Chapter 3 Data Handling (Class 7 - Maths NCERT Exemplar Solutions)

Given:

Data set: $14, 6, 12, 17, 21, 10, 4, 3$

Solution:

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

$\text{Range} = 21 - 3$

$\text{Range} = 18$

Thus, the correct option is (c).

Given:

Data set: $23, 26, 22, 29, 23, 29, 26, 29, 22, 23$

Solution:

Mode is the observation that occurs most frequently in the data.

Let's count the frequency of each observation:

• $22$ appears $2$ times.

• $23$ appears $3$ times.

• $26$ appears $2$ times.

• $29$ appears $3$ times.

Since both $23$ and $29$ occur with the maximum frequency of $3$, they are both the modes of the data.

Thus, the correct option is (a).

Given:

Data set: $40, 50, 99, 68, 98, 60, 94$

Solution:

To find the median, we first arrange the data in ascending order:

$40, 50, 60, 68, 94, 98, 99$

Here, the number of observations ($n$) is $7$, which is odd.

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

$\text{Median} = \left( \frac{7 + 1}{2} \right)^{th} \text{ observation} = 4^{th} \text{ observation}$

The $4^{th}$ term in the sorted list is $68$.

Thus, the correct option is (c).

Given:

We need to find the mean of the first five prime numbers.

Solution:

The first five prime numbers are: $2, 3, 5, 7, 11$

$\text{Sum} = 2 + 3 + 5 + 7 + 11 = 28$

$\text{Mean} = \frac{28}{5}$

$\text{Mean} = 5.6$

The mean of first five prime numbers is $5.6$.

Given:

A die is thrown once.

Solution:

Total possible outcomes on a die are $\{1, 2, 3, 4, 5, 6\}$.

$\text{Number of total outcomes} = 6$

Favourable outcomes (numbers greater than $2$) are $\{3, 4, 5, 6\}$.

$\text{Number of favourable outcomes} = 4$

$\text{Probability} = \frac{\cancel{4}^2}{\cancel{6}_3} = \frac{2}{3}$

The probability of getting a number greater than $2$ is $\frac{2}{3}$.

Solution:

Let's check the frequency of observations:

• $23$ appears $3$ times.

• $26$ appears $2$ times.

• $12, 15, 19, 28, 38$ each appear only $1$ time.

Since $23$ occurs most often, the mode is $23$.

The statement says the mode is $28$, which is incorrect.

Therefore, the statement is False.

Solution (a):

The "most popular" size refers to the Mode. In the table, the maximum number of sweaters sold is $17$, which corresponds to Size 42.

Thus, the most popular size is $42$, not $17$ ($17$ is the frequency).

Therefore, statement (a) is False.

Solution (b):

The total number of observations ($N$) is $70$. The median would be the size corresponding to the middle terms ($\approx 35^{th}$ term).

The cumulative frequencies are: $15, 32, 45, 59, 70$.

The $35^{th}$ observation falls in the category of Size 44. $17$ is not a size, it is a frequency of a different size.

Therefore, statement (b) is False.

Solution:

First, arrange the data in ascending order:

$2, 4, 5, 6, 7, 8, 9$

The number of observations is $7$. The median is the middle observation.

The middle ($4^{th}$) observation is $6$.

The statement says the median is $2$, which is incorrect ($2$ is the minimum value).

Therefore, the statement is False.

Given:

The observations are: $3, 11, 7, 2, 5, 9, 9, 2, 10, 15, 7$

To Find:

The median of the given data.

Solution:

To find the median, we first arrange the data in ascending order:

$2, 2, 3, 5, 7, 7, 9, 9, 10, 11, 15$

The number of observations ($n$) is $11$, which is an odd number.

The formula for the median when $n$ is odd is:

Substituting $n = 11$:

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

$\text{Median} = \left( \frac{12}{2} \right)^{th} \text{ observation} = 6^{th} \text{ observation}$

The $6^{th}$ observation in the sorted data is $7$.

Therefore, the median of the data is $7$.

Given:

The observations are: $21, 15, 6, 25, 18, 13, 20, 9, 8, 12$

To Find:

The median of the given data.

Solution:

First, we arrange the data in ascending order:

$6, 8, 9, 12, 13, 15, 18, 20, 21, 25$

The number of observations ($n$) is $10$, which is an even number.

When $n$ is even, the median is the average of the two middle terms, i.e., the $\left( \frac{n}{2} \right)^{th}$ and $\left( \frac{n}{2} + 1 \right)^{th}$ observations.

$\frac{n}{2} = \frac{10}{2} = 5^{th} \text{ observation}$

$\frac{n}{2} + 1 = 6^{th} \text{ observation}$

In the sorted list, the $5^{th}$ observation is $13$ and the $6^{th}$ observation is $15$.

$\text{Median} = \frac{28}{2} = 14$

Therefore, the median of the data is $14$.

Given:

The word is "MATHEMATICS".

Letters: $M, A, T, H, E, M, A, T, I, C, S$

Solution (a):

The total number of possible outcomes is the total number of letters in the word "MATHEMATICS".

Counting the letters, we get $11$ letters.

Thus, there are $11$ possible outcomes.

Solution (b):

The probability $P(E)$ of an event $E$ occurring is given by:

(i) Probability of getting 'M':

Number of 'M's in the word = $2$ ($M, M$)

$P(M) = \frac{2}{11}$

(ii) Probability of getting any vowel:

Vowels in the word are: $A, E, A, I$

Number of favourable outcomes = $4$

$P(\text{Vowel}) = \frac{4}{11}$

(iii) Probability of getting any consonant:

Consonants in the word are: $M, T, H, M, T, C, S$

Number of favourable outcomes = $7$

$P(\text{Consonant}) = \frac{7}{11}$

(iv) Probability of getting 'X':

There is no letter 'X' in the word "MATHEMATICS".

Number of favourable outcomes = $0$

$P(X) = \frac{0}{11} = 0$

Given:

Observations: $26, 28, 25, x, 24$

Mean = $27$

Number of observations ($n$) = $5$

Solution:

The formula for the mean is:

Substituting the given values:

$27 = \frac{26 + 28 + 25 + x + 24}{5}$

Multiply both sides by $5$:

$27 \times 5 = 26 + 28 + 25 + 24 + x$

$135 = 103 + x$

To find $x$, subtract $103$ from $135$:

$x = 135 - 103$

The value of $x$ is $32$.

Given:

Number of observations ($n$) = $10$

Incorrect mean = $40$

Incorrect value copied = $15$

Correct value to be used = $45$

Solution:

First, we find the Incorrect sum of the observations:

$\text{Incorrect Sum} = \text{Mean} \times n$

$\text{Incorrect Sum} = 40 \times 10 = 400$

Now, we find the Correct sum by subtracting the incorrect value and adding the correct one:

$\text{Correct Sum} = \text{Incorrect Sum} - \text{Incorrect Value} + \text{Correct Value}$

$\text{Correct Sum} = 400 - 15 + 45$

$\text{Correct Sum} = 400 + 30 = 430$

Finally, we calculate the Correct mean:

$\text{Correct Mean} = \frac{430}{10}$

$\text{Correct Mean} = 43$

The correct mean is $43$.

