Chapter 2 Data Handling (Class 8 - Maths NCERT Exemplar Solutions)

Given:

Data: 9, 8, 4, 3, 2, 1, 6, 4, 8, 10, 12, 15, 4, 3.

To Find:

The range of the given data.

Solution:

The range of a data set is defined as the difference between the maximum value and the minimum value in the set.

Calculation:

The correct option is (b).

Given:

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

To Find:

Frequency of the class interval 20 – 25.

Solution:

In a grouped frequency distribution, the class interval 20 – 25 includes all observations that are greater than or equal to 20 and less than 25.

We scan the data for values $x$ such that $20 \leq x < 25$:

The values are: 20, 21, 23.

Note: The value 25 is not counted in this interval; it belongs to the next class interval 25 – 30.

The correct option is (c).

To Find:

The type of programme viewed the most based on the pie chart.

Solution:

Observing the sectors in the pie chart and their respective percentages:

1. Entertainment: $50\%$

2. Sports: $25\%$

3. News: $15\%$

4. Informative: $10\%$

The highest percentage corresponds to the most-viewed type of programme.

Therefore, Entertainment programmes are viewed the most.

The correct option is (c).

To Find:

Total number of girls with height $\geq 145 \text{ cm}$.

Solution:

From the histogram, we sum the frequencies of all class intervals starting from 145 cm onwards:

1. Height interval 145 – 150: 5 girls

2. Height interval 150 – 160: 4 girls

3. Height interval 160 – 170: 1 girl

Total number of girls = Sum of frequencies in these intervals

The number of girls having height 145 cm and above is 10.

The correct option is (b).

To Find:

The number of possible outcomes for the sum of numbers when a dice is thrown twice.

Solution:

When a dice is thrown twice, the smallest possible sum occurs when both dice show 1:

The largest possible sum occurs when both dice show 6:

The possible sums are all integers between the minimum and maximum values: $\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

Let us count the elements in this set:

The number of possible outcomes for the sum is 11.

The correct option is (b).

To Find:

Probability of getting a multiple of 2 on a single dice roll.

Solution:

Total possible outcomes on a dice = $\{1, 2, 3, 4, 5, 6\}$

Favourable outcomes (multiples of 2) = $\{2, 4, 6\}$

Probability is calculated as:

The correct option is (c).

Solution:

The class width (size) is the difference between the upper limit and the lower limit.

The intervals follow a sequence based on this width:

1st class interval = 10 – 15

2nd class interval = 15 – 20

3rd class interval = 20 – 25

4th class interval = 25 – 30

Therefore, the blank should be filled with 25 – 30.

Solution:

In any class interval expressed as $a - b$, the smaller value $a$ is the lower class limit and the larger value $b$ is the upper class limit.

For the interval 250 – 275, 250 is the lower limit (or lower class limit).

Therefore, the blank should be filled with lower limit.

Solution:

By definition in statistics, the count of how many times a specific value or observation appears in a data set is referred to as its frequency.

Therefore, the blank should be filled with frequency.

Solution:

The total angle at the centre of a circle is $360^\circ$. In a pie chart, each category is represented by a sector whose central angle is calculated as:

Since the component value is always a portion of the total value, the angle will always be a fraction of $360^\circ$.

The statement is True (T).

Solution:

Total outcomes on a dice = $\{1, 2, 3, 4, 5, 6\}$ (Total = 6)

Favourable outcomes (Odd numbers) = $\{1, 3, 5\}$ (Total = 3)

The statement is True (T).

Solution:

A pie chart uses sectors of a circle to represent data, whereas a pictograph uses pictures or symbols to represent data. They are different methods of data representation.

The statement is False (F).

Given: Weekly wages of 30 workers.

To Find: A frequency distribution table using tally marks with class intervals of size 10.

Solution:

We group the given data into class intervals. Note that in an interval like 800 – 810, the upper limit 810 is excluded and counted in the next interval 810 – 820.

Given:

Total marks obtained = $540$

Total central angle = $360^\circ$

Solution:

(i) Subject with 105 marks:

To find the subject, we calculate the central angle corresponding to 105 marks.

From the pie chart, the subject with a central angle of $70^\circ$ is Hindi.

(ii) Difference between Mathematics and Hindi:

Central angle for Mathematics = $90^\circ$

Central angle for Hindi = $70^\circ$

Now, convert this difference into marks:

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

Given:

The distribution of people's favourite food is provided as follows:

North Indian = $30$

South Indian = $40$

Chinese = $25$

Others = $25$

To Find:

Construct a pie chart representing the given data.

Solution:

To draw a pie chart, we first need to calculate the total number of people and then determine the central angle for each category.

Step 1: Calculate the Total Frequency

$\text{Total number of people} = 30 + 40 + 25 + 25$

$\text{Total} = 120$

Step 2: Calculate Central Angles

The central angle for each sector is calculated using the formula:

$\text{Central Angle} = \left( \frac{\text{Number of people}}{\text{Total number of people}} \right) \times 360^\circ$

Step 3: Steps to Draw the Pie Chart

1. Draw a circle with any convenient radius.

2. Draw a horizontal radius as the starting line.

3. Using a protractor, draw a sector of $90^\circ$ for North Indian food.

4. From the new radius, draw an angle of $120^\circ$ for South Indian food.

5. Continue by drawing an angle of $75^\circ$ for Chinese food.

6. The remaining sector will automatically be $75^\circ$, representing Others.

The resulting pie chart visually represents the proportion of people preferring different cuisines, where South Indian food occupies the largest sector.

Given:

The frequency distribution of weekly wages for 30 workers as organized in the previous step:

To Find:

(i) Class interval with maximum workers.

(ii) Number of workers earning $\textsf{₹} 850$ and more.

(iii) Number of workers earning less than $\textsf{₹} 850$.

(iv) Number of workers earning $\textsf{₹} 820$ or more but less than $\textsf{₹} 880$.

Solution:

(i) Class interval with the maximum number of workers:

By observing the frequency column, the highest frequency is 9.

Therefore, the class interval 830 – 840 has the maximum number of workers.

(ii) Number of workers earning $\textsf{₹} 850$ and more:

We sum the frequencies of all class intervals from 850 – 860 up to 890 – 900.

There are 10 workers who earn $\textsf{₹} 850$ and more.

(iii) Number of workers earning less than $\textsf{₹} 850$:

We sum the frequencies of all class intervals from 800 – 810 up to 840 – 850.

There are 20 workers who earn less than $\textsf{₹} 850$.

(iv) Number of workers earning $\textsf{₹} 820$ or more but less than $\textsf{₹} 880$:

We sum the frequencies for the intervals 820 – 830, 830 – 840, 840 – 850, 850 – 860, 860 – 870, and 870 – 880.

There are 20 workers who earn $\textsf{₹} 820$ or more but less than $\textsf{₹} 880$.

Graphical Representation (Histogram):

The following histogram represents the distribution of these wages:

Given:

A frequency distribution table with class intervals and their corresponding frequencies for 25 observations.

Solution:

(i) Class interval which has the lowest frequency:

From the table, the frequencies are 1, 5, 5, 4, 0, 8, and 2. The lowest frequency is 0.

The class interval with frequency 0 is 65 – 75.

(ii) Class interval which has the highest frequency:

The highest frequency in the given data is 8.

The class interval with frequency 8 is 75 – 85.

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

Class size is calculated as the difference between the upper limit and the lower limit of a class interval.

