Chapter 3 Data Handling

Solution:

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Given:

Hours studied on the first day = 4 hours.

Hours studied on the second day = 5 hours.

Hours studied on the third day = 3 hours.

Number of days = 3.

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To Find:

The average hours Ashish studies daily.

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Solution:

The average is calculated by dividing the sum of the values by the number of values.

Average hours of study = $\frac{\text{Sum of hours studied on three days}}{\text{Number of days}}$

Sum of hours studied = $4 + 5 + 3 = 12$ hours.

Average hours of study = $\frac{12 \text{ hours}}{3}$

Average hours of study = $4$ hours.

Thus, Ashish studies $\mathbf{4}$ hours daily on an average.

Solution:

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Given:

Runs scored in six innings: 36, 35, 50, 46, 60, 55.

Number of innings = 6.

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To Find:

The mean runs scored by the batsman.

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Solution:

The mean (average) runs is the sum of the runs scored in all innings divided by the number of innings.

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

Sum of runs scored = $36 + 35 + 50 + 46 + 60 + 55$

Sum of runs scored = $282$.

Mean runs = $\frac{282}{6}$

Mean runs = $47$.

Thus, the mean runs scored by the batsman in an inning is $\mathbf{47}$.

Solution:

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Given:

The ages in years of 10 teachers are: $32, 41, 28, 54, 35, 26, 23, 33, 38, 40$.

Number of teachers = $10$.

---

To Find:

(i) Age of the oldest and youngest teacher.

(ii) Range of the ages.

(iii) Mean age of the teachers.

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Solution:

First, let's arrange the given ages in ascending order:

$23, 26, 28, 32, 33, 35, 38, 40, 41, 54$.

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(i) Age of the oldest and youngest teacher:

The oldest teacher has the maximum age, which is the highest value in the data set.

Maximum age = $54$ years.

The youngest teacher has the minimum age, which is the lowest value in the data set.

Minimum age = $23$ years.

The age of the oldest teacher is $\mathbf{54}$ years.

The age of the youngest teacher is $\mathbf{23}$ years.

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(ii) Range of the ages of the teachers:

The range is the difference between the maximum age and the minimum age.

Range = Maximum age - Minimum age

Range = $54 - 23 = 31$ years.

The range of the ages of the teachers is $\mathbf{31}$ years.

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(iii) Mean age of these teachers:

The mean age is the sum of all the ages divided by the number of teachers.

Sum of ages = $23 + 26 + 28 + 32 + 33 + 35 + 38 + 40 + 41 + 54$

Sum of ages = $350$ years.

Number of teachers = $10$.

Mean age = $\frac{\text{Sum of ages}}{\text{Number of teachers}}$

Mean age = $\frac{350}{10}$

Mean age = $35$ years.

The mean age of these teachers is $\mathbf{35}$ years.

Solution:

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To find the range of the heights of any ten students, we first need to collect the data, i.e., the heights of ten students of the class. The range is defined as the difference between the highest value and the lowest value in a data set.

The steps to find the range are:

1. Collect the heights of ten students.

2. Identify the highest height among these ten students.

3. Identify the lowest height among these ten students.

4. Subtract the lowest height from the highest height to find the range.

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Since the actual heights of the students are not provided, let's assume a hypothetical set of heights (in cm) for ten students in a class:

Hypothetical heights: 135, 140, 138, 145, 132, 141, 136, 143, 139, 140.

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Arranging the hypothetical heights in ascending order:

132, 135, 136, 138, 139, 140, 140, 141, 143, 145.

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From the ordered list:

The highest hypothetical height = $145$ cm.

The lowest hypothetical height = $132$ cm.

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Now, calculate the range:

Range = Highest height - Lowest height

Range = $145 \text{ cm} - 132 \text{ cm}$

Range = $13$ cm.

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Based on this hypothetical data, the range of heights of the ten students is $\mathbf{13}$ cm.

Note: To get the actual answer, you should collect the real heights of ten students from your class and perform the calculation as shown above.

Solution:

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The given marks in the class assessment are: 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7.

Total number of marks (data points) = 20.

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Organising the marks in a tabular form (frequency distribution table):

Mark	Tally Marks	Frequency (Number of students)
1	$|$	1
2	$||$	2
3	$|$	1
4	$|||$	3
5	$\bcancel{||||}$	5
6	$||||$	4
7	$||$	2
8	$|$	1
9	$|$	1

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(i) Which number is the highest?

Looking at the given marks, the highest mark is the largest value among them.

The marks are: 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7.

The highest mark is $\mathbf{9}$.

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(ii) Which number is the lowest?

Looking at the given marks, the lowest mark is the smallest value among them.

The marks are: 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7.

The lowest mark is $\mathbf{1}$.

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(iii) What is the range of the data?

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

Range = Highest mark - Lowest mark

Range = $9 - 1 = 8$.

The range of the data is $\mathbf{8}$.

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(iv) Find the arithmetic mean.

The arithmetic mean (or simply mean) is the sum of all the observations divided by the total number of observations.

Mean = $\frac{\text{Sum of all marks}}{\text{Total number of marks}}$

Sum of all marks = $4+6+7+5+3+5+4+5+2+6+2+5+1 \ $$ +9+6+5 \ $$ +8+4 \ $$+6+7$

Sum of all marks = $(1 \times 1) + (2 \times 2) + (3 \times 1) + (4 \times 3) + (5 \times 5) \ $$ + (6 \times 4) + (7 \times 2) \ $$ + (8 \times 1) + (9 \times 1)$

Sum of all marks = $1 + 4 + 3 + 12 + 25 + 24 + 14 + 8 + 9 = 100$.

