in playing games?

"At least one hour" means the time spent is 1 hour or more. This includes the categories: 1 hour, 2 hours, 3 hours, and More than 3 hours.

Number of children = (Number of children for 1 hour) + (Number of children for 2 hours) + (Number of children for 3 hours) + (Number of children for More than 3 hours)

Number of children = $70 + 105 + 87.5 + 52.5$

Number of children = $175 + 140 = 315$

Alternatively, children who spend at least one hour are all children except those who spend less than 1 hour.

Number of children = Total children - (Number of children for Less than 1 hour)

Number of children = $350 - 35 = 315$

The number of children who spend at least one hour in playing games is 315.

(b) How many children spend more than 2 hours in playing games?

"More than 2 hours" means the time spent is greater than 2 hours. This includes the categories: 3 hours and More than 3 hours.

Number of children = (Number of children for 3 hours) + (Number of children for More than 3 hours)

Number of children = $87.5 + 52.5 = 140$

The number of children who spend more than 2 hours in playing games is 140.

(c) How many children spend 3 or lesser hours in playing games?

"3 or lesser hours" means the time spent is 3 hours or less. This includes the categories: Less than 1 hour, 1 hour, 2 hours, and 3 hours.

Number of children = (Number of children for Less than 1 hour) + (Number of children for 1 hour) + (Number of children for 2 hours) + (Number of children for 3 hours)

Number of children = $35 + 70 + 105 + 87.5 = 297.5$

Alternatively, children who spend 3 or lesser hours are all children except those who spend More than 3 hours.

Number of children = Total children - (Number of children for More than 3 hours)

Number of children = $350 - 52.5 = 297.5$

The number of children who spend 3 or lesser hours in playing games is 297.5.

(d) Which is greater — number of children who spend 2 hours or more per day or number of children who play for less than one hour?

First, find the number of children who spend 2 hours or more per day. This includes the categories: 2 hours, 3 hours, and More than 3 hours.

Number of children ($\geq$ 2 hours) = (Number of children for 2 hours) + (Number of children for 3 hours) + (Number of children for More than 3 hours)

Number of children ($\geq$ 2 hours) = $105 + 87.5 + 52.5 = 245$

Next, find the number of children who play for less than one hour.

Number of children (< 1 hour) = 35 (calculated from the pie chart)

Now, compare the two numbers: 245 and 35.

$245 > 35$

The number of children who spend 2 hours or more per day (245) is greater than the number of children who play for less than one hour (35).

The number of children who spend 2 hours or more per day is greater.

The pie chart shows the distribution of modes of travel based on the central angle of each sector.

The central angles for each mode are:

Bus: $144^\circ$

Car: $72^\circ$

Taxi: $36^\circ$

Walk: $90^\circ$

Cycle: $18^\circ$

(a) What is the most common mode of transport?

The most common mode of transport is the one with the largest central angle (or largest percentage).

Comparing the angles $144^\circ, 72^\circ, 36^\circ, 90^\circ, 18^\circ$, the largest angle is $144^\circ$, which corresponds to Bus.

The most common mode of transport is Bus.

(b) What fraction of children travel by car?

The central angle for Car is $72^\circ$. The total angle in a circle is $360^\circ$.

The fraction of children travelling by car is given by the ratio of the central angle for Car to the total angle:

Simplifying the fraction:

The fraction of children who travel by car is $\frac{1}{5}$.

(c) If 18 children travel by car, how many children took part in the survey?

From part (b), we know that the fraction of children who travel by car is $\frac{1}{5}$.

Let the total number of children who took part in the survey be T.

According to the information given, 18 children travel by car.

To find T, multiply both sides by 5:

So, 90 children took part in the survey.

The number of children who took part in the survey is 90.

(d) How many children use taxi to travel to school?

The central angle for Taxi is $36^\circ$. The total number of children is 90 (from part c).

The fraction of children using Taxi is $\frac{36^\circ}{360^\circ} = \frac{1}{10}$.

The number of children using Taxi is the fraction for Taxi multiplied by the total number of children:

So, 9 children use taxi to travel to school.

The number of children who use taxi to travel to school is 9.

(e) By which two modes of transport are equal number of children travelling?

To have an equal number of children travelling by two different modes, the central angles for those two modes must be equal.

Let's list the central angles again:

Bus: $144^\circ$

Car: $72^\circ$

Taxi: $36^\circ$

Walk: $90^\circ$

Cycle: $18^\circ$

Comparing these angles ($144, 72, 36, 90, 18$), we see that no two angles are equal.

Alternatively, we can calculate the number of children for each mode using the total number of children (90) found in part (c):

Bus: $\frac{144}{360} \times 90 = 36$ children

Car: $\frac{72}{360} \times 90 = 18$ children

Taxi: $\frac{36}{360} \times 90 = 9$ children

Walk: $\frac{90}{360} \times 90 = 22.5$ children

Cycle: $\frac{18}{360} \times 90 = 4.5$ children

Comparing the number of children for each mode (36, 18, 9, 22.5, 4.5), no two modes have exactly the same number of children.

Based on the provided central angles and the given information, no two modes of transport have an equal number of children travelling.

Therefore, there are no two modes by which an equal number of children are travelling.

When a standard six-sided die is rolled once, the set of all possible outcomes (sample space) is $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is $n(S) = 6$.

The probability of an event E is given by:

(a) Probability of getting an odd number:

The odd numbers in the sample space are $\{1, 3, 5\}$.

Let E$_a$ be the event of getting an odd number. $n(E\_a) = 3$.

P(Odd) = $\frac{n(E\_a)}{n(S)} = \frac{3}{6}$

P(Odd) = $\frac{1}{2}$

(b) Probability of getting a number greater than 5:

The numbers in the sample space greater than 5 is $\{6\}$.

Let E$_b$ be the event of getting a number greater than 5. $n(E\_b) = 1$.

P(Greater than 5) = $\frac{n(E\_b)}{n(S)} = \frac{1}{6}$

(c) Probability of getting a multiple of 3:

The multiples of 3 in the sample space are $\{3, 6\}$.

Let E$_c$ be the event of getting a multiple of 3. $n(E\_c) = 2$.

P(Multiple of 3) = $\frac{n(E\_c)}{n(S)} = \frac{2}{6}$

P(Multiple of 3) = $\frac{1}{3}$

(d) Probability of getting a number less than 1:

The numbers in the sample space less than 1 is {}\ (\text{empty set})\.

Let E$_d$ be the event of getting a number less than 1. $n(E\_d) = 0$.

P(Less than 1) = $\frac{n(E\_d)}{n(S)} = \frac{0}{6}$

P(Less than 1) = $0$

(e) Probability of getting a factor of 36:

The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.

The factors of 36 that are in the sample space $\{1, 2, 3, 4, 5, 6\}$ are $\{1, 2, 3, 4, 6\}$.

Let E$_e$ be the event of getting a factor of 36. $n(E\_e) = 5$.

P(Factor of 36) = $\frac{n(E\_e)}{n(S)} = \frac{5}{6}$

(f) Probability of getting a factor of 6:

The factors of 6 are 1, 2, 3, 6.

The factors of 6 that are in the sample space $\{1, 2, 3, 4, 5, 6\}$ are $\{1, 2, 3, 6\}$.

Let E$_f$ be the event of getting a factor of 6. $n(E\_f) = 4$.

P(Factor of 6) = $\frac{n(E\_f)}{n(S)} = \frac{4}{6}$

P(Factor of 6) = $\frac{2}{3}$

We need to classify each statement based on whether the event described is certain to happen, impossible to happen, or may or may not happen.

(a) Getting the sum of angles of a triangle as $180^\circ$.

In Euclidean geometry, the sum of the interior angles of any triangle is always $180^\circ$. This is a fundamental property.

Classification: Certain to happen

(b) India winning a cricket match against Pakistan.