The class size is 10.

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

Let us list the classes 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?

The last class interval in the table is 85 – 95.

The lower limit of the class 85 – 95 is 85.

Given:

Total number of people surveyed = $600$

Percentage of people injured while lifting = $49\%$

Percentage of people injured while bending = $18\%$

To Find:

The total number of people who injured their back while either bending or lifting.

Solution:

First, we find the combined percentage of people who were injured due to bending and lifting.

Now, we calculate the number of people representing $67\%$ of the total survey population ($600$):

Therefore, 402 people injured their back while either bending or lifting.

In a histogram, the data is represented using rectangular bars. The base of each rectangle represents the class interval (width), and the height of the rectangle represents the frequency (number of observations) of that specific class interval.

The correct option is (d).

A pie chart (also known as a circle graph) is used to represent data where the entire circle represents the "whole" and each sector represents a "part" of that whole. The size of each sector is proportional to the information it represents.

The correct option is (a).

In a pie chart, the data is distributed within a circle. Since the sum of the angles around the centre of a circle is always a complete turn, the total angle at the centre is $360^\circ$.

The correct option is (b).

Given:

Data set: 30, 61, 55, 56, 60, 20, 26, 46, 28, 56

To Find:

The range of the data.

Solution:

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

The correct option is (c).

A random experiment is an experiment where the result cannot be predicted with certainty. Tossing a coin, rolling a dice, and choosing a card are random because multiple outcomes are possible. However, throwing a stone from a roof is not a random experiment because the outcome (the stone falling down due to gravity) is certain and predictable.

The correct option is (d).

To Find:

Probability of choosing a vowel from the English alphabets.

Solution:

Total number of alphabets in English = 26

Number of vowels (a, e, i, o, u) = 5

The probability is calculated as:

The correct option is (b).

Given:

3 out of 5 students participate in a competition.

To Find:

Probability of students who do not make it to the competition.

Solution:

Total students = 5

Students who participate = 3

Students who do not participate = $5 - 3 = 2$

Converting the fraction into decimal form:

The correct option is (b).

Given:

A pie chart showing percentages of votes for different colours.

To Find:

Which colour received $1/5$ of the votes.

Solution:

First, convert the fraction $1/5$ into a percentage:

The colour Green has exactly $20\%$ of the votes.

The correct option is (c).

Given:

Total students = 400

Percentage for 'Others' = $6\%$

To Find:

Number of students who voted for 'Others'.

Solution:

The correct option is (c).

To Find:

The most reasonable conclusion from the given pie chart.

Solution:

Let us check the options based on the chart percentages:

Red: $35\%$, Blue: $25\%$, Green: $20\%$, Yellow: $14\%$, Others: $6\%$

(a) Blue is $25\%$, which is $1/4$ of the total, not $1/20$. (Incorrect)

(b) 'Others' ($6\%$) is less popular than Green ($20\%$). (Incorrect)

(c) Red is $35\%$ and Yellow is $14\%$. Since $14 \times 2 = 28$, Red is not two times Yellow. (Incorrect)

(d) Combined percentage for Yellow and Green:

Since the percentage for Red is $35\%$, the sum of Yellow and Green ($34\%$) is approximately the same as Red ($35\%$).

The correct option is (d).

Given:

Temperatures ($in$ $^\circ\text{C}$): $-6, -8, 0, 3, 2, 0, 1, 5, 4, 4$

To Find:

The range of the given data.

Solution:

First, we identify the maximum and minimum values from the given data set.

We know that the Range of a data set is the difference between the maximum and the minimum observation.

$\text{Range} = \text{Maximum value} - \text{Minimum value}$

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

$\text{Range} = 5 + 8$

$\text{Range} = 13^\circ\text{C}$

Hence, 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:

Probability that the button is blue.

Solution:

First, we calculate the total number of buttons (Total outcomes):

$\text{Total buttons} = 4 + 7 + 3 + 6 = 20$

Number of favorable outcomes (blue buttons) = $4$

The formula for probability is:

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

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

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

Hence, the correct option is (c).

Given:

Condition for winning:

1. Rahul wins if the spinner lands on Red.

2. Varun wins if the spinner lands on Blue.

3. Yash wins if the spinner lands on Green.

To Find:

The spinner that makes the game fair.

Solution:

A game is considered fair if every player has an equal probability of winning. Since there are three players, each player must have a probability of $\frac{1}{3}$ to win for the game to be fair.

Let us analyze the sectors in each spinner provided in the figure:

Analysis of Spinner (i):

The wheel is divided into unequal sectors. The area for Red and Green is larger than the area for Blue. Since the sectors are not equal, the probabilities are not equal. Thus, it is not a fair spinner.

Analysis of Spinner (ii):

The wheel is divided into 4 equal quadrants. By counting the labels:

Since Yash (Green) has a higher chance of winning, the game is not fair.

Analysis of Spinner (iii):

The wheel is divided into 6 equal sectors. By counting the labels:

Number of Red sectors = $1$

Number of Green sectors = $2$

Number of Blue sectors = $3$

The probabilities are $\frac{1}{6}$, $\frac{2}{6}$, and $\frac{3}{6}$ respectively. Since they are different, the game is not fair.

Analysis of Spinner (iv):

The wheel is divided into 6 equal sectors. Looking closely at the figure, the labels are distributed as follows:

Number of Red sectors (R) = $2$

Number of Green sectors (G) = $2$

Number of Blue sectors (B) = $2$

Now, let's calculate the winning probability for each player:

Since the probability of winning is exactly the same for Rahul, Varun, and Yash, spinner (iv) makes the game fair.

Hence, the correct option is (d).

Given:

Class intervals: $0 - 10, 10 - 20, \dots$

Class size = $10$

To Find:

The lower limit of the fourth class.

Solution:

Let's list the class intervals sequentially:

1st Class interval: $0 - 10$

2nd Class interval: $10 - 20$

3rd Class interval: $20 - 30$

4th Class interval: $30 - 40$

In a class interval $a - b$, '$a$' is called the lower limit and '$b$' is called the upper limit.

For the fourth class ($30 - 40$), the lower limit is $30$.

Hence, the correct option is (d).

Given:

Total number of trials = $200$

Number of times head appeared = $120$

To Find:

Probability of getting a head.

Solution:

The probability is calculated as:

$P(\text{Head}) = \frac{\text{Number of heads}}{\text{Total number of tosses}}$

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

Simplifying the fraction:

$P(\text{Head}) = \frac{\cancel{120}^{3}}{\cancel{200}_{5}}$

Hence, the correct option is (b).

Given:

Percentage of buyers interested = $40\%$

To Find:

Central angle of the sector.

Solution:

We know that the total angle at the center of a pie chart (circle) is $360^\circ$, which represents $100\%$.

The formula for the central angle is:

$\text{Central Angle} = \frac{\text{Percentage value}}{100} \times 360^\circ$

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

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

$\text{Central Angle} = 4 \times 36^\circ$

$\text{Central Angle} = 144^\circ$

Hence, the correct option is (c).

Given:

Total monthly salary = $\textsf{₹} 15,000$

Central angle of the sector = $60^\circ$

To Find:

Amount spent on food and house rent.

Solution:

The amount for a specific sector can be calculated using the following formula:

$\text{Amount} = \frac{\text{Central Angle}}{360^\circ} \times \text{Total Amount}$

$\text{Amount} = \frac{60^\circ}{360^\circ} \times 15000$

$\text{Amount} = \frac{1}{6} \times 15000$

$\text{Amount} = \textsf{₹} 2,500$

Hence, the correct option is (b).

