Chapter 3 Data Handling

To find the range of the data, we need to identify the maximum and minimum values in the given set.

The given data is: $14, 6, 12, 17, 21, 10, 4, 3$.

Let's arrange the data in ascending order (this step is not strictly necessary to find the range, but it can help identify the minimum and maximum values easily):

$3, 4, 6, 10, 12, 14, 17, 21$

The maximum value in the data set is the largest number, which is $21$.

The minimum value in the data set is the smallest number, which is $3$.

The range of the data is the difference between the maximum value and the minimum value.

Range = Maximum Value - Minimum Value

Range = $21 - 3$

Range = $18$

The range of the given data is $18$.

Comparing this result with the given options:

(a) $21$

(b) $17$

(c) $18$

(d) $11$

The correct option is (c).

The mode of a data set is the value that appears most frequently.

The given data set is: $23, 26, 22, 29, 23, 29, 26, 29, 22, 23$.

Let's count the frequency of each value in the data set:

Value 22 appears $2$ times.

Value 23 appears $3$ times.

Value 26 appears $2$ times.

Value 29 appears $3$ times.

The value(s) that appear most frequently are those with the highest frequency.

The highest frequency in this data set is $3$.

The values that appear $3$ times are $23$ and $29$.

Therefore, the mode of the data is $23$ and $29$. This is a bimodal data set.

Comparing this result with the given options:

(a) $23$ and $29$

(b) $23$ only

(c) $29$ only

(d) $26$ only

The correct option is (a).

To find the median of a data set, we first need to arrange the data in either ascending or descending order.

The given data set is: $40, 50, 99, 68, 98, 60, 94$.

Arranging the data in ascending order:

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

Count the number of observations in the data set. Let $n$ be the number of observations.

In this case, $n = 7$.

Since the number of observations ($n=7$) is an odd number, the median is the middle value.

The position of the median for an odd number of observations is given by the formula $\left(\frac{n+1}{2}\right)^{\text{th}}$ term.

Position of the median = $\left(\frac{7+1}{2}\right)^{\text{th}}$ term

Position of the median = $\left(\frac{8}{2}\right)^{\text{th}}$ term

Position of the median = $4^{\text{th}}$ term

Looking at the ordered data set ($40, 50, 60, 68, 94, 98, 99$), the $4^{\text{th}}$ term is $68$.

Therefore, the median of the data is $68$.

Comparing this result with the given options:

(a) $40$

(b) $60$

(c) $68$

(d) $99$

The correct option is (c).

The mean of a set of numbers is the sum of the numbers divided by the count of the numbers.

We need to find the first five prime numbers.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The first prime number is $2$.

The second prime number is $3$.

The third prime number is $5$.

The fourth prime number is $7$.

The fifth prime number is $11$.

So, the first five prime numbers are $2, 3, 5, 7,$ and $11$.

Now, we calculate the sum of these five prime numbers:

Sum = $2 + 3 + 5 + 7 + 11$

Sum = $28$

The number of observations (prime numbers) is $5$.

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

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

Mean = $5.6$

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

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

When a standard six-sided die is thrown once, the possible outcomes are the numbers $1, 2, 3, 4, 5,$ and $6$.

The set of all possible outcomes is called the sample space.

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

The total number of outcomes is the number of elements in the sample space.

Total number of outcomes $= 6$

We are interested in the probability of getting a number greater than $2$.

The numbers in the sample space that are greater than $2$ are $3, 4, 5,$ and $6$.

The set of favourable outcomes is $\{3, 4, 5, 6\}$.

The number of favourable outcomes is the number of elements in this set.

Number of favourable outcomes $= 4$

The probability of an event is calculated using the formula:

Probability (Event) $= \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

In this case, the event is getting a number greater than $2$.

Probability (getting a number greater than 2) $= \frac{4}{6}$

We can simplify the fraction $\frac{4}{6}$ by dividing both the numerator and the denominator by their greatest common divisor, which is $2$.

$\frac{4}{6} = \frac{4 \div 2}{6 \div 2} = \frac{2}{3}$

So, the probability of getting a number greater than 2 on throwing a die once is $\frac{2}{3}$.

The statement is: The probability of getting a number greater than 2 on throwing a die once is $\frac{2}{3}$.

The mode of a set of observations is the value that occurs most frequently in the set.

The given observations are: $23, 26, 15, 12, 28, 38, 19, 23, 26, 23$.

Let's count the frequency of each distinct observation:

• $12$: occurs $1$ time
• $15$: occurs $1$ time
• $19$: occurs $1$ time
• $23$: occurs $3$ times
• $26$: occurs $2$ times
• $28$: occurs $1$ time
• $38$: occurs $1$ time

The observation with the highest frequency is $23$, which occurs $3$ times.

Therefore, the mode of the given observations is $23$.

The statement claims that the mode is $28$. This is incorrect.

So, the statement "The mode of the observations 23, 26, 15, 12, 28, 38, 19, 23, 26, 23 is 28" is False.

The given table shows the size of sweaters sold and the number of sweaters sold for each size. The number of sweaters sold represents the frequency of each size.

Let's analyse statement (a): "The most popular size is 17."

The most popular size is the size that has the highest frequency (number of sweaters sold).

From the table, the frequencies are 15, 17, 13, 14, and 11 for sizes 40, 42, 44, 46, and 48 respectively.

The highest frequency is $17$, which corresponds to the size $42$.

Therefore, the most popular size is $42$, not $17$.

Statement (a) is False.

Let's analyse statement (b): "17 is the median for above data."

The data set consists of the sizes of the sweaters sold. Since each size is listed with its frequency, the full data set is: $15$ times $40$, $17$ times $42$, $13$ times $44$, $14$ times $46$, and $11$ times $48$.

The total number of observations ($n$) is the sum of the frequencies:

$n = 15 + 17 + 13 + 14 + 11 = 70$

To find the median, we need to arrange the data in ascending order. The sizes are already in ascending order: $40, 42, 44, 46, 48$.

Since the number of observations ($n=70$) is even, the median is the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2}+1\right)^{\text{th}}$ terms.

The positions of the middle terms are the $\left(\frac{70}{2}\right)^{\text{th}} = 35^{\text{th}}$ term and the $\left(\frac{70}{2}+1\right)^{\text{th}} = 36^{\text{th}}$ term.

Let's find the values at these positions by looking at the cumulative frequencies:

• Size $40$ occurs $15$ times (terms 1 to 15).
• Size $42$ occurs $17$ times. Cumulative frequency up to $42$ is $15 + 17 = 32$ (terms 16 to 32).
• Size $44$ occurs $13$ times. Cumulative frequency up to $44$ is $32 + 13 = 45$ (terms 33 to 45).

The $35^{\text{th}}$ term falls within the range of terms for size $44$ (which are terms 33 to 45). So, the $35^{\text{th}}$ term is $44$.

The $36^{\text{th}}$ term also falls within the range of terms for size $44$. So, the $36^{\text{th}}$ term is $44$.

The median is the average of the $35^{\text{th}}$ and $36^{\text{th}}$ terms:

Median $= \frac{44 + 44}{2} = \frac{88}{2} = 44$

The median of the data is $44$, not $17$.

Statement (b) is False.

To find the median of a data set, we first need to arrange the data in either ascending or descending order.

The given data set is: $4, 5, 9, 2, 6, 8, 7$.

Arranging the data in ascending order:

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

Count the number of observations in the data set. Let $n$ be the number of observations.

In this case, $n = 7$.

Since the number of observations ($n=7$) is an odd number, the median is the middle value.

The position of the median for an odd number of observations is given by the formula $\left(\frac{n+1}{2}\right)^{\text{th}}$ term.

Position of the median = $\left(\frac{7+1}{2}\right)^{\text{th}}$ term

Position of the median = $\left(\frac{8}{2}\right)^{\text{th}}$ term

Position of the median = $4^{\text{th}}$ term

Looking at the ordered data set ($2, 4, 5, 6, 7, 8, 9$), the $4^{\text{th}}$ term is $6$.

Therefore, the median of the data is $6$.

The statement claims that the median of the data is $2$. This is incorrect.

So, the statement "Median of the data: 4, 5, 9, 2, 6, 8, 7 is 2" is False.

To find the median of a data set, we first need to arrange the data in either ascending or descending order.

The given data set is: $3, 11, 7, 2, 5, 9, 9, 2, 10, 15, 7$.

Count the number of observations in the data set. Let $n$ be the number of observations.

$n = 11$

Arranging the data in ascending order:

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

Since the number of observations ($n=11$) is an odd number, the median is the middle value.

The position of the median for an odd number of observations is given by the formula $\left(\frac{n+1}{2}\right)^{\text{th}}$ term.

Position of the median = $\left(\frac{11+1}{2}\right)^{\text{th}}$ term

Position of the median = $\left(\frac{12}{2}\right)^{\text{th}}$ term

Position of the median = $6^{\text{th}}$ term

Looking at the ordered data set ($2, 2, 3, 5, 7, 7, 9, 9, 10, 11, 15$), the $6^{\text{th}}$ term is $7$.

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

To find the median of a data set, we first need to arrange the data in either ascending or descending order.

The given data set is: $21, 15, 6, 25, 18, 13, 20, 9, 8, 12$.

Count the number of observations in the data set. Let $n$ be the number of observations.

$n = 10$

Arranging the data in ascending order:

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

Since the number of observations ($n=10$) is an even number, the median is the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2}+1\right)^{\text{th}}$ terms.

The positions of the middle terms are the $\left(\frac{10}{2}\right)^{\text{th}} = 5^{\text{th}}$ term and the $\left(\frac{10}{2}+1\right)^{\text{th}} = 6^{\text{th}}$ term.

Looking at the ordered data set ($6, 8, 9, 12, 13, 15, 18, 20, 21, 25$), the $5^{\text{th}}$ term is $13$ and the $6^{\text{th}}$ term is $15$.

The median is the average of the $5^{\text{th}}$ and $6^{\text{th}}$ terms:

Median $= \frac{5^{\text{th}} \text{ term} + 6^{\text{th}} \text{ term}}{2}$

Median $= \frac{13 + 15}{2}$

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

Median $= 14$

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

The word is "MATHEMATICS".

Let's list the letters in the word: M, A, T, H, E, M, A, T, I, C, S.

(a) How many outcomes are possible when a letter is taken out of the bag at random?

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

Counting the letters, we have $11$ letters.

Total number of outcomes = $11$.

(b) What is the probability of getting

To calculate the probability of an event, we use the formula:

Probability (Event) $= \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

From part (a), the total number of outcomes is $11$.

(i) M?

The favourable outcome is getting the letter 'M'.

Let's count the number of times the letter 'M' appears in the word "MATHEMATICS": M, A, T, H, E, M, A, T, I, C, S.

The letter 'M' appears $2$ times.

Number of favourable outcomes (getting M) = $2$.

Probability of getting M = $\frac{\text{Number of M's}}{\text{Total number of letters}} = \frac{2}{11}$.

(ii) Any vowel?

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

Let's identify the vowels in the word "MATHEMATICS": M, A, T, H, E, M, A, T, I, C, S.

The vowels are A, E, A, I.

Counting the vowels, we have $4$ vowels.

Number of favourable outcomes (getting a vowel) = $4$.

Probability of getting any vowel = $\frac{\text{Number of vowels}}{\text{Total number of letters}} = \frac{4}{11}$.

(iii) Any consonant?

The consonants are the letters that are not vowels.

The letters in "MATHEMATICS" are M, A, T, H, E, M, A, T, I, C, S.

The vowels are A, E, A, I ($4$ letters).

The total number of letters is $11$.

Number of consonants = Total number of letters - Number of vowels

Number of consonants = $11 - 4 = 7$.

Alternatively, identify the consonants: M, A, T, H, E, M, A, T, I, C, S.

The consonants are M, T, H, M, T, C, S. Counting them, we get $7$.

Number of favourable outcomes (getting a consonant) = $7$.

Probability of getting any consonant = $\frac{\text{Number of consonants}}{\text{Total number of letters}} = \frac{7}{11}$.

(iv) X?

The favourable outcome is getting the letter 'X'.

Let's check if the letter 'X' appears in the word "MATHEMATICS".

The word is M, A, T, H, E, M, A, T, I, C, S. The letter 'X' is not present.

The number of times the letter 'X' appears is $0$.

Number of favourable outcomes (getting X) = $0$.

Probability of getting X = $\frac{\text{Number of X's}}{\text{Total number of letters}} = \frac{0}{11} = 0$.

Given:

The given data are $26, 28, 25, x, 24$.

The mean of the data is $27$.

To Find:

The value of $x$.

Solution:

The formula for the mean of a data set is:

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

In the given data set, the observations are $26, 28, 25, x, 24$.

The number of observations is $5$.

The sum of the observations is:

Sum $= 26 + 28 + 25 + x + 24$

Sum $= (26 + 28 + 25 + 24) + x$

Sum $= 103 + x$

We are given that the mean is $27$.

Using the formula for the mean, we can write the equation:

Now, we solve this equation for $x$.