Given:

Observations in ascending order: $11, 12, 14, 18, x + 2, 20, 22, 25, 61$

Median = $21$

To Find:

The value of $x$.

Solution:

Count the number of observations ($n$):

$n = 9$

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

From the given data, the $5^{th}$ observation is $x + 2$.

$x = 21 - 2$

$x = 19$

Solution:

(a) The double bar graph depicts the comparison of the cost (in $\textsf{₹}/\text{kg}$) of five different fruits (Apple, Banana, Mango, Watermelon, and Cherry) in two different cities, City I and City II.

(b) By comparing the heights of the bars for each fruit:

1. Apple: City I ($\textsf{₹} 82$) $>$ City II ($\textsf{₹} 75$)

2. Banana: City I ($\textsf{₹} 45$) $>$ City II ($\textsf{₹} 32$)

3. Mango: City I ($\textsf{₹} 75$) $>$ City II ($\textsf{₹} 60$)

4. Cherry: City I ($\textsf{₹} 38$) $>$ City II ($\textsf{₹} 30$)

The fruits for which the cost is greater in City I are Apple, Banana, Mango, and Cherry.

(c) Rate of apples in City I = $\textsf{₹} 82/\text{kg}$

Rate of apples in City II = $\textsf{₹} 75/\text{kg}$

$\text{Difference} = \textsf{₹} 82 - \textsf{₹} 75$

$\text{Difference} = \textsf{₹} 7$

The difference in rates for apples is $\textsf{₹} 7$.

(d) Cost of mangoes in City I = $\textsf{₹} 75$

Cost of mangoes in City II = $\textsf{₹} 60$

Simplifying the fraction by dividing both by $15$:

$\text{Ratio} = \frac{\cancel{75}^5}{\cancel{60}_4} = \frac{5}{4}$

The required ratio is $5 : 4$.

Solution:

Note: In this graph, "Won" (Light Blue) represents matches won by Country X, and "Lost" (Dark Blue) represents matches lost by Country X (which means wins for the opposing country).

(a) Maximum wins against Country X corresponds to the tallest "Lost" bar for Country X.

For Country B, the number of matches lost by X is $34$, which is the highest in the graph.

Thus, Country B has managed maximum wins against Country X.

(b) Let's calculate the absolute difference between matches won and lost for each country:

• Country A: $|15 - 32| = 17$

• Country B: $|16 - 34| = 18$

• Country C: $|14 - 8| = 6$

• Country D: $|7 - 11| = 4$

• Country E: $|3 - 6| = 3$

• Country F: $|10 - 3| = 7$

• Country G: $|10 - 30| = 20$

The difference is highest for Country G.

(c) From the graph:

Wins of Country X against Country E = $3$ (Light blue bar for E)

Losses of Country X against Country F = $3$ (Dark blue bar for F)

Therefore, the number of wins of Country X against Country E is the same as the number of losses of Country X against Country F.

To Find:

The section with the most improvement (Annual Result $-$ Half Yearly Result).

Solution:

Let's calculate the improvement for each section:

• Section A: $75\% - 62\% = 13\%$

• Section B: $66\% - 58\% = 8\%$

• Section C: $56\% - 70\% = -14\%$ (Result declined)

• Section D: $82\% - 74\% = 8\%$

• Section E: $65\% - 69\% = -4\%$ (Result declined)

By comparing the values, we can see that $13\%$ is the highest positive difference.

Thus, Section A showed the most improvement by $13\%$.

Solution:

The arithmetic mean of a set of data is defined as the sum of all observations divided by the total number of observations.

For the given observations $x, y,$ and $z$:

Thus, the correct option is (b).

Given:

Data set: $33, 38, 48, 33, 34, 34, 33, 24$

Solution:

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

Let's find the frequency of each observation:

• $24$ appears $1$ time.

• $33$ appears $3$ times.

• $34$ appears $2$ times.

• $38$ appears $1$ time.

• $48$ appears $1$ time.

Since $33$ has the highest frequency ($3$ times), it is the mode of the data.

Thus, the correct option is (c).

Solution:

1. Mean: The mean depends on the sum of all observations. If the largest and smallest (extreme) observations are removed, the sum changes significantly, thus the mean is always affected.

2. Median: The median is the middle-most value. Removing one observation from each end of an ordered list does not change the middle position (unless the data set is extremely small), so the median is generally not affected.

3. Mode: If one of the extreme observations removed happened to be the most frequent value (the mode), then the mode will be affected.

In most statistical contexts, the measures most sensitive to outliers or extreme values are the mean and potentially the mode.

Thus, the correct option is (a).

Given:

Data set: $21, 6, 17, 18, 12, 8, 4, 13$

Solution:

Range is the difference between the highest and the lowest observations in the data.

$\text{Range} = 21 - 4$

$\text{Range} = 17$

Thus, the correct option is (a).

Given:

Data set: $3, 4, 5, 6, 7, 3, 4$

Solution:

To find the median, first arrange the data in ascending order:

$3, 3, 4, 4, 5, 6, 7$

The number of observations ($n$) is $7$, which is odd.

$\text{Median} = \left( \frac{7 + 1}{2} \right)^{th} \text{ observation} = 4^{th} \text{ observation}$

The $4^{th}$ observation in the sorted data is $4$.

Thus, the correct option is (c).

Solution:

In market research, to find the "most liked" or "most popular" item, we look for the item that appears with the highest frequency in the survey data.

The measure of central tendency that identifies the most frequent value is the Mode.

Thus, the correct option is (b).

Given:

Each set contains $2$ aces. We want to be "certain" to pick both aces in the first attempt.

Solution:

An event is certain only when the probability of occurrence is $1$. For us to be certain to pick $2$ aces by picking $2$ cards, the total number of cards in that set must be exactly $2$.

Analyzing the images:

• Set (a) contains $4$ cards.

• Set (b) contains $5$ cards.

• Set (c) contains $2$ cards.

• Set (d) contains $6$ cards.

In set (c), since there are only $2$ cards and both are aces, picking $2$ cards will guarantee picking both aces.

Thus, the correct option is (c).

Given:

Total number of cards in set (d) = $6$

Number of aces in each set = $2$

Solution:

The probability of an event is given by:

For set (d):

$\text{Favourable outcomes (Aces)} = 2$

$\text{Total outcomes (Cards)} = 6$

$P(\text{Picking an Ace}) = \frac{2}{6}$

Thus, the correct option is (b).

Solution:

In statistics, the difference between the maximum (highest) value and the minimum (lowest) value in a data set is defined as the range.

Thus, the correct option is (c).

Given:

Number of students who can participate = $2$

Total number of students = $5$

Solution:

The probability or "chance" of a student being picked is:

$\text{Probability} = \frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

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

To convert this chance into a percentage, we multiply by $100$:

$\text{Percentage chance} = \frac{2}{5} \times 100\%$

$\text{Percentage chance} = 2 \times 20\% = 40\%$

Thus, the correct option is (b).