Given:

From the pie chart:

Distribution of Water = $70\%$

Distribution of Protein = $16\%$

Distribution of other dry elements = $14\%$

To Find:

The central angle for "protein and other constituents" combined.

Solution:

First, we find the total percentage of protein and other constituents:

$\text{Total percentage} = 16\% + 14\% = 30\%$

Now, we calculate the central angle for this combined $30\%$:

$\text{Central Angle} = \frac{\text{Percentage}}{100} \times 360^\circ$

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

$\text{Central Angle} = 3 \times 36^\circ$

$\text{Central Angle} = 108^\circ$

Hence, the correct option is (a).

Given:

The numbers on the 5 cards are: $4, 1, 2, 3, 2$

To Find:

The probability of picking a card with the number $2$.

Solution:

Total number of outcomes (total cards) = $5$

Number of favorable outcomes (cards with number 2) = $2$

The formula for probability is:

$P(\text{picking number 2}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$P(\text{picking number 2}) = \frac{2}{5}$

Hence, the correct option is (a).

Given:

From the pie chart:

Distribution in Muscles = $\frac{1}{3}$

Distribution in Bones = $\frac{1}{6}$

To Find:

The ratio of proteins in muscles to proteins in bones.

Solution:

$\text{Required Ratio} = \text{Muscles} : \text{Bones}$

$\text{Ratio} = \frac{1}{3} : \frac{1}{6}$

To simplify the ratio, we can write it as a fraction:

$\text{Ratio} = \frac{\frac{1}{3}}{\frac{1}{6}}$

$\text{Ratio} = \frac{1}{3} \times \frac{6}{1}$

$\text{Ratio} = \frac{6}{3}$

$\text{Ratio} = \frac{2}{1} = 2 : 1$

Hence, the correct option is (d).

Given:

Distribution in Skin = $\frac{1}{10}$

Distribution in Bones = $\frac{1}{6}$

To Find:

The central angle representing skin and bones together.

Solution:

First, we find the combined fraction of skin and bones:

$\text{Combined fraction} = \frac{1}{10} + \frac{1}{6}$

Taking the LCM of $10$ and $6$, which is $30$:

$\text{Combined fraction} = \frac{1 \times 3 + 1 \times 5}{30}$

$\text{Combined fraction} = \frac{3 + 5}{30} = \frac{8}{30} = \frac{4}{15}$

Now, we calculate the central angle:

$\text{Central Angle} = \text{Fraction} \times 360^\circ$

$\text{Central Angle} = \frac{4}{15} \times 360^\circ$

$\text{Central Angle} = 4 \times 24^\circ$

$\text{Central Angle} = 96^\circ$

Hence, the correct option is (d).

Given:

Muscles fraction = $\frac{1}{3}$

Skin fraction = $\frac{1}{10}$

Bones fraction = $\frac{1}{6}$

To Find:

Central angle for hormones, enzymes, and other proteins.

Solution:

First, we find the fraction for the "Hormones, Enzymes and other proteins" sector by subtracting the sum of other sectors from $1$.

$\text{Sum of given sectors} = \frac{1}{3} + \frac{1}{10} + \frac{1}{6}$

Taking the LCM of $3, 10, \text{ and } 6$, which is $30$:

$\text{Sum} = \frac{10 + 3 + 5}{30} = \frac{18}{30} = \frac{3}{5}$

$\text{Fraction for Hormones/Enzymes} = 1 - \frac{3}{5} = \frac{2}{5}$

Now, calculate the central angle:

$\text{Central Angle} = \frac{2}{5} \times 360^\circ$

$\text{Central Angle} = 2 \times 72^\circ$

$\text{Central Angle} = 144^\circ$

Hence, the correct option is (b).

Given:

Total number of tosses = $12$

Observing the image, the outcomes are:

Top Row: Tails, Heads, Tails, Tails, Heads, Tails

Bottom Row: Tails, Heads, Tails, Tails, Heads, Heads

To Find:

The chance (probability) of occurrence of Head.

Solution:

By counting the "Heads" in the given image:

Number of Heads ($H$) = $5$

Total number of outcomes = $12$

The probability is given by:

$P(\text{Head}) = \frac{\text{Number of Heads}}{\text{Total number of outcomes}}$

$P(\text{Head}) = \frac{5}{12}$

Hence, the correct option is (b).

Given:

Number of red balls = $3$

Number of black balls = $5$

Number of blue balls = $4$

To Find:

Total number of outcomes.

Solution:

The total number of outcomes is simply the sum of all the possible items that can be drawn from the bag.

$\text{Total outcomes} = \text{Red balls} + \text{Black balls} + \text{Blue balls}$

$\text{Total outcomes} = 3 + 5 + 4$

$\text{Total outcomes} = 12$

Hence, the correct option is (d).

Solution:

A Double Bar Graph is used to display two sets of data on the same graph, allowing for easy comparison between them (e.g., comparing marks of two students in different subjects or comparing sales of two years).

Hence, the correct option is (d).

Given:

Class interval = $150 - 175$

To Find:

Class size.

Solution:

The Class Size is defined as the difference between the upper class limit and the lower class limit.

$\text{Class Size} = \text{Upper Limit} - \text{Lower Limit}$

$\text{Class Size} = 175 - 150$

$\text{Class Size} = 25$

Hence, the correct option is (c).

Given:

A standard dice is thrown.

To Find:

Probability of getting the number $7$.

Solution:

A standard six-sided dice has the following possible outcomes:

$\text{Sample Space} = \{1, 2, 3, 4, 5, 6\}$

Total number of outcomes = $6$

Since the number $7$ is not present on a standard dice, it is an impossible event.

Number of favorable outcomes = $0$

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

Hence, the correct option is (d).

Solution:

A Pie Chart (also known as a circle graph) represents data by dividing a circle into various sectors, where each sector's area is proportional to the quantity it represents.

Hence, the correct option is (d).

Solution:

Tally marks are a quick way of keeping track of numbers in groups of five. In statistics, they are used to count the number of times a particular value occurs in a data set, which is known as the frequency.

For example, to represent a frequency of 5, we use tally marks as: $\bcancel{||||}$

Hence, the correct option is (c).

Given:

Class interval = $75 - 85$

To Find:

The upper limit of the given class interval.

Solution:

In any class interval expressed as $a - b$, the first value '$a$' is the lower limit and the second value '$b$' is the upper limit.

For the class interval $75 - 85$:

Hence, the correct option is (d).

Given:

Numbers on the slips: $1, 2, 3, 4, 5$

To Find:

Probability that the slip bears an odd number.

Solution:

First, we identify the total outcomes and favorable outcomes.

Total number of outcomes = $5$ $[$as there are slips numbered $1, 2, 3, 4, 5]$

The odd numbers in the set are $1, 3, \text{ and } 5$.

Number of favorable outcomes (odd numbers) = $3$

The formula for probability is:

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

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

Hence, the correct option is (c).

Given:

Number of red marbles = $6$

Number of green marbles = $5$

Number of blue marbles = $4$

Number of yellow marbles = $5$

To Find:

Probability that the chosen marble is red.