Multiply both sides of the equation (i) by $5$:

Calculate the product of $27$ and $5$:

$27 \times 5 = 135$

To isolate $x$, subtract $103$ from both sides of the equation:

Calculate the difference:

$135 - 103 = 32$

Thus, the value of $x$ is $32$.

Verification:

Substitute $x = 32$ into the data set: $26, 28, 25, 32, 24$.

Sum of the data $= 26 + 28 + 25 + 32 + 24 = 135$.

Number of observations $= 5$.

Mean $= \frac{135}{5} = 27$.

This matches the given mean, so the value of $x=32$ is correct.

Given:

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

Incorrect mean = $40$.

Incorrectly copied value = $15$.

Correct value = $45$.

To Find:

The correct mean.

Solution:

The formula for the mean is:

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

Using the given incorrect mean and the number of observations, we can find the incorrect sum of observations.

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

To find the Incorrect Sum, multiply the incorrect mean by the number of observations:

The incorrect sum includes the wrongly copied value ($15$) instead of the correct value ($45$).

To find the correct sum of observations, we need to subtract the incorrect value from the incorrect sum and add the correct value.

Substitute the values from the given information and equation (i):

Now, we calculate the correct mean using the correct sum (from equation (ii)) and the number of observations.

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

Divide the correct sum by $10$:

The correct mean of the 10 observations is $43$.

Given:

The observations arranged in ascending order are: $11, 12, 14, 18, x + 2, 20, 22, 25, 61$.

The median of the data is $21$.

To Find:

The value of $x$.

Solution:

The given data is already arranged in ascending order.

Count the number of observations in the data set. Let $n$ be the number of observations.

The observations are $11, 12, 14, 18, x + 2, 20, 22, 25, 61$.

There are $9$ observations.

Since the number of observations ($n=9$) is an odd number, the median is the middle value.

The position of the median for an odd number of observations is given by the formula $\left(\frac{n+1}{2}\right)^{\text{th}}$ term.

Position of the median = $\left(\frac{9+1}{2}\right)^{\text{th}}$ term

Position of the median = $\left(\frac{10}{2}\right)^{\text{th}}$ term

Position of the median = $5^{\text{th}}$ term

Looking at the ordered data set ($11, 12, 14, 18, x + 2, 20, 22, 25, 61$), the $5^{\text{th}}$ term is $x+2$.

We are given that the median of the data is $21$.

Since the median is the $5^{\text{th}}$ term and the $5^{\text{th}}$ term is $x+2$, we can set up the equation:

Now, we solve this equation for $x$.

Subtract $2$ from both sides of the equation (i):

So, $x = 19$.

Thus, the value of $x$ is $19$.

Verification:

If $x = 19$, the $5^{\text{th}}$ term is $x+2 = 19+2 = 21$.

The ordered data set becomes: $11, 12, 14, 18, 21, 20, 22, 25, 61$.

Wait, the data must be in ascending order. Let's recheck the ordered data set with $x=19$.

The terms are $11, 12, 14, 18, x+2, 20, 22, 25, 61$. With $x=19$, the term is $19+2=21$.

The sequence is $11, 12, 14, 18, 21, 20, 22, 25, 61$. This sequence is NOT in ascending order because $21 > 20$.

Let's re-read the question carefully. "arranged in ascending order is 21". This implies that $11, 12, 14, 18, x + 2, 20, 22, 25, 61$ is indeed the ordered list, and its median (which is the $5^{\text{th}}$ term) is $21$. This interpretation leads to $x+2=21$, so $x=19$. However, for this list to be in ascending order, the terms must satisfy the inequalities:

$18 \leq x+2 \leq 20$

Subtracting 2 from all parts:

$18 - 2 \leq x \leq 20 - 2$

$16 \leq x \leq 18$

This contradicts the result $x=19$ obtained from the median value $21$.

There seems to be an inconsistency in the question statement as the list given is claimed to be in ascending order, but if the median is 21, the $5^{\text{th}}$ term must be 21, which is $x+2$. If $x+2 = 21$, then $x=19$. However, for the list $11, 12, 14, 18, x + 2, 20, 22, 25, 61$ to be in ascending order, we must have $18 \le x+2 \le 20$, which means $16 \le x \le 18$. A value of $x=19$ makes the sequence $11, 12, 14, 18, 21, 20, 22, 25, 61$, which is not in ascending order.

Assuming the median value (21) and the $5^{\text{th}}$ term ($x+2$) are correctly identified as the median, the calculation for $x$ is as shown above:

If we strictly follow the instruction that the median is 21 and the data is as listed (with $x+2$ being the $5^{\text{th}}$ term in the ordered list of 9 terms), then $x+2$ must equal 21, leading to $x=19$. The inconsistency arises from the additional claim that the list is "arranged in ascending order", which would require $16 \le x \le 18$.

Given the phrasing, it is most likely that the median value provided is correct and corresponds to the term $x+2$. The phrase "arranged in ascending order" confirms that the median is indeed the $5^{\text{th}}$ term.

Therefore, assuming the median value given (21) is correct for the $5^{\text{th}}$ term in the sequence as listed:

Solving for $x$ from equation (i):

The value of $x$ is $19$. Note that this makes the $5^{th}$ term $21$. The list would be $11, 12, 14, 18, 21, 20, 22, 25, 61$, which is stated to be in ascending order. This is where the inconsistency lies; $21$ is not less than or equal to $20$. However, based on standard median problems of this type, the value of the median is usually provided correctly, and the value involving $x$ is the term at the median position.

Assuming the median value of $21$ is correct and corresponds to the $5^{\text{th}}$ term, the value of $x$ is $19$. The statement about the list being strictly in ascending order has an issue.

Understanding the Double Bar Graph:

The double bar graph shows the cost per kilogram of four different fruits (Apple, Mango, Orange, and Grapes) in two different cities (City I and City II).

Answering the questions:

(a) What information does the above double graph depict?

The double bar graph depicts the comparison of the cost per kilogram of different fruits in two cities, City I and City II.

(b) Name the fruits for which cost of 1 kg is greater in City I as compared to City II.

We need to look for the fruits where the blue bar (City I) is taller than the red bar (City II).

• Apple: Cost in City I (\textsf{₹} 30) > Cost in City II (\textsf{₹} 25). So, for Apple, the cost is greater in City I.
• Mango: Cost in City I (\textsf{₹} 20) < Cost in City II (\textsf{₹} 30). So, for Mango, the cost is less in City I.
• Orange: Cost in City I (\textsf{₹} 40) > Cost in City II (\textsf{₹} 35). So, for Orange, the cost is greater in City I.
• Grapes: Cost in City I (\textsf{₹} 25) < Cost in City II (\textsf{₹} 30). So, for Grapes, the cost is less in City I.

The fruits for which the cost of 1 kg is greater in City I as compared to City II are Apple and Orange.

(c) What is the difference of rates for apples in both the cities?

Cost of apples in City I = \textsf{₹} 30 per kg.

Cost of apples in City II = \textsf{₹} 25 per kg.

Difference in rates = Cost in City I - Cost in City II

Difference = \textsf{₹} $30$ - \textsf{₹} $25$

Difference = \textsf{₹} $5$

The difference of rates for apples in both the cities is \textsf{₹} 5.

(d) Find the ratio of the cost of mangoes per kg in City I to the cost of mangoes per kg in City II.

Cost of mangoes in City I = \textsf{₹} 20 per kg.

Cost of mangoes in City II = \textsf{₹} 30 per kg.

Ratio of cost in City I to City II = $\frac{\text{Cost in City I}}{\text{Cost in City II}}$

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

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

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 10.

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

The ratio of the cost of mangoes per kg in City I to the cost of mangoes per kg in City II is $\frac{2}{3}$ or $2:3$.

Understanding the Double Bar Graph:

The double bar graph represents the test match results summary for the Cricket Team of country X against five different countries (A, B, C, D, and E). The blue bars show the number of matches won by Team X, and the red bars show the number of matches lost by Team X against each country.

Data from the Bar Graph:

• Against Country A: Team X won 6 matches, Lost 4 matches.
• Against Country B: Team X won 5 matches, Lost 3 matches.
• Against Country C: Team X won 6 matches, Lost 2 matches.
• Against Country D: Team X won 4 matches, Lost 3 matches.
• Against Country E: Team X won 5 matches, Lost 1 match.

Answering the questions:

(a) Which country has managed maximum wins against country X?

The number of wins of a country against Team X is equal to the number of losses of Team X against that country. We need to find the country against which Team X has the maximum number of losses (i.e., the tallest red bar).

• Losses against A: 4
• Losses against B: 3
• Losses against C: 2
• Losses against D: 3
• Losses against E: 1

The maximum number of losses for Team X is 4, which occurred against Country A.

Therefore, Country A has managed maximum wins against country X.

(b) The difference between the number of matches won and lost is highest for which country against country X?

We calculate the absolute difference between the number of matches won by Team X and the number of matches lost by Team X for each country.

• Against Country A: Difference = $|6 - 4| = 2$
• Against Country B: Difference = $|5 - 3| = 2$
• Against Country C: Difference = $|6 - 2| = 4$
• Against Country D: Difference = $|4 - 3| = 1$
• Against Country E: Difference = $|5 - 1| = 4$

The highest difference is 4, which is observed for both Country C and Country E.

The difference is highest for Country C and Country E against country X.

(c) Number of wins of country E is the same as number of losses of which country against country X?

First, find the number of wins of Country E against country X. This is the number of losses of Team X against Country E, which is 1 (from the red bar for Country E).

Number of wins of Country E = 1.

Now, we need to find the country against which Team X has the same number of losses (which is 1). We look for a red bar with a height of 1.

• Losses against A: 4
• Losses against B: 3
• Losses against C: 2
• Losses against D: 3
• Losses against E: 1

The number of losses is 1 against Country E.

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

Understanding the Double Bar Graph:

The double bar graph compares the class averages in the half-yearly and annual examinations for five sections (A, B, C, D, E) of Class VII.

Extracting Data from the Graph:

Let's note down the class averages for each section in both examinations:

• Section A: Half-yearly Average = 75, Annual Average = 80
• Section B: Half-yearly Average = 60, Annual Average = 65
• Section C: Half-yearly Average = 85, Annual Average = 85
• Section D: Half-yearly Average = 70, Annual Average = 75
• Section E: Half-yearly Average = 65, Annual Average = 70

Calculating the Improvement:

Improvement for each section is the difference between the Annual Average and the Half-yearly Average (Annual Average - Half-yearly Average).

• Section A: Improvement = $80 - 75 = 5$
• Section B: Improvement = $65 - 60 = 5$
• Section C: Improvement = $85 - 85 = 0$
• Section D: Improvement = $75 - 70 = 5$
• Section E: Improvement = $70 - 65 = 5$

Finding the Section with Most Improvement:

We compare the improvements calculated for each section: 5, 5, 0, 5, 5.

The highest improvement is 5.

Sections A, B, D, and E all showed an improvement of 5 marks.

The question asks which section showed the most improvement. Since multiple sections showed the same highest improvement, we list all of them.

Sections A, B, D and E showed the most improvement.

Determining the Amount of Improvement:

The amount of improvement for these sections is 5 marks.

Conclusion:

Sections A, B, D, and E showed the most improvement, and the amount of improvement is 5 marks.

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

Given three observations: $x$, $y$, and $z$.

The number of observations is 3.

The sum of the observations is $x + y + z$.

Therefore, the mean of these observations is calculated as:

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

$Mean = \frac{x + y + z}{3}$

Now, we look at the given options:

(a) $\frac{x \times y \times z}{4}$

(b) $\frac{x + y + z}{4}$

(c) $\frac{x − y − z}{4}$

(d) $\frac{x \times y + z}{4}$

Comparing the correct formula for the mean of three observations ($\frac{x+y+z}{3}$) with the given options, we see that none of the options exactly match the correct result.

However, option (b) $\frac{x+y+z}{4}$ has the sum of the observations in the numerator and is divided by a constant in the denominator. This structure is consistent with the definition of a mean, although the denominator (4) is incorrect for the mean of three observations (where it should be 3).

Options (a), (c), and (d) involve multiplication or subtraction in ways that do not correspond to the standard definition of the mean.

Assuming there is a typographical error in the denominator of option (b) and that it was intended to represent the mean, or that among the given incorrect options, (b) is the closest in structure to the correct formula for the mean, we choose option (b).

The correct option is (b).

To Find: The mode of the given data.

Solution:

The given data represents the number of trees in different parks of a city: 33, 38, 48, 33, 34, 34, 33, 24.

The mode of a data set is the value that appears most frequently in the set.

To find the mode, we first arrange the data in ascending order or count the frequency of each observation.

Let's count the frequency of each unique value in the data:

• 24 appears 1 time.
• 33 appears 3 times.
• 34 appears 2 times.
• 38 appears 1 time.
• 48 appears 1 time.

We can also represent the frequencies in a table:

The observation with the highest frequency is 33, which appears 3 times.

Therefore, the mode of the data is 33.