Total number of marks = 20.

Mean = $\frac{100}{20} = 5$.

The arithmetic mean of the marks is $\mathbf{5}$.

Solution:

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

Number of observations = 5.

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To Find:

The mean of these numbers.

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Solution:

The mean is the sum of the observations divided by the number of observations.

Mean = $\frac{\text{Sum of the first five whole numbers}}{\text{Number of whole numbers}}$

Sum of the first five whole numbers = $0 + 1 + 2 + 3 + 4 = 10$.

Mean = $\frac{10}{5}$

Mean = $2$.

The mean of the first five whole numbers is $\mathbf{2}$.

Solution:

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Given:

Runs scored by a cricketer in eight innings are: 58, 76, 40, 35, 46, 45, 0, 100.

Number of innings = 8.

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To Find:

The mean score.

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Solution:

The mean score is the sum of the runs scored in all innings divided by the number of innings.

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

Sum of runs scored = $58 + 76 + 40 + 35 + 46 + 45 + 0 + 100$

Sum of runs scored = $400$.

Mean score = $\frac{400}{8}$

Mean score = $50$.

The mean score of the cricketer is $\mathbf{50}$.

Solution:

The mean (average) is calculated as: Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$.

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(i) Find the mean to determine A’s average number of points scored per game.

Points scored by Player A in four games: 14, 16, 10, 10.

Number of games played by A = 4.

Sum of points scored by A = $14 + 16 + 10 + 10 = 50$.

Mean score of A = $\frac{\text{Sum of points}}{\text{Number of games}} = \frac{50}{4}$.

Mean score of A = $12.5$.

A's average number of points scored per game is $\mathbf{12.5}$.

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(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?

Player C played in Game 1 (8 points), Game 2 (11 points), and Game 4 (13 points). Player C did not play in Game 3.

To find the mean number of points per game for C, we would divide the total points by the number of games C actually played.

Number of games played by C = 3.

Therefore, we would divide the total points by 3.

Reason: The mean is calculated over the number of instances the event occurred. Since C only played in 3 games, we consider only those 3 games for the average.

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

Points scored by Player B in four games: 0, 8, 6, 4.

Number of games played by B = 4.

To find the mean number of points per game for B, we would sum the points scored in all four games and divide by the total number of games B played (which is 4).

Sum of points scored by B = $0 + 8 + 6 + 4 = 18$.

Mean score of B = $\frac{\text{Sum of points}}{\text{Number of games}} = \frac{18}{4}$.

Mean score of B = $4.5$.

The mean score for B is $\mathbf{4.5}$.

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(iv) Who is the best performer?

To determine the best performer, we compare the mean scores of the three players.

Mean score of A = 12.5

Mean score of B = 4.5

Mean score of C: We use the calculation from part (ii), dividing by 3 games.

Points scored by C = 8 + 11 + 13 = 32.

Mean score of C = $\frac{32}{3} \approx 10.67$.

Comparing the mean scores: 12.5 (A), 4.5 (B), 10.67 (C).

The highest mean score is 12.5, which is A's mean score.

Therefore, Player A is the best performer based on the average points per game played.

Solution:

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Given:

Marks obtained by a group of students in a science test are: 85, 76, 90, 85, 39, 48, 56, 95, 81, 75.

Number of students = 10.

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To Find:

(i) Highest and lowest marks.

(ii) Range of the marks.

(iii) Mean marks.

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Solution:

First, let's arrange the given marks in ascending order:

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

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(i) Highest and the lowest marks obtained by the students:

The highest mark is the maximum value in the data set.

Highest mark = 95.

The lowest mark is the minimum value in the data set.

Lowest mark = 39.

The highest mark obtained is $\mathbf{95}$.

The lowest mark obtained is $\mathbf{39}$.

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(ii) Range of the marks obtained:

The range is the difference between the highest mark and the lowest mark.

Range = Highest mark - Lowest mark

Range = $95 - 39 = 56$.

The range of the marks obtained is $\mathbf{56}$.

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(iii) Mean marks obtained by the group:

The mean marks is the sum of all the marks divided by the number of students.

Sum of marks = $85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75$

Sum of marks = $730$.

Number of students = 10.

Mean marks = $\frac{\text{Sum of marks}}{\text{Number of students}}$

Mean marks = $\frac{730}{10} = 73$.

The mean marks obtained by the group is $\mathbf{73}$.

Solution:

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Given:

The enrolment in a school during six consecutive years: 1555, 1670, 1750, 2013, 2540, 2820.

Number of years = 6.

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To Find:

The mean enrolment for this period.

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Solution:

The mean enrolment is the sum of the enrolment figures for each year divided by the number of years.

Mean enrolment = $\frac{\text{Sum of enrolment for six years}}{\text{Number of years}}$

Sum of enrolment = $1555 + 1670 + 1750 + 2013 + 2540 + 2820$

Sum of enrolment = $12348$.

Mean enrolment = $\frac{12348}{6}$

Divide 12348 by 6:

$\begin{array}{r} 2058 \\ 6{\overline{\smash{\big)}\,12348}} \\ \underline{-~12\phantom{000}} \\ 034\phantom{0} \\ \underline{-~30\phantom{0}} \\ 48 \\ \underline{-~48} \\ 0 \end{array}$

Mean enrolment = $2058$.