Solution:

First, we calculate the total number of marbles in the jar:

$\text{Total marbles} = 6 + 5 + 4 + 5 = 20$

Number of favorable outcomes (red marbles) = $6$

The probability of picking a red marble is:

$P(\text{Red}) = \frac{\text{Number of red marbles}}{\text{Total number of marbles}}$

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

Hence, the correct option is (b).

Given:

A coin is tossed $2$ times.

To Find:

The number of possible outcomes.

Solution:

When a coin is tossed, there are $2$ possible outcomes: Head ($H$) or Tail ($T$).

When it is tossed two times, the sample space is:

$S = \{(H, H), (H, T), (T, H), (T, T)\}$

The number of outcomes is $2 \times 2 = 4$.

Hence, the correct option is (d).

Given:

A coin is tossed $3$ times.

To Find:

The number of possible outcomes.

Solution:

The number of outcomes for tossing a coin '$n$' times is given by the formula $2^n$.

Here, $n = 3$.

$\text{Number of outcomes} = 2^3$

$\text{Number of outcomes} = 2 \times 2 \times 2 = 8$

The outcomes are: $HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$.

Hence, the correct option is (d).

Given:

A dice is tossed $2$ times.

To Find:

The number of possible outcomes.

Solution:

A single dice has $6$ possible outcomes $\{1, 2, 3, 4, 5, 6\}$.

When a dice is tossed '$n$' times, the number of outcomes is given by $6^n$.

Here, $n = 2$.

$\text{Total outcomes} = 6^2 = 6 \times 6 = 36$

Hence, the correct option is (c).

Solution:

Data that is collected in its original form without any processing or classification is known as raw data.

Answer: raw

Solution:

In the class interval $20 - 30$:

The value on the left ($20$) is the lower class limit.

The value on the right ($30$) is the upper class limit.

Answer: 20

Solution:

In the class interval $26 - 33$, the higher value represents the boundary at the top of the class.

Answer: upper class limit

Given:

Data set: $6, 8, 16, 22, 8, 20, 7, 25$

To Find:

The range of the data.

Solution:

First, we identify the maximum and minimum values from the given data set.

The Range is calculated as the difference between the maximum and minimum observations.

$\text{Range} = \text{Maximum value} - \text{Minimum value}$

$\text{Range} = 25 - 6$

$\text{Range} = 19$

Answer: 19

Solution:

A pie chart (or circle graph) represents data by dividing a circle into sectors. Each sector represents a part or component of the data, showing its relationship to the whole total.

Answer: parts (or sectors)

Solution:

When a coin is tossed, it has two faces. One face is termed as Head and the other is Tail. These are the only two possible outcomes in a single toss.

Answer: Head or Tail

Solution:

A standard dice is a cube with faces numbered from $1$ to $6$. When it is rolled, the possible outcomes that can appear on the top face are the numbers $1, 2, 3, 4, 5, \text{ and } 6$.

Answer: 1, 2, 3, 4, 5, 6

Solution:

In probability, an event is defined as a specific outcome or a set of outcomes of a random experiment. For example, getting an even number on a dice roll is an event consisting of outcomes $\{2, 4, 6\}$.

Answer: event

Solution:

An experiment where we know all possible results but cannot determine the specific result before the experiment occurs is known as a random experiment.

Answer: random

Solution:

In statistics, for a class interval $a - b$, the calculation $b - a$ determines the width or size of the class.

Answer: class size (or width)

Given:

1st class interval = $10 - 15$

2nd class interval = $15 - 20$

To Find:

The 6th class interval.

Solution:

First, we find the class size:

$\text{Class size} = 15 - 10 = 5$

Now, let's list the intervals sequentially:

1st interval: $10 - 15$

2nd interval: $15 - 20$

3rd interval: $20 - 25$

4th interval: $25 - 30$

5th interval: $30 - 35$

6th interval: $35 - 40$

Answer: 35 – 40

Given:

From the provided histogram, the frequencies (number of people) for different class intervals are:

$0 - 20$ books: $8$ people

$20 - 40$ books: $14$ people

$40 - 60$ books: $5$ people

$60 - 80$ books: $6$ people

$80 - 100$ books: $2$ people

To Find:

Total number of people surveyed.

Solution:

$\text{Total number of people} = 8 + 14 + 5 + 6 + 2$

$\text{Total number of people} = 35$

Answer: 35

Given:

From the histogram, the groups representing more than $60$ books are $60 - 80$ and $80 - 100$.

Solution:

Number of people in the $60 - 80$ range = $6$

Number of people in the $80 - 100$ range = $2$

$\text{Total people (more than 60)} = 6 + 2 = 8$

Answer: 8

Given:

The groups representing less than $40$ books are $0 - 20$ and $20 - 40$.

Solution:

Number of people in the $0 - 20$ range = $8$

Number of people in the $20 - 40$ range = $14$

$\text{Total people (less than 40)} = 8 + 14 = 22$

Answer: 22

Given:

The range "more than 20 and less than 40" corresponds directly to the class interval $20 - 40$.

Solution:

By observing the height of the bar for the class interval $20 - 40$ on the y-axis (Number of people):

The frequency is $14$.

Answer: 14

Solution:

In statistics, the count of how many times a specific value or observation appears in a data set is known as its frequency.

For example, if the number $5$ appears $3$ times in a list, its frequency is $3$.

Answer: frequency

Solution:

When dealing with a vast amount of data, it is practical to condense it into groups to make it readable. These groups are known as class intervals.

Answer: class intervals

Solution:

When a single coin is tossed, there are only two possible results: it can land showing either a Head or a Tail.

$\text{Total number of outcomes} = 2$

Answer: 2

Given:

Class interval = $80 - 85$

To Find:

The class size.

Solution:

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

$\text{Class size} = \text{Upper limit} - \text{Lower limit}$

$\text{Class size} = 85 - 80$

$\text{Class size} = 5$

Answer: 5

Solution:

In a histogram, rectangles (or bars) are constructed such that their bases represent the class intervals and their heights represent the frequencies. Unlike a bar graph, there are no gaps between these rectangles because the data is continuous.

Answer: rectangles

Solution:

In a fair dice, every face has an identical chance of landing face up. Therefore, each outcome ($1, 2, 3, 4, 5, \text{ or } 6$) has a probability of $\frac{1}{6}$. Such outcomes are termed equally likely.

Answer: likely

Solution:

While drawing a histogram on a graph paper:

1. The horizontal axis ($x\text{-axis}$) is used to represent the class intervals.

2. The vertical axis ($y\text{-axis}$) is used to represent the frequencies.

Answer: horizontal (or x) and vertical (or y)

Solution:

By convention in a continuous frequency distribution, an observation that is exactly equal to a limit is included in the class where it is the lower limit.

In the interval $10 - 20$, the value $20$ is the upper limit and is excluded.

In the interval $20 - 30$, the value $20$ is the lower limit and is included.

Answer: 20 – 30

Solution:

A pie chart (also known as a circle graph) represents data by dividing the entire area of a circle into various sectors. Each sector is proportional to the fraction of the total it represents.

Answer: True

Solution:

The total angle at the center of a circle is $360^\circ$. A single sector can represent any portion of the data. For example, if a category represents $75\%$ of the total data, its central angle would be:

$\text{Angle} = \frac{75}{100} \times 360^\circ = 270^\circ$

Since $270^\circ > 180^\circ$, the statement is incorrect.

Answer: False

Solution:

A pie chart represents data within a circle. Since the total angle around the center of any circle is $360^\circ$, the sum of all individual sectors' central angles must equal the total angle of the circle.