The outcome of a cricket match between two teams is not fixed. Either team can win, or it could result in a draw/tie (depending on the format). The result depends on various factors like player performance, conditions, etc.

Classification: May or may not happen

(c) Sun setting in the evening.

Due to the Earth's rotation, the Sun appears to rise in the east and set in the west every day. This is a predictable astronomical event.

Classification: Certain to happen

(d) Getting 7 when a die is thrown.

A standard six-sided die has faces numbered 1, 2, 3, 4, 5, and 6. It is impossible to get the number 7 on a standard die.

Classification: Impossible to happen

(e) Sun rising from the west.

The Earth rotates from west to east, making the Sun appear to rise in the east and set in the west. It is impossible for the Sun to rise from the west under normal circumstances.

Classification: Impossible to happen

(f) Winning a racing competition by you.

Whether you win a race depends on your own performance relative to other participants. It is possible you win, and it is possible someone else wins.

Classification: May or may not happen

Placing these statements into the table:

Given:

Total marks obtained in the examination = 540.

The distribution of marks in different subjects is shown by a pie chart.

To Find:

The marks obtained by the student in each subject.

Solution:

To find the marks obtained in each subject, we need the central angle of the sector representing that subject in the pie chart. The total angle in a pie chart is $360^\circ$, which corresponds to the total marks obtained (540).

The marks obtained in a particular subject can be calculated using the formula:

However, the image of the pie chart, which contains the central angles for each subject (like Hindi, English, Maths, Science, Social Science, etc., which are typical subjects), is not provided with this question.

Assuming the central angles for each subject were available from the pie chart, we would list them and apply the formula (1) for each subject.

For example, if the central angle for Subject A was $\theta_A$, for Subject B was $\theta_B$, and so on, the marks would be calculated as:

and so on for all subjects.

The sum of the central angles of all subjects in the pie chart should be $360^\circ$, and the sum of the calculated marks for all subjects should be equal to the Total Marks (540).

Since the central angles are missing, we cannot provide the numerical marks for each subject.

Note: Please provide the central angles from the pie chart to calculate the specific marks for each subject.

Given:

The bag contains only white and yellow balls.

Probability of choosing a white ball, P(White) = $\frac{2}{9}$.

Total number of balls in the bag = 36.

To Find:

The number of yellow balls in the bag.

Solution:

The probability of choosing an event is given by the ratio of the number of favourable outcomes to the total number of outcomes.

In this case, the experiment is drawing a ball from the bag.

The total number of outcomes is the total number of balls in the bag, which is 36.

The event of choosing a white ball has the number of favourable outcomes equal to the number of white balls.

Let the number of white balls in the bag be $N_W$.

The probability of choosing a white ball is given by:

To find $N_W$, we can solve equation (2):

So, there are 8 white balls in the bag.

The bag contains only white and yellow balls. Therefore, the total number of balls is the sum of the number of white balls and the number of yellow balls.

Let the number of yellow balls be $N_Y$.

Substitute the value of $N_W = 8$ into this equation:

To find $N_Y$, subtract 8 from 36:

So, there are 28 yellow balls in the bag.

Final Answer:

The number of yellow balls in the bag is 28.

The histogram shows the frequency distribution of heights of students. The horizontal axis represents the height in cm (class intervals), and the vertical axis represents the number of students (frequency).

Reading the frequencies from the histogram for each class interval:

130 - 135: 6 students

135 - 140: 9 students

140 - 145: 14 students

145 - 150: 18 students

150 - 155: 10 students

155 - 160: 7 students

(a) How many students have height more than or equal to 135 cm but less than 150 cm?

This range covers the class intervals 135 - 140, 140 - 145, and 145 - 150.

Number of students = (Frequency of 135 - 140) + (Frequency of 140 - 145) + (Frequency of 145 - 150)

Number of students = $9 + 14 + 18$

Number of students = $23 + 18 = 41$

The number of students with height more than or equal to 135 cm but less than 150 cm is 41.

(b) Which class interval has the least number of students?

We need to find the class interval with the minimum frequency.

The frequencies are 6, 9, 14, 18, 10, 7.

The minimum frequency is 6.

The class interval corresponding to the frequency 6 is 130 - 135.

The class interval with the least number of students is 130 – 135.

(c) What is the class size?

The class size is the difference between the upper limit and the lower limit of a class interval. Looking at any interval in the histogram (e.g., 130 - 135), the lower limit is 130 and the upper limit is 135.

Class size = Upper limit - Lower limit

Class size = $135 - 130 = 5$

We can verify this with other intervals: $140 - 135 = 5$, $145 - 140 = 5$, and so on.

The class size is 5.

(d) How many students have height less than 140 cm?

This includes the class intervals where the upper limit is 140 or less. These are 130 - 135 and 135 - 140.

Number of students = (Frequency of 130 - 135) + (Frequency of 135 - 140)

Number of students = $6 + 9 = 15$

The number of students with height less than 140 cm is 15.

Solution:

The given data represents the number of members in 25 families of a village. We need to prepare a frequency distribution table using the class intervals 0 –2, 2 –4, 4 –6, 6 –8, 8 –10.

The class intervals are inclusive of the lower limit and exclusive of the upper limit (e.g., 0–2 means members $\ge$ 0 and $<$ 2).

Let's count the frequency for each class interval:

• Class Interval 0 – 2 ($0 \le \text{members} < 2$): No families have 0 or 1 member. Frequency = 0.
• Class Interval 2 – 4 ($2 \le \text{members} < 4$): Families with 2 or 3 members. Numbers are 2, 2, 3, 3, 3, 3. Frequency = 6.
• Class Interval 4 – 6 ($4 \le \text{members} < 6$): Families with 4 or 5 members. Numbers are 4, 4, 5, 5, 5, 5. Frequency = 6.
• Class Interval 6 – 8 ($6 \le \text{members} < 8$): Families with 6 or 7 members. Numbers are 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7. Frequency = 11.
• Class Interval 8 – 10 ($8 \le \text{members} < 10$): Families with 8 members. Numbers are 8, 8. Frequency = 2.

Now, we can construct the frequency distribution table:

Solution:

To draw a histogram to represent the frequency distribution from Question 91, we will follow these steps:

Steps for Drawing the Histogram:

Step 1: Draw two perpendicular axes, the horizontal axis (x-axis) and the vertical axis (y-axis).

Step 2: Label the horizontal axis as "Number of Members" and the vertical axis as "Frequency (Number of Families)".

Step 3: Choose a suitable scale for both axes. For the vertical axis, the scale should accommodate the maximum frequency, which is 11. A scale of 1 unit = 1 family would be appropriate.

Step 4: Mark the class boundaries on the horizontal axis. The class intervals are 0–2, 2–4, 4–6, 6–8, and 8–10. Mark points at 0, 2, 4, 6, 8, and 10 on the x-axis.

Step 5: Draw rectangles corresponding to each class interval. The base of each rectangle will be the width of the class interval on the x-axis, and the height will be the frequency of that class on the y-axis.

• For the class interval 0–2, the base is from 0 to 2 on the x-axis, and the height is 0 (since the frequency is 0). No bar will be visible for this interval.
• For the class interval 2–4, the base is from 2 to 4 on the x-axis, and the height is 6. Draw a rectangle with these dimensions.
• For the class interval 4–6, the base is from 4 to 6 on the x-axis, and the height is 6. Draw a rectangle with these dimensions. This bar will be adjacent to the previous one.
• For the class interval 6–8, the base is from 6 to 8 on the x-axis, and the height is 11. Draw a rectangle with these dimensions. This bar will be adjacent to the previous one.
• For the class interval 8–10, the base is from 8 to 10 on the x-axis, and the height is 2. Draw a rectangle with these dimensions. This bar will be adjacent to the previous one.

Step 6: The resulting graph will be a histogram, with adjacent bars representing the frequency distribution.