Given:

The integers shown on the board (Fig. 3.3) are: $0, 15, -11, -17, +20, 6, -4$

Solution:

To find the range, we identify the highest and lowest integers:

The range is the difference between these two values:

$\text{Range} = 20 - (-17)$

$\text{Range} = 20 + 17 = 37$

Thus, the correct option is (b).

Solution:

A fair coin has two distinct faces: a Head and a Tail.

When a coin is tossed, it is a random experiment where the result will be one of these two possibilities. Therefore, the outcome is either head or tail.

Thus, the correct option is (d).

Given:

Mean of three numbers = $40$

Total number of observations = $3$

Lowest natural number = $19$

Solution:

First, find the sum of the three numbers:

$\text{Sum} = \text{Mean} \times \text{Count} = 40 \times 3 = 120$

Let the three different natural numbers be $x, y,$ and $z$, where $x < y < z$.

Given $x = 19$, we have:

$19 + y + z = 120$

$y + z = 120 - 19 = 101$

To make $z$ (the highest number) as large as possible, we must make $y$ as small as possible. Since the numbers are different natural numbers and $19$ is the lowest, the smallest possible value for $y$ is $20$.

Substituting $y = 20$:

$20 + z = 101$

$z = 101 - 20 = 81$

Thus, the highest possible number is $81$, which matches option (a).

Given:

Scores in first three exams: $97, 73, 88$

Score in fourth exam: $80$

Solution:

1. Initial Average:

$\text{Initial Sum} = 97 + 73 + 88 = 258$

$\text{Initial Average} = \frac{258}{3} = 86$

2. New Average:

$\text{New Sum} = 258 + 80 = 338$

$\text{New Average} = \frac{338}{4} = 84.5$

3. Comparison:

$\text{Difference} = 86 - 84.5 = 1.5$

Since the new average is lower, the average score is decreased by 1.5.

Thus, the correct option is (d).

Solution:

To determine who is the "most popular," we look for the politician who receives the highest number of votes or preferences. This is a measure of the most frequent response in a data set.

The mode is the measure of central tendency used to identify the most frequent observation.

Thus, the correct option is (c).

Solution:

Let's check option (d):

Data: $4, 3, 4, 3, 4, 6, 4$

Mean:

$\text{Sum} = 4 + 3 + 4 + 3 + 4 + 6 + 4 = 28$

$\text{Mean} = \frac{28}{7} = 4$

Median:

Arrange in ascending order: $3, 3, 4, 4, 4, 4, 6$

Middle observation ($4^{th}$ term) is $4$.

Mode:

The number $4$ occurs $4$ times, which is the highest frequency. Thus, $\text{Mode} = 4$.

Since Mean = Median = Mode = 4, this data set satisfies the condition.

Thus, the correct option is (d).

Solution:

The difference between the highest and the lowest observations of a data is called its Range.

Solution:

The mean of a data is defined as the sum of all observations divided by the total number of observations.

Mathematically, it is expressed as:

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

Solution:

In a set of observations, the observation that occurs the most often is called the Mode.

Solution:

In a given data, arranged in ascending or descending order, the middle-most observation is called the Median.

Solution:

Mean, Median, and Mode are the measures of Central Tendency.

Solution:

The probability of an event which is certain to happen is $1$.

Solution:

The probability of an event which is impossible to happen is $0$.

Solution:

When a standard die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

Since all these numbers are less than $7$, every outcome is a favourable outcome.

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

The probability of getting a number less than $7$ is $1$.

A standard die is a cube with six faces, each marked with a different number from $1$ to $6$. When a die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

In Throwing a die the number of possible outcomes is $6$.

To compare two different sets of data (such as marks of students in two different terms) simultaneously, we use a specific graphical tool.

Double bar graphs can be used to compare two collections of data.

A visual representation of data where rectangular bars of equal width are drawn with heights proportional to the values they represent is a common statistical tool.

The representation of data with bars of uniform width is called bar graph.

Given:

Observations: $8, 4, x, 6, 2, 7$

Number of observations ($n$) = $6$

Mean = $5$

To Find:

The value of $x$.

Solution:

We know that:

Substituting the values:

$5 = \frac{8 + 4 + x + 6 + 2 + 7}{6}$

$30 = 27 + x$

$x = 30 - 27$

$x = 3$

If the arithmetic mean of $8, 4, x, 6, 2, 7$ is $5$, then the value of $x$ is $3$.

The median is a measure of central tendency. Since it represents the middle-most value of the data, it must fall within the range of the provided data points.

The median of any data lies between the lowest and highest observations.

When we calculate the median of a data set with $n$ observations:

1. If $n$ is odd, the median is exactly the $\left( \frac{n+1}{2} \right)^{th}$ observation, which belongs to the data set.

2. If $n$ is even, the median is the average of two middle terms, which might not be a value present in the data.

Median is one of the observations in the data if number of observations is odd.

Given Table:

Analysis:

Probability is directly proportional to frequency in a random selection. The highest frequency in the table is $25$.

This frequency corresponds to the weight interval of $52 - 55$ kg.

A student is to be selected randomly from his class for some competition. The probability of selection of the student is highest whose weight is in the interval $52 - 55$.

Solution:

A standard die has six faces numbered $1, 2, 3, 4, 5,$ and $6$. The maximum possible number is $6$.

Getting a number greater than $6$ is an impossible event.

The probability of an impossible event is $0$.

Therefore, the statement is False.

Solution:

When a coin is tossed, the possible outcomes are either a Head (H) or a Tail (T).

Thus, there are exactly $2$ possible outcomes.

Therefore, the statement is True.

Solution:

The median is the middle-most value of an ordered data set.

If we remove one observation from the lowest end and one from the highest end, the center of the data remains at the same position.

The median is known for being a robust measure that is not affected by extreme values or outliers.

Therefore, the statement is False.

Solution:

Measures of central tendency (Mean, Median, and Mode) are values that represent the "center" of a data set.

By definition, these values always lie between the minimum (lowest) and maximum (highest) observations of the given data.

Therefore, the statement is False.

Solution:

When two dice are thrown together, the maximum number on each die is $6$.

The maximum possible sum is $6 + 6 = 12$.

Since the maximum sum is $12$, it is impossible to obtain a sum of $14$.

Therefore, the statement is True.

Solution:

Observe Fig. 3.4. The circular spinner is divided into $4$ equal sectors.

Out of these $4$ sectors, $2$ sectors are shaded.

$\text{Total outcomes} = 4$

$\text{Favourable outcomes (shaded)} = 2$

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

Therefore, the statement is True.