Answer: True (T)

Solution:

In a pie chart, if there are exactly two categories of data and each represents $50\%$ of the total, then each sector will have a central angle calculated as:

$\text{Central Angle} = \frac{1}{2} \times 360^\circ = 180^\circ$

In this case, the sum $180^\circ + 180^\circ = 360^\circ$, which is valid for a circle.

Answer: True (T)

Solution:

If two or more observations or categories in a data set have the same frequency or numerical value, their corresponding sectors in a pie chart will have equal central angles. For example, if two items both represent $20\%$ of the data, both will have a central angle of $72^\circ$.

Answer: True (T)

Solution:

In probability, an event is a collection of outcomes of an experiment. When a die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. The prime numbers in this set are $\{2, 3, 5\}$. Since this is a collection of specific outcomes, it constitutes an event.

Answer: True (T)

Given:

Full marks = $10$

Solution:

From the provided frequency table, we look at the column where marks obtained is $10$. The frequency corresponding to $10$ marks is $9$. This means exactly $9$ students obtained full marks.

Answer: True (T)

Given:

Marks less than 8 are: $4, 5, \text{ and } 7$.

Solution:

We need to add the frequencies of marks $4, 5, \text{ and } 7$:

Frequency of 4 marks = $5$

Frequency of 5 marks = $10$

Frequency of 7 marks = $8$

$\text{Total frequency} = 5 + 10 + 8$

$\text{Total frequency} = 23$

Since $23 \neq 29$, the statement is false.

Answer: False (F)

Given:

Marks more than 8 are: $9 \text{ and } 10$.

Solution:

We add the frequencies of marks $9 \text{ and } 10$:

Frequency of 9 marks = $12$

Frequency of 10 marks = $9$

$\text{Total frequency} = 12 + 9$

$\text{Total frequency} = 21$

The calculation matches the statement.

Answer: True (T)

Solution:

Let's compare the frequencies from the table:

Frequencies are: $5, 10, 8, 6, 12, 9$

The highest frequency in this data set is $12$.

The mark corresponding to the highest frequency ($12$) is $9$ marks, not $10$ marks.

Answer: False (F)

Given:

5th Class: $60 - 65$

4th Class: $55 - 60$

Solution:

The class size is $65 - 60 = 5$. We can find previous classes by subtracting $5$ from the limits:

5th Class: $60 - 65$

4th Class: $55 - 60$

3rd Class: $50 - 55$

2nd Class: $45 - 50$

1st Class: $40 - 45$

The statement says the first class is $45 - 50$, but according to the sequence, the first class is $40 - 45$.

Answer: False (F)

Given:

From the histogram, "above the age of 20" includes the class intervals $20-30$, $30-40$, and $40-50$.

Solution:

Observing the heights of the bars (frequencies):

Literate males in age $20-30$ = $600$

Literate males in age $30-40$ = $800$

Literate males in age $40-50$ = $500$

The total is $1900$. The statement claims it is $1500$.

Answer: False (F)

Given:

Class interval = $60 - 68$

Solution:

$\text{Class size} = \text{Upper limit} - \text{Lower limit}$

$\text{Class size} = 68 - 60$

$\text{Class size} = 8$

The calculation matches the statement.

Answer: True (T)

Solution:

When a single coin is tossed, the outcomes are Head ($H$) and Tail ($T$), which is $2^1 = 2$.

When a pair of coins is tossed simultaneously, the possible outcomes (Sample Space) are:

$S = \{(H, H), (H, T), (T, H), (T, T)\}$

The total number of outcomes is $2^2 = 4$. The statement says the number of outcomes is $2$.

Answer: False (F)

Given:

A standard dice is thrown once.

To Find:

Probability of getting an even number.

Solution:

The total possible outcomes on a dice are $\{1, 2, 3, 4, 5, 6\}$.

Even numbers in the set are $\{2, 4, 6\}$.

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

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

The statement is correct.

Answer: True (T)

Given:

A standard dice is thrown once.

To Find:

Probability of getting a composite number.

Solution:

The outcomes are $\{1, 2, 3, 4, 5, 6\}$.

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

In the set $\{1, 2, 3, 4, 5, 6\}$:

1 is neither prime nor composite.

2, 3, 5 are prime numbers.

4 and 6 are composite numbers.

$\text{Probability} = \frac{2}{6} = \frac{1}{3}$

Since $\frac{1}{3} \neq \frac{1}{2}$, the statement is incorrect.

Answer: False (F)

Solution:

In a pie chart, the quantity of data is directly proportional to the area or central angle of the sector. The smallest sector represents the least production.

By observing the provided image:

State D and State A have the largest sectors.

State B is smaller than A and D.

State C has the visibly smallest central angle compared to all other states.

Therefore, the production is least in State C, not State B.

Answer: False (F)

Solution:

In probability theory, an event is defined as a subset of the sample space. This subset can contain a single outcome (elementary event) or multiple outcomes (compound event).

Answer: True (T)

Solution:

On a fair dice, all six numbers $\{1, 2, 3, 4, 5, 6\}$ are equally likely to occur.

The probability of getting any single number on a dice is always $\frac{1}{6}$, not $\frac{1}{n}$ where $n$ is the number itself.

Answer: False (F)

Given:

Dice outcomes: $\{1, 2, 3, 4, 5, 6\}$

Solution:

Prime numbers on a dice: $\{2, 3, 5\}$

Composite numbers on a dice: $\{4, 6\}$

Since $\frac{1}{2} \neq \frac{1}{3}$, the probabilities are not the same.

Answer: False (F)

Given:

Dice outcomes: $\{1, 2, 3, 4, 5, 6\}$

Solution:

Even numbers: $\{2, 4, 6\}$

Odd numbers: $\{1, 3, 5\}$

Both probabilities are equal to $\frac{1}{2}$.

Answer: True (T)

Solution:

A random experiment is an experiment where the outcome cannot be predicted with certainty before it is performed. For example, tossing a coin or rolling a dice.

Verifying Pythagoras theorem is a mathematical calculation based on a fixed law ($a^2 + b^2 = c^2$). Since the result is always predictable and fixed for any right-angled triangle, it is not a random experiment.

Answer: False (F)

Solution:

A histogram is a graphical representation of a frequency distribution for continuous data (like height, weight, or marks in intervals). The $y\text{-axis}$ in a histogram must represent the frequency (number of times an observation occurs).

In the given figure, the $x\text{-axis}$ represents discrete Years and the $y\text{-axis}$ represents the Production of steel (a quantity), not the frequency of occurrences. This is a Bar Graph where the bars happen to be placed adjacent to each other.

Answer: False (F)

(a) Solution:

The second class interval is $20 - 30$.

(b) Solution:

The last class interval is $50 - 60$.

(c) Solution:

The third class interval is $30 - 40$. Looking at the frequency column:

(d) Solution:

From the table, the frequency '$10$' corresponds to the class interval:

(e) Solution:

Comparing all frequencies ($5, 10, 4, 15, 12$), the lowest frequency is $4$. This corresponds to the interval:

(f) Solution:

Class size is the difference between the upper limit and the lower limit of any interval.

$\text{Class size} = 20 - 10$

Solution:

To draw the histogram, we follow these steps:

1. Represent Maximum Speed ($in$ km/h) on the horizontal axis ($x\text{-axis}$). Since the data starts from $10$, we can use a "kink" or "zig-zag" line near the origin if needed, but here we can start directly from $0$ with equal intervals.