The mean enrolment of the school for this period is $\mathbf{2058}$.

Solution:

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Given:

Rainfall (in mm) on 7 days: 0.0, 12.2, 2.1, 0.0, 20.5, 5.5, 1.0.

Number of days = 7.

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(i) Find the range of the rainfall in the above data.

Arrange the rainfall data in ascending order:

0.0, 0.0, 1.0, 2.1, 5.5, 12.2, 20.5.

Highest rainfall = 20.5 mm.

Lowest rainfall = 0.0 mm.

Range = Highest rainfall - Lowest rainfall

Range = $20.5 \text{ mm} - 0.0 \text{ mm} = 20.5 \text{ mm}$.

The range of the rainfall is $\mathbf{20.5}$ mm.

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(ii) Find the mean rainfall for the week.

Mean rainfall = $\frac{\text{Sum of rainfall on 7 days}}{\text{Number of days}}$

Sum of rainfall = $0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0$

Sum of rainfall = $41.3$ mm.

Mean rainfall = $\frac{41.3}{7}$

Divide 41.3 by 7:

$\begin{array}{r} 5.9 \\ 7{\overline{\smash{\big)}\,41.3}} \\ \underline{-~35\phantom{.0}} \\ 63\phantom{0} \\ \underline{-~63\phantom{0}} \\ 0\phantom{0} \end{array}$

Mean rainfall = $5.9$ mm.

The mean rainfall for the week is $\mathbf{5.9}$ mm.

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(iii) On how many days was the rainfall less than the mean rainfall.

The mean rainfall is 5.9 mm.

We look at the daily rainfall data and count the number of days where the rainfall was less than 5.9 mm.

• Monday: 0.0 mm ($0.0 < 5.9$) - Yes
• Tuesday: 12.2 mm ($12.2 \not< 5.9$) - No
• Wednesday: 2.1 mm ($2.1 < 5.9$) - Yes
• Thursday: 0.0 mm ($0.0 < 5.9$) - Yes
• Friday: 20.5 mm ($20.5 \not< 5.9$) - No
• Saturday: 5.5 mm ($5.5 < 5.9$) - Yes
• Sunday: 1.0 mm ($1.0 < 5.9$) - Yes

The days with rainfall less than the mean are Monday, Wednesday, Thursday, Saturday, and Sunday.

Counting these days, we find there are 5 days.

On 5 days, the rainfall was less than the mean rainfall.

Solution:

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Given:

The heights of 10 girls in cm are: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141.

Number of girls = 10.

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To Find:

(i) Height of the tallest girl.

(ii) Height of the shortest girl.

(iii) Range of the data.

(iv) Mean height of the girls.

(v) Number of girls with heights more than the mean height.

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Solution:

First, let's arrange the given heights in ascending order:

128, 132, 135, 139, 141, 143, 146, 149, 150, 151.

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(i) Height of the tallest girl:

The height of the tallest girl is the maximum value in the data set.

Maximum height = 151 cm.

The height of the tallest girl is $\mathbf{151}$ cm.

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(ii) Height of the shortest girl:

The height of the shortest girl is the minimum value in the data set.

Minimum height = 128 cm.

The height of the shortest girl is $\mathbf{128}$ cm.

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(iii) Range of the data:

The range of the data is the difference between the highest height and the lowest height.

Range = Maximum height - Minimum height

Range = $151 \text{ cm} - 128 \text{ cm} = 23$ cm.

The range of the data is $\mathbf{23}$ cm.

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(iv) Mean height of the girls:

The mean height is the sum of all the heights divided by the number of girls.

Sum of heights = $135 + 150 + 139 + 128 + 151 + 132 + 146 + 149 + 143 + 141$

Sum of heights = $1414$ cm.

Number of girls = 10.

Mean height = $\frac{\text{Sum of heights}}{\text{Number of girls}}$

Mean height = $\frac{1414}{10} = 141.4$ cm.

The mean height of the girls is $\mathbf{141.4}$ cm.

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

The mean height is 141.4 cm.

We look at the heights of the girls and count how many are greater than 141.4 cm.

The heights are: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141.

Heights greater than 141.4 cm are:

150 ($150 > 141.4$)

151 ($151 > 141.4$)

146 ($146 > 141.4$)

149 ($149 > 141.4$)

143 ($143 > 141.4$)

141 ($141 \not> 141.4$)

Counting these heights, we find there are 5 girls whose heights are more than the mean height.

5 girls have heights more than the mean height.

Solution:

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Given set of numbers: $1, 1, 2, 4, 3, 2, 1, 2, 2, 4$.

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To Find:

The mode of the given set of numbers.

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Solution:

The mode of a data set is the observation that occurs most frequently.

Let's count the frequency of each distinct number in the set:

• Number 1 appears 3 times.
• Number 2 appears 4 times.
• Number 3 appears 1 time.
• Number 4 appears 2 times.

Comparing the frequencies (3, 4, 1, 2), the highest frequency is 4, which corresponds to the number 2.

Since the number 2 occurs most frequently (4 times), the mode of the given set of numbers is 2.

The mode of the given set of numbers is $\mathbf{2}$.