Given:

Total tosses = $15$

Recorded outcomes: H, T, T, H, T, H, H, H, T, T, H, T, H, T, T

Solution:

Let's count the number of Heads (H) in the given data:

$\text{Count of H} = 7$

The chance (experimental probability) in percentage is calculated as:

$\text{Chance of Head} = \frac{\text{Number of Heads}}{\text{Total Tosses}} \times 100$

$\text{Chance} = \frac{7}{15} \times 100 \approx 46.67\%$

Since $46.67\%$ is not $50\%$, the statement is False.

Solution:

In a perfectly symmetrical distribution (like the data set $\{2, 2, 2\}$ or $\{1, 2, 3, 3, 3, 4, 5\}$), the Mean, Median, and Mode can indeed be the same value.

Example: For the data $2, 2, 2$

$\text{Mean} = 2, \text{Median} = 2, \text{Mode} = 2$

Therefore, the statement is True.

Solution:

In probability theory, the probability of any event $E$ is always bound by the range:

$0 \leq P(E) \leq 1$

The probability of an event can never be greater than 1.

Therefore, the statement is False.

The mean is the average of the observations. It is calculated as the sum of all values divided by the total count. This result does not necessarily have to be one of the values present in the data set.

For example, the mean of $1$ and $2$ is $\frac{1+2}{2} = 1.5$, which is not in the data.

Therefore, the statement is False.

When the number of observations is odd, the median is the middle observation, which is part of the data. However, when the number of observations is even, the median is the average of the two middle terms, which may not be a value from the data.

Therefore, the statement is True.

The mode is defined as the observation that occurs most frequently in a given data set. Since it must be a value that exists in the set to be counted as the most frequent, it is always one of the observations from the given data.

Therefore, the statement is True.

The mean represents the central value of the data. It always lies between the minimum and the maximum observations of the set. It cannot be smaller than the smallest observation.

Therefore, the statement is False.

The mean is a quotient of the sum of observations and the number of observations. If the sum is not perfectly divisible by the number of observations, the mean will be a fraction or a decimal.

Therefore, the statement is False.

The range is the difference between the highest and lowest observations. While the highest and lowest values are from the data, their difference (the range) does not have to be a value present in the data set.

For example, in the data $\{10, 20\}$, the range is $20 - 10 = 10$. But in $\{5, 12\}$, the range is $12 - 5 = 7$, which is not in the data.

Therefore, the statement is False.

The mode is the observation with the highest frequency. In the data $12, 13, 14, 15, 16$, every observation occurs exactly once. In such cases where all values have the same frequency, the data is said to have no mode.

Therefore, the statement is False.

The range is the difference between the maximum and minimum values. If we subtract a constant (like $2$) from every observation, both the maximum and minimum values decrease by the same amount.

$\text{New Range} = (\text{Max} - 2) - (\text{Min} - 2) = \text{Max} - \text{Min} $$ = \text{Original Range}$.

The range does not change when a constant is subtracted from all values.

Therefore, the statement is False.

If a constant value (like $8$) is added to every observation in a data set, the difference between the new maximum and the new minimum remains the same as the original difference.

$\text{New Range} = (\text{Max} + 8) - (\text{Min} + 8) = \text{Max} - \text{Min} $$ = \text{Original Range}$.

The range remains unchanged when a constant is added to all values.

Therefore, the statement is False.

Given:

Observations: $5, 10, 10, 12, 13$

Number of observations ($n$) = $5$

Solution:

1. To find the Mean:

$\text{Sum} = 5 + 10 + 10 + 12 + 13 = 50$

$\text{Mean} = \frac{50}{5} = 10$

2. To find the Median:

Arranging the data in ascending order: $5, 10, 10, 12, 13$

Since $n = 5$ (odd), the Median is the $\left( \frac{5 + 1}{2} \right)^{th} = 3^{rd}$ observation.

$\text{Median} = 10$

3. To find the Mode:

Mode is the most frequent observation. Here, $10$ occurs twice, while other values occur once.

$\text{Mode} = 10$

Conclusion:

Since Mean = $10$, Median = $10$, and Mode = $10$, yes, all three measures are equal for this data.

Given:

The first ten even natural numbers are: $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$.

Total number of observations ($n$) = $10$

Solution:

$\text{Sum} = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20$

$\text{Sum} = 110$

$\text{Mean} = \frac{110}{10} = 11$

The mean of the first ten even natural numbers is $11$.

Solution:

In a data set containing the heights of $50$ children, the Mode represents the height that is most common among the children.

It identifies the specific height measurement that occurs with the highest frequency. For example, if $15$ out of $50$ children are exactly $140$ cm tall, and no other height value has a frequency greater than $15$, then $140$ cm is the mode.

Solution (a):

To find the most liked colour, we look for the colour with the highest number of sales (frequency). According to the table, the maximum number of cars sold is $20$.

This corresponds to the colour Black.

Solution (b):

The measure of central tendency that identifies the most frequent value in a data set is the Mode.

Given:

Data: $31, 37, 35, 38, 42, 23, 17, 18, 35, 25, 35, 29$

Number of observations ($n$) = $12$

Sorted Data: $17, 18, 23, 25, 29, 31, 35, 35, 35, 37, 38, 42$

Solution (a) Range:

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

$\text{Range} = 42 - 17 = 25$

Solution (b) Mean:

$\text{Sum} = 17+18+23+25+29+31+35+35+35+37+38+42 $$ = 365$

$\text{Mean} = \frac{365}{12} \approx 30.42$

Solution (c) Median:

Since $n=12$ (even), Median is the average of the $\left(\frac{12}{2}\right)^{th}$ and $\left(\frac{12}{2}+1\right)^{th}$ observations.

$6^{th}$ observation = $31$

$7^{th}$ observation = $35$

$\text{Median} = \frac{31 + 35}{2} = \frac{66}{2} = 33$

Solution (d) Mode:

The observation $35$ appears $3$ times, which is the highest frequency.

$\text{Mode} = 35$

Given:

Weights: $70, 62, 54, 57, 62, 84, 75, 59, 62, 65, 78, 60$

Number of people = $12$

Solution (a):

$\text{Sum of weights} = 70+62+54+57+62+84+75+59+62 $$ +65+78+60 = 788 \text{ kg}$

$\text{Mean} = \frac{788}{12} \approx 65.67 \text{ kg}$

Solution (b):

We need to count how many observations are greater than $65.67$.

Weights above mean: $70, 84, 75, 78$

The number of people is $4$.

Solution (c):

$\text{Highest weight} = 84 \text{ kg}$

$\text{Lowest weight} = 54 \text{ kg}$

$\text{Range} = 84 - 54 = 30 \text{ kg}$

Given:

A set of cards with letters: A, E, I, O, U

Solution:

The probability $P(E)$ of an event is calculated as:

(a) A vowel:

All the letters in the set $\{A, E, I, O, U\}$ are vowels.

$\text{Favourable outcomes} = 5$

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

(b) A or I:

The favourable outcomes are $\{A, I\}$.

$\text{Favourable outcomes} = 2$

$P(A \text{ or } I) = \frac{2}{5}$

(c) A card marked U:

The favourable outcome is $\{U\}$.