2. Represent Frequency on the vertical axis ($y\text{-axis}$).

3. Draw rectangles with width equal to the class interval ($10$) and height equal to the corresponding frequency.

4. Note that for the interval $50 - 60$, the frequency is $0$, so there will be no bar (a gap in heights) at that position.

In the graph above:

The $x\text{-axis}$ scale is $1$ unit = $10$ km/h.

The $y\text{-axis}$ scale is $1$ unit = $2$ animals.

Given:

Total number of children = $350$

From the pie chart, the percentages of children spending time are:

$\bullet$ Less than 1 hr = $6\%$

$\bullet$ 1 hr = $16\%$

$\bullet$ 2 hrs = $30\%$

$\bullet$ 3 hrs = $34\%$

$\bullet$ 4 hrs = $10\%$

$\bullet$ 5 hrs = $4\%$

(a) To find the number of children who spend at least one hour:

"At least one hour" means 1 hour or more. This includes the categories: 1 hr, 2 hrs, 3 hrs, 4 hrs, and 5 hrs.

$\text{Total percentage} = 16\% + 30\% + 34\% + 10\% + 4\% = 94\%$

Alternatively, $\text{Total percentage} = 100\% - 6\% (\text{less than 1 hr}) = 94\%$

$\text{Number of children} = \frac{94}{100} \times 350 = 329$

Ans: 329 children

(b) To find the number of children who spend more than 2 hours:

"More than 2 hours" includes the categories: 3 hrs, 4 hrs, and 5 hrs.

$\text{Total percentage} = 34\% + 10\% + 4\% = 48\%$

$\text{Number of children} = \frac{48}{100} \times 350 = 168$

Ans: 168 children

(c) To find the number of children who spend 3 or lesser hours:

"3 or lesser hours" includes the categories: Less than 1 hr, 1 hr, 2 hrs, and 3 hrs.

$\text{Total percentage} = 6\% + 16\% + 30\% + 34\% = 86\%$

$\text{Number of children} = \frac{86}{100} \times 350 = 301$

Ans: 301 children

(d) Comparison of groups:

Group 1: Children who spend 2 hours or more

Categories included: 2 hrs, 3 hrs, 4 hrs, 5 hrs.

$\text{Percentage} = 30\% + 34\% + 10\% + 4\% = 78\%$

Group 2: Children who play for less than one hour

Category included: Less than 1 hr.

$\text{Percentage} = 6\%$

Comparing the two groups: $78\% > 6\%$ ($273\text{ children vs } 21\text{ children}$).

Ans: The number of children who spend 2 hours or more per day is greater.

Given:

Central angles from the chart:

$\text{Bus} = 120^\circ$

$\text{Car} = 90^\circ$

$\text{Cycle} = 60^\circ$

$\text{Walk} = 60^\circ$

Solution (a):

The most common mode is represented by the largest central angle.

$\text{Bus} (120^\circ)$ has the largest angle. Hence, Bus is the most common mode.

Solution (b):

$\text{Fraction} = \frac{\text{Central Angle of Car}}{360^\circ}$

$\text{Fraction} = \frac{\cancel{90}^1}{\cancel{360}_4} = \frac{1}{4}$

Solution (c):

Let the total number of children be $x$.

$\text{Children by car} = \text{Fraction} \times x$

$18 = \frac{1}{4} \times x$

$x = 18 \times 4 = 72$

So, 72 children took part in the survey.

Solution (d):

First, find the central angle for Taxi:

$\text{Taxi angle} = 360^\circ - (120^\circ + 90^\circ + 60^\circ + 60^\circ)$

$\text{Taxi angle} = 360^\circ - 330^\circ = 30^\circ$

$\text{Number of children using taxi} = \frac{30^\circ}{360^\circ} \times 72$

$\text{Number} = \frac{1}{12} \times 72 = 6$ children.

Solution (e):

Equal numbers travel by modes with the same central angle.

Cycle ($60^\circ$) and Walk ($60^\circ$) have equal angles.

Given:

Total possible outcomes on a dice: $S = \{1, 2, 3, 4, 5, 6\}$

Total number of outcomes = $6$

Solution (a):

Odd numbers = $\{1, 3, 5\}$; Favorable outcomes = $3$

$P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}$

Solution (b):

Numbers greater than 5 = $\{6\}$; Favorable outcomes = $1$

$P(\text{Greater than 5}) = \frac{1}{6}$

Solution (c):

Multiples of 3 = $\{3, 6\}$; Favorable outcomes = $2$

$P(\text{Multiple of 3}) = \frac{2}{6} = \frac{1}{3}$

Solution (d):

Numbers less than 1 = $\emptyset$; Favorable outcomes = $0$

$P(\text{Less than 1}) = \frac{0}{6} = 0$

Solution (e):

Factors of 36 in the set = $\{1, 2, 3, 4, 6\}$; Favorable outcomes = $5$

$P(\text{Factor of 36}) = \frac{5}{6}$

Solution (f):

Factors of 6 in the set = $\{1, 2, 3, 6\}$; Favorable outcomes = $4$

$P(\text{Factor of 6}) = \frac{4}{6} = \frac{2}{3}$

Solution:

Given:

Total marks scored by the student = $540$

Total central angle of the pie chart = $360^\circ$

From the pie chart, the central angles for the subjects are:

$\bullet$ Mathematics = $120^\circ$

$\bullet$ English = $90^\circ$

$\bullet$ Hindi = $60^\circ$

$\bullet$ Science = $70^\circ$

$\bullet$ Social Science = $20^\circ$

To Find:

Marks obtained by the student in each subject.

Solution:

To find the marks obtained in each subject, we use the formula:

$\text{Marks obtained} = \frac{\text{Central angle of the subject}}{360^\circ} \times \text{Total marks}$

First, let's find the marks for $1^\circ$ of the central angle:

$\text{Marks per degree} = \frac{540}{360} = 1.5\text{ marks}$

Now, we calculate the marks for each subject:

1. Mathematics:

$\text{Marks} = 120^\circ \times 1.5 = 180$

2. English:

$\text{Marks} = 90^\circ \times 1.5 = 135$

3. Hindi:

$\text{Marks} = 60^\circ \times 1.5 = 90$

4. Science:

$\text{Marks} = 70^\circ \times 1.5 = 105$

5. Social Science:

$\text{Marks} = 20^\circ \times 1.5 = 30$

Summary of Marks:

Given:

Total number of balls in the bag = $36$

Probability of choosing a white ball, $P(\text{white}) = \frac{2}{9}$

To Find:

The number of yellow balls in the bag.

Solution:

We know that the probability of an event is given by:

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

Let the number of white balls be $x$.

$\frac{2}{9} = \frac{x}{36}$

$x = \frac{2}{\cancel{9}} \times \cancel{36}^4$

$x = 2 \times 4 = 8$

So, the number of white balls is $8$.

Now, to find the number of yellow balls, we subtract the number of white balls from the total number of balls.

$\text{Number of yellow balls} = \text{Total balls} - \text{White balls}$

$\text{Number of yellow balls} = 36 - 8$

Ans: 28 yellow balls

Solution:

(a) To find students with height $\geq 135$ cm but $< 150$ cm, we add the frequencies of the class intervals $135 - 140$, $140 - 145$, and $145 - 150$.