Solution:

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Given data represents the margins of victory in football matches:

1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 1, 1, 2, 3, 2, 6, 4, 3, 2, 1, 1, 4, 2, 1, 5, 3, 3, 2, 3, 2, 4, 2, 1, 2.

Total number of observations = 40.

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To Find:

The mode of this data.

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Solution:

The mode is the observation that appears most frequently in the data set.

Let's organize the data by counting the frequency of each distinct margin of victory:

• Margin 1: Appears 9 times. ($\bcancel{||||} \bcancel{||||} |$)
• Margin 2: Appears 14 times. ($\bcancel{||||} \bcancel{||||} \bcancel{||||} ||||$)
• Margin 3: Appears 7 times. ($\bcancel{||||} ||$)
• Margin 4: Appears 5 times. ($\bcancel{||||}$)
• Margin 5: Appears 3 times. ($|||$)
• Margin 6: Appears 2 times. ($||$)

Summarizing the frequencies:

Margin of Victory	Frequency
1	9
2	14
3	7
4	5
5	3
6	2

The highest frequency is 14, which corresponds to the margin of victory 2.

Since the margin of victory 2 occurs most frequently (14 times), the mode of the data is 2.

The mode of this data is $\mathbf{2}$.

Solution:

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Given set of numbers: 2, 2, 2, 3, 3, 4, 5, 5, 5, 6, 6, 8.

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To Find:

The mode of the given set of numbers.

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Solution:

The mode of a data set is the observation that occurs most frequently.

Let's count the frequency of each distinct number in the set:

• Number 2 appears 3 times.
• Number 3 appears 2 times.
• Number 4 appears 1 time.
• Number 5 appears 3 times.
• Number 6 appears 2 times.
• Number 8 appears 1 time.

Comparing the frequencies (3, 2, 1, 3, 2, 1), we observe that the highest frequency is 3, which occurs for both the number 2 and the number 5.

Since two numbers (2 and 5) have the same highest frequency, this data set has two modes.

The modes of the given set of numbers are $\mathbf{2}$ and $\mathbf{5}$.

Note: A data set can have one mode (unimodal), two modes (bimodal), or more than two modes (multimodal). If all observations occur with the same frequency, the data set has no mode.

Solution:

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Given data: 24, 36, 46, 17, 18, 25, 35.

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To Find:

The median of the data.

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Solution:

The median is the middle observation when the data is arranged in ascending or descending order.

First, arrange the data in ascending order:

17, 18, 24, 25, 35, 36, 46.

The number of observations, $n$, is 7.

Since the number of observations (7) is odd, the median is the $\left(\frac{n+1}{2}\right)$-th observation.

Median = $\left(\frac{7+1}{2}\right)$-th observation = $\left(\frac{8}{2}\right)$-th observation = 4th observation.

Looking at the ordered data (17, 18, 24, 25, 35, 36, 46), the 4th observation is 25.

The median of the data is $\mathbf{25}$.

Solution:

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Given scores in mathematics test (out of 25) for 15 students:

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

Number of students (observations), $n = 15$.

---

To Find:

The mode and median of the data, and if they are the same.

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Solution:

Finding the Mode:

The mode is the observation that occurs most frequently.

Let's count the frequency of each score:

• Score 5: 1 time
• Score 9: 1 time
• Score 10: 1 time
• Score 12: 1 time
• Score 15: 1 time
• Score 16: 1 time
• Score 19: 1 time
• Score 20: 4 times
• Score 23: 1 time
• Score 24: 1 time
• Score 25: 2 times

The score that appears most frequently is 20, which appears 4 times.

The mode of the data is $\mathbf{20}$.

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Finding the Median:

The median is the middle observation when the data is arranged in ascending or descending order.

Arrange the scores in ascending order:

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

The number of observations, $n = 15$, is odd.

The median is the $\left(\frac{n+1}{2}\right)$-th observation.

Median = $\left(\frac{15+1}{2}\right)$-th observation = $\left(\frac{16}{2}\right)$-th observation = 8th observation.

Looking at the ordered data, the 8th observation is 20.

The median of the data is $\mathbf{20}$.

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Comparison of Mode and Median:

Mode = 20

Median = 20

Since both the mode and the median are 20, they are the same.

The mode and median of this data are 20. Yes, they are the same.

Solution:

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Given:

The runs scored by 11 players in a cricket match are: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15.

Number of players (observations), $n = 11$.

---

To Find:

The mean, mode, and median of this data, and if they are the same.

---

Solution:

Finding the Mean:

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

Mean = $\frac{\text{Sum of runs scored}}{\text{Number of players}}$

Sum of runs scored = $6 + 15 + 120 + 50 + 100 + 80 + 10 \ $$ + 15 + 8 \ $$ + 10 + 15$

Sum of runs scored = $429$.

Mean = $\frac{429}{11}$.

Mean = $39$.

The mean score is $\mathbf{39}$.

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Finding the Mode:

The mode is the observation that occurs most frequently.

Let's count the frequency of each score:

• Score 6: 1 time
• Score 8: 1 time
• Score 10: 2 times
• Score 15: 3 times
• Score 50: 1 time
• Score 80: 1 time
• Score 100: 1 time
• Score 120: 1 time

The score that appears most frequently is 15, which appears 3 times.

The mode of the data is $\mathbf{15}$.

---

Finding the Median:

The median is the middle observation when the data is arranged in ascending or descending order.