Comparing this with the given options:

(a) 24

(b) 34

(c) 33

(d) 48

The correct option is (c).

The correct option is (c) 33.

Solution:

We need to determine which measures of central tendency (Mean, Median, and Mode) are affected when the extreme observations (the largest and the smallest values) are removed from a dataset.

Let the data arranged in descending order be $x_1, x_2, ..., x_n$, where $x_1$ is the largest observation and $x_n$ is the smallest observation.

The extreme observations on both ends are $x_1$ and $x_n$. Removing these two observations leaves the data set $x_2, x_3, ..., x_{n-1}$. This new data set has $n-2$ observations.

Let's consider each measure of central tendency:

1. Mean:

The mean of the original data is given by the sum of all observations divided by the number of observations:

$Mean_{original} = \frac{\sum_{i=1}^{n} x_i}{n}$

The mean of the new data (after removing $x_1$ and $x_n$) is the sum of the remaining observations divided by the new number of observations:

$Mean_{new} = \frac{\sum_{i=2}^{n-1} x_i}{n-2} = \frac{(\sum_{i=1}^{n} x_i) - x_1 - x_n}{n-2}$

In general, for $n > 2$, $Mean_{original} \neq Mean_{new}$. The mean is calculated using the values of all observations. Removing the largest and smallest values will change the sum and the count, thus changing the mean value in most cases.

Therefore, the Mean is affected.

2. Median:

The median is the middle value of the data arranged in order. Its position depends on the number of observations.

If the original data has $n$ observations, the median is found based on the values at the center of the ordered list $x_1, x_2, ..., x_n$.

After removing $x_1$ and $x_n$, the new ordered data is $x_2, x_3, ..., x_{n-1}$, with $n-2$ observations.

The median of the new data set is calculated based on the values at the center of this new list.

For $n > 2$, removing the single largest and single smallest values does not affect the values of the observations located in the middle of the dataset. The values used to calculate the median of the original dataset (if $n$ is even, the two middle values; if $n$ is odd, the single middle value) will still be the middle values (or involved in the calculation of the median) in the new dataset, although their positions/indices relative to the start of the list will shift.

For example, if $n$ is odd ($n=2k+1$), the original median is $x_{k+1}$. The new list has $2k-1$ elements, and the median is the $k$-th element of the new list, which is $x_{k+1}$ from the original list. The value remains the same.

If $n$ is even ($n=2k$), the original median is $(x_k + x_{k+1})/2$. The new list has $2k-2$ elements. The new median is the average of the $(k-1)$-th and $k$-th elements of the new list, which are $x_k$ and $x_{k+1}$ from the original list. The value remains the same.

Thus, the value of the Median is generally not affected by removing the single largest and smallest observations, provided $n > 2$. The median is known for its resistance to extreme values (outliers).

3. Mode:

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

Removing $x_1$ and $x_n$ reduces the frequency count for these specific values by one. The frequencies of all other values remain unchanged.

The mode can be affected if:

• The original mode was one of the extreme values ($x_1$ or $x_n$), and removing it causes its frequency to drop below the highest frequency among the remaining values.
• The removal of $x_1$ or $x_n$ changes the relative frequencies such that a different value now has the highest frequency, or the set of values with the highest frequency changes (e.g., from unimodal to bimodal or vice versa).

For example, in the data {1, 2, 2, 3, 4, 10, 10}, the modes are 2 and 10 (each with frequency 2). Removing 1 and 10 gives {2, 2, 3, 4, 10}. The mode is now 2 (frequency 2), while 10 has frequency 1. The set of modes changed from {2, 10} to {2}.

In other cases, the mode might not be affected (e.g., {1, 2, 3, 3, 3, 10}, mode is 3. Removing 1, 10 gives {2, 3, 3, 3}, mode is still 3).

Therefore, the Mode can be affected.

Based on the analysis, the Mean is generally affected, the Median value is generally not affected (for $n>2$), and the Mode can be affected.

The question asks which measures "get affected", implying those that are generally or can be affected.

The measures that are affected are Mean and Mode.

Comparing with the options:

(a) Mean and mode

(b) Mean and Median

(c) Mode and Median

(d) Mean, Median and Mode

The correct option is (a) as the Mean is always (or almost always for $n>2$) affected, and the Mode can be affected, while the Median value is generally not affected.

The correct option is (a) Mean and mode.

To Find: The range of the given data.

Solution:

The given data set is: 21, 6, 17, 18, 12, 8, 4, 13.

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

$Range = Maximum \ Value - Minimum \ Value$

First, let's identify the observations in the data set:

21, 6, 17, 18, 12, 8, 4, 13.

Next, we find the highest observation in this data set.

The observations are: 21, 6, 17, 18, 12, 8, 4, 13.

The highest observation is 21.

Maximum Value = 21.

Now, we find the lowest observation in this data set.

The observations are: 21, 6, 17, 18, 12, 8, 4, 13.

The lowest observation is 4.

Minimum Value = 4.

Finally, we calculate the range using the formula:

$Range = Maximum \ Value - Minimum \ Value$

$Range = 21 - 4$

$Range = 17$

Comparing this result with the given options:

(a) 17

(b) 12

(c) 8

(d) 15

The calculated range, 17, matches option (a).

The correct option is (a) 17.

To Find: The median of the given data.

Solution:

The given data set is: 3, 4, 5, 6, 7, 3, 4.

To find the median, we first need to arrange the data in ascending or descending order.

Arranging the data in ascending order, we get:

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

Next, we count the number of observations in the data set. Let $n$ be the number of observations.

In this case, $n = 7$.

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

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

Looking at the ordered data (3, 3, 4, 4, 5, 6, 7), the 4th observation is 4.

Therefore, the median of the data is 4.

Comparing this result with the given options:

(a) 5

(b) 3

(c) 4

(d) 6

The calculated median, 4, matches option (c).

The correct option is (c) 4.

To Determine: The most appropriate measure of central tendency to find the chocolate brand most liked by children.

Solution:

The problem asks to find the chocolate brand that is most liked by children. This means we are looking for the brand that occurs most frequently in a dataset representing the preferences of children.

Let's consider what each measure of central tendency represents:

Mean: The average value of a dataset. It is typically used for numerical data where the sum and division are meaningful.

Median: The middle value of a dataset when arranged in order. It represents the central value, separating the higher half from the lower half.

Mode: The value that appears most frequently in a dataset. It is used to identify the most common category or value in the data.

In this scenario, the data would likely consist of the brand preferred by each child surveyed (e.g., Brand A, Brand B, Brand C, etc.). This type of data is categorical or qualitative, although we could represent it with numbers if needed for counting frequencies.

We want to find the brand that has the highest frequency, i.e., the brand chosen by the largest number of children.

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

The mean and median are generally not appropriate for finding the most popular item or category in a set of preferences, especially when the data is not numerical in nature where averaging or ordering makes sense in the context of the preference itself.

Therefore, the most appropriate measure of central tendency to determine the most liked brand is the Mode.

Comparing this with the given options:

(a) Mean

(b) Mode

(c) Median

(d) Any of the three

The correct option is (b).

The correct option is (b) Mode.

To Determine: From which set it is certain to pick the two aces in the first go.

Solution:

The question asks from which set we are certain to pick the two aces in the first go. Being "certain" means the probability of the event is 1.

Let's interpret "in the first go" as picking a certain number of cards from the set. Since we want to pick "the two aces", the number of cards we intend to pick is 2.

We need to find the set where, if we pick 2 cards, we are guaranteed that both cards are aces.

Let's look at the contents of each set as shown in the image:

• Set (a) contains 4 cards: 2 Aces and 2 other cards.
• Set (b) contains 4 cards: 3 Aces and 1 other card.
• Set (c) contains 3 cards: 2 Aces and 1 other card.
• Set (d) contains 2 cards: 2 Aces.

(Note: The initial statement "There are 2 aces in each of the given set of cards" seems to be a general description, but the image shows the specific contents of each option. We will use the image content for each option.)

Now, let's consider picking 2 cards from each set and determine the probability of picking exactly two aces.

For Set (a):

Total number of cards = 4.

Number of Aces = 2.

Number of other cards = 2.

Total number of ways to pick 2 cards from 4 is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.

Number of ways to pick 2 Aces from 2 Aces is $\binom{2}{2} = 1$.

Probability of picking 2 Aces from Set (a) = $\frac{\text{Ways to pick 2 Aces}}{\text{Total ways to pick 2 cards}} = \frac{1}{6}$.

Since the probability is $1/6 \neq 1$, we are not certain to pick the two aces from Set (a).

For Set (b):

Total number of cards = 4.

Number of Aces = 3.

Number of other cards = 1.

Total number of ways to pick 2 cards from 4 is $\binom{4}{2} = 6$.

Number of ways to pick 2 Aces from 3 Aces is $\binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3$.

Probability of picking 2 Aces from Set (b) = $\frac{\text{Ways to pick 2 Aces}}{\text{Total ways to pick 2 cards}} = \frac{3}{6} = \frac{1}{2}$.

Since the probability is $1/2 \neq 1$, we are not certain to pick the two aces from Set (b).

For Set (c):

Total number of cards = 3.

Number of Aces = 2.

Number of other cards = 1.

Total number of ways to pick 2 cards from 3 is $\binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3$.

Number of ways to pick 2 Aces from 2 Aces is $\binom{2}{2} = 1$.

Probability of picking 2 Aces from Set (c) = $\frac{\text{Ways to pick 2 Aces}}{\text{Total ways to pick 2 cards}} = \frac{1}{3}$.

Since the probability is $1/3 \neq 1$, we are not certain to pick the two aces from Set (c).

For Set (d):

Total number of cards = 2.

Number of Aces = 2.

Number of other cards = 0.

Total number of ways to pick 2 cards from 2 is $\binom{2}{2} = 1$.

Number of ways to pick 2 Aces from 2 Aces is $\binom{2}{2} = 1$.

Probability of picking 2 Aces from Set (d) = $\frac{\text{Ways to pick 2 Aces}}{\text{Total ways to pick 2 cards}} = \frac{1}{1} = 1$.

Since the probability is 1, we are certain to pick the two aces from Set (d) when picking 2 cards.

Therefore, the set from which you are certain to pick the two aces in the first go is Set (d).

The correct option is (d).

To Find: The probability of picking up an ace from set (d).

Solution:

Based on the description and image in the previous question (Question 7), Set (d) contains 2 cards, and both of these cards are Aces.

The question asks for the probability of picking up "an ace" from set (d). We assume this means picking one card from the set and it being an ace.

Total number of cards in Set (d) = 2.

Number of Aces in Set (d) = 2.

The probability of picking an ace when picking one card from Set (d) is given by:

$P(\text{picking an ace}) = \frac{\text{Number of Aces}}{\text{Total number of cards}}$

$P(\text{picking an ace}) = \frac{2}{2} = 1$

The probability of picking an ace from set (d) is 1.

Now let's compare this result with the given options:

(a) $\frac{1}{6}$

(b) $\frac{2}{6}$

(c) $\frac{3}{6}$

(d) $\frac{4}{6}$

The result, 1, is not among the provided options.

Let's consider if the question meant picking "the two aces" from set (d). In this case, we are picking 2 cards from Set (d).

Total number of ways to pick 2 cards from 2 is $\binom{2}{2} = 1$.

Number of ways to pick 2 Aces from 2 Aces is $\binom{2}{2} = 1$.

The probability of picking the two aces from set (d) (when picking 2 cards) is $\frac{1}{1} = 1$. Again, this is not among the options.

Given the provided options are fractions based on parts of the previous question, it is highly likely that there is a typographical error in the question and it was intended to ask about a different set from the previous question, specifically set (c), and the probability of picking the two aces.

Let's calculate the probability of picking the two aces from set (c).

From the previous question, Set (c) contains 3 cards: 2 Aces and 1 other card.

Total number of cards in Set (c) = 3.

Number of Aces in Set (c) = 2.

Number of other cards in Set (c) = 1.

We assume the question implies picking 2 cards to get "the two aces".

Total number of ways to pick 2 cards from 3 is $\binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3$.

Number of ways to pick the two aces (which are 2) from the 2 available aces is $\binom{2}{2} = 1$.

The probability of picking the two aces from set (c) is $\frac{\text{Ways to pick 2 Aces}}{\text{Total ways to pick 2 cards}} = \frac{1}{3}$.

We can write $\frac{1}{3}$ as $\frac{2}{6}$.

Comparing this probability ($\frac{2}{6}$) with the given options, we find that it matches option (b).

Assuming the question intended to ask for the probability of picking the two aces from set (c), the correct answer is (b).

Based on the provided options and the likely intended question, the correct option is (b) $\frac{2}{6}$.

Solution:

We are asked to identify the term that describes the difference between the highest and the lowest observations in a data set.

Let's consider the definitions of the terms provided in the options:

Frequency: Frequency is the number of times a particular observation or value appears in a data set.

Width: Width usually refers to the size of a class interval in grouped data, or the distance between points on a scale, not specifically the difference between the highest and lowest observations of the entire data set.