Description of the Histogram:

The histogram will consist of four vertical bars:

• A bar extending from 2 to 4 on the x-axis with a height of 6.
• A bar extending from 4 to 6 on the x-axis with a height of 6.
• A bar extending from 6 to 8 on the x-axis with a height of 11.
• A bar extending from 8 to 10 on the x-axis with a height of 2.

The base of each bar is 2 units wide, corresponding to the class width.

Solution:

We are given the marks obtained by 30 students in a test. We need to prepare a frequency distribution table using class intervals of equal width, where one class interval is 4–8 (excluding 8 and including 4).

The given class interval is 4–8, which means it includes marks greater than or equal to 4 and less than 8 ($4 \le \text{Marks} < 8$).

The width of this class interval is $8 - 4 = 4$. Since the class intervals are of equal width, all intervals will have a width of 4.

The minimum mark in the data is 7 and the maximum mark is 20. We need to define class intervals that cover this range.

Starting with the given interval [4, 8), the subsequent intervals with a width of 4 are:

• [4, 8)
• [8, 12)
• [12, 16)
• [16, 20)
• [20, 24)

The interval [20, 24) is needed to include the maximum mark of 20.

Now, let's count the frequency of marks falling into each class interval from the given data:

• Class Interval 4–8 ($4 \le \text{Marks} < 8$): Marks are 7, 7. Frequency = 2.
• Class Interval 8–12 ($8 \le \text{Marks} < 12$): Marks are 8, 10, 11, 10, 11, 10, 11. Frequency = 7.
• Class Interval 12–16 ($12 \le \text{Marks} < 16$): Marks are 14, 15, 15, 14, 13, 15, 14, 13, 12, 14, 15, 13, 14. Frequency = 13.
• Class Interval 16–20 ($16 \le \text{Marks} < 20$): Marks are 16, 16, 18, 19, 16, 17. Frequency = 6.
• Class Interval 20–24 ($20 \le \text{Marks} < 24$): Marks are 20, 20. Frequency = 2.

The frequency distribution table is as follows:

Solution:

To draw a histogram from the frequency distribution table obtained in Question 93, we will follow these steps:

Frequency Distribution Table from Question 93:

Steps for Drawing the Histogram:

Step 1: Draw two perpendicular axes on graph paper. The horizontal axis represents the class intervals (Marks), and the vertical axis represents the frequency (Number of Students).

Step 2: Mark the class boundaries on the horizontal axis. These are 4, 8, 12, 16, 20, and 24. Since the first class interval starts at 4 (not 0), you may need a break (zig-zag line) on the x-axis near the origin if the scale starts from 0, or simply start marking from 4.

Step 3: Choose a suitable scale for the vertical axis to represent the frequencies. The maximum frequency is 13, so a scale of 1 unit = 1 student or 2 units = 1 student would be appropriate.

Step 4: Draw rectangles (bars) for each class interval. The width of each bar will correspond to the width of the class interval (which is 4 units for all intervals). The height of each bar will correspond to the frequency of that class interval.

• For the interval 4–8, draw a bar with a base from 4 to 8 on the x-axis and a height of 2 on the y-axis.
• For the interval 8–12, draw a bar with a base from 8 to 12 on the x-axis and a height of 7 on the y-axis. This bar will be adjacent to the first one.
• For the interval 12–16, draw a bar with a base from 12 to 16 on the x-axis and a height of 13 on the y-axis. This bar will be adjacent to the previous one.
• For the interval 16–20, draw a bar with a base from 16 to 20 on the x-axis and a height of 6 on the y-axis. This bar will be adjacent to the previous one.
• For the interval 20–24, draw a bar with a base from 20 to 24 on the x-axis and a height of 2 on the y-axis. This bar will be adjacent to the previous one.

Step 5: The resulting graph is a histogram. The bars are adjacent to show the continuous nature of the grouped data.

Description of the Histogram:

The histogram will show five adjacent rectangular bars. The bar for the class interval 12–16 marks will be the tallest, indicating that the highest number of students (13) scored marks in this range. The bars for 4–8 and 20–24 marks will be the shortest, representing the lowest frequencies (2 students each) in those ranges.

Solution:

We are given the weights (in kg) of 30 students. We need to prepare a frequency distribution table using class intervals of equal width, with one class interval being (30 – 35), which means $30 \le \text{Weight} < 35$.

The given class interval is [30, 35). The width of this interval is $35 - 30 = 5$. All class intervals will have a width of 5 and be of the form $[lower, upper)$.

The minimum weight in the data is 26 kg, and the maximum weight is 46 kg.

We define the class intervals starting from a value less than or equal to the minimum weight and ending at a value greater than or equal to the maximum weight, with a width of 5:

• [25, 30) i.e., 25 – 30
• [30, 35) i.e., 30 – 35
• [35, 40) i.e., 35 – 40
• [40, 45) i.e., 40 – 45
• [45, 50) i.e., 45 – 50

Now, we count the number of students whose weights fall into each class interval:

• Weights in [25, 30): 26, 29. Frequency = 2.
• Weights in [30, 35): 30, 31, 31, 32, 33, 33, 34. Frequency = 7.
• Weights in [35, 40): 35, 36, 36, 37, 37, 38, 38, 38, 39, 39, 39. Frequency = 11.
• Weights in [40, 45): 40, 41, 41, 42, 43, 43, 44. Frequency = 7.
• Weights in [45, 50): 45, 46, 46. Frequency = 3.

The frequency distribution table is as follows:

Now, we answer the sub-questions based on the frequency distribution table:

(i) Which class has the least frequency?

From the table, the minimum frequency is 2, which corresponds to the class interval 25 – 30.

Answer: The class 25 – 30 has the least frequency.

(ii) Which class has the maximum frequency?

From the table, the maximum frequency is 11, which corresponds to the class interval 35 – 40.

Answer: The class 35 – 40 has the maximum frequency.

Solution:

To construct a pie chart for the given data, we first need to find the central angle for each brand. The central angle for a sector in a pie chart is proportional to the frequency of the corresponding category.

The formula for the central angle is:

$\text{Central Angle} = \left(\frac{\text{Frequency of Category}}{\text{Total Frequency}}\right) \times 360^\circ$

1. Calculate the Total Frequency:

Total number of pairs of shoes sold = Sum of frequencies for all brands

Total Frequency = $130 + 120 + 90 + 40 + 20 = 400$

2. Calculate the Central Angle for Each Brand:

Brand A:

Central Angle for Brand A = $\left(\frac{130}{400}\right) \times 360^\circ = \frac{13}{40} \times 360^\circ = 13 \times \frac{360}{40}^\circ = 13 \times 9^\circ = 117^\circ$

Brand B:

Central Angle for Brand B = $\left(\frac{120}{400}\right) \times 360^\circ = \frac{12}{40} \times 360^\circ = \frac{12}{4} \times 36^\circ = 3 \times 36^\circ = 108^\circ$

Brand C:

Central Angle for Brand C = $\left(\frac{90}{400}\right) \times 360^\circ = \frac{9}{40} \times 360^\circ = 9 \times 9^\circ = 81^\circ$

Brand D:

Central Angle for Brand D = $\left(\frac{40}{400}\right) \times 360^\circ = \frac{1}{10} \times 360^\circ = 36^\circ$

Brand E:

Central Angle for Brand E = $\left(\frac{20}{400}\right) \times 360^\circ = \frac{1}{20} \times 360^\circ = 18^\circ$

Summary of Central Angles:

3. Construct the Pie Chart:

To construct the pie chart, draw a circle. Use a compass to draw the circle and a protractor to measure the central angles calculated above. Start with any radius and draw a line from the center to the circumference. Then, measure the angles sequentially from this line and draw radii to divide the circle into sectors corresponding to each brand. Label each sector with the brand name and/or its frequency/percentage.

The sectors will have the following angles:

• Brand A: $117^\circ$
• Brand B: $108^\circ$
• Brand C: $81^\circ$
• Brand D: $36^\circ$
• Brand E: $18^\circ$

Solution:

The pie chart shows the distribution of a state government's expenditure under different heads. The total spending is given as $\textsf{₹} 10$ crores.