Arrange the scores in ascending order:

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

The number of observations, $n = 11$, is odd.

The median is the $\left(\frac{n+1}{2}\right)$-th observation.

Median = $\left(\frac{11+1}{2}\right)$-th observation = $\left(\frac{12}{2}\right)$-th observation = 6th observation.

Looking at the ordered data (6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120), the 6th observation is 15.

The median of the data is $\mathbf{15}$.

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Comparison of Mean, Mode, and Median:

Mean = 39

Mode = 15

Median = 15

The mean is 39, while the mode and median are both 15.

Are the three same? No, the mean, mode, and median are not all the same.

Solution:

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Given weights (in kg) of 15 students of a class:

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

Number of students (observations), $n = 15$.

---

To Find:

(i) The mode and median of this data.

(ii) If there is more than one mode.

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Solution:

(i) Finding the Mode and Median:

First, let's arrange the weights in ascending order:

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

Number of observations, $n = 15$, is odd.

Mode: The mode is the observation that occurs most frequently.

Count the frequency of each weight:

• Weight 32: 1 time
• Weight 35: 1 time
• Weight 36: 1 time
• Weight 37: 1 time
• Weight 38: 3 times
• Weight 40: 1 time
• Weight 42: 1 time
• Weight 43: 3 times
• Weight 45: 1 time
• Weight 47: 1 time
• Weight 50: 1 time

The highest frequency is 3, which occurs for both weight 38 and weight 43.

The modes of the data are $\mathbf{38}$ and $\mathbf{43}$.

Median: The median is the middle observation when the data is arranged in order.

Since $n = 15$ (odd), the median is the $\left(\frac{n+1}{2}\right)$-th observation.

Median = $\left(\frac{15+1}{2}\right)$-th observation = $\left(\frac{16}{2}\right)$-th observation = 8th observation.

Looking at the ordered data (32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50), the 8th observation is 40.

The median of the data is $\mathbf{40}$.

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(ii) Is there more than one mode?

From the frequency count in part (i), we found that both the weight 38 and the weight 43 appear with the highest frequency (3 times).

Since there are two distinct values (38 and 43) that occur with the highest frequency, there is more than one mode.

Yes, there is more than one mode. The modes are 38 and 43.

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Answers:

(i) The mode is $\mathbf{38}$ and $\mathbf{43}$. The median is $\mathbf{40}$.

(ii) Yes, there is more than one mode.

Solution:

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Given data: 13, 16, 12, 14, 19, 12, 14, 13, 14.

Number of observations, $n = 9$.

---

To Find:

The mode and median of the data.

---

Solution:

Finding the Mode:

The mode is the observation that occurs most frequently.

Let's count the frequency of each number:

• Number 12: 2 times
• Number 13: 2 times
• Number 14: 3 times
• Number 16: 1 time
• Number 19: 1 time

The number that appears most frequently is 14, which appears 3 times.

The mode of the data is $\mathbf{14}$.

---

Finding the Median:

The median is the middle observation when the data is arranged in ascending or descending order.

Arrange the data in ascending order:

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

The number of observations, $n = 9$, is odd.

The median is the $\left(\frac{n+1}{2}\right)$-th observation.

Median = $\left(\frac{9+1}{2}\right)$-th observation = $\left(\frac{10}{2}\right)$-th observation = 5th observation.

Looking at the ordered data (12, 12, 13, 13, 14, 14, 14, 16, 19), the 5th observation is 14.

The median of the data is $\mathbf{14}$.

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Answers:

The mode of the data is $\mathbf{14}$.

The median of the data is $\mathbf{14}$.

Solution:

---

Let's analyze each statement:

(i) The mode is the most frequently occurring observation in a data set. By definition, the mode is a value that is present in the data set itself. Therefore, the mode is always one of the numbers in a data.

This statement is True.

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(ii) The mean is the average of the data, calculated by summing all observations and dividing by the number of observations. The mean does not necessarily have to be one of the numbers in the data. For example, the mean of 1, 2, and 3 is $\frac{1+2+3}{3} = \frac{6}{3} = 2$, which is in the data. However, the mean of 1 and 2 is $\frac{1+2}{2} = \frac{3}{2} = 1.5$, which is not in the data.

This statement is False.

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(iii) The median is the middle observation when the data is arranged in order. If the number of observations is odd, the median is the single middle value, which is always one of the numbers in the data. If the number of observations is even, the median is the average of the two middle values, which may or may not be one of the numbers in the data. For example, in the data set {1, 2, 3, 4}, the median is $\frac{2+3}{2} = 2.5$, which is not in the data.

This statement is False.

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(iv) The data is 6, 4, 3, 8, 9, 12, 13, 9. The number of observations is 8.

Let's calculate the mean:

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

Sum of observations = $6 + 4 + 3 + 8 + 9 + 12 + 13 + 9 = 64$.

Mean = $\frac{64}{8} = 8$.

The mean of the data is 8, not 9. Therefore, the statement is incorrect.

This statement is False.

Solution:

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The given data can be tabulated as follows:

Favourite Colour	Number of Students
Red	43
Green	19
Blue	55
Yellow	49
Orange	34

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Bar Graph Representation:

To represent this data on a bar graph, we follow these steps:

1. Draw two perpendicular lines on a graph paper. The horizontal line is the x-axis (representing Favourite Colour) and the vertical line is the y-axis (representing Number of Students).