Range: The range of a data set is defined as the difference between the maximum (highest) observation and the minimum (lowest) observation in the data set.

$Range = Maximum \ Value - Minimum \ Value$

Mode: The mode is the value that appears most frequently in a data set.

Based on the definitions, the term that represents the difference between the highest and the lowest observations is the Range.

Comparing this with the given options:

(a) frequency

(b) width

(c) range

(d) mode

The correct option is (c).

The correct option is (c) range.

To Find: The chance that a student picked at random makes it to the competition.

Solution:

The chance of an event occurring is the probability of that event.

Probability is calculated as:

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

In this case, the event is "a student picked at random makes it to the competition".

The total number of students is 5.

The number of students who can participate in the quiz (favorable outcomes) is 2.

So, the probability that a student picked at random makes it to the competition is:

$P(\text{student makes it}) = \frac{\text{Number of students who can participate}}{\text{Total number of students}}$

$P(\text{student makes it}) = \frac{2}{5}$

The question asks for the chance as a percentage. To convert a probability (fraction or decimal) to a percentage, we multiply by 100%.

$Chance = P(\text{student makes it}) \times 100\%$

$Chance = \frac{2}{5} \times 100\%$

$Chance = (0.4) \times 100\%$

$Chance = 40\%$

Comparing this result with the given options:

(a) 20%

(b) 40%

(c) 50%

(d) 30%

The calculated chance, 40%, matches option (b).

The correct option is (b) 40%.

To Find: The range of the integers marked on the board.

Solution:

The integers marked on the board, as shown in the image (number line), are -25, -17, -6, 0, 3, and 12.

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

$Range = Maximum \ Value - Minimum \ Value$

From the marked integers:

Highest integer = 12

Lowest integer = -25

Now, we calculate the range:

$Range = 12 - (-25)$

$Range = 12 + 25$

$Range = 37$

Comparing this result with the given options:

(a) 31

(b) 37

(c) 20

(d) 3

The calculated range, 37, matches option (b).

The correct option is (b) 37.

To Determine: The possible outcome when tossing a coin.

Solution:

When a standard, fair coin is tossed, there are two possible outcomes: a Head or a Tail.

Let's consider the given options:

(a) only head: This is incorrect, as a tail can also occur.

(b) only tail: This is incorrect, as a head can also occur.

(c) neither head nor tail: This is incorrect in the context of a standard coin toss, which typically results in either a head or a tail (excluding rare cases like landing on the edge, which is usually disregarded in basic probability). The outcome is one of the two defined sides.

(d) either head or tail: This means the outcome is either a head or a tail, which correctly describes the possible results of tossing a standard coin.

Therefore, on tossing a coin, the outcome is either head or tail.

The correct option is (d) either head or tail.

To Find: The highest possible value of one of the remaining two numbers.

Given:

• There are three different natural numbers.
• The mean of the three numbers is 40.
• The lowest number is 19.

Solution:

Let the three different natural numbers be $a, b,$ and $c$.

The mean of these three numbers is given by:

$Mean = \frac{a + b + c}{3}$

We are given that the mean is 40. So,

Multiplying both sides by 3, we get the sum of the three numbers:

$a + b + c = 40 \times 3$

$a + b + c = 120$

We are given that the lowest number is 19.

Since the three numbers are different natural numbers, the other two numbers must be natural numbers greater than 19.

Let the three numbers be 19, $x$, and $y$, where $19 < x$ and $19 < y$, and $x \neq y$, $x \neq 19$, $y \neq 19$.

The sum of these numbers is 120:

$19 + x + y = 120$

$x + y = 120 - 19$

$x + y = 101$

We want to find the highest possible value of one of the remaining two numbers (either $x$ or $y$). Let's say we want to maximize $x$. To maximize $x$, we need to minimize $y$, subject to the conditions that $y$ is a natural number, $y > 19$, and $y \neq x$.

Since the numbers must be different natural numbers and $y > 19$, the smallest possible natural number greater than 19 is 20.

So, the minimum possible value for $y$ is 20.

Let $y = 20$.

We must check if this choice makes $x$ different from $y$ and 19. If $y=20$, then $x = 101 - 20 = 81$. The numbers are 19, 20, and 81. These are three different natural numbers, and the lowest is 19. This combination is valid.

With $y_{min} = 20$, the maximum possible value for $x$ is:

$x_{max} = 101 - y_{min}$

$x_{max} = 101 - 20$

$x_{max} = 81$

So, the highest possible number among the remaining two numbers is 81.

Let's verify with the options:

(a) 81

(b) 40

(c) 100

(d) 71

The calculated highest possible value is 81, which matches option (a).

The correct option is (a) 81.

Given:

• Scores in first three examinations: 97, 73, 88.
• Score in the fourth examination: 80.

To Find: The change in her average score after the fourth examination.

Solution:

First, let's calculate the average score of the first three examinations.

Sum of scores in the first three examinations = $97 + 73 + 88$

Sum = $170 + 88 = 258$

Number of examinations = 3

Average score for the first three examinations = $\frac{\text{Sum of scores}}{\text{Number of examinations}}$

Average (3 exams) = $\frac{258}{3}$

Average (3 exams) = 86

Now, let's calculate the average score after the fourth examination.

Scores in the first four examinations: 97, 73, 88, 80.

Sum of scores in the first four examinations = $97 + 73 + 88 + 80$

Sum = $258 + 80 = 338$

Number of examinations = 4

Average score for the first four examinations = $\frac{\text{Sum of scores}}{\text{Number of examinations}}$

Average (4 exams) = $\frac{338}{4}$

Average (4 exams) = 84.5

To find the change in the average score, we subtract the average of the first three exams from the average of the first four exams.

Change in average = Average (4 exams) - Average (3 exams)

Change in average = $84.5 - 86$

Change in average = $-1.5$

Since the change is $-1.5$, the average score decreased by 1.5.

Comparing this result with the given options:

(a) increased by 1

(b) increased by 1.5

(c) decreased by 1

(d) decreased by 1.5

The calculated change corresponds to option (d).

The correct option is (d) decreased by 1.5.

To Determine: The most appropriate measure of central tendency to represent the most popular politician.

Solution:

We are looking for the measure of central tendency that best represents the "most popular" politician after a debate. Popularity in this context refers to which politician is preferred or favored by the largest number of people.

Let's consider what each measure of central tendency represents:

Mean: The average value of a dataset. It is suitable for numerical data where summing and dividing are meaningful, such as average score or average height.

Median: The middle value of a dataset when arranged in order. It is useful for ordered numerical data, especially when the data might be skewed by extreme values.

Mode: The value or category that appears most frequently in a dataset. It is used to find the most common or most popular item in a set of data, which can be numerical or categorical.

In the context of popularity among politicians, the data would likely involve collecting preferences from a group of people (e.g., which politician they liked best). The result would be a count of how many people preferred each politician. The "most popular" politician is the one with the highest count (frequency).

The measure of central tendency that directly identifies the value or category with the highest frequency is the Mode.

Therefore, the Mode is the most appropriate measure of central tendency to represent the data of the most popular politician.

Comparing this with the given options:

(a) Mean

(b) Median

(c) Mode

(d) Any of the above

The correct option is (c).

The correct option is (c) Mode.

To Find: The data set among the given options which has the same mean, median, and mode.

Solution:

We need to calculate the mean, median, and mode for each data set provided in the options and check if they are equal.

Recall:

• Mean: The sum of all observations divided by the number of observations.
• Median: The middle value of the data when arranged in ascending or descending order. If the number of observations is odd, it is the single middle value. If the number of observations is even, it is the average of the two middle values.
• Mode: The value that appears most frequently in the data set.

Let's examine each option:

(a) Data: 6, 2, 5, 4, 3, 4, 1

Arranged data: 1, 2, 3, 4, 4, 5, 6

Number of observations, $n = 7$.

Mean = $\frac{1+2+3+4+4+5+6}{7} = \frac{25}{7} \approx 3.57$

Median = Middle value (4th value) = 4

Mode = Most frequent value = 4 (appears twice)

Mean $\approx 3.57$, Median = 4, Mode = 4. They are not all equal.

(b) Data: 4, 2, 2, 1, 3, 2, 3

Arranged data: 1, 2, 2, 2, 3, 3, 4

Number of observations, $n = 7$.

Mean = $\frac{1+2+2+2+3+3+4}{7} = \frac{17}{7} \approx 2.43$

Median = Middle value (4th value) = 2

Mode = Most frequent value = 2 (appears three times)

Mean $\approx 2.43$, Median = 2, Mode = 2. They are not all equal.

(c) Data: 2, 3, 7, 3, 8, 3, 2

Arranged data: 2, 2, 3, 3, 3, 7, 8

Number of observations, $n = 7$.

Mean = $\frac{2+2+3+3+3+7+8}{7} = \frac{28}{7} = 4$

Median = Middle value (4th value) = 3

Mode = Most frequent value = 3 (appears three times)

Mean = 4, Median = 3, Mode = 3. They are not all equal.

(d) Data: 4, 3, 4, 3, 4, 6, 4

Arranged data: 3, 3, 4, 4, 4, 4, 6

Number of observations, $n = 7$.

Mean = $\frac{3+3+4+4+4+4+6}{7} = \frac{28}{7} = 4$

Median = Middle value (4th value) = 4

Mode = Most frequent value = 4 (appears four times)

Mean = 4, Median = 4, Mode = 4. They are all equal.

Thus, the data set in option (d) has the same mean, median, and mode.

The correct option is (d) 4, 3, 4, 3, 4, 6, 4.

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

$Range = Maximum \ Value - Minimum \ Value$

The blank should be filled with the term "range".

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

The mean of a data set is a measure of central tendency. It is calculated by summing all the observations in the data set and then dividing the sum by the total number of observations.

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

The blank can be filled with "the sum of all observations divided by the total number of observations".

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

In statistics, the measure of central tendency that represents the most frequently occurring value in a data set is called the mode.

The blank should be filled with the term "mode".

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

The measure of central tendency that represents the middle value of a data set when the data is arranged in ascending or descending order is called the median.

If the number of observations is odd, the median is the single middle observation.

If the number of observations is even, the median is the average of the two middle observations.

The blank should be filled with the term "median".

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

Mean, Median, and Mode are statistical measures used to describe the central position of a frequency distribution for a set of data.

They are commonly referred to as measures of central tendency.

The blank should be filled with "central tendency".

Mean, Median, Mode are the measures of central tendency.

An event that is certain to happen is called a sure event or a certain event.

The probability of a sure event is defined as 1.

For example, the probability that the sun will rise tomorrow is a sure event, and its probability is 1.

The blank should be filled with "1".

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

An event that is impossible to happen is called an impossible event.

The probability of an impossible event is defined as 0.

For example, the probability of rolling a 7 on a standard six-sided die is an impossible event, and its probability is 0.

The blank should be filled with "0".

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

To Find: The probability of getting a number less than 7 when a die is thrown.

Solution:

When a standard six-sided die is thrown, the possible outcomes are the numbers on its faces: 1, 2, 3, 4, 5, and 6.

The total number of possible outcomes is 6.

We are interested in the event of getting a number less than 7.

The numbers in the set of possible outcomes that are less than 7 are 1, 2, 3, 4, 5, and 6.

All the possible outcomes (1, 2, 3, 4, 5, 6) are less than 7.

The number of favorable outcomes (getting a number less than 7) is 6.

The probability of an event is given by:

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

$P(\text{number less than 7}) = \frac{6}{6}$

$P(\text{number less than 7}) = 1$

Getting a number less than 7 when throwing a standard die is a sure event.

The blank should be filled with "1".

When a die is thrown, the probability of getting a number less than 7 is 1.

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

These outcomes are the integers from 1 to 6, inclusive.

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

The number of possible outcomes is the count of distinct values in this set.

Number of possible outcomes = 6.

The blank should be filled with "6".

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

When we want to visually compare two different sets of data, especially when they relate to the same categories, a specific type of graph is commonly used.

A double bar graph is a graphical tool designed to present two sets of data side-by-side, allowing for easy comparison of the values for each category across the two collections.

Other methods like comparing means, medians, or ranges are numerical comparisons, but a graphical representation like a double bar graph provides a direct visual comparison.

The blank should be filled with "Double bar graph".

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

A bar graph is a chart or graph that presents categorical data with rectangular bars with heights or lengths proportional to the values that they represent.

The bars are typically drawn such that they have uniform width and are spaced equally apart.

The blank should be filled with the term "bar graph".

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

To Find: The value of $x$.

Given:

• The data set is: 8, 4, x, 6, 2, 7.
• The arithmetic mean of the data is 5.

Solution:

The arithmetic mean (or mean) of a data set is calculated by summing all the observations and dividing by the number of observations.

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

The observations in the data set are 8, 4, $x$, 6, 2, and 7.

The total number of observations is 6.

The sum of the observations is $8 + 4 + x + 6 + 2 + 7$.

Sum = $(8+4+6+2+7) + x$

Sum = $(12+6+2+7) + x$

Sum = $(18+2+7) + x$

Sum = $(20+7) + x$

Sum = $27 + x$

We are given that the mean is 5.