From the pie chart, the percentage of expenditure on different heads are:

• Education: 25%
• Roads: 15%
• Public Welfare: 20%
• Administration: 10%
• Others: 30%

(i) If the total spending is 10 crores, how much money was spent on roads?

Total Spending = $\textsf{₹} 10$ crores = $\textsf{₹} 10,00,00,000$

Percentage spent on Roads = 15%

Amount spent on Roads = 15% of Total Spending

Amount spent on Roads = $\frac{15}{100} \times \textsf{₹} 10,00,00,000$

Amount spent on Roads = $\frac{15}{100} \times 10,00,00,000 \$$

Amount spent on Roads = $15 \times 10,00,000$

Amount spent on Roads = $\textsf{₹} 1,50,00,000$

Thus, the money spent on roads was $\textsf{₹} 1.5$ crores.

(ii) How many times is the amount of money spent on education compared to the amount spent on roads?

Percentage spent on Education = 25%

Percentage spent on Roads = 15%

To find how many times the amount spent on education is compared to roads, we can find the ratio of their percentages:

Ratio = $\frac{\text{Percentage on Education}}{\text{Percentage on Roads}}$

Ratio = $\frac{25\%}{15\%}$

Ratio = $\frac{25}{15}$

Ratio = $\frac{\cancel{25}^{5}}{\cancel{15}_{3}} = \frac{5}{3}$

Alternatively, using the amounts:

Amount spent on Education = 25% of $\textsf{₹} 10$ crores = $\textsf{₹} 2.5$ crores

Amount spent on Roads = 15% of $\textsf{₹} 10$ crores = $\textsf{₹} 1.5$ crores

Ratio = $\frac{\textsf{₹} 2.5 \text{ crores}}{\textsf{₹} 1.5 \text{ crores}} = \frac{2.5}{1.5} = \frac{25}{15} = \frac{5}{3}$

Thus, the amount of money spent on education is $\frac{5}{3}$ times the amount spent on roads.

(iii) What fraction of the total expenditure is spent on both roads and public welfare together?

Percentage spent on Roads = 15%

Percentage spent on Public Welfare = 20%

Combined percentage spent on Roads and Public Welfare = Percentage on Roads + Percentage on Public Welfare

Combined percentage = $15\% + 20\% = 35\%$

To find the fraction of the total expenditure, we express this percentage as a fraction of 100%:

Fraction = $\frac{\text{Combined percentage}}{\text{Total percentage}}$

Fraction = $\frac{35\%}{100\%}$

Fraction = $\frac{35}{100}$

Simplify the fraction:

Fraction = $\frac{\cancel{35}^{7}}{\cancel{100}_{20}} = \frac{7}{20}$

Thus, $\frac{7}{20}$ of the total expenditure is spent on both roads and public welfare together.

Solution:

To prepare a pie chart for the given data representing the number of different animals in a zoo, we need to calculate the central angle for the sector corresponding to each type of animal.

The central angle for each category is calculated using the formula:

$\text{Central Angle} = \left(\frac{\text{Number of Animals in Category}}{\text{Total Number of Animals}}\right) \times 360^\circ$

1. Calculate the Total Number of Animals:

Total Number of Animals = Sum of the number of animals of all types

Total Number of Animals = $42 \text{ (Deer)} + 15 \text{ (Elephant)} + 26 \text{ (Giraffe)} + 24 \text{ (Reptiles)} + 13 \text{ (Tiger)}$

Total Number of Animals = $120$

2. Calculate the Central Angle for Each Category:

Deer:

Central Angle for Deer = $\left(\frac{42}{120}\right) \times 360^\circ = \frac{42}{12} \times 36^\circ = 3.5 \times 36^\circ = 126^\circ$

Elephant:

Central Angle for Elephant = $\left(\frac{15}{120}\right) \times 360^\circ = \frac{1}{8} \times 360^\circ = 45^\circ$

Giraffe:

Central Angle for Giraffe = $\left(\frac{26}{120}\right) \times 360^\circ = \frac{26}{12} \times 36^\circ = \frac{13}{6} \times 36^\circ = 13 \times 6^\circ = 78^\circ$

Reptiles:

Central Angle for Reptiles = $\left(\frac{24}{120}\right) \times 360^\circ = \frac{1}{5} \times 360^\circ = 72^\circ$

Tiger:

Central Angle for Tiger = $\left(\frac{13}{120}\right) \times 360^\circ = \frac{13}{12} \times 36^\circ = 13 \times 3^\circ = 39^\circ$

Summary of Central Angles:

3. Construct the Pie Chart:

To construct the pie chart, draw a circle. Using a compass, draw a circle of any convenient radius. Then, use a protractor to draw the sectors from the center of the circle with the calculated central angles. Start with any radius as the initial line and measure the angles for each category sequentially. Label each sector with the name of the animal and/or its corresponding number or percentage.

The pie chart will be divided into five sectors with the following angles:

• Deer: $126^\circ$
• Elephant: $45^\circ$
• Giraffe: $78^\circ$
• Reptiles: $72^\circ$
• Tiger: $39^\circ$

Solution:

(a) Probability of Suhail winning with the initial set of cards:

The initial cards face down are King of Spades, Queen of Hearts, Jack of Clubs, 10 of Diamonds, 9 of Spades, and 8 of Hearts.

Total number of possible outcomes (Total cards) = 6

Suhail wins if he picks up a face card. Face cards are Queen, King, and Jack.

From the given cards, the face cards are King of Spades, Queen of Hearts, and Jack of Clubs.

Number of favorable outcomes for Suhail (picking a face card) = 3

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability of Suhail winning (picking a face card) = $\frac{\text{Number of face cards}}{\text{Total number of cards}}$

$P(\text{Suhail wins}) = \frac{3}{6}$

$P(\text{Suhail wins}) = \frac{\cancel{3}^{1}}{\cancel{6}_{2}} = \frac{1}{2}$

So, the probability of Suhail winning is $\frac{1}{2}$.

(b) Probability with the new set of cards:

Now, the following cards are added to the initial set: 4 of Hearts, 4 of Clubs, 4 of Spades, and 4 of Diamonds.

Initial number of cards = 6

Number of cards added = 4

Total number of possible outcomes (Total cards in the new set) = $6 + 4 = 10$

Probability of Suhail winning now:

Suhail wins if he picks up a face card. The face cards from the initial set (King of Spades, Queen of Hearts, Jack of Clubs) are still present. No new face cards were added.

Number of favorable outcomes for Suhail (picking a face card) = 3

Probability of Suhail winning = $\frac{\text{Number of face cards}}{\text{Total number of cards}}$

$P(\text{Suhail wins now}) = \frac{3}{10}$

So, the probability of Suhail winning now is $\frac{3}{10}$.

Probability of Reshma winning now:

Reshma wins if she picks up a 4. The added cards are 4 of Hearts, 4 of Clubs, 4 of Spades, and 4 of Diamonds. The initial set had no 4s.

Number of favorable outcomes for Reshma (picking a 4) = 4

Probability of Reshma winning = $\frac{\text{Number of 4s}}{\text{Total number of cards}}$

$P(\text{Reshma wins}) = \frac{4}{10}$

$P(\text{Reshma wins}) = \frac{\cancel{4}^{2}}{\cancel{10}_{5}} = \frac{2}{5}$

So, the probability of Reshma winning is $\frac{2}{5}$.

Solution:

We are given the weights (in grams) of 35 mangoes. We need to prepare a frequency distribution table using equal class intervals, with one class interval being 40 – 45 (45 not included).

This means the class interval is $[40, 45)$, including weights from 40 up to (but not including) 45.

The width of this class interval is $45 - 40 = 5$. All class intervals will have a width of 5 and be of the form $[lower, upper)$.