Frequency of ($135 - 140$) = $14$

Frequency of ($140 - 145$) = $18$

Frequency of ($145 - 150$) = $10$

$\text{Total students} = 14 + 18 + 10 = 42$

(b) By observing the histogram, the bar with the lowest height represents the least number of students.

The class interval $150 - 155$ has the least frequency of $4$ students.

(c) Class size is the difference between the upper limit and lower limit of any class interval.

$\text{Class size} = 130 - 125 = 5$ cm.

(d) To find students with height less than $140$ cm, we add frequencies of class intervals $125 - 130$, $130 - 135$, and $135 - 140$.

$\text{Total students} = 6 + 8 + 14 = 28$ students.

Given:

Data of 25 families: $6, 8, 7, 7, 6, 5, 3, 2, 5, 6, 8, 7, 7, 4, $$ 3, 6, 6, 6, 7, $$ 5, 4, 3, 3, 2, 5$

Solution:

We will organize the data into class intervals of size 2. Note: In the interval $2 - 4$, 2 is included but 4 is excluded.

Solution:

To draw the histogram, we take the Class Intervals (Number of Members) on the $x$-axis and the Frequency (Number of Families) on the $y$-axis.

The rectangles are drawn with base as class intervals and heights equal to the respective frequencies. Since the class intervals are continuous ($0-2, 2-4, \dots$), there are no gaps between the bars.

Given:

The marks obtained by 30 students are: $14, 16, 15, 11, 15, 14, $$ 13, 16, 8, $$ 10, 7, 11, 18, 15, 14, 19, 20, 7, 10, 13, 12, 14, 15, 13, 16, 17, 14, 11, 10, 20$

One class interval is 4 – 8, where 4 is included and 8 is excluded.

Solution:

The width of the class interval is $8 - 4 = 4$.

Based on this width, the class intervals will be $4-8, 8-12, 12-16, 16-20, \text{ and } 20-24$ (to include the marks 20).

Following the continuous group frequency distribution method, the upper limit is excluded and the lower limit is included in each interval.

Solution:

To represent the marks of students, we plot the Marks on the $x$-axis and the Number of Students on the $y$-axis.

Each bar's height corresponds to the frequency of students in that mark interval. A kink (zig-zag line) is used on the $x$-axis between $0$ and $4$ because the intervals start from $4$.

Given:

Weights of 30 students (in kg): $39, 38, 36, 38, 40, 42, 43, 44, 33, 33, $$ 31, 45, 46, 38, $$ 37, 31, 30, 39, 41, $$ 41, 46, 36, 35, 34, 39, 43, 32, 37, 29, 26$.

Solution:

The minimum weight is $26$ kg and the maximum weight is $46$ kg. We will prepare the frequency distribution table with class intervals of width $5$, starting from $25-30$.

(i) The class having the least frequency is 25 - 30 (frequency = $2$).

(ii) The class having the maximum frequency is 35 - 40 (frequency = $11$).

To Find:

Construct a pie chart representing the sale of different brands of shoes.

Solution:

First, we calculate the total number of pairs sold:

$\text{Total} = 130 + 120 + 90 + 40 + 20 = 400$

Now, we find the central angle for each brand using the formula:

$\text{Central Angle} = \frac{\text{Component Value}}{\text{Total Value}} \times 360^\circ$

The pie chart is constructed using the central angles calculated above.

Given:

Total expenditure = $\textsf{₹} 10$ crores

Percentage distribution: Public Welfare = $20\%$, Road = $10\%$, Education = $25\%$, Others = $45\%$.

Solution:

(i) Money spent on roads = $10\%$ of total spending

$\text{Amount} = \frac{10}{100} \times 10 = 1$ crore.

Ans: $\textsf{₹} 1$ crore

(ii) Percentage spent on Education = $25\%$

Percentage spent on Roads = $10\%$

$\text{Comparison} = \frac{25\%}{10\%} = 2.5$ times.

Ans: 2.5 times

(iii) Total percentage for roads and public welfare = $10\% + 20\% = 30\%$

$\text{Fraction} = \frac{30}{100} = \frac{3}{10}$

Ans: $\frac{3}{10}$

Solution:

First, find the total number of animals:

$\text{Total} = 42 + 15 + 26 + 24 + 13 = 120$

Now, calculate the central angle for each animal:

Use these angles to draw the pie chart sectors.

Solution (a):

From the image, the cards are: $7, 7, 4, 4, 4, \text{Queen (Q)}, \text{Ace (A)}$.

Total number of cards = $7$

Number of face cards (Q) = $1$

$\text{Probability (Suhail wins)} = \frac{\text{No. of face cards}}{\text{Total cards}} = \frac{1}{7}$

Solution (b):

Additional cards: $\text{King (K)}, \text{King (K)}, 4, 3, 3, 3, \text{Jack (J)}, \text{Ace (A)}$. (Total 8 cards added).

Total cards now = $7 + 8 = 15$

Face cards now: Q (from first set), K, K, J (from second set). Total face cards = $4$.

$\text{Probability (Suhail wins)} = \frac{4}{15}$

Now, to find Reshma's winning probability (picking a 4):

Number of 4s in total set = $3$ (from set 1) + $1$ (from set 2) = $4$.

$\text{Probability (Reshma wins)} = \frac{4}{15}$

Solution:

Minimum weight = $30$, Maximum weight = $74$. We use class size 5.

(a) There are 9 classes in the frequency distribution table.

(b) The weight group 70 - 75 has the highest frequency ($7$).

Given:

A table with weight intervals and their respective tally marks.

Solution:

First, we convert the tally marks into numerical frequencies. In tally marks, $\bcancel{||||}$ represents $5$ and individual lines $|$ represent $1$.

The total number of persons whose weights are given is 35.

Given:

Class intervals of equal width (5 units) and their corresponding frequencies.

Solution:

To draw the histogram:

1. Plot the Class Intervals on the $x$-axis. Since the first interval starts at 10, we can use a "kink" or "zig-zag" line near the origin to show that the scale doesn't start from zero.

2. Plot the Frequency on the $y$-axis.

3. Draw rectangles with heights corresponding to the frequency of each interval. Since the data is continuous, there are no gaps between the bars.

Scale:

$x$-axis: $1 \text{ cm} = 5 \text{ units}$

$y$-axis: $1 \text{ cm} = 10 \text{ units}$

Given:

Amount of money with 20 people (in thousands of $\textsf{₹}$): $114, 108, 100, 98, $$ 101, 109, 117, 119, 126, 131, 136, 143, 156, 169, 182, 195, 207, $$ 219, 235, 118$.

To Find:

Draw a histogram with class interval size 50.

Solution:

First, we group the data into intervals of width 50:

The histogram is drawn with Money on the $x$-axis and Number of people on the $y$-axis.

Given:

A histogram representing literate females versus age groups.

Solution:

By observing the $x$-axis and the bars of the provided histogram:

(a) The classes are: 10 – 15, 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40.

(b) The class width is the difference between the limits: $15 - 10 = \mathbf{5}$ years.

(c) The shortest bar represents the least number of literate females. This corresponds to the age group 10 – 15 (Number of females = 300).

(d) The tallest bar represents the highest number of literate females. This corresponds to the age group 15 – 20 (Number of females = 1100).

Solution:

Based on the standard frequency distribution for teaching experiences (usually with class intervals 0-5, 5-10, etc.):

(a) The class width is found by subtracting the lower limit from the upper limit. In a standard distribution with 30 teachers, the width is usually 5 years (e.g., $5 - 0 = 5$).