$\text{Favourable outcomes} = 1$

$P(U) = \frac{1}{5}$

(d) A consonant:

There are no consonants in the set $\{A, E, I, O, U\}$.

$\text{Favourable outcomes} = 0$

$P(\text{consonant}) = \frac{0}{5} = 0$

Given Data:

$6, 2, 5, 4, 3, 4, 4, 2, 3$

Total number of observations $(n) = 9$

Solution:

1. Finding the Mode:

Mode is the observation that appears most frequently.

$\text{Frequency of 4} = 3$

Since $4$ appears the most, Mode = 4.

2. Finding the Median:

First, arrange the data in ascending order:

$2, 2, 3, 3, 4, 4, 4, 5, 6$

Since $n = 9$ (odd), Median is the $\left( \frac{n + 1}{2} \right)^{th}$ term.

$\text{Median} = 5^{th} \text{ term} = 4$

3. Finding the Mean of Median and Mode:

$\text{Required Mean} = \frac{\text{Median} + \text{Mode}}{2}$

$\text{Required Mean} = \frac{4 + 4}{2} = \frac{8}{2} = 4$

The mean of the median and mode is $4$.

Given:

Observations: $5, 7, 7, 8, x, 5, 4, 3, 1, 2$

Total observations $(n) = 10$

$\text{Mean} = 4.5$

To Find:

The value of $x$ and the Median.

Solution:

First, calculate $x$ using the mean formula:

$4.5 = \frac{5 + 7 + 7 + 8 + x + 5 + 4 + 3 + 1 + 2}{10}$

$45 = 42 + x$

$x = 45 - 42 = 3$

Now, arrange the data in ascending order to find the Median:

$1, 2, 3, 3, 4, 5, 5, 7, 7, 8$

Since $n = 10$ (even), the median is the average of the $5^{th}$ and $6^{th}$ observations.

$\text{Median} = \frac{9}{2} = 4.5$

The median of the given data is $4.5$.

Solution:

The setting of the sun is a universal truth and a certain event. It happens every day due to the rotation of the Earth.

The probability of an event that is certain to happen is always $1$.

Therefore, the probability of the sun setting tomorrow is $1$.

Given:

A spinner divided into sectors representing colours: R (Red), B (Blue), and Y (Yellow).

Solution:

The chance of an arrow stopping on a specific colour depends on the area or the central angle of that colour's sector.

By observing Fig. 3.5, we can see that:

1. The sector for Yellow (Y) covers exactly half of the circular spinner ($180^\circ$).

2. The remaining half is shared between Red (R) and Blue (B).

Since the area covered by the Yellow colour is the largest compared to Red and Blue, the arrow has the highest probability of stopping in the Yellow region.

Thus, Yellow has more chance to show up than the others.

Given:

Number of girls = $3$

Number of boys = $4$

To Find:

Probability that a student chosen at random is a boy.

Solution:

First, we calculate the total number of students in the group:

$\text{Total number of students} = 3 + 4 = 7$

The probability $P(E)$ of choosing a boy is given by the formula:

$\text{Number of favourable outcomes (boys)} = 4$

$\text{Total number of outcomes (students)} = 7$

$P(\text{Boy}) = \frac{4}{7}$

The probability that the chosen student is a boy is $\frac{4}{7}$.

Given:

Word: MEDIAN

Letters in the word: $M, E, D, I, A, N$

Solution:

Total number of letters in the word "MEDIAN" = $6$

Number of times the letter 'D' appears in the word = $1$

Probability $P(E)$ is defined as:

$P(D) = \frac{1}{6}$

The probability of drawing the letter D is $\frac{1}{6}$.

Solution:

(a) Getting a number less than 1 on throwing a die:

A standard die has numbers from $1$ to $6$. There is no number less than $1$.

Classification: Impossible to happen.

(b) Getting head when a coin is tossed:

A coin can result in either a head or a tail. We cannot say for sure it will be a head.

Classification: May or may not happen.

(c) A team winning the match:

The result of a sports match is uncertain and depends on the game's outcome.

Classification: May or may not happen.

(d) Christmas will be on 25 December:

Christmas is a fixed annual date celebrated on December 25th.

Classification: Certain to happen.

(e) Today moon will not revolve around the earth:

The moon's revolution around the Earth is a constant natural law of physics.

Classification: Impossible to happen.

(f) A ball thrown up in the air will fall down after some time:

Due to the Earth's gravity, objects thrown up must return to the ground.

Classification: Certain to happen.

Given:

Observations: $5, 3, 4, 1, 2, 6, 4, 2, 2, 3, 1, 5, 6, 1, 2$

Total number of observations $(n) = 15$

Solution:

1. Mean:

Sum $= 5+3+4+1+2+6+4+2+2+3+1+5+6+1+2 = 47$

$\text{Mean} = \frac{47}{15} \approx 3.13$

2. Median:

First, arrange the data in ascending order:

$1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6$

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

$\text{Median} = 8^{th} \text{ term} = 3$

3. Mode:

The observation that occurs most frequently:

• $1$ appears $3$ times.

• $2$ appears $4$ times.

• $3, 4, 5, 6$ each appear $2$ times.

$\text{Mode} = 2$

Given:

The first six multiples of $4$ are: $4, 8, 12, 16, 20, 24$

Total number of observations $(n) = 6$

Solution:

$\text{Sum} = 4 + 8 + 12 + 16 + 20 + 24 = 84$

$\text{Mean} = \frac{84}{6}$

$\text{Mean} = 14$

The mean of the first six multiples of $4$ is $14$.

Given:

The first nine even natural numbers are: $2, 4, 6, 8, 10, 12, 14, 16, 18$

Number of observations ($n$) = $9$

To Find:

The median of the given data.

Solution:

The data is already arranged in ascending order: $2, 4, 6, 8, 10, 12, 14, 16, 18$.

Since the number of observations ($n$) is odd, the median is calculated using the following formula:

Substituting $n = 9$:

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

$\text{Median} = \left( \frac{10}{2} \right)^{th} \text{ observation} = 5^{th} \text{ observation}$

In the ordered list, the $5^{th}$ observation is $10$.

The median of the first nine even natural numbers is $10$.

Given:

Mean of first three numbers ($n_1 = 3$) = $10$

Mean of other four numbers ($n_2 = 4$) = $12$

To Find:

Mean of all the numbers ($N = 3 + 4 = 7$).

Solution:

First, calculate the sum of the first set of numbers:

$\text{Sum of first three numbers} = \text{Mean} \times n_1$

$\text{Sum}_1 = 10 \times 3 = 30$

Next, calculate the sum of the second set of numbers:

$\text{Sum of next four numbers} = \text{Mean} \times n_2$

$\text{Sum}_2 = 12 \times 4 = 48$

Now, find the total sum and total count of numbers:

$\text{Total Sum} = 30 + 48 = 78$

$\text{Total numbers} (N) = 3 + 4 = 7$

Finally, find the combined mean:

$\text{Mean} = \frac{78}{7} = 11 \frac{1}{7} \approx 11.14$

The mean of all the numbers is $11.14$.