Using the formula for the mean:

$5 = \frac{27 + x}{6}$

To solve for $x$, we multiply both sides of the equation by 6:

$5 \times 6 = 27 + x$

$30 = 27 + x$

Subtract 27 from both sides of the equation:

$30 - 27 = x$

$3 = x$

So, the value of $x$ is 3.

We can verify this by calculating the mean with $x=3$:

Data: 8, 4, 3, 6, 2, 7

Sum = $8+4+3+6+2+7 = 12+3+6+2+7 = 15+6+2+7 = 21+2+7 = 23+7 = 30$

Mean = $\frac{30}{6} = 5$. This matches the given mean.

The blank should be filled with "3".

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

The median of a data set is the middle value when the data is arranged in ascending or descending order.

Let the data set be $x_1, x_2, ..., x_n$ arranged in ascending order, where $x_1$ is the lowest observation and $x_n$ is the highest observation.

If $n$ is odd, the median is the single middle value, which is $x_{(n+1)/2}$. This value is always between the lowest observation ($x_1$) and the highest observation ($x_n$), inclusive.

If $n$ is even, the median is the average of the two middle values, $\frac{x_{n/2} + x_{n/2 + 1}}{2}$. Since $x_{n/2} \geq x_1$ and $x_{n/2 + 1} \leq x_n$, their average will also be between $x_1$ and $x_n$, inclusive.

Therefore, the median of any data set always lies between the lowest and highest observations (inclusive).

The blanks should be filled with "lowest" and "highest".

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

The median is the middle value of a data set when the data is arranged in ascending or descending order.

If the number of observations ($n$) is odd, there is a single middle term at the position $\frac{n+1}{2}$. This middle term is one of the observations in the original data set.

If the number of observations ($n$) is even, there are two middle terms at positions $\frac{n}{2}$ and $\frac{n}{2}+1$. The median is the average of these two terms. Unless the two middle terms are equal, their average will not be one of the original observations.

Therefore, the median is one of the observations in the data set specifically when the number of observations is odd.

The blank should be filled with "odd".

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

To Find: The weight interval in which the probability of selecting a student is highest.

Solution:

The probability of selecting a student from a particular weight interval is proportional to the number of students in that interval, assuming random selection from the entire class.

$P(\text{selecting a student from an interval}) = \frac{\text{Number of students in that interval}}{\text{Total number of students}}$

To find the interval where the probability of selection is highest, we need to find the interval that contains the largest number of students (highest frequency).

From the given table:

Weight (in kg): 44 - 47, Number of Students: 3

Weight (in kg): 48 - 51, Number of Students: 5

Weight (in kg): 52 - 55, Number of Students: 25

Weight (in kg): 56 - 60, Number of Students: 7

The frequencies (number of students) for the given intervals are 3, 5, 25, and 7.

The highest frequency is 25.

The interval corresponding to the highest frequency (25 students) is 52 - 55 kg.

Therefore, the probability of selecting a student is highest for the weight interval 52 - 55 kg.

The blank should be filled with "52 - 55 kg" or "52 - 55". Using the interval as presented in the question options format (like in MCQs if this were one) or the table heading seems appropriate.

The probability of selection of the student is highest whose weight is in the interval 52 - 55.

To Determine: Whether the statement is True or False.

Statement: If a die is thrown, the probability of getting a number greater than 6 is 1.

Solution:

When a standard six-sided die is thrown, the possible outcomes are the numbers 1, 2, 3, 4, 5, and 6.

The total number of possible outcomes is 6.

We are interested in the event of getting a number greater than 6.

Let's look at the possible outcomes: $\{1, 2, 3, 4, 5, 6\}$.

There are no numbers in this set that are greater than 6.

The number of favorable outcomes (getting a number greater than 6) is 0.

The probability of an event is given by:

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

$P(\text{number greater than 6}) = \frac{0}{6} = 0$

The statement says that the probability of getting a number greater than 6 is 1.

Our calculation shows the probability is 0.

An event with a probability of 1 is a sure event (certain to happen), whereas getting a number greater than 6 on a standard die is an impossible event (probability 0).

Therefore, the statement is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: When a coin is tossed, there are 2 possible outcomes.

Solution:

When a standard, fair coin is tossed, there are two distinct sides it can land on: one is called Head, and the other is called Tail.

The possible outcomes of a single coin toss are Head (H) and Tail (T).

The set of possible outcomes is $\{H, T\}$.

The number of possible outcomes is the count of distinct results, which is 2.

The statement says there are 2 possible outcomes when a coin is tossed. This matches our understanding of the possible results (Head and Tail).

Therefore, the statement is true.

The statement is True.

To Determine: Whether the statement is True or False.

Statement: If the extreme observations on both the ends of a data arranged in ascending order are removed, the median gets affected.

Solution:

Let the data arranged in ascending order be $x_1, x_2, ..., x_n$, where $x_1$ is the lowest observation and $x_n$ is the highest observation.

The extreme observations on both ends are $x_1$ and $x_n$. When these are removed, the new data set is $x_2, x_3, ..., x_{n-1}$. The number of observations in the new data set is $n-2$ (assuming $n \geq 2$).

The median is the middle value of the ordered data.

• If the original number of observations ($n$) is odd, say $n = 2k+1$, the median is the $(k+1)$-th observation, which is $x_{k+1}$. After removing the two extreme values, the new number of observations is $(2k+1) - 2 = 2k - 1$, which is odd. The median of the new data is the middle observation at position $\frac{(2k-1)+1}{2} = \frac{2k}{2} = k$. This corresponds to the $k$-th term in the new ordered list $x_2, x_3, ..., x_{2k}$. The $k$-th term in this list is $x_{k+1}$ from the original list. So, the median remains $x_{k+1}$.
• If the original number of observations ($n$) is even, say $n = 2k$, the median is the average of the $k$-th and $(k+1)$-th observations, which are $x_k$ and $x_{k+1}$. Median $= \frac{x_k + x_{k+1}}{2}$. After removing the two extreme values, the new number of observations is $2k - 2$, which is even. The median of the new data is the average of the middle two observations. The positions of the middle two observations in the new list $x_2, x_3, ..., x_{2k-1}$ are $\frac{2k-2}{2} = k-1$ and $(k-1)+1 = k$. These positions in the new list correspond to the values $x_{k-1+1} = x_k$ and $x_{k+1}$ from the original list. So, the new median is $\frac{x_k + x_{k+1}}{2}$.

In both cases (for $n \geq 2$), the value of the median remains the same after removing the lowest and highest observations. The median is resistant to extreme values.

Therefore, the statement that the median gets affected is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: The measures of central tendency may not lie between the maximum and minimum values of data.

Solution:

Let's examine the common measures of central tendency: Mean, Median, and Mode.

Let the data set be $x_1, x_2, ..., x_n$. Let the minimum value be $x_{min}$ and the maximum value be $x_{max}$. By definition, for every observation $x_i$ in the data set, $x_{min} \leq x_i \leq x_{max}$.

1. Mean:

The mean is $\frac{\sum x_i}{n}$. Since $x_{min} \leq x_i \leq x_{max}$ for all $i$, we have:

$\sum x_{min} \leq \sum x_i \leq \sum x_{max}$

$n \cdot x_{min} \leq \sum x_i \leq n \cdot x_{max}$

Dividing by $n$ (assuming $n > 0$):

$x_{min} \leq \frac{\sum x_i}{n} \leq x_{max}$

Thus, the mean always lies between the minimum and maximum values (inclusive).

2. Median:

The median is the middle value (or the average of the two middle values) after the data is arranged in order. Let the ordered data be $y_1 \leq y_2 \leq ... \leq y_n$, where $y_1 = x_{min}$ and $y_n = x_{max}$. The median is either $y_{(n+1)/2}$ (if $n$ is odd) or $\frac{y_{n/2} + y_{n/2 + 1}}{2}$ (if $n$ is even).

Since $y_1 \leq y_i \leq y_n$ for all $i$, the middle value(s) $y_{(n+1)/2}$ or $y_{n/2}$ and $y_{n/2 + 1}$ are always between $y_1$ and $y_n$. The average of two values between $y_1$ and $y_n$ also lies between $y_1$ and $y_n$.

Thus, the median always lies between the minimum and maximum values (inclusive).

3. Mode:

The mode is the observation that occurs most frequently in the data set. By definition, the mode must be one of the values present in the data set.

Since every value in the data set is greater than or equal to the minimum value ($x_{min}$) and less than or equal to the maximum value ($x_{max}$), the mode, being one of these values, must also lie between the minimum and maximum values (inclusive).

Since the Mean, Median, and Mode all always lie between the minimum and maximum values of the data (inclusive), the statement "The measures of central tendency may not lie between the maximum and minimum values of data" is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: It is impossible to get a sum of 14 of the numbers on both dice when a pair of dice is thrown together.

Solution:

When a pair of standard six-sided dice is thrown, each die can show a number from 1 to 6.

Let the outcome of the first die be $D_1$ and the outcome of the second die be $D_2$. The possible values for $D_1$ and $D_2$ are $\{1, 2, 3, 4, 5, 6\}$.

We are interested in the sum of the numbers on both dice, which is $S = D_1 + D_2$.

To find the maximum possible sum, we take the maximum value for each die and add them together.

Maximum value on one die = 6.

Maximum sum = Maximum value on die 1 + Maximum value on die 2 = $6 + 6 = 12$.

The possible sums when throwing two dice range from the minimum sum ($1+1=2$) to the maximum sum ($6+6=12$).

The possible sums are the integers from 2 to 12.

The statement says it is impossible to get a sum of 14. Since the maximum possible sum is 12, it is indeed impossible to get a sum of 14.

An impossible event has a probability of 0.

Therefore, the statement is true.

The statement is True.

To Determine: Whether the statement is True or False.

Statement: The probability of the spinning arrow stopping in the shaded region (Fig. 3.4) is $\frac{1}{2}$.

Solution:

The image shows a circle divided into several equally sized sectors with a spinning arrow at the center.

We need to count the total number of sectors and the number of shaded sectors.

Total number of sectors = 8

Number of shaded sectors = 4

Assuming the spinning arrow is equally likely to stop in any of the sectors, the probability of stopping in the shaded region is given by:

$P(\text{shaded region}) = \frac{\text{Number of shaded sectors}}{\text{Total number of sectors}}$

$P(\text{shaded region}) = \frac{4}{8}$

Simplifying the fraction:

$P(\text{shaded region}) = \frac{\cancel{4}^{1}}{\cancel{8}_{2}} = \frac{1}{2}$

The calculated probability of the spinning arrow stopping in the shaded region is $\frac{1}{2}$.

The statement says that the probability is $\frac{1}{2}$. This matches our calculation.

Therefore, the statement is true.

The statement is True.

To Determine: Whether the statement is True or False based on the given experimental data.

Statement: The chance of occurence of a head is 50 per cent.

Solution:

The given data shows the outcomes of tossing a coin 15 times:

H, T, T, H, T, H, H, H, T, T, H, T, H, T, T

We need to find the experimental probability (chance) of getting a head based on this data.

Total number of coin tosses = 15

Number of times a Head (H) occurred = Count the occurrences of 'H' in the data.

The outcomes are: H, T, T, H, T, H, H, H, T, T, H, T, H, T, T

Number of Heads = 7

The experimental probability of an event is given by:

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

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

$P(\text{Head}) = \frac{7}{15}$

To express this probability as a percentage (chance), we multiply by 100%:

$Chance(\text{Head}) = \frac{7}{15} \times 100\%$

$Chance(\text{Head}) \approx 0.4667 \times 100\%$

$Chance(\text{Head}) \approx 46.67\%$

The statement claims that the chance of occurrence of a head is 50 per cent.

Our calculated chance based on the experiment is approximately 46.67%, which is not equal to 50%.

While the theoretical probability of getting a head with a fair coin is 50%, the experimental probability based on a limited number of trials may differ from the theoretical probability.

The statement refers to "The chance of occurence of a head" in the context of the provided experimental results. Based on these results, the chance is 7 out of 15, which is not 50 per cent.

Therefore, the statement is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: Mean, Median and Mode may be the same for some data.

Solution:

We need to check if there exists a data set for which the calculated mean, median, and mode are all equal.

Consider the data set: 4, 4, 4, 4, 4.

Mean = $\frac{4+4+4+4+4}{5} = \frac{20}{5} = 4$.

Arranged data: 4, 4, 4, 4, 4. Median = Middle value = 4.

Mode = Most frequent value = 4.

In this data set, Mean = Median = Mode = 4.

Consider the data set: 3, 3, 4, 4, 4, 4, 6 (from Question 16(d)).

Mean = 4.

Arranged data: 3, 3, 4, 4, 4, 4, 6. Median = Middle value = 4.

Mode = Most frequent value = 4.

In this data set, Mean = Median = Mode = 4.

Data sets that are symmetric and unimodal (like a normal distribution) often have the mean, median, and mode equal or very close to each other.