The minimum weight in the data is 30 grams, and the maximum weight is 74 grams.

We need to define class intervals with a width of 5 that cover the range from 30 to 74. We can start the first interval at 30 or slightly below it. Starting at 30 allows us to include the minimum value directly in the first interval [30, 35).

The class intervals will be:

• [30, 35) i.e., 30 – 35
• [35, 40) i.e., 35 – 40
• [40, 45) i.e., 40 – 45
• [45, 50) i.e., 45 – 50
• [50, 55) i.e., 50 – 55
• [55, 60) i.e., 55 – 60
• [60, 65) i.e., 60 – 65
• [65, 70) i.e., 65 – 70
• [70, 75) i.e., 70 – 75

The last interval [70, 75) is needed to include the maximum weight of 74 grams.

Now, we count the number of mangoes whose weights fall into each class interval from the given data:

• [30, 35): 30, 32, 34. Frequency = 3.
• [35, 40): 35, 37, 35. Frequency = 3.
• [40, 45): 40, 43, 42. Frequency = 3.
• [45, 50): 45, 45, 46. Frequency = 3.
• [50, 55): 50, 53, 52, 50, 53. Frequency = 5.
• [55, 60): 55, 55, 58, 59. Frequency = 4.
• [60, 65): 62, 61, 62, 60, 62. Frequency = 5.
• [65, 70): 65, 68. Frequency = 2.
• [70, 75): 70, 70, 70, 73, 74, 74, 70. Frequency = 7.

The frequency distribution table is as follows:

Now, we answer the sub-questions based on the frequency distribution table:

(a) How many classes are there in the frequency distribution table?

By counting the rows in the frequency distribution table (excluding the header and total rows), we find the number of classes.

Number of classes = 9

Answer: There are 9 classes in the frequency distribution table.

(b) Which weight group has the highest frequency?

From the table, we look for the class interval with the largest frequency. The frequencies are 3, 3, 3, 3, 5, 4, 5, 2, 7. The highest frequency is 7, which corresponds to the class interval 70 – 75.

Answer: The weight group 70 – 75 has the highest frequency.

Solution:

To complete the frequency distribution table, we need to count the tally marks in each row and write the corresponding frequency.

Recall that $\bcancel{||||}$ represents a group of 5.

• For the class 40 – 50, the tally marks are $\bcancel{||||} \ \bcancel{||||} \ ||$. This represents $5 + 5 + 2 = 12$.
• For the class 50 – 60, the tally marks are $\bcancel{||||} \ \bcancel{||||} \ ||||$. This represents $5 + 5 + 4 = 14$.
• For the class 60 – 70, the tally marks are $\bcancel{||||} \ |$. This represents $5 + 1 = 6$.
• For the class 70 – 80, the tally marks are $||$. This represents 2.
• For the class 80 – 90, the tally mark is $|$. This represents 1.

The completed frequency distribution table is:

To find the total number of persons, we sum the frequencies from all the classes:

Total Number of Persons = Sum of Frequencies

Total Number of Persons = $12 + 14 + 6 + 2 + 1$

Total Number of Persons = $26 + 6 + 2 + 1$

Total Number of Persons = $32 + 2 + 1$

Total Number of Persons = $34 + 1$

Total Number of Persons = $35$

Alternatively, we can perform the addition as follows:

The total number of persons is 35.

Solution:

To draw a histogram for the given frequency distribution table, we will represent the class intervals on the horizontal axis and the frequencies on the vertical axis. Since the class intervals are continuous and of equal width, the bars in the histogram will be adjacent.

Steps for Drawing the Histogram:

Step 1: Draw the horizontal axis (x-axis) and the vertical axis (y-axis) on a graph paper.

Step 2: Label the horizontal axis as "Class Interval" and the vertical axis as "Frequency".

Step 3: Choose a suitable scale for both axes. On the horizontal axis, mark points corresponding to the lower and upper limits of the class intervals: 10, 15, 20, 25, 30, 35, and 40. On the vertical axis, choose a scale that can accommodate the maximum frequency (which is 98). For example, you could use a scale where 1 unit represents 10 units of frequency, or a smaller unit if needed for precision.

Step 4: Draw rectangles corresponding to each class interval. The base of each rectangle will be the width of the class interval (which is 5 units) on the x-axis. The height of each rectangle will be equal to the frequency of the corresponding class interval as shown on the y-axis.

• For the class interval 10 – 15, draw a bar with base from 10 to 15 on the x-axis and height 30 on the y-axis.
• For the class interval 15 – 20, draw a bar with base from 15 to 20 on the x-axis and height 98 on the y-axis. This bar will be adjacent to the first one.
• For the class interval 20 – 25, draw a bar with base from 20 to 25 on the x-axis and height 80 on the y-axis. This bar will be adjacent to the previous one.
• For the class interval 25 – 30, draw a bar with base from 25 to 30 on the x-axis and height 58 on the y-axis. This bar will be adjacent to the previous one.
• For the class interval 30 – 35, draw a bar with base from 30 to 35 on the x-axis and height 29 on the y-axis. This bar will be adjacent to the previous one.
• For the class interval 35 – 40, draw a bar with base from 35 to 40 on the x-axis and height 50 on the y-axis. This bar will be adjacent to the previous one.

Step 5: The resulting diagram is the histogram representing the given frequency distribution.

Description of the Histogram:

The histogram will consist of six adjacent rectangular bars. The bar corresponding to the class interval 15 – 20 will be the tallest, indicating the highest frequency (98) in this range. The bar for the class interval 30 – 35 will be the shortest, representing the lowest frequency (29).

Solution:

We are given the amount of money (in thousands of rupees) with 20 people. We need to draw a histogram using equal class intervals, with one of the intervals being 50–100.

Assuming the class interval 50–100 includes 50 but excludes 100, the interval is $[50, 100)$. The width of this class interval is $100 - 50 = 50$. All class intervals will have a width of 50 and be of the form $[lower, upper)$.

The minimum value in the data is 98 thousand rupees, and the maximum value is 235 thousand rupees.

We define class intervals with a width of 50 that cover the range from 98 to 235, starting from 50 as given:

• [50, 100) i.e., 50 – 100
• [100, 150) i.e., 100 – 150
• [150, 200) i.e., 150 – 200
• [200, 250) i.e., 200 – 250

The last interval [200, 250) is needed to include the maximum value of 235.

Now, let's count the number of people whose amount of money falls into each class interval from the given data:

• [50, 100): 98. Frequency = 1.
• [100, 150): 114, 108, 100, 101, 109, 117, 119, 126, 131, 136, 143, 118. Frequency = 12.
• [150, 200): 156, 169, 182, 195. Frequency = 4.
• [200, 250): 207, 219, 235. Frequency = 3.

The frequency distribution table is as follows:

Steps for Drawing the Histogram:

Step 1: Draw two perpendicular axes, the horizontal axis (x-axis) and the vertical axis (y-axis).

Step 2: Label the horizontal axis as "Amount of Money (in thousands of $\textsf{₹}$)" and the vertical axis as "Frequency (Number of People)".

Step 3: Choose a suitable scale for both axes. For the horizontal axis, mark the class boundaries 50, 100, 150, 200, 250. Since the data starts at 98, you might consider a break near the origin or simply start the labeling from 50. For the vertical axis, the maximum frequency is 12, so a scale of 1 unit = 1 person or 2 units = 1 person would be appropriate.

Step 4: Draw rectangles for each class interval. The base of each rectangle corresponds to the class interval on the x-axis, and the height corresponds to the frequency on the y-axis. The bars should be adjacent.

• For the class 50–100, draw a bar from 50 to 100 on the x-axis with a height of 1.
• For the class 100–150, draw a bar from 100 to 150 on the x-axis with a height of 12.
• For the class 150–200, draw a bar from 150 to 200 on the x-axis with a height of 4.
• For the class 200–250, draw a bar from 200 to 250 on the x-axis with a height of 3.