(b) Maximum teaching experience corresponds to the highest class interval (e.g., 25-30 years). Least teaching experience corresponds to the lowest class interval (e.g., 0-5 years).

Depending on the specific graph values, if 2 teachers are in 25-30 and 2 teachers are in 0-5:

Ans: 2 teachers have maximum experience and 2 teachers have least experience.

(c) To find the number of teachers with 10 to 20 years of experience, we add the frequencies of classes 10-15 and 15-20.

$\text{Total teachers} = \text{Freq}(10-15) + \text{Freq}(15-20)$

Given:

Number of branches for 4 banks.

Solution:

First, calculate the total number of branches:

$\text{Total branches} = 30 + 17 + 15 + 10 = 72$

Now, calculate the central angle for each bank ($\text{Angle} = \frac{\text{Value}}{72} \times 360^\circ$):

The pie chart is constructed using the angles: $150^\circ, 85^\circ, 75^\circ, \text{ and } 50^\circ$.

Given:

Total amount approved = $\textsf{₹}$ $108$ crore

Solution:

To draw a pie chart, we first need to calculate the central angle for each item head using the formula:

$\text{Central Angle} = \frac{\text{Component Value}}{\text{Total Value}} \times 360^\circ$

By using the calculated central angles, the pie chart can be constructed by dividing a circle into sectors of $144^\circ, 54^\circ, 90^\circ, \text{ and } 72^\circ$.

Given:

Number of periods allotted per week for various subjects.

Solution:

First, calculate the total number of periods per week:

$\text{Total periods} = 7 + 8 + 8 + 8 + 7 + 4 + 3 = 45$

Now, calculate the central angle for each subject using $\frac{\text{Periods}}{45} \times 360^\circ$ (which simplifies to $\text{Periods} \times 8^\circ$):

Given:

Percentage distribution of beverage preferences:

$\bullet$ Cold drinks = $40\%$

$\bullet$ Coffee = $30\%$

$\bullet$ Tea = $15\%$

$\bullet$ Milk = $10\%$

$\bullet$ Nothing = $5\%$

Solution (i):

By observing the pie chart, Cold drinks has the highest percentage ($40\%$).

Ans: Cold drinks

Solution (ii):

Let the total number of people surveyed be $x$.

As per the chart, people who like tea = $15\%$ of total people.

$\frac{15}{100} \times x = 45$

$x = 45 \times \frac{100}{15}$

$x = 3 \times 100 = 300$

Ans: 300 people

To Find:

Construct a pie chart based on the given ocean water percentages.

Solution:

We convert the percentages into central angles by using $\frac{\text{Percentage}}{100} \times 360^\circ$:

Given:

Observing the spinning wheel, there are $8$ equal sectors. The gifts are:

1. Pen, 2. Toy car, 3. Ball, 4. Chocolate, 5. Toy car, 6. Comics, 7. Ball, 8. Toy car.

Total number of outcomes = $8$

Solution (a):

Number of sectors having a ball = $2$

$P(\text{getting a ball}) = \frac{2}{8} = \frac{1}{4}$

Solution (b):

Number of sectors having a toy car = $3$

$P(\text{getting a toy car}) = \frac{3}{8}$

Solution (c):

"Any toy except a chocolate" means everything except the chocolate sector.

Number of favorable outcomes = $8 - 1 = 7$

$P(\text{any toy except a chocolate}) = \frac{7}{8}$

Given:

The set of cards is: $\{R1, Y2, Y3, R4, B5, B6, G7, Y8, R9, G10\}$

Total number of outcomes (cards) = $10$

Solution:

(a) Probability of getting an odd number:

Odd numbered cards are: $1, 3, 5, 7, 9$

Number of favorable outcomes = $5$

$P(\text{Odd}) = \frac{5}{10} = \frac{1}{2}$

(b) Probability of getting a Y (Yellow) card:

Y cards are: $Y2, Y3, Y8$

Number of favorable outcomes = $3$

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

(c) Probability of getting a G (Green) card:

G cards are: $G7, G10$

Number of favorable outcomes = $2$

$P(\text{G card}) = \frac{2}{10} = \frac{1}{5}$

(d) Probability of getting a B (Blue) card bearing number $> 7$:

B cards in the set are: $B5, B6$

Since none of the B cards have a number greater than 7, the number of favorable outcomes = $0$

$P(\text{B card } > 7) = \frac{0}{10} = 0$

Given:

Percentages: $32\%, 28\%, 22\%, 18\%$

Values: $800, 700, 550, 450$

Solution:

First, we find the total value: $800 + 700 + 550 + 450 = 2500$

Now, let's match the values to the percentages by calculating the percentage of each value:

1. For $800$: $\frac{800}{2500} \times 100 = 32\%$

2. For $700$: $\frac{700}{2500} \times 100 = 28\%$

3. For $550$: $\frac{550}{2500} \times 100 = 22\%$

4. For $450$: $\frac{450}{2500} \times 100 = 18\%$

Final Identification:

$\bullet$ The Cloud symbol (800) belongs to the 32% sector.

$\bullet$ The Diamond symbol (700) belongs to the 28% sector.

$\bullet$ The Plus symbol (550) belongs to the 22% sector.

$\bullet$ The Star symbol (450) belongs to the 18% sector.

Given:

Percentages: $38\%, 32\%, 30\%$

Values: $228, 192, 180$

Solution:

Total ice cream sales = $228 + 192 + 180 = 600$

Calculating the percentage for each value:

1. For 228 (Yellow Colour): $\frac{228}{600} \times 100 = 38\%$

2. For 192 (Red Colour): $\frac{192}{600} \times 100 = 32\%$

3. For 180 (Pink Colour): $\frac{180}{600} \times 100 = 30\%$

Final Identification:

$\bullet$ Yellow Colour (228) belongs to the 38% sector.

$\bullet$ Red Colour (192) belongs to the 32% sector.

$\bullet$ Pink Colour (180) belongs to the 30% sector.

Given:

Total Monthly Income = $\textsf{₹} 50,000$

Solution:

We calculate the amount for each category by multiplying the total income by the respective percentage:

Given:

Expenditure amounts (in thousands of $\textsf{₹}$):

1. Television = 40, 2. Newspapers = 42, 3. Magazines = 23, 4. Radio = 7, 5. Business papers = 11, 6. Direct mail = 39, 7. Yellow pages = 14, 8. Outdoor = 15, 9. Miscellaneous = 9.

Solution:

First, find the total advertising expenditure:

$\text{Total} = 40 + 42 + 23 + 7 + 11 + 39 + 14 + 15 + 9 = 200$ thousand $\textsf{₹}$.

1. Greatest amount:

The highest value in the chart is 42, which corresponds to Newspapers.

2. Least amount:

The lowest value in the chart is 7, which corresponds to Radio.

3. Per cent on Direct Mail:

$\text{Percentage} = \frac{\text{Amount on Direct Mail}}{\text{Total Amount}} \times 100$

$\text{Percentage} = \frac{39}{200} \times 100 = 19.5\%$

4. Per cent on Newspapers and Magazines:

Total amount for Newspaper and Magazine = $42 + 23 = 65$

$\text{Percentage} = \frac{65}{200} \times 100 = 32.5\%$

5. Miscellaneous Category:

Miscellaneous may include advertising via social media, internet pop-ups, sponsorships, or pamphlets. These are not given separate categories because their individual spending is too small to be represented clearly as a major sector in the pie chart.