Since we have found examples where the mean, median, and mode are the same, the statement is true.

The statement is True.

To Determine: Whether the statement is True or False.

Statement: The probability of getting an ace out of a deck of cards is greater than 1.

Solution:

In probability theory, the probability of any event must be a number between 0 and 1, inclusive.

$0 \leq P(\text{Event}) \leq 1$

A probability of 0 indicates an impossible event.

A probability of 1 indicates a sure event (certain to happen).

Probabilities cannot be less than 0 or greater than 1.

The statement claims that the probability of getting an ace out of a deck of cards is greater than 1.

Let's consider a standard deck of 52 playing cards. There are 4 aces in a deck (Ace of Spades, Ace of Hearts, Ace of Diamonds, Ace of Clubs).

Total number of outcomes (drawing one card) = 52.

Number of favorable outcomes (drawing an ace) = 4.

The probability of getting an ace is $\frac{\text{Number of Aces}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}$.

The value $\frac{1}{13}$ is between 0 and 1.

Even if we were talking about a non-standard deck or a different scenario, the fundamental principle of probability dictates that the probability value itself must be between 0 and 1 (inclusive).

Since the probability of any event cannot be greater than 1, the statement that the probability of getting an ace out of a deck of cards is greater than 1 is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: Mean of the data is always from the given data.

Solution:

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

Let's consider some examples.

Example 1: Data set = {2, 4, 6}

Mean = $\frac{2+4+6}{3} = \frac{12}{3} = 4$.

In this case, the mean (4) is one of the observations in the data set.

Example 2: Data set = {1, 2, 3, 4}

Mean = $\frac{1+2+3+4}{4} = \frac{10}{4} = 2.5$.

In this case, the mean (2.5) is not one of the observations in the data set.

Example 3: Data set = {5, 10}

Mean = $\frac{5+10}{2} = \frac{15}{2} = 7.5$.

In this case, the mean (7.5) is not one of the observations in the data set.

From these examples, we can see that the mean of a data set is not always one of the values present in the data set. It is only necessarily equal to a data point if all the data points are identical, or in specific cases where the calculation results in one of the values.

Therefore, the statement that the mean of the data is always from the given data is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: Median of the data may or may not be from the given data.

Solution:

The median is the middle value of a data set when the data is arranged in ascending or descending order.

Let $n$ be the number of observations in the data set.

Case 1: $n$ is odd.

If $n$ is odd, the median is the value at the $\left(\frac{n+1}{2}\right)$-th position in the ordered data. This value is always one of the original observations in the data set.

Example: Data = {1, 3, 5}. Ordered Data = {1, 3, 5}. $n=3$. Median is the $\frac{3+1}{2} = 2$nd term, which is 3. 3 is in the data set.

Case 2: $n$ is even.

If $n$ is even, the median is the average of the values at the $\frac{n}{2}$-th and $\left(\frac{n}{2}+1\right)$-th positions in the ordered data.

Median = $\frac{x_{n/2} + x_{n/2 + 1}}{2}$.

If the two middle values ($x_{n/2}$ and $x_{n/2 + 1}$) are different, their average will generally not be one of the original observations. However, if the two middle values are the same, their average is equal to that value, which is in the data set.

Example 1 (Middle values different): Data = {1, 2, 4, 5}. Ordered Data = {1, 2, 4, 5}. $n=4$. Median = $\frac{2+4}{2} = \frac{6}{2} = 3$. 3 is not in the data set.

Example 2 (Middle values same): Data = {1, 3, 3, 5}. Ordered Data = {1, 3, 3, 5}. $n=4$. Median = $\frac{3+3}{2} = \frac{6}{2} = 3$. 3 is in the data set.

Since the median is always one of the observations when $n$ is odd, but may or may not be one of the observations when $n$ is even, the statement that the median of the data may or may not be from the given data is true.

The statement is True.

To Determine: Whether the statement is True or False.

Statement: Mode of the data is always from the given data.

Solution:

The mode of a data set is defined as the value or values that appear most frequently in the data set.

By definition, the mode is determined by counting the frequencies of each distinct value in the data set and identifying the value(s) with the highest frequency.

Therefore, the mode must be one of the values that are present in the original data set. It cannot be a value that does not exist in the data.

Examples:

• Data: {1, 2, 2, 3, 4}. Mode is 2, which is in the data.
• Data: {5, 5, 6, 7, 7}. Modes are 5 and 7, both of which are in the data.
• Data: {10, 20, 30}. Each value appears once. There might be no mode, or all values can be considered modes depending on the definition used, but if a mode exists, it must be one of the values from the data.

Thus, the mode of a data set is always one of the observations in the given data.

The statement is True.

To Determine: Whether the statement is True or False.

Statement: Mean of the observations can be lesser than each of the observations.

Solution:

Let the observations in the data set be $x_1, x_2, ..., x_n$.

The mean ($\overline{x}$) is given by $\overline{x} = \frac{x_1 + x_2 + ... + x_n}{n}$.

The statement says that the mean can be lesser than each of the observations. This means $\overline{x} < x_i$ for all $i=1, 2, ..., n$.

Let's assume this is true for a data set. Then, for each observation $x_i$, we have $\overline{x} < x_i$.

Summing these inequalities for all $n$ observations, we get:

$\underbrace{\overline{x} + \overline{x} + ... + \overline{x}}_{n \ times} < x_1 + x_2 + ... + x_n$

$n \cdot \overline{x} < \sum_{i=1}^{n} x_i$

We know that $\overline{x} = \frac{\sum x_i}{n}$, which means $n \cdot \overline{x} = \sum x_i$.

So the inequality becomes:

$\sum x_i < \sum x_i$

This is a contradiction, as a quantity cannot be strictly less than itself.

This implies that our assumption that the mean can be lesser than each of the observations must be false, unless the data set is empty (which is usually not considered in these calculations). If the data set has at least one observation, the mean must be less than or equal to at least one of the observations.

More formally, if the mean were strictly less than every observation ($\overline{x} < x_i$ for all $i$), then $\overline{x}$ would be strictly less than the minimum observation ($x_{min}$). However, we know that $\overline{x} \geq x_{min}$. The only way this is possible is if the data set is empty.

If the data set is not empty ($n \geq 1$), then the mean is always greater than or equal to the minimum value and less than or equal to the maximum value:

This means that the mean cannot be lesser than every single observation unless all observations are equal, in which case the mean is equal to each observation, not strictly lesser.

The statement "Mean of the observations can be lesser than each of the observations" implies it is possible. Our analysis shows it is not possible for a non-empty data set.

Therefore, the statement is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: Mean can never be a fraction.

Solution:

The mean of a data set is calculated by summing all the observations and dividing by the total number of observations.

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

Let's consider an example:

Data set = {1, 2, 3, 4}

Sum of observations = $1 + 2 + 3 + 4 = 10$

Number of observations = 4

Mean = $\frac{10}{4} = \frac{5}{2} = 2.5$

In this example, the mean is 2.5, which is a decimal or a fraction ($\frac{5}{2}$).

The result of dividing the sum of observations by the number of observations is not necessarily a whole number. It can be a fraction or a decimal, depending on the values in the data set.

Therefore, the statement that the mean can never be a fraction is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: Range of the data is always from the data.

Solution:

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

$Range = Maximum \ Value - Minimum \ Value$

Let's consider some examples to see if the calculated range is always one of the values present in the original data.

Example 1: Data set = {2, 5, 8}

Highest Value = 8

Lowest Value = 2

Range = $8 - 2 = 6$. Is 6 in the data set {2, 5, 8}? No.

Example 2: Data set = {10, 12, 15, 20}

Highest Value = 20

Lowest Value = 10

Range = $20 - 10 = 10$. Is 10 in the data set {10, 12, 15, 20}? Yes, in this specific case, the range is equal to the lowest value which is in the data.

Example 3: Data set = {7, 7, 7}

Highest Value = 7

Lowest Value = 7

Range = $7 - 7 = 0$. Is 0 in the data set {7, 7, 7}? No (unless 0 was originally included in the data, which is not the case here).

The range is a value calculated from the data (specifically the difference between the maximum and minimum values), but the resulting difference value itself is not necessarily one of the original data points.

Since we have found examples where the range is not present in the data (Examples 1 and 3), the statement that the range of the data is *always* from the data is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: The data 12, 13, 14, 15, 16 has every observation as mode.

Solution:

The given data set is: 12, 13, 14, 15, 16.

The mode of a data set is the value or values that appear most frequently.

Let's count the frequency of each observation in the given data:

• 12 appears 1 time.
• 13 appears 1 time.
• 14 appears 1 time.
• 15 appears 1 time.
• 16 appears 1 time.

Each observation in the data set appears exactly once.

In such a case, where all observations have the same frequency (and that frequency is 1), the concept of mode can be interpreted in different ways depending on the specific definition used. Some definitions state that there is no mode in such a dataset, while others state that every observation is a mode (as they all have the highest frequency). In the context of elementary statistics, where the mode is the value(s) with the highest frequency, and all values share the same highest frequency (1), it is often considered that every observation is a mode.

If every observation is considered a mode because they all appear with the same highest frequency (which is 1), then the statement is true.

If the definition requires at least one value to have a frequency greater than other values for it to be a mode, then there would be no mode in this dataset, and the statement would be false.

However, the phrasing "every observation as mode" suggests the interpretation where all elements with the highest frequency (even if that frequency is 1) are considered modes. Given that all values appear once, they all share the highest frequency of 1.

Under the interpretation where any value with the highest frequency is a mode, and all values have the same frequency of 1, every observation is a mode.

Therefore, the statement is true under this common interpretation.

The statement is True.

To Determine: Whether the statement is True or False.

Statement: The range of the data 2, –5, 4, 3, 7, 6 would change if 2 was subtracted from each value in the data.

Solution:

First, let's find the range of the original data set: 2, –5, 4, 3, 7, 6.

Arranged data: -5, 2, 3, 4, 6, 7

Highest observation = 7

Lowest observation = -5

Range = Highest - Lowest = $7 - (-5) = 7 + 5 = 12$.

Now, let's subtract 2 from each value in the data set and find the range of the new data set.

Original data: 2, -5, 4, 3, 7, 6

Subtracting 2 from each value: $2-2=0$, $-5-2=-7$, $4-2=2$, $3-2=1$, $7-2=5$, $6-2=4$.

New data set: 0, -7, 2, 1, 5, 4.

Now, find the range of the new data set.

Arranged new data: -7, 0, 1, 2, 4, 5.

Highest observation in new data = 5

Lowest observation in new data = -7

Range of new data = Highest - Lowest = $5 - (-7) = 5 + 7 = 12$.

The range of the original data was 12.

The range of the new data (after subtracting 2 from each value) is also 12.

The range did not change.

In general, subtracting a constant value ($c$) from each observation in a data set shifts the entire data set by that constant amount. If the original data is $x_1, x_2, ..., x_n$, the new data is $x_1-c, x_2-c, ..., x_n-c$. The new highest value is $x_{max}-c$, and the new lowest value is $x_{min}-c$. The new range is $(x_{max}-c) - (x_{min}-c) = x_{max} - c - x_{min} + c = x_{max} - x_{min}$, which is the same as the original range. Adding or subtracting a constant to each observation does not change the range.

The statement says the range would change. Our calculation shows it does not change.

Therefore, the statement is false.

The statement is False.

To Determine: Whether the statement is True or False.

Statement: The range of the data 3, 7, 1, –2, 2, 6, –3, –5 would change if 8 was added to each value in the data.

Solution:

First, let's find the range of the original data set: 3, 7, 1, –2, 2, 6, –3, –5.

To find the range, we need the highest and lowest observations.

Highest observation = 7

Lowest observation = -5

Range = Highest - Lowest = $7 - (-5) = 7 + 5 = 12$.

Now, let's add 8 to each value in the data set and find the range of the new data set.

Original data: 3, 7, 1, -2, 2, 6, -3, -5

Adding 8 to each value:

• $3 + 8 = 11$
• $7 + 8 = 15$
• $1 + 8 = 9$
• $-2 + 8 = 6$
• $2 + 8 = 10$
• $6 + 8 = 14$
• $-3 + 8 = 5$
• $-5 + 8 = 3$

New data set: 11, 15, 9, 6, 10, 14, 5, 3.

Now, find the range of the new data set.

Highest observation in new data = 15

Lowest observation in new data = 3

Range of new data = Highest - Lowest = $15 - 3 = 12$.

The range of the original data was 12.

The range of the new data (after adding 8 to each value) is also 12.

The range did not change.

Adding or subtracting a constant value to every observation in a data set shifts the entire data set by that constant amount but does not change the difference between the highest and lowest values. Therefore, the range remains unchanged.

The statement says the range would change. Our calculation shows it does not change.

Therefore, the statement is false.

The statement is False.

Given Data:

The given data set is: 5, 10, 10, 12, 13

The number of observations is $n = 5$.