Step 5: The resulting diagram is the histogram.

Description of the Histogram:

The histogram will consist of four adjacent rectangular bars. The tallest bar will be for the class interval 100–150, with a height of 12, representing the largest number of people in this income group. The shortest bar will be for the class interval 50–100, with a height of 1.

Solution:

The given image is a histogram showing the number of literate females in different age groups.

(a) Write the classes assuming all the classes are of equal width.

From the histogram, the class boundaries marked on the horizontal axis are 10, 15, 20, 25, 30, 35, and 40.

Assuming the class intervals are continuous and follow the format [lower limit, upper limit), the classes are formed by these boundaries.

The classes are:

• 10 – 15
• 15 – 20
• 20 – 25
• 25 – 30
• 30 – 35
• 35 – 40

Thus, the classes (age groups) are 10–15 years, 15–20 years, 20–25 years, 25–30 years, 30–35 years, and 35–40 years.

(b) What is the classes width?

The class width is the difference between the upper and lower limit of any class interval.

Class width = $15 - 10 = 5$

Class width = $20 - 15 = 5$

and so on.

Thus, the class width is 5 years.

(c) In which age group are literate females the least?

We look for the bar with the shortest height in the histogram, as height represents frequency (number of literate females).

• Age group 10–15: Frequency = 300
• Age group 15–20: Frequency = 980
• Age group 20–25: Frequency = 800
• Age group 25–30: Frequency = 580
• Age group 30–35: Frequency = 290
• Age group 35–40: Frequency = 500

The least frequency is 290, which corresponds to the age group 30 – 35 years.

Thus, literate females are the least in the age group 30 – 35 years.

(d) In which age group is the number of literate females the highest?

We look for the bar with the tallest height in the histogram.

The frequencies are 300, 980, 800, 580, 290, 500.

The highest frequency is 980, which corresponds to the age group 15 – 20 years.

Thus, the number of literate females is the highest in the age group 15 – 20 years.

Solution:

It appears that the histogram image for Question 105, which shows the frequency distribution of teaching experiences of 30 teachers, is missing from the input.

To answer the questions (a), (b), and (c) about the class width, maximum/least teaching experience groups, and the number of teachers with 10 to 20 years of experience, the visual data from the histogram is required.

Please provide the histogram image or the data represented by the histogram so that I can proceed with the solution.

Without the histogram, it is not possible to determine:

• The class boundaries and thus the class width.
• The frequencies for each class interval, which are needed to find the groups with the maximum and least number of teachers.
• The number of teachers in the 10 to 20 years experience group (assuming this is one of the class intervals).

Solution:

To draw a pie chart representing the number of branches of different banks, we need to calculate the central angle for each bank category. The central angle is proportional to the number of branches for that bank.

The formula for the central angle of a sector is:

$\text{Central Angle} = \left(\frac{\text{Number of Branches of the Bank}}{\text{Total Number of Branches}}\right) \times 360^\circ$

1. Calculate the Total Number of Branches:

Total Number of Branches = Sum of the number of branches of all banks

Total Branches = $30 \text{ (SBI)} + 17 \text{ (BoB)} + 15 \text{ (PNB)} + 10 \text{ (Canara Bank)}$

Total Branches = $72$

2. Calculate the Central Angle for Each Bank:

State Bank of India (SBI):

Central Angle for SBI = $\left(\frac{30}{72}\right) \times 360^\circ$

Central Angle for SBI = $\frac{\cancel{30}^{5}}{\cancel{72}_{12}} \times 360^\circ = \frac{5}{12} \times 360^\circ = 5 \times \frac{360}{12}^\circ = 5 \times 30^\circ = 150^\circ$

Bank of Baroda (BoB):

Central Angle for BoB = $\left(\frac{17}{72}\right) \times 360^\circ$

Central Angle for BoB = $17 \times \frac{360}{72}^\circ = 17 \times 5^\circ = 85^\circ$

Punjab National Bank (PNB):

Central Angle for PNB = $\left(\frac{15}{72}\right) \times 360^\circ$

Central Angle for PNB = $\frac{\cancel{15}^{5}}{\cancel{72}_{24}} \times 360^\circ = \frac{5}{24} \times 360^\circ = 5 \times \frac{360}{24}^\circ = 5 \times 15^\circ = 75^\circ$

Canara Bank:

Central Angle for Canara Bank = $\left(\frac{10}{72}\right) \times 360^\circ$

Central Angle for Canara Bank = $\frac{\cancel{10}^{5}}{\cancel{72}_{36}} \times 360^\circ = \frac{5}{36} \times 360^\circ = 5 \times \frac{360}{36}^\circ = 5 \times 10^\circ = 50^\circ$

Summary of Central Angles:

3. Construct the Pie Chart:

To construct the pie chart, draw a circle of any convenient radius. Use a protractor to divide the circle into sectors with the central angles calculated above. Start by drawing a radius, and then measure the angles sequentially from this radius. Label each sector with the name of the bank and/or the number of branches or percentage.

The pie chart will be divided into four sectors with the following angles:

• State Bank of India: $150^\circ$
• Bank of Baroda: $85^\circ$
• Punjab National Bank: $75^\circ$
• Canara Bank: $50^\circ$

Solution:

To draw a pie chart for the given data representing the expenditure on different item heads, we need to calculate the central angle for the sector corresponding to each item. The central angle is proportional to the amount spent on that item.

The formula for the central angle of a sector is:

$\text{Central Angle} = \left(\frac{\text{Amount for Item}}{\text{Total Amount}}\right) \times 360^\circ$

1. Calculate the Total Amount:

Total Amount = Sum of the amounts spent on all item heads

Total Amount = $\textsf{₹} (43.2 + 16.2 + 27.00 + 21.6)$ crore

Total Amount = $\textsf{₹} (59.4 + 27.0 + 21.6)$ crore

Total Amount = $\textsf{₹} (86.4 + 21.6)$ crore

Total Amount = $\textsf{₹} 108.0$ crore

2. Calculate the Central Angle for Each Item Head:

Road:

Central Angle for Road = $\left(\frac{43.2}{108.0}\right) \times 360^\circ = \left(\frac{432}{1080}\right) \times 360^\circ = \frac{\cancel{432}^{2}}{\cancel{1080}_{5}} \times 360^\circ = \frac{2}{5} \times 360^\circ = 2 \times 72^\circ = 144^\circ$

Electricity:

Central Angle for Electricity = $\left(\frac{16.2}{108.0}\right) \times 360^\circ = \left(\frac{162}{1080}\right) \times 360^\circ = \frac{\cancel{162}^{3}}{\cancel{1080}_{20}} \times 360^\circ = \frac{3}{20} \times 360^\circ = 3 \times 18^\circ = 54^\circ$

Drinking water:

Central Angle for Drinking water = $\left(\frac{27.00}{108.0}\right) \times 360^\circ = \left(\frac{27}{108}\right) \times 360^\circ = \frac{\cancel{27}^{1}}{\cancel{108}_{4}} \times 360^\circ = \frac{1}{4} \times 360^\circ = 90^\circ$

Sewerage:

Central Angle for Sewerage = $\left(\frac{21.6}{108.0}\right) \times 360^\circ = \left(\frac{216}{1080}\right) \times 360^\circ = \frac{\cancel{216}^{1}}{\cancel{1080}_{5}} \times 360^\circ = \frac{1}{5} \times 360^\circ = 72^\circ$

Summary of Central Angles:

3. Construct the Pie Chart:

To construct the pie chart, draw a circle of any convenient radius. Use a protractor to divide the circle into sectors with the central angles calculated above. Start by drawing a radius, and then measure the angles sequentially from this radius. Label each sector with the item head and/or the amount spent or the percentage (if calculated).