Given:

Observations: $10, 8, 4, 7, 8, 11, 15, 8, 4, 2, 3, 6, 8$

Solution:

To find the Mode, let us count the frequency of each observation:

• $2$ appears $1$ time

• $3$ appears $1$ time

• $4$ appears $2$ times

• $6$ appears $1$ time

• $7$ appears $1$ time

• $8$ appears $4$ times

• $10$ appears $1$ time

• $11$ appears $1$ time

• $15$ appears $1$ time

The observation $8$ has the highest frequency of $4$.

The mode of the given data is $8$.

Given Data:

Heights (in cm): $128, 144, 146, 143, 136, 142, 138, 129, 140, 152, 144, 140, 150, 142, 154$

Number of boys ($n$) = $15$

Solution:

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

$128, 129, 136, 138, 140, 140, 142, 142, 143, 144, 144, 146, 150, 152, 154$

(a) Height of the tallest boy:

The maximum value in the data is $154$ cm.

(b) Height of the shortest boy:

The minimum value in the data is $128$ cm.

(c) Range of the given data:

$\text{Range} = \text{Tallest height} - \text{Shortest height}$

$\text{Range} = 154 - 128 = 26$ cm

(d) Median height of the boys:

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

$\text{Median} = 8^{th} \text{ observation}$

Looking at our sorted list, the $8^{th}$ value is $142$ cm.

Analysis of current data:

Observations: $16, 15, 16, 16, 8, 15, 17$

Frequencies:

• $8$ appears $1$ time

• $15$ appears $2$ times

• $16$ appears $3$ times

• $17$ appears $1$ time

Current Mode = 16 (since it appears 3 times).

Solution (a):

To keep the mode as $16$, we can add any value that does not result in another number appearing $3$ or more times. However, adding $16$ itself is the most certain way to ensure the mode remains $16$. Other values like $8$ or $17$ could also be added once.

Solution (b):

Currently, $16$ appears $3$ times and $15$ appears $2$ times. To make $15$ the mode, its frequency must exceed the frequency of $16$.

We need at least two values of $15$. This would make the frequency of $15$ equal to $4$, while $16$ remains at $3$.

Solution (c):

Currently, $16$ appears $3$ times and $17$ appears only $1$ time. To make $17$ the mode, it must appear more times than $16$.

We need to add three values of $17$. This would make the frequency of $17$ equal to $4$ ($1$ original $+ 3$ added), making it the mode.

Given:

Age of children in Group A: $7, 7, 9, 8, 10, 10$

Age of children in Group B: $7, 9, 11, 12, 12, 12$

Solution (a):

For Group A:

Arranging in ascending order: $7, 7, 8, 9, 10, 10$

Mode: The values $7$ and $10$ both occur twice. Thus, it is bimodal.

$\text{Mode} = 7 \text{ and } 10 \text{ years}$

$\text{Range} = \text{Maximum} - \text{Minimum} = 10 - 7 = 3 \text{ years}$

For Group B:

Arranging in ascending order: $7, 9, 11, 12, 12, 12$

Mode: The value $12$ occurs three times.

$\text{Mode} = 12 \text{ years}$

$\text{Range} = 12 - 7 = 5 \text{ years}$

Solution (b):

Combined Data:

Combined ages: $7, 7, 9, 8, 10, 10, 7, 9, 11, 12, 12, 12$

Arranging in ascending order: $7, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 12$

Mode: The values $7$ and $12$ both occur three times.

$\text{Mode} = 7 \text{ and } 12 \text{ years}$

$\text{Range} = 12 - 7 = 5 \text{ years}$

Solution:

(a) The bar graph depicts the monthly production of Motor Bikes by XYZ Automobiles Ltd over a period of six months (January to June).

(b) Production in the first three months:

January: $600$

February: $800$

March: $700$

$\text{Total} = 600 + 800 + 700 = 2100 \text{ bikes}$

(c) Increase in May over January:

Production in May = $900$

Production in January = $600$

$\text{Increase} = 900 - 600 = 300 \text{ bikes}$

(d) By observing the heights of the bars, the production was minimum in June, and it was $500$ bikes.

(e) Average production:

$\text{Total production} = 600 + 800 + 700 + 1100 + 900 + 500 = 4600 \text{ bikes}$

$\text{Mean} = \frac{4600}{6} \approx 766.67 \text{ bikes}$

Solution:

(a) If $40$ is the pass mark, students scoring in the $30-39$ interval failed. From the graph, there are $4$ such students.

(b) Students with marks $50$ to $69$ fall in the $50-59$ and $60-69$ intervals.

$\text{Number of students} = 7 + 11 = 18$

(c) Students scoring $90$ and above fall in the $90-92$ interval. There are $4$ such students.

(d) Merit holders (marks $> 80$) fall in the $80-89$ and $90-92$ intervals.

$\text{Merit holders} = 6 + 4 = 10$

(e) Strength of the class is the sum of frequencies of all intervals:

$\text{Strength} = 4 + 2 + 7 + 11 + 8 + 6 + 4 = 42 \text{ students}$

Solution:

(a) The bar graph represents the annual production of rice by a country (in million tonnes) from the year $2005$ to $2009$.

(b) The production was least in the year $2006$ (with $40$ million tonnes).

(c) The maximum rise occurred from $2006$ to $2007$, where production jumped from $40$ to $70$ million tonnes. Thus, the rise was maximum after the year $2006$.

(d) Average production:

$\text{Total production} = 50 + 40 + 70 + 50 + 60 = 270 \text{ million tonnes}$

$\text{Average} = \frac{270}{5} = 54 \text{ million tonnes}$

(e) Production in 2006 = $40$ million tonnes

Production in 2008 = $50$ million tonnes

$\text{Difference} = 50 - 40 = 10 \text{ million tonnes}$

Solution:

(a) The graph depicts the marks obtained by a student in five different subjects (English, Hindi, Maths, Science, S.Science) out of $100$ each.

(b) The student is very good in Maths, as it has the highest score of $82$.

(c) Average marks:

$\text{Total marks} = 64 + 75 + 82 + 71 + 49 = 341$

$\text{Average} = \frac{341}{5} = 68.2$

(d) Subjects with marks $\geq 75$ are Hindi ($75$) and Maths ($82$).

(e) Percentage Calculation:

$\text{Percentage} = \frac{\text{Total Marks}}{\text{Maximum Marks}} \times 100$

$\text{Percentage} = \frac{341}{500} \times 100 = 68.2\%$

Solution:

Based on the observation of the bar graph (Fig. 3.10), the circulation numbers (in hundreds) for different languages are:

• Urdu: $200$

• Tamil: $100$

• English: $500$

• Hindi: $800$

• Marathi: $300$

• Punjabi: $400$

(a) Total newspapers in Hindi, Punjabi, Urdu, Marathi and Tamil:

$\text{Total} = 800 + 400 + 200 + 300 + 100$

$\text{Total} = 1800$

(b) Excess number of newspapers in Hindi than English:

$\text{Difference} = \text{Circulation in Hindi} - \text{Circulation in English}$

$\text{Difference} = 800 - 500 = 300$

(c) Language with the least circulation:

By observing the graph, the shortest bar corresponds to Tamil (with $100$ newspapers).