Calculation of Mean:

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

Formula for Mean ($\bar{x}$) = $\frac{\sum x_i}{n}$

Sum of observations ($\sum x_i$) = $5 + 10 + 10 + 12 + 13 = 50$

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

Mean = $10$

Calculation of Median:

First, arrange the data in ascending order:

5, 10, 10, 12, 13

The number of observations is $n = 5$, which is an odd number.

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

Median position = $\left(\frac{5+1}{2}\right)^{\text{th}} = \left(\frac{6}{2}\right)^{\text{th}} = 3^{\text{rd}}$ observation.

The $3^{\text{rd}}$ observation in the sorted data is 10.

Median = $10$

Calculation of Mode:

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

Let's look at the frequency of each observation:

• 5 occurs 1 time
• 10 occurs 2 times
• 12 occurs 1 time
• 13 occurs 1 time

The observation with the highest frequency is 10, which occurs 2 times.

Mode = $10$

Comparison:

We found the following values:

Mean = 10

Median = 10

Mode = 10

Yes, the Mean, Median, and Mode for the given data are all equal.

To Find:

The mean of the first ten even natural numbers.

Solution:

The first ten even natural numbers are:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20.

The number of observations is $n = 10$.

The sum of these numbers ($\sum x_i$) is:

$\sum x_i = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20$

We can group these for easier addition:

$\sum x_i = (2+18) + (4+16) + (6+14) + (8+12) + 10 + 20$

$\sum x_i = 20 + 20 + 20 + 20 + 10 + 20$

$\sum x_i = 4 \times 20 + 10 + 20$

$\sum x_i = 80 + 10 + 20$

$\sum x_i = 110$

Alternatively, the sum of the first $k$ even numbers is $k(k+1)$. Here $k=10$.

Sum = $10(10+1) = 10 \times 11 = 110$.

The formula for Mean ($\bar{x}$) is:

$\bar{x} = \frac{\sum x_i}{n}$

Substitute the values:

$\bar{x} = \frac{110}{10}$

$\bar{x} = 11$

Thus, the mean of the first ten even natural numbers is 11.

In a data set consisting of the heights (in cm) of 50 children, the mode represents the height that occurs most frequently among the 50 children.

Specifically:

1. The mode is a measure of central tendency.

2. It identifies the value that appears with the highest frequency in the data set.

3. If we calculate the mode for the heights of the 50 children, the resulting value (in cm) will be the height that the largest number of children in that group possess.

For example, if the mode is found to be 125 cm, it means that 125 cm is the height that is repeated more often than any other height among the 50 children.

Given Data:

The table shows the number of cars sold for different colours.

(a) Which colour of the car is most liked?

To find the most liked colour, we need to look for the colour with the highest number of cars sold.

• Red: 15 cars
• Black: 20 cars
• White: 17 cars
• Silver: 12 cars
• Others: 9 cars

Comparing the numbers, the highest number of cars sold is 20, which corresponds to the colour Black.

Therefore, the colour of the car that is most liked is Black.

(b) Which measure of central tendency was used in (a)?

In part (a), we identified the category (colour) that appeared most frequently (had the highest count) in the data.

The measure of central tendency that represents the value (or category) that occurs most frequently in a data set is called the Mode.

Therefore, the measure of central tendency used in (a) is the Mode.

Given Data:

The marks of 12 students are: 31, 37, 35, 38, 42, 23, 17, 18, 35, 25, 35, 29.

The number of observations is $n = 12$.

(a) Range:

The range is the difference between the highest and the lowest observation in the data.

First, let's arrange the data in ascending order to easily find the minimum and maximum values:

17, 18, 23, 25, 29, 31, 35, 35, 35, 37, 38, 42

Lowest value = 17

Highest value = 42

Range = Highest value - Lowest value

Range = $42 - 17$

Range = 25

(b) Mean:

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

Mean ($\bar{x}$) = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

Sum of observations ($\sum x_i$) = $31 + 37 + 35 + 38 + 42 + 23 + 17 + 18 + 35 + 25 + 35 + 29$

Sum of observations ($\sum x_i$) = 355

Number of observations ($n$) = 12

Mean = $\frac{355}{12}$

Mean $\approx 29.58$

(c) Median:

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

Arranging the data in ascending order:

17, 18, 23, 25, 29, 31, 35, 35, 35, 37, 38, 42

Since the number of observations ($n = 12$) is even, the median is the average of the two middle values.

The middle values are the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ observations.

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

$\left(\frac{12}{2} + 1\right)^{\text{th}} = 7^{\text{th}}$ observation

The 6th observation is 31.

The 7th observation is 35.

Median = $\frac{6^{\text{th}} \text{ observation} + 7^{\text{th}} \text{ observation}}{2}$

Median = $\frac{31 + 35}{2}$

Median = $\frac{66}{2}$

Median = 33

(d) Mode:

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

Let's count the frequency of each observation in the sorted data (17, 18, 23, 25, 29, 31, 35, 35, 35, 37, 38, 42):

• 17: 1 time
• 18: 1 time
• 23: 1 time
• 25: 1 time
• 29: 1 time
• 31: 1 time
• 35: 3 times
• 37: 1 time
• 38: 1 time
• 42: 1 time

The observation 35 occurs most frequently (3 times).

Mode = 35

Given Data:

The weights (in kg) of 12 people are: 70, 62, 54, 57, 62, 84, 75, 59, 62, 65, 78, 60.

Number of people, $n = 12$.

(a) Find the mean of the weights of the people.

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

Formula for Mean ($\bar{x}$) = $\frac{\sum x_i}{n}$

Sum of weights ($\sum x_i$) = $70 + 62 + 54 + 57 + 62 + 84 + 75 + 59 + 62 + 65 + 78 + 60$

Sum of weights ($\sum x_i$) = 708 kg

Number of observations ($n$) = 12

Mean = $\frac{708}{12}$

Mean = 59 kg

(b) How many people weigh above the mean weight?

The mean weight is 59 kg.

We need to count the number of people whose weight is greater than 59 kg from the given data:

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

Looking through the data, the weights greater than 59 are:

70, 62, 62, 84, 75, 62, 65, 78, 60.

Count the number of values in this list.

There are 9 weights that are greater than 59.

Therefore, 9 people weigh above the mean weight.

(c) Find the range of the given data.

The range is the difference between the highest and the lowest observation in the data.

Given data: 70, 62, 54, 57, 62, 84, 75, 59, 62, 65, 78, 60

Identify the lowest weight in the data set: 54 kg.

Identify the highest weight in the data set: 84 kg.

Range = Highest weight - Lowest weight

Range = $84 - 54$

Range = 30 kg

Given:

The cards are marked with the letters: M, A, T, H, E, M, A, T, I, C, S.

To Find:

The probability (chance) of drawing out:

(a) a vowel

(b) A or I

(c) a card marked U

(d) a consonant

Solution:

The letters on the cards are: M, A, T, H, E, M, A, T, I, C, S.

The total number of possible outcomes (total number of cards) is $N = 11$.

The formula for the probability of an event E is given by:

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

(a) Drawing a vowel:

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

In the given set of letters {M, A, T, H, E, M, A, T, I, C, S}, the vowels are A, E, I.

Counting the occurrences:

• A appears 2 times.
• E appears 1 time.
• I appears 1 time.

The total number of vowels (favorable outcomes) is $2 + 1 + 1 = 4$.

The probability of drawing a vowel is:

$P(\text{vowel}) = \frac{\text{Number of vowels}}{\text{Total number of cards}} = \frac{4}{11}$

(b) Drawing A or I:

The favorable outcomes are the cards marked 'A' or 'I'.

Counting the occurrences:

• A appears 2 times.
• I appears 1 time.

The total number of cards marked A or I (favorable outcomes) is $2 + 1 = 3$.

The probability of drawing A or I is:

$P(\text{A or I}) = \frac{\text{Number of cards marked A or I}}{\text{Total number of cards}} = \frac{3}{11}$

(c) Drawing a card marked U:

We need to check if the letter 'U' is present in the set {M, A, T, H, E, M, A, T, I, C, S}.

The letter U is not present in this set.

The number of cards marked U (favorable outcomes) is 0.

The probability of drawing a card marked U is:

$P(\text{U}) = \frac{\text{Number of cards marked U}}{\text{Total number of cards}} = \frac{0}{11}$

$P(\text{U}) = 0$

(d) Drawing a consonant:

The consonants in the given set of letters {M, A, T, H, E, M, A, T, I, C, S} are M, T, H, C, S.

Counting the occurrences:

• M appears 2 times.
• T appears 2 times.
• H appears 1 time.
• C appears 1 time.
• S appears 1 time.

The total number of consonants (favorable outcomes) is $2 + 2 + 1 + 1 + 1 = 7$.

The probability of drawing a consonant is:

$P(\text{consonant}) = \frac{\text{Number of consonants}}{\text{Total number of cards}} = \frac{7}{11}$

Given Data:

The given data set is: 6, 2, 5, 4, 3, 4, 4, 2, 3

The number of observations is $n = 9$.

Calculation of Median:

First, arrange the data in ascending order:

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

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

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

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

The $5^{\text{th}}$ observation in the sorted data is 4.

Median = 4

Calculation of Mode:

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

Let's count the frequency of each observation:

• 2 occurs 2 times.
• 3 occurs 2 times.
• 4 occurs 3 times.
• 5 occurs 1 time.
• 6 occurs 1 time.

The observation with the highest frequency is 4 (occurs 3 times).

Mode = 4

Calculation of the Mean of Median and Mode:

Median = 4

Mode = 4

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

Mean = $\frac{4 + 4}{2}$

Mean = $\frac{8}{2}$

Mean = 4

Thus, the mean of the median and mode of the given data is 4.

Given Data:

The given data set is: 5, 7, 7, 8, x, 5, 4, 3, 1, 2.

The mean of the data is 4.5.

The number of observations is $n = 10$.

To Find:

The median of the given data.

Solution:

First, we need to find the value of 'x' using the given mean.

The formula for Mean ($\bar{x}$) is:

$\bar{x} = \frac{\text{Sum of observations}}{\text{Number of observations}}$

Sum of observations = $5 + 7 + 7 + 8 + x + 5 + 4 + 3 + 1 + 2$

Sum of observations = $42 + x$

Given Mean ($\bar{x}$) = 4.5

Number of observations ($n$) = 10

Substitute the values into the mean formula:

Multiply both sides by 10:

$4.5 \times 10 = 42 + x$

$45 = 42 + x$

Solve for x:

$x = 45 - 42$

$x = 3$

Now substitute the value of x back into the data set:

The data set is: 5, 7, 7, 8, 3, 5, 4, 3, 1, 2.

Calculating the Median:

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

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

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

For an even number of observations, the median is the average of the two middle values.

The middle values are the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ observations.

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

$\left(\frac{10}{2} + 1\right)^{\text{th}} = 6^{\text{th}}$ observation

From the ordered data (1, 2, 3, 3, 4, 5, 5, 7, 7, 8):

The 5th observation is 4.

The 6th observation is 5.

Median = $\frac{5^{\text{th}} \text{ observation} + 6^{\text{th}} \text{ observation}}{2}$

Median = $\frac{4 + 5}{2}$

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

Median = 4.5

The median of the given data is 4.5.

The event "the sun setting tomorrow" is a certain event based on our understanding of the Earth's rotation and its orbit around the sun under normal circumstances.

A certain event is an event that is sure to happen.

The probability of a certain event is always 1 (or 100%).

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

Given:

A spinner with three colours: Red, Blue, and Green, as shown in Fig. 3.5.

To Find:

Which colour has a greater chance of showing up with the arrow.

Solution:

We need to determine how many sections are colored with each colour by looking at the figure.

The spinner is divided into sections.

Number of Red sections = 2

Number of Blue sections = 1

Number of Green sections = 1

Total number of sections on the spinner = $2 + 1 + 1 = 4$.

Assuming that each section is of equal size, the chance or probability of the spinner landing on a certain colour is proportional to the number of sections assigned to that colour.

The probability of an event E is given by:

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

For the given spinner:

Total possible outcomes = Total number of sections = 4.

Probability of landing on Red = $\frac{\text{Number of Red sections}}{\text{Total sections}} = \frac{2}{4} = \frac{1}{2}$

Probability of landing on Blue = $\frac{\text{Number of Blue sections}}{\text{Total sections}} = \frac{1}{4}$

Probability of landing on Green = $\frac{\text{Number of Green sections}}{\text{Total sections}} = \frac{1}{4}$

Comparing the probabilities:

$P(\text{Red}) = \frac{1}{2} = 0.5$

$P(\text{Blue}) = \frac{1}{4} = 0.25$

$P(\text{Green}) = \frac{1}{4} = 0.25$

The probability of landing on Red ($0.5$) is greater than the probability of landing on Blue ($0.25$) and Green ($0.25$).

This is because the Red colour occupies a larger portion (2 out of 4 sections) of the spinner compared to Blue (1 section) and Green (1 section).

Therefore, the colour with more chance to show up with the arrow is Red.

Given:

Number of girls = 3

Number of boys = 4

To Find:

The probability that a student chosen at random is a boy.

Solution:

First, find the total number of students.