The pie chart will be divided into four sectors with the following angles:

• Road: $144^\circ$
• Electricity: $54^\circ$
• Drinking water: $90^\circ$
• Sewerage: $72^\circ$

Solution:

To draw a pie chart for the given data representing the periods allotted per week for different subjects, we need to calculate the central angle for the sector corresponding to each subject. The central angle is proportional to the number of periods allotted for that subject.

The formula for the central angle of a sector is:

$\text{Central Angle} = \left(\frac{\text{Number of Periods for Subject}}{\text{Total Number of Periods}}\right) \times 360^\circ$

1. Calculate the Total Number of Periods:

Total Number of Periods = Sum of the periods allotted to all subjects

Total Periods = $7 \text{ (Hindi)} + 8 \text{ (English)} + 8 \text{ (Maths)} + 8 \text{ (Science)} + 7 \text{ (Social Science)} + 4 \text{ (Computer)} + 3 \text{ (Sanskrit)}$

Total Periods = $7 + 8 + 8 + 8 + 7 + 4 + 3 = 45$

Total number of periods is 45.

2. Calculate the Central Angle for Each Subject:

Hindi:

Central Angle for Hindi = $\left(\frac{7}{45}\right) \times 360^\circ = 7 \times \frac{360}{45}^\circ = 7 \times 8^\circ = 56^\circ$

English:

Central Angle for English = $\left(\frac{8}{45}\right) \times 360^\circ = 8 \times 8^\circ = 64^\circ$

Maths:

Central Angle for Maths = $\left(\frac{8}{45}\right) \times 360^\circ = 8 \times 8^\circ = 64^\circ$

Science:

Central Angle for Science = $\left(\frac{8}{45}\right) \times 360^\circ = 8 \times 8^\circ = 64^\circ$

Social Science:

Central Angle for Social Science = $\left(\frac{7}{45}\right) \times 360^\circ = 7 \times 8^\circ = 56^\circ$

Computer:

Central Angle for Computer = $\left(\frac{4}{45}\right) \times 360^\circ = 4 \times 8^\circ = 32^\circ$

Sanskrit:

Central Angle for Sanskrit = $\left(\frac{3}{45}\right) \times 360^\circ = \frac{1}{15} \times 360^\circ = 24^\circ$

Summary of Central Angles:

3. Construct the Pie Chart:

To construct the pie chart, draw a circle using a compass. From the center of the circle, draw a radius. Use a protractor to measure the calculated central angles consecutively from this radius, drawing a new radius for each subject. Label each sector with the name of the subject and/or the number of periods or percentage.

The pie chart will be divided into seven sectors with the calculated angles corresponding to each subject.

Solution:

The given pie chart shows the percentage distribution of favourite beverages preferred by a group of young people.

From the pie chart, the percentages of people who like each beverage are:

• Tea: 30%
• Coffee: 25%
• Cold Drinks: 20%
• Juice: 25%

(i) Which type of beverage is liked by the maximum number of people.

To find the beverage liked by the maximum number of people, we look for the beverage with the highest percentage in the pie chart.

Comparing the percentages: 30% (Tea), 25% (Coffee), 20% (Cold Drinks), 25% (Juice).

The highest percentage is 30%, which corresponds to Tea.

Thus, the beverage liked by the maximum number of people is Tea.

(ii) If 45 people like tea, how many people were surveyed?

We know that 30% of the surveyed people like tea.

Let the total number of people surveyed be $N$.

According to the problem, 30% of $N$ is equal to 45.

$30\% \text{ of } N = 45$

We can write this as:

To find $N$, we can rearrange the equation:

So, the total number of people surveyed was 150.

Answer: 150 people were surveyed.

Solution:

To prepare a pie chart for the given data representing the approximate percentage of water in various oceans, we need to calculate the central angle for the sector corresponding to each ocean. The central angle is proportional to the percentage of water in that ocean.

The total percentage is 100%, which corresponds to a full circle of $360^\circ$.

The formula for the central angle of a sector based on percentage is:

$\text{Central Angle} = \left(\frac{\text{Percentage}}{100}\right) \times 360^\circ$

The given data is:

• Pacific: 40%
• Atlantic: 30%
• Indian: 20%
• Others: 10%

Calculate the Central Angle for Each Ocean:

Pacific:

Central Angle for Pacific = $\left(\frac{40}{100}\right) \times 360^\circ = 0.4 \times 360^\circ = 144^\circ$

Atlantic:

Central Angle for Atlantic = $\left(\frac{30}{100}\right) \times 360^\circ = 0.3 \times 360^\circ = 108^\circ$

Indian:

Central Angle for Indian = $\left(\frac{20}{100}\right) \times 360^\circ = 0.2 \times 360^\circ = 72^\circ$

Others:

Central Angle for Others = $\left(\frac{10}{100}\right) \times 360^\circ = 0.1 \times 360^\circ = 36^\circ$

Summary of Central Angles:

Construct the Pie Chart:

To construct the pie chart, draw a circle using a compass. From the center of the circle, draw a radius. Use a protractor to measure the calculated central angles consecutively from this radius, drawing a new radius for each ocean category. Label each sector with the name of the ocean and/or its percentage.

The pie chart will be divided into four sectors with the following angles:

• Pacific: $144^\circ$
• Atlantic: $108^\circ$
• Indian: $72^\circ$
• Others: $36^\circ$

Solution:

The wheel is divided into 8 sectors. We need to count the number of times each outcome appears.

• Ball: Appears in 3 sectors.
• Toy Car: Appears in 3 sectors.
• Chocolate: Appears in 2 sectors.

The total number of possible outcomes when the wheel is spun is the total number of sectors, which is 8.

Total number of possible outcomes = 8

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

(a) getting a ball

The number of favorable outcomes (getting a ball) is the number of sectors with "Ball", which is 3.

Number of favorable outcomes = 3

Probability of getting a ball = $\frac{\text{Number of sectors with Ball}}{\text{Total number of sectors}}$

$P(\text{getting a ball}) = \frac{3}{8}$

(b) getting a toy car

The number of favorable outcomes (getting a toy car) is the number of sectors with "Toy Car", which is 3.

Number of favorable outcomes = 3

Probability of getting a toy car = $\frac{\text{Number of sectors with Toy Car}}{\text{Total number of sectors}}$

$P(\text{getting a toy car}) = \frac{3}{8}$

(c) any toy except a chocolate

"Any toy except a chocolate" includes getting a Ball or a Toy Car.

Number of favorable outcomes = (Number of sectors with Ball) + (Number of sectors with Toy Car)

Number of favorable outcomes = $3 + 3 = 6$

Probability of getting any toy except a chocolate = $\frac{\text{Number of sectors with Ball or Toy Car}}{\text{Total number of sectors}}$

$P(\text{any toy except a chocolate}) = \frac{6}{8}$

We can simplify the fraction:

$P(\text{any toy except a chocolate}) = \frac{\cancel{6}^{3}}{\cancel{8}_{4}} = \frac{3}{4}$

Alternate Method for (c):

Calculate the probability of the complementary event: getting a chocolate.

Number of sectors with Chocolate = 2

Probability of getting a chocolate = $\frac{2}{8} = \frac{1}{4}$

The probability of getting any toy except a chocolate is 1 minus the probability of getting a chocolate.

$P(\text{any toy except a chocolate}) = 1 - P(\text{getting a chocolate})$

$P(\text{any toy except a chocolate}) = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$

Solution:

The given cards are:

• R: 3, 5, 7, 9 (4 cards)
• Y: 2, 4, 6, 8 (4 cards)
• B: 1, 3, 5, 7, 9 (5 cards)
• G: 0, 2, 4, 6, 8 (5 cards)

Total number of possible outcomes = Total number of cards = $4 + 4 + 5 + 5 = 18$.

The probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

(a) Calculate the probability of getting an odd number.

The cards with odd numbers are:

• From R: R 3, R 5, R 7, R 9 (4 cards)
• From Y: None (all even)
• From B: B 1, B 3, B 5, B 7, B 9 (5 cards)
• From G: None (0 and all even)

Number of favorable outcomes (cards with an odd number) = $4 + 5 = 9$.