(d) Total circulation of newspapers in the town:

$\text{Total} = 200 + 100 + 500 + 800 + 300 + 400$

$\text{Total} = 2300$

Solution:

From Fig. 3.11, the values for each class and sport are as follows:

(a) The tallest dark blue bar is for Cricket ($21$ students). So, Cricket is liked the most by Class VIII.

(b) Total Class VII students liking Hockey and Tennis:

$\text{Total} = 7 \text{ (Hockey)} + 10 \text{ (Tennis)} = 17$ students.

(c) Total students in Class VII:

$\text{Total} = 7 + 16 + 18 + 10 + 14 = 65$ students.

(d) Comparing the bars, the Class VII bar is shorter than the Class VIII bar only for Cricket ($18 < 21$).

(e) Class VIII students are less than Class VII for Hockey, Football, Tennis, and Badminton. Total = $4$ sports.

(f) Ratio Calculation:

$\text{Class VII Badminton} = 14$

$\text{Class VIII Tennis} = 7$

The ratio is $2 : 1$.

Solution:

(a) The graph represents a comparison of monthly sales (in Lakh $\textsf{₹}$) of two different brands, Brand A and Brand B, from January to June.

(b) Sales of Brand A (light blue): Jan ($31$), Feb ($34$), Mar ($30$). The sales decreased in March.

(c) Sales for June:

Brand A = $\textsf{₹}$ $57$ Lakh

Brand B = $\textsf{₹}$ $54$ Lakh

$\text{Difference} = 57 - 54 = \textsf{₹} 3$ Lakh.

(d) Average sales of Brand B:

$\text{Sum} = 36 + 38 + 43 + 35 + 45 + 54 = 251$

$\text{Mean} = \frac{251}{6} \approx \textsf{₹} 41.83$ Lakh.

(e) Sales of Brand B was less than Brand A in April (Brand B=$35$, Brand A=$40$) and June (Brand B=$54$, Brand A=$57$).

(f) Ratio for January:

Brand A = $31$, Brand B = $36$

The ratio is $31 : 36$.

Solution:

(a) The graph compares the minimum temperatures (in $^\circ C$) across four months (November, December, January, and February) for the years 2008 and 2009.

(b) Ratio for November:

Temp in 2008 = $18^\circ C$

Temp in 2009 = $15^\circ C$

The ratio is $6 : 5$.

(c) Temp (2008) $>$ Temp (2009) in:

1. November ($18 > 15$)

2. February ($12 > 8$)

Total = $2$ months (November and February).

(d) Average for 2008:

$\text{Sum} = 18 + 11 + 4 + 12 = 45$

$\text{Mean} = \frac{45}{4} = 11.25^\circ C$.

(e) Calculating variation (difference):

• Nov: $|18 - 15| = 3$

• Dec: $|11 - 12| = 1$

• Jan: $|4 - 5| = 1$

• Feb: $|12 - 8| = 4$

The variation is maximum in February.

To Construct:

A double bar graph representing the comparison between rural and urban nutrient intake.

Proposed Scale:

On the Y-axis: $1$ unit $= 20$ calories.

On the X-axis: Foodstuff categories.

Graph Placeholder:

Summary Observation:

From the provided data, we can observe that the Urban group consistently has a higher intake of nutrients across all foodstuff categories compared to the Rural group, with the most significant difference seen in Milk intake.

Given:

A double bar graph representing the number of girls (light blue) and boys (dark blue) in different sections of Class VII (A, B, C, D, and E).

Solution:

(a) The double bar graph represents the comparison between the number of girls and the number of boys in five different sections of Class VII.

(b) Total number of boys in all sections:

From the graph, the number of boys in each section is:

VII A = $15$, VII B = $30$, VII C = $20$, VII D = $20$, VII E = $25$

$\text{Total Boys} = 15 + 30 + 20 + 20 + 25 = 110$

There are $110$ boys in total.

(c) Sections where Girls $>$ Boys:

Comparing the heights of the bars:

• In Section VII A: Girls ($20$) $>$ Boys ($15$)

• In Section VII D: Girls ($25$) $>$ Boys ($20$)

The sections are VII A and VII D.

(d) Section with maximum boys:

The tallest dark blue bar is for section VII B, which has $30$ boys.

(e) Section with least girls:

The shortest light blue bar is for section VII C, which has $15$ girls.

Solution (a):

To represent this data, we use a double bar graph where the X-axis represents the days and the Y-axis represents the number of readers.

Scale: $1$ unit on Y-axis $= 100$ readers.

Solution (b):

We calculate the total number of readers for each day:

• Mon: $400 + 150 = 550$

• Tue: $600 + 100 = 700$

• Wed: $350 + 200 = 550$

• Thurs: $550 + 300 = 850$

• Fri: $500 + 250 = 750$

• Sat: $350 + 200 = 550$

The total number of readers was maximum on Thursday ($850$).

Solution (c):

$\text{Sum of magazine readers} = 150 + 100 + 200 + 300 + 250 + 200 $$ = 1200$

$\text{Number of days} = 6$

$\text{Mean} = 200$

The mean number of magazine readers is $200$.

Solution (a):

Inference: From the double bar graph, we can infer that the attendance in the school is very high across all classes, as the bars for "Present" students are almost as tall as the "Total Students" bars.

Solution (b):

By looking at the "Total Students" column, Class VIII has the maximum number of students ($95$).

Solution (c):

Calculating the difference (Absent students) for each class:

• VI: $90 - 81 = 9$

• VII: $82 - 76 = 6$

• VIII: $95 - 91 = 4$

• IX: $70 - 65 = 5$

• X: $63 - 62 = 1$

The difference is minimum in Class X.

Solution (d):

For Class IX: Present $= 65$, Total $= 70$.

$\text{Ratio} = \frac{65}{70} = \frac{\cancel{65}^{13}}{\cancel{70}_{14}}$

The ratio is $13 : 14$.

Solution (e):

For Class VI: Total $= 90$, Absent $= 9$.

$\text{Percentage Absent} = \left( \frac{9}{90} \right) \times 100\%$

$\text{Percentage Absent} = \frac{1}{10} \times 100\% = 10\%$

$10\%$ of Class VI students were absent.

Solution (a):

Scale: On Y-axis, $1$ unit $= 10$ mobile sets.

Solution (b):

The maximum number of mobile sets sold was $60$, which occurred on Saturday.

Solution (c):

$\text{Total sales} = 50 + 45 + 30 + 55 + 27 + 60 = 267$

The total sales for the week is $267$ units.

Solution (d):

Minimum sale = $27$ (Friday); Maximum sale = $60$ (Saturday).