2. Along the x-axis, mark points at equal intervals and label them with the names of the colours: Red, Green, Blue, Yellow, Orange.

3. Choose a suitable scale for the y-axis. Since the maximum value is 55, we can choose a scale where 1 unit = 10 students.

4. Draw bars of equal width for each colour, with the height of each bar corresponding to the number of students for that colour on the chosen scale.

• Height of Red bar = 43 units
• Height of Green bar = 19 units
• Height of Blue bar = 55 units
• Height of Yellow bar = 49 units
• Height of Orange bar = 34 units

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Answers to the questions:

(i) Which is the most preferred colour and which is the least preferred?

By observing the bar graph, the tallest bar corresponds to the most preferred colour and the shortest bar corresponds to the least preferred colour.

The tallest bar is for Blue, with 55 students.

The shortest bar is for Green, with 19 students.

Therefore, the most preferred colour is Blue and the least preferred colour is Green.

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(ii) How many colours are there in all? What are they?

By counting the categories on the x-axis (or in the table), we can find the total number of colours.

There are 5 colours in all.

They are: Red, Green, Blue, Yellow, and Orange.

Solution:

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The given data is tabulated below:

Students	Marks Obtained (out of 600)
Ajay	450
Bali	500
Dipti	300
Faiyaz	360
Geetika	400
Hari	540

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Bar Graph Representation:

To represent this data on a bar graph, we follow these steps:

1. Draw a horizontal x-axis (for Students) and a vertical y-axis (for Marks Obtained).

2. On the x-axis, mark points at equal intervals for each student: Ajay, Bali, Dipti, Faiyaz, Geetika, and Hari.

3. Choose a suitable scale for the y-axis. The marks range from 300 to 540. A suitable scale would be 1 unit = 100 marks.

4. Draw bars of equal width for each student, with the height corresponding to their marks.

• Height of Ajay's bar = 450 units (4.5 on the scale)
• Height of Bali's bar = 500 units (5 on the scale)
• Height of Dipti's bar = 300 units (3 on the scale)
• Height of Faiyaz's bar = 360 units (3.6 on the scale)
• Height of Geetika's bar = 400 units (4 on the scale)
• Height of Hari's bar = 540 units (5.4 on the scale)

Solution:

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The given data shows the scores of 5 students in Quarterly and Half-yearly tests:

Students	Quarterly Score (out of 25)	Half-yearly Score (out of 25)
Ashish	10	15
Arun	15	18
Kavish	12	16
Maya	20	21
Rita	9	15

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To check the effectiveness of the new teaching technique, we need to compare the performance in both tests. A double bar graph is the most suitable way to represent this data as it allows for a direct visual comparison of the scores for each student.

---

Double Bar Graph Representation:

1. Draw the x-axis (for Students) and y-axis (for Scores).

2. On the x-axis, mark the names of the students.

3. Choose a scale for the y-axis, for example, 1 unit = 5 marks, going up to 25.

4. For each student, draw two adjacent bars. One bar represents the Quarterly score and the other represents the Half-yearly score.

5. Use different colours or patterns for the Quarterly and Half-yearly bars and provide a legend to indicate what each colour/pattern represents.

• For Ashish: Draw a bar of height 10 (Quarterly) and an adjacent bar of height 15 (Half-yearly).
• For Arun: Draw a bar of height 15 (Quarterly) and an adjacent bar of height 18 (Half-yearly).
• For Kavish: Draw a bar of height 12 (Quarterly) and an adjacent bar of height 16 (Half-yearly).
• For Maya: Draw a bar of height 20 (Quarterly) and an adjacent bar of height 21 (Half-yearly).
• For Rita: Draw a bar of height 9 (Quarterly) and an adjacent bar of height 15 (Half-yearly).

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Conclusion:

From the double bar graph, we can clearly see that for every student, the bar representing the Half-yearly score is taller than the bar representing the Quarterly score. This indicates that every student has shown improvement.

Therefore, we can conclude that the new teaching technique was effective.

Solution:

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The bar graph shows the favourite pets of students. The horizontal axis represents the pets, and the vertical axis represents the number of students.

From the bar graph, we can read the number of students for each pet:

• Dogs: 8 students
• Cats: 10 students
• Rabbits: 2 students
• Hamsters: 5 students
• Others: 3 students

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(a) Which is the most popular pet?

The most popular pet is the one with the highest number of students. Looking at the number of students for each pet, the highest number is 10, which corresponds to Cats.

The most popular pet is Cats.

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(b) How many students have dog as a pet?

Looking at the bar for 'Dogs', its height corresponds to the number of students who have a dog as a pet.

From the bar graph, the bar for Dogs reaches up to 8 on the vertical axis.

So, 8 students have a dog as a pet.

Solution:

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The bar graph shows the number of books sold by a bookstore from 1989 to 1993. The horizontal axis represents the Year, and the vertical axis represents the Number of books sold. The scale on the vertical axis is 1 unit = 100 books.

Let's read the approximate number of books sold each year from the bar graph:

• 1989: The bar is a little less than halfway between 100 and 200. Let's estimate it as 175.
• 1990: The bar is slightly below the 500 mark. Let's estimate it as 475.
• 1991: The bar reaches the 300 mark.
• 1992: The bar is a little more than halfway between 200 and 300. Let's estimate it as 225.
• 1993: The bar is between 400 and 500, reaching the 450 mark.