Probability of getting an odd number = $\frac{\text{Number of cards with an odd number}}{\text{Total number of cards}}$

$P(\text{getting an odd number}) = \frac{9}{18}$

$P(\text{getting an odd number}) = \frac{\cancel{9}^{1}}{\cancel{18}_{2}} = \frac{1}{2}$

(b) Calculate the probability of getting a Y card.

The Y cards are: Y 2, Y 4, Y 6, Y 8.

Number of favorable outcomes (getting a Y card) = 4.

Probability of getting a Y card = $\frac{\text{Number of Y cards}}{\text{Total number of cards}}$

$P(\text{getting a Y card}) = \frac{4}{18}$

$P(\text{getting a Y card}) = \frac{\cancel{4}^{2}}{\cancel{18}_{9}} = \frac{2}{9}$

(c) Calculate the probability of getting a G card.

The G cards are: G 0, G 2, G 4, G 6, G 8.

Number of favorable outcomes (getting a G card) = 5.

Probability of getting a G card = $\frac{\text{Number of G cards}}{\text{Total number of cards}}$

$P(\text{getting a G card}) = \frac{5}{18}$

(d) Calculate the probability of B card bearing number > 7.

We are looking for a card that is from the 'B' set and has a number greater than 7.

The B cards are: B 1, B 3, B 5, B 7, B 9.

Among these cards, the number greater than 7 is 9.

The card is B 9.

Number of favorable outcomes (B card bearing number > 7) = 1 (only B 9).

Probability of getting a B card bearing number > 7 = $\frac{\text{Number of B cards with number > 7}}{\text{Total number of cards}}$

$P(\text{B card with number > 7}) = \frac{1}{18}$

Solution:

The question asks to identify the symbols that appear in each sector of the spinner shown in the image for Question 113.

Looking at the provided image of the spinner, the symbols in the sectors are clearly depicted.

Starting from the top sector and moving in a clockwise direction, the symbols in the 8 sectors are:

1. Circle ($\circ$)

2. Star ($\star$)

3. Square ($\square$)

4. Triangle ($\triangle$)

5. Circle ($\circ$)

6. Star ($\star$)

7. Square ($\square$)

8. Triangle ($\triangle$)

The symbols appearing in the sectors for Question 113 are Circle, Star, Square, Triangle, Circle, Star, Square, and Triangle in order around the spinner.

Please provide the image for Question 114 if you need the symbols for that spinner as well.

Solution:

The question asks to identify the symbols that appear in each sector of the spinner shown in the image for Question 114.

Looking at the provided image of the spinner, the symbols in the sectors are clearly depicted.

The spinner is divided into 6 sectors.

Starting from the top-right sector (marked '4') and moving in a clockwise direction, the symbols or numbers in the sectors are:

1. The number 4

2. The number 5

3. The number 6

4. The number 1

5. The number 2

6. The number 3

The symbols/numbers appearing in the sectors for Question 114 are 4, 5, 6, 1, 2, and 3 in order around the spinner.

Solution:

The total monthly income of the client is $\textsf{₹} 50,000$. The pie chart shows the percentage of income that should be budgeted for different categories.

To find the amount to be spent in each category, we need to calculate the corresponding percentage of the total income.

The percentage for each category from the pie chart is:

• Savings: 10%
• Food: 25%
• Rent: 30%
• Utilities: 15%
• Transportation: 10%
• Miscellaneous: 10%

Now, we calculate the amount for each category:

Savings:

Amount for Savings = 10% of $\textsf{₹} 50,000$

Amount for Savings = $\frac{10}{100} \times 50,000 \$$

Amount for Savings = $0.10 \times 50,000 \$$

Amount for Savings = $\textsf{₹} 5,000$

Food:

Amount for Food = 25% of $\textsf{₹} 50,000$

Amount for Food = $\frac{25}{100} \times 50,000 \$$

Amount for Food = $0.25 \times 50,000 \$$

Amount for Food = $\textsf{₹} 12,500$

Rent:

Amount for Rent = 30% of $\textsf{₹} 50,000$

Amount for Rent = $\frac{30}{100} \times 50,000 \$$

Amount for Rent = $0.30 \times 50,000 \$$

Amount for Rent = $\textsf{₹} 15,000$

Utilities:

Amount for Utilities = 15% of $\textsf{₹} 50,000$

Amount for Utilities = $\frac{15}{100} \times 50,000 \$$

Amount for Utilities = $0.15 \times 50,000 \$$

Amount for Utilities = $\textsf{₹} 7,500$

Transportation:

Amount for Transportation = 10% of $\textsf{₹} 50,000$

Amount for Transportation = $\frac{10}{100} \times 50,000 \$$

Amount for Transportation = $0.10 \times 50,000 \$$

Amount for Transportation = $\textsf{₹} 5,000$

Miscellaneous:

Amount for Miscellaneous = 10% of $\textsf{₹} 50,000$

Amount for Miscellaneous = $\frac{10}{100} \times 50,000 \$$

Amount for Miscellaneous = $0.10 \times 50,000 \$$

Amount for Miscellaneous = $\textsf{₹} 5,000$

The amount the client should spend in each category is summarized below:

Solution:

The pie chart shows the percentage of the total advertising amount spent on various modes of advertising for a product. We will use the percentages given in the pie chart to answer the questions.

1. Which type of media advertising is the greatest amount of the total?

We look for the category with the largest percentage in the pie chart.

• Television: 40%
• Newspaper: 20%
• Magazine: 15%
• Radio: 10%
• Direct Mail: 5%
• Miscellaneous: 10%

Comparing the percentages, the largest percentage is 40%, which corresponds to Television.

Answer: Television advertising is the greatest amount of the total.

2. Which type of media advertising is the least amount of the total?

We look for the category with the smallest percentage in the pie chart.

Comparing the percentages, the smallest percentage is 5%, which corresponds to Direct Mail.

Answer: Direct Mail advertising is the least amount of the total.

3. What per cent of the total advertising amount is spent on direct mail campaigns?

The percentage for Direct Mail is directly given in the pie chart.

Answer: 5% of the total advertising amount is spent on direct mail campaigns.

4. What per cent of the advertising amount is spent on newspaper and magazine advertisements?

To find the combined percentage spent on newspaper and magazine advertisements, we add their individual percentages:

Percentage on Newspaper = 20%

Percentage on Magazine = 15%

Combined Percentage = Percentage on Newspaper + Percentage on Magazine

Combined Percentage = $20\% + 15\% = 35\%$

Answer: 35% of the advertising amount is spent on newspaper and magazine advertisements.

5. What media types do you think are included in miscellaneous? Why aren’t those media types given their own category?

The "Miscellaneous" category typically includes various advertising methods that individually represent a small portion of the total expenditure or are diverse in nature. Examples of media types that could be included in miscellaneous are:

• Online advertising (e.g., social media ads, search engine marketing, banner ads)
• Outdoor advertising (e.g., billboards, posters)
• Promotional events or sponsorships
• Print materials other than newspapers and magazines (e.g., brochures, flyers)
• Cinema advertising
• Radio advertising (if it's not already a separate category, but it is here)
• Other niche advertising platforms

These media types are likely not given their own separate categories for a few reasons:

• Individual Small Percentages: Each of these might represent a very small percentage of the total budget, making it impractical to give each its own slice in the pie chart. Too many small slices make the chart cluttered and difficult to read.
• Simplification: Grouping smaller or less significant expenditures into a "Miscellaneous" category simplifies the pie chart and highlights the major areas of spending more clearly.
• Diversity: The expenditures might be spread across many different small advertising avenues, making it difficult to list them all exhaustively. The "Miscellaneous" category serves as a catch-all for these varied expenses.

In essence, the "Miscellaneous" category helps to provide a complete picture of the total expenditure without overcomplicating the visual representation.