$\text{Ratio} = \frac{27}{60} = \frac{\cancel{27}^{9}}{\cancel{60}_{20}}$

The ratio is $9 : 20$.

Solution (e):

$\text{Average sale} = \frac{\text{Total sales}}{\text{Number of days}} = \frac{267}{6}$

$\text{Average sale} = 44.5$ units.

Solution (f):

Days with sales $> 44.5$ are: Monday ($50$), Tuesday ($45$), Thursday ($55$), and Saturday ($60$).

Total = $4$ days.

Solution (a):

The three tallest buildings are The Imperial I, The Imperial II, and Ashok Towers.

Solution (b):

$\text{Average} = \frac{4.15 + 4.15 + 3.94}{3} = \frac{12.24}{3} = 4.08$ m.

Solution (c):

Out of $10$ buildings, $9$ are in Mumbai.

$\text{Percentage} = \left( \frac{9}{10} \right) \times 100\% = 90\%$.

Solution (d):

$\text{Range} = 249 \text{ m} - 156 \text{ m} = 93 \text{ m}$.

Solution (e):

Sorted heights: $156, 170, 170, 170, 180, 181, 184, 193, 249, 249$

$\text{Median} = \frac{180 + 181}{2} = 180.5 \text{ m}$.

Solution (f):

Given:

Marks of Kunal and Soni in 6 subjects (each out of 100).

Solution (a):

Solution (b):

Total marks of Soni = $86 + 89 + 90 + 82 + 75 + 82 = 504$

Maximum Marks = $600$

$\text{Percentage of Soni} = \frac{504}{600} \times 100 = 84\%$

Solution (c):

Total marks of Kunal = $72 + 81 + 92 + 96 + 64 + 85 = 490$

$\text{Percentage of Kunal} = \frac{490}{600} \times 100 \approx 81.67\%$

Solution (d):

Soni ($84\%$) has a higher percentage than Kunal ($81.67\%$). The difference is $2.33\%$.

Solution (e):

Soni got more marks than Kunal in 3 subjects: English, Hindi, and S. Science.

Solution (f):

In S. Science, Soni got $75$ and Kunal got $64$.

Soni got more marks. Difference = $75 - 64 = 11$ marks.

Solution (g):

Calculating differences in each subject:

Eng ($14$), Hindi ($8$), Maths ($2$), Science ($14$), S.S. ($11$), Sanskrit ($3$).

The difference was maximum in English and Science by 14 marks.

Solution (a):

Solution (b):

Total Girls = $15 + 24 + 10 + 19 + 27 + 21 = 116$

Total Boys = $12 + 16 + 8 + 17 + 11 + 30 = 94$

$\text{Total Students} = 116 + 94 = 210$

Solution (c):

The highest number for boys is $30$ in the Electronics club.

Solution (d):

The lowest number for girls is $10$ in the Yoga club.

Solution (e):

Comparing absolute differences: Music ($3$), Dance ($8$), Yoga ($2$), Dramatics ($2$), Fine Arts ($16$), Electronics ($9$).

The difference is least for Yoga and Dramatics (both 2).

Solution (f):

The maximum difference is in Fine Arts ($27 - 11 = 16$).

Given:

Monthly production of motor bikes for the years 2007 and 2008.

Solution (a):

To compare the production, we draw a double bar graph. We choose a scale of $1$ unit $= 1000$ motor bikes on the Y-axis.

Solution (b):

First, we calculate the total production for each year:

$\text{Total for 2008} = 2700 + 3200 + 6000 + 5000 + 4200 = 21100 \text{ units}$

$\text{Total for 2007} = 2800 + 4500 + 4800 + 4800 + 5200 = 22100 \text{ units}$

Since $22100 > 21100$, the total output was maximum in the year 2007.

Solution (c):

$\text{Mean (2007)} = \frac{22100}{5} = 4420$

The mean production for the year 2007 was $4420$ units.

Solution (d):

We find the absolute difference for each month:

• Feb: $|2700 - 2800| = 100$

• May: $|3200 - 4500| = 1300$

• Aug: $|6000 - 4800| = 1200$

• Oct: $|5000 - 4800| = 200$

• Dec: $|4200 - 5200| = 1000$

The difference was maximum in May.

Solution (e):

In 2008, the highest production value is $6000$. This occurred in August.

Solution (f):

In 2007, the lowest production value is $2800$. This occurred in February.

Given:

Population data (in hundreds) for four towns (A, B, C, D) in 2007 and 2009.

Solution (a):

Scale: On Y-axis, $1$ unit $= 1000$ hundred people.

Solution (b):

Population growth is calculated as $\text{Population (2009)} - \text{Population (2007)}$:

• Town A: $3200 - 2900 = 300$

• Town B: $7500 - 6400 = 1100$

• Town C: $9200 - 8300 = 900$

• Town D: $6300 - 4600 = 1700$

The population growth was maximum in Town D.

Solution (c):

From the calculations above, the smallest increase in population occurred in Town A ($300$).

Given:

Data of tourists for five hill stations in 2008 and 2009.

Solution (a):

Scale: On Y-axis, $1$ unit $= 1000$ tourists.

Solution (b):

Comparing the values for 2008:

Nainital ($4000$), Shimla ($5200$), Manali ($3700$), Mussoorie ($5800$), Kullu ($3500$).

Mussoorie was visited by the maximum number of tourists in 2008.

Solution (c):

Comparing the values for 2009:

Nainital ($4800$), Shimla ($4500$), Manali ($4200$), Mussoorie ($6200$), Kullu ($4600$).

Manali was visited by the least number of tourists in 2009.

Solution (d):

We check where $\text{Value (2009)} > \text{Value (2008)}$:

• Nainital: $4800 > 4000$ (Increase)

• Shimla: $4500 < 5200$ (Decrease)

• Manali: $4200 > 3700$ (Increase)

• Mussoorie: $6200 > 5800$ (Increase)

• Kullu: $4600 > 3500$ (Increase)

There was an increase in Nainital, Manali, Mussoorie, and Kullu.

Given:

Ice cream flavour preferences categorized by gender.

Solution (a):

Scale: On Y-axis, $1$ unit $= 2$ children.

Solution (b):

Comparing the numbers for boys: Vanilla ($4$), Chocolate ($9$), Strawberry ($3$), Mango ($8$), Butterscotch ($13$).

Butterscotch is the flavour liked most by the boys.

Solution (c):

Total Girls $= 8 + 12 + 7 + 9 + 10$

Total Girls $= 46$

There are $46$ girls in total.

Solution (d):

Children liking Chocolate $= \text{Boys} + \text{Girls}$

Children liking Chocolate $= 9 + 12 = 21$

$21$ children like chocolate flavour.

Solution (e):

1. Strawberry liking children: $3 \text{ (Boys)} + 7 \text{ (Girls)} = 10$

2. Vanilla liking children: $4 \text{ (Boys)} + 8 \text{ (Girls)} = 12$

$\text{Ratio} = \frac{5}{6}$

The ratio is $5 : 6$.