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

From our estimations:

In 1989, about $\mathbf{175}$ books were sold.

In 1990, about $\mathbf{475}$ books were sold.

In 1992, about $\mathbf{225}$ books were sold.

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

From our estimations:

About 475 books were sold in the year 1990 (the bar for 1990 is slightly below 500, consistent with 475).

About 225 books were sold in the year 1992 (the bar for 1992 is a little above 200, consistent with 225).

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

We need to check the years where the number of books sold is less than 250.

• 1989: About 175 books (175 < 250) - Yes
• 1990: About 475 books (475 $\not<$ 250) - No
• 1991: 300 books (300 $\not<$ 250) - No
• 1992: About 225 books (225 < 250) - Yes
• 1993: 450 books (450 $\not<$ 250) - No

The years in which fewer than 250 books were sold are 1989 and 1992.

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

To estimate the number of books sold in 1989 using the bar graph:

1. Locate the bar corresponding to the year 1989 on the horizontal axis.

2. Look at the top of this bar and find the point on the vertical axis that aligns with the top of the bar.

3. Observe the scale on the vertical axis (1 unit = 100 books).

4. The top of the bar for 1989 is between the mark for 100 books and the mark for 200 books. It appears to be approximately halfway between 100 and 200, or slightly below halfway. This visually corresponds to a value like 150, 175, or similar.

5. Based on the position relative to the 100 and 200 marks, we make an estimate. A reasonable estimate based on the visual representation is about 175 books.

Explanation: The top of the bar for 1989 lies between 100 and 200 on the vertical scale. By observing its position relative to these two marks and the space between them, we can make an approximate reading. In this case, it appears to be roughly $\frac{3}{4}$ of the way from 100 to 200, which is $100 + \frac{3}{4}(200-100) = 100 + \frac{3}{4}(100) = 100 + 75 = 175$. Thus, we estimate the number of books sold in 1989 to be about 175.

Solution:

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The given data is as follows:

Class	Number of Children
Fifth	135
Sixth	120
Seventh	95
Eighth	100
Ninth	90
Tenth	80

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Bar Graph Representation:

A bar graph for the given data is created by representing 'Class' on the horizontal axis and 'Number of Children' on the vertical axis.

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(a) How would you choose a scale?

To choose a scale, we observe the range of the data. The minimum value is 80 and the maximum is 135. A scale should be chosen to comfortably accommodate this range.

A convenient scale would be 1 unit = 20 children on the vertical axis. This would make the graph of a reasonable size. The axis would be marked at 0, 20, 40, 60, 80, 100, 120, 140.

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(b) Answer the following questions:

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

By observing the data table or the heights of the bars in the graph:

The maximum number of children is 135, which is in the Fifth class.

The minimum number of children is 80, which is in the Tenth class.

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

Number of students in Class Sixth = 120.

Number of students in Class Eighth = 100.

Ratio = $\frac{\text{Number of students in Class Sixth}}{\text{Number of students in Class Eighth}} = \frac{120}{100}$

To simplify the ratio, we divide both numbers by their greatest common divisor, which is 20.

$\frac{120 \div 20}{100 \div 20} = \frac{6}{5}$

Therefore, the required ratio is 6:5.

Solution:

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The given data shows the marks of a student in 1st Term and 2nd Term tests:

Subject	1st Term Marks (out of 100)	2nd Term Marks (out of 100)
English	67	70
Hindi	72	65
Maths	88	95
Science	81	85
S. Science	73	75

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Double Bar Graph Representation:

A double bar graph is drawn with 'Subject' on the horizontal axis and 'Marks' on the vertical axis. An appropriate scale for the vertical axis would be 1 unit = 10 marks. For each subject, two adjacent bars are drawn representing the marks in the 1st and 2nd terms respectively.

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To answer the questions, let's find the difference in marks (2nd Term - 1st Term) for each subject:

• English: $70 - 67 = 3$ marks improvement.
• Hindi: $65 - 72 = -7$ marks (performance went down).
• Maths: $95 - 88 = 7$ marks improvement.
• Science: $85 - 81 = 4$ marks improvement.
• S. Science: $75 - 73 = 2$ marks improvement.

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

Comparing the improvements (3, 7, 4, 2), the highest improvement is 7 marks.

The child has improved his performance the most in Maths.

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

Looking only at the subjects where performance improved, the improvements are 3, 7, 4, and 2. The smallest of these is 2.

The improvement is the least in S. Science.

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

Yes, the difference in marks for Hindi is negative (-7), which means the marks decreased from the 1st Term to the 2nd Term.

The performance has gone down in Hindi.

Solution:

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The given data from the survey is:

Favourite Sport	Watching	Participating
Cricket	1240	620
Basket Ball	470	320
Swimming	510	320
Hockey	430	250
Athletics	250	105

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

To draw the double bar graph, we represent 'Favourite Sport' on the horizontal axis and 'Number of People' on the vertical axis. The maximum value is 1240, so an appropriate scale would be 1 unit = 200 people.

Inference from the bar graph:

From the bar graph, we can infer that:

• For all the given sports, the number of people who prefer watching is greater than the number of people who prefer participating.
• Cricket is the most liked sport, while Athletics is the least liked.

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(ii) Which sport is most popular?

The popularity of a sport can be judged by the number of people watching or participating. In the graph, the bars for Cricket are the tallest for both categories.

Total people interested in Cricket = 1240 (Watching) + 620 (Participating) = 1860.

This total is the highest among all sports.

Therefore, Cricket is the most popular sport.

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

To determine this, we find the total number of people for each preference.

Total number of people who prefer watching:

$1240 + 470 + 510 + 430 + 250 = 2900$

Total number of people who prefer participating:

$620 + 320 + 320 + 250 + 105 = 1615$

Since $2900 > 1615$, it is clear that watching sports is more preferred.

Solution:

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First, we create a double bar graph for the given temperature data. We'll use 'City' for the horizontal axis and 'Temperature in °C' for the vertical axis. A suitable scale would be 1 unit = 5°C.

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Now, we answer the questions by analyzing the data. Let's calculate the difference between the maximum and minimum temperature for each city:

• Ahmedabad: $38 - 29 = 9^\circ C$
• Amritsar: $37 - 26 = 11^\circ C$
• Bangalore: $28 - 21 = 7^\circ C$
• Chennai: $36 - 27 = 9^\circ C$
• Delhi: $38 - 28 = 10^\circ C$
• Jaipur: $39 - 29 = 10^\circ C$
• Jammu: $41 - 26 = 15^\circ C$
• Mumbai: $32 - 27 = 5^\circ C$

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

From the calculations above, the largest difference is $15^\circ C$.

The city with the largest temperature difference is Jammu.

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

The hottest city is the one with the highest maximum temperature. The highest maximum temperature is $41^\circ C$.

The hottest city is Jammu.

The coldest city is the one with the lowest minimum temperature. The lowest minimum temperature is $21^\circ C$.

The coldest city is Bangalore.

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

We need to find two cities, City A and City B, such that Maximum Temperature of City A < Minimum Temperature of City B.

Let's consider Bangalore, which has a maximum temperature of $28^\circ C$. Now we look for a city with a minimum temperature greater than $28^\circ C$.

Ahmedabad has a minimum temperature of $29^\circ C$.

Since $28^\circ C$ (Max of Bangalore) < $29^\circ C$ (Min of Ahmedabad), one such pair is Bangalore and Ahmedabad.

Another example: Jaipur has a minimum temperature of $29^\circ C$. So, another pair is Bangalore and Jaipur.

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

From the temperature differences calculated earlier, the smallest difference is $5^\circ C$.

The city with the least temperature difference is Mumbai.

Solution:

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(i) You are older today than yesterday.

This statement is always true because time passes continuously. Every person ages by a day from yesterday to today.

Classification: Certain to happen.

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(ii) A tossed coin will land heads up.

When a fair coin is tossed, there are two possible outcomes: heads or tails. While landing heads up is a possible outcome, it is not guaranteed to happen. It could also land tails up.

Classification: Can happen but not certain.

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

A standard die has faces numbered from 1 to 6. It does not have a face with the number 8.

Classification: Impossible.

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

Traffic lights typically cycle through red, yellow, and green. While seeing a green light is a possible outcome, it is not guaranteed. You might see a red light or a yellow light first.

Classification: Can happen but not certain.

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(v) Tomorrow will be a cloudy day.

Weather is unpredictable. Tomorrow might be cloudy, sunny, rainy, or something else. A cloudy day is a possible weather condition, but it is not certain to happen.

Classification: Can happen but not certain.

Solution:

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Given:

There are 6 marbles in a box, marked with numbers from 1 to 6.

The numbers on the marbles are {1, 2, 3, 4, 5, 6}.

Total number of possible outcomes (marbles in the box) = 6.

---

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

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

The favourable outcome is drawing a marble with the number 2.

There is only one marble with the number 2 on it.

Number of favourable outcomes = 1.

Total number of possible outcomes = 6.

Probability of drawing a marble with number 2 = $\frac{\text{Number of marbles with 2}}{\text{Total number of marbles}}$

Probability = $\frac{1}{6}$.

The probability of drawing a marble with number 2 is $\mathbf{\frac{1}{6}}$.

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

The favourable outcome is drawing a marble with the number 5.

There is only one marble with the number 5 on it.

Number of favourable outcomes = 1.

Total number of possible outcomes = 6.

Probability of drawing a marble with number 5 = $\frac{\text{Number of marbles with 5}}{\text{Total number of marbles}}$

Probability = $\frac{1}{6}$.

The probability of drawing a marble with number 5 is $\mathbf{\frac{1}{6}}$.

Solution:

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Given:

A coin is flipped to decide which team starts the game.

When a coin is flipped, there are two possible outcomes:

1. Heads (H)

2. Tails (T)

Total number of possible outcomes = 2.

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To Find:

The probability that your team will start.

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Solution:

Before flipping the coin, one team usually chooses 'Heads' and the other chooses 'Tails'. The team whose call matches the outcome of the flip gets to start the game.

Assume your team calls either 'Heads' or 'Tails'. There is only one outcome out of the two possibilities (Heads or Tails) that will result in your team starting.

If your team calls Heads, the favourable outcome is Heads (1 outcome).

If your team calls Tails, the favourable outcome is Tails (1 outcome).

In either case, the number of favourable outcomes for your team to start is 1.

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

Probability that your team will start = $\frac{\text{Number of outcomes where your team starts}}{\text{Total number of possible outcomes}}$

Probability = $\frac{1}{2}$.

The probability that your team will start is $\mathbf{\frac{1}{2}}$ or $\mathbf{0.5}$.