Total number of students = Number of girls + Number of boys

Total number of students = $3 + 4 = 7$

The total number of possible outcomes when choosing one student is the total number of students, which is 7.

The favorable outcome is choosing a boy.

The number of favorable outcomes (number of boys) is 4.

The formula for the probability of an event E is given by:

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

In this case, the event E is choosing a boy.

$P(\text{boy}) = \frac{\text{Number of boys}}{\text{Total number of students}}$

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

Therefore, the probability that a student chosen at random is a boy is $\frac{4}{7}$.

Given:

The letters on the paper slips are from the word "MEDIAN".

To Find:

The probability that a randomly drawn slip bears the letter D.

Solution:

The letters in the word MEDIAN are: M, E, D, I, A, N.

Count the total number of letters in the word "MEDIAN".

Total number of letters = 6.

This is the total number of possible outcomes when drawing one slip.

Identify the number of times the letter 'D' appears in the word "MEDIAN".

The letter 'D' appears 1 time.

This is the number of favorable outcomes.

The formula for the probability of an event E is given by:

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

In this case, the event E is drawing the letter D.

$P(\text{drawing D}) = \frac{\text{Number of times D appears}}{\text{Total number of letters}}$

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

Therefore, the probability that the slip bears the letter D is $\frac{1}{6}$.

Here is the classification of the given events:

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

A standard die has faces marked with numbers 1, 2, 3, 4, 5, and 6. There is no number less than 1 on a standard die.

Classification: Impossible to happen.

(b) Getting head when a coin is tossed.

When a coin is tossed, the outcome can be either a head or a tail. Getting a head is a possible outcome, but it is not guaranteed.

Classification: May or may not happen.

(c) A team winning the match.

In a match, a team can win, lose, or draw. Winning is a possible outcome for a team, but whether a specific team wins depends on various factors and is not certain beforehand.

Classification: May or may not happen.

(d) Christmas will be on 25 December.

Christmas is a fixed date in the calendar, always celebrated on the 25th of December.

Classification: Certain to happen.

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

The Moon revolves around the Earth due to gravitational forces. This is a continuous astronomical process. For the Moon to stop revolving today is contrary to fundamental physics and highly improbable under normal circumstances.

Classification: Impossible to happen.

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

Due to Earth's gravity, any object thrown upwards will be pulled back towards the Earth and will fall down after some time, assuming it is thrown within the atmosphere and does not reach escape velocity.

Classification: Certain to happen.

Given Data:

The outcomes of throwing a die 15 times are: 5, 3, 4, 1, 2, 6, 4, 2, 2, 3, 1, 5, 6, 1, 2.

The number of observations is $n = 15$.

Calculation of Mean:

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

Formula for Mean ($\bar{x}$) = $\frac{\sum x_i}{n}$

Sum of observations ($\sum x_i$) = $5 + 3 + 4 + 1 + 2 + 6 + 4 + 2 + 2 + 3 + 1 + 5 + 6 + 1 + 2$

Sum of observations ($\sum x_i$) = 47

Number of observations ($n$) = 15

Mean = $\frac{47}{15}$

Mean $\approx 3.13$ (rounded to two decimal places)

Calculation of Median:

First, arrange the data in ascending order:

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

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

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

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

The $8^{\text{th}}$ observation in the sorted data (1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6) is 3.

Median = 3

Calculation of Mode:

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

Let's count the frequency of each outcome from the sorted data (1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6):

• 1 appears 3 times.
• 2 appears 4 times.
• 3 appears 2 times.
• 4 appears 2 times.
• 5 appears 2 times.
• 6 appears 2 times.

The outcome with the highest frequency is 2 (occurs 4 times).

Mode = 2

To Find:

The mean of the first six multiples of 4.

Solution:

The first six multiples of 4 are:

$4 \times 1 = 4$

$4 \times 2 = 8$

$4 \times 3 = 12$

$4 \times 4 = 16$

$4 \times 5 = 20$

$4 \times 6 = 24$

So, the data set is: 4, 8, 12, 16, 20, 24.

The number of observations is $n = 6$.

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

Formula for Mean ($\bar{x}$) = $\frac{\sum x_i}{n}$

Sum of observations ($\sum x_i$) = $4 + 8 + 12 + 16 + 20 + 24$

Sum of observations ($\sum x_i$) = $12 + 12 + 16 + 20 + 24$

Sum of observations ($\sum x_i$) = $24 + 16 + 20 + 24$

Sum of observations ($\sum x_i$) = $40 + 20 + 24$

Sum of observations ($\sum x_i$) = $60 + 24$

Sum of observations ($\sum x_i$) = 84

Number of observations ($n$) = 6

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

Mean = 14

Thus, the mean of the first six multiples of 4 is 14.

To Find:

The median of the first nine even natural numbers.

Solution:

The first nine even natural numbers are:

2, 4, 6, 8, 10, 12, 14, 16, 18.

The data is already arranged in ascending order.

The number of observations is $n = 9$.

Since the number of observations ($n=9$) is an odd number, the median is the value of the $\left(\frac{n+1}{2}\right)^{\text{th}}$ observation.

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

From the ordered data (2, 4, 6, 8, 10, 12, 14, 16, 18), the $5^{\text{th}}$ observation is 10.

Median = 10

Thus, the median of the first nine even natural numbers is 10.

Given:

Mean of the first three numbers = 10

Mean of the next four numbers = 12

To Find:

The mean of all the seven numbers.

Solution:

The formula for the mean of a set of numbers is:

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

This can be rearranged to find the sum of observations:

Sum of observations = Mean $\times$ Number of observations

For the first three numbers:

Number of observations ($n_1$) = 3

Mean ($\bar{x}_1$) = 10

Sum of the first three numbers = $\bar{x}_1 \times n_1$

Sum$_1 = 30$

For the next four numbers:

Number of observations ($n_2$) = 4

Mean ($\bar{x}_2$) = 12

Sum of the next four numbers = $\bar{x}_2 \times n_2$

Sum$_2 = 48$

For all the numbers:

Total number of observations ($N$) = $n_1 + n_2 = 3 + 4 = 7$

Total sum of observations ($\sum X$) = Sum$_1$ + Sum$_2$

$\sum X = 30 + 48$

$\sum X = 78$

The mean of all the numbers ($\bar{X}$) is:

$\bar{X} = \frac{\text{Total sum of observations}}{\text{Total number of observations}}$

$\bar{X} = \frac{78}{7}$

The mean of all the numbers is $\frac{78}{7}$. This can also be expressed as approximately $11.14$ (rounded to two decimal places).

Given Data:

The given data set is: 10, 8, 4, 7, 8, 11, 15, 8, 4, 2, 3, 6, 8.

To Find:

The mode of the given data.

Solution:

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

To find the mode, we count the frequency of each observation in the data set:

Data: 10, 8, 4, 7, 8, 11, 15, 8, 4, 2, 3, 6, 8.

Let's list the unique values and their counts:

• 10: appears 1 time
• 8: appears 4 times
• 4: appears 2 times
• 7: appears 1 time
• 11: appears 1 time
• 15: appears 1 time
• 2: appears 1 time
• 3: appears 1 time
• 6: appears 1 time

The observation that occurs most frequently is 8, which appears 4 times.

Therefore, the mode of the given data is 8.

Given Data:

The heights (in cm) of 15 boys are: 128, 144, 146, 143, 136, 142, 138, 129, 140, 152, 144, 140, 150, 142, 154.

The number of boys (observations) is $n = 15$.

(a) The height of the tallest boy:

To find the height of the tallest boy, we need to find the maximum value in the given data set.

The heights are: 128, 144, 146, 143, 136, 142, 138, 129, 140, 152, 144, 140, 150, 142, 154.

Comparing the values, the highest height is 154 cm.

The height of the tallest boy is 154 cm.

(b) The height of the shortest boy:

To find the height of the shortest boy, we need to find the minimum value in the given data set.

The heights are: 128, 144, 146, 143, 136, 142, 138, 129, 140, 152, 144, 140, 150, 142, 154.

Comparing the values, the lowest height is 128 cm.

The height of the shortest boy is 128 cm.

(c) The range of the given data:

The range is the difference between the highest and the lowest observation in the data.

Highest height = 154 cm

Lowest height = 128 cm

Range = Highest height - Lowest height

Range = $154 - 128$

Range = 26 cm

(d) The median height of the boys:

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

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

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

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

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

From the ordered data (128, 129, 136, 138, 140, 140, 142, 142, 143, 144, 144, 146, 150, 152, 154), the $8^{\text{th}}$ observation is 142.

Median = 142 cm

Given Data:

The given data set is: 16, 15, 16, 16, 8, 15, 17.

First, let's find the initial mode of the data by counting the frequency of each value:

• 8: 1 time
• 15: 2 times
• 16: 3 times
• 17: 1 time

The value with the highest frequency is 16. So, the current mode is 16.

(a) Which data value can be put in the data so that the mode remains the same?

The current mode is 16, with a frequency of 3.

To keep the mode as 16, the new value(s) added must not increase the frequency of any other value to 3 or more, unless it's also 16.

If we add 16, the frequency of 16 becomes 4, and it remains the mode.

If we add 8, 15, or 17, their frequencies will become 2, 3, or 2 respectively. Adding 15 would make its frequency 3, which ties with 16 (resulting in two modes, 15 and 16, if only one 15 is added). Adding 8 or 17 does not change the mode.

To ensure the mode remains 16, the added value should either be 16 (increasing its frequency) or any value other than 15, 8, or 17 if adding just one value.

However, the question asks for "which data value can be put in". Adding 16 definitely keeps 16 as the mode.

A simple and direct answer is to add the current mode.

Therefore, putting in the value 16 will ensure the mode remains the same.

(b) At least how many and which value(s) must be put in to change the mode to 15?

The current frequency of 16 is 3.

The current frequency of 15 is 2.

To make 15 the mode, its frequency must be greater than the frequency of 16.

If we add one 15, the frequency of 15 becomes 3. The frequency of 16 is also 3. This results in two modes (15 and 16). To make 15 the *sole* mode, its frequency must be strictly greater than 3.

If we add two 15s, the frequency of 15 becomes $2 + 2 = 4$. The frequency of 16 is 3. Now 15 has the highest frequency.

So, we must add at least two values of 15.

(c) What is the least number of data values that must be put in to change the mode to 17? Name them.

The current frequency of 16 is 3.

The current frequency of 17 is 1.

To make 17 the mode, its frequency must be greater than the frequency of 16 (which is 3).

If we add one 17, its frequency becomes 2 (not greater than 3).

If we add two 17s, its frequency becomes 3 (equal to 16's frequency, resulting in two modes). To make 17 the sole mode, its frequency must be at least 4.

To reach a frequency of 4 for 17, we need to add $4 - 1 = 3$ values of 17.

If we add three 17s, the frequency of 17 becomes $1 + 3 = 4$. The frequency of 16 is 3. Now 17 has the highest frequency.

The least number of data values that must be put in to change the mode to 17 is three.

The values are 17, 17, and 17.

Given Data:

Group A ages: 7, 7, 9, 8, 10, 10 (Number of children $n_A = 6$)

Group B ages: 7, 9, 11, 12, 12, 12 (Number of children $n_B = 6$)

(a) Find the mode and range for each group.

For Group A:

Data for Group A: 7, 7, 9, 8, 10, 10

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

Mode (Group A): The value that appears most frequently.

• 7 appears 2 times.
• 8 appears 1 time.
• 9 appears 1 time.
• 10 appears 2 times.

Both 7 and 10 appear with the highest frequency (2 times). This data set is bimodal.

Mode for Group A is 7 and 10.

Range (Group A): Highest value - Lowest value.

Highest age in Group A = 10

Lowest age in Group A = 7

Range for Group A = $10 - 7 = 3$

Range for Group A is 3.

For Group B:

Data for Group B: 7, 9, 11, 12, 12, 12

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

Mode (Group B): The value that appears most frequently.

• 7 appears 1 time.
• 9 appears 1 time.
• 11 appears 1 time.
• 12 appears 3 times.

The value 12 appears most frequently (3 times).

Mode for Group B is 12.

Range (Group B): Highest value - Lowest value.

Highest age in Group B = 12

Lowest age in Group B = 7

Range for Group B = $12 - 7 = 5$

Range for Group B is 5.

(b) Find the range and mode if the two groups are combined together.

Combined data set: 7, 7, 9, 8, 10, 10, 7, 9, 11, 12, 12, 12

Total number of children $n = n_A + n_B = 6 + 6 = 12$.

Arranging the combined data in ascending order:

7, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 12

Mode (Combined Data): The value that appears most frequently.

• 7 appears 3 times.
• 8 appears 1 time.
• 9 appears 2 times.
• 10 appears 2 times.
• 11 appears 1 time.
• 12 appears 3 times.

Both 7 and 12 appear with the highest frequency (3 times). The combined data set is bimodal.

Mode for the combined data is 7 and 12.

Range (Combined Data): Highest value - Lowest value.

From the ordered combined data (7, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 12):

Highest age in combined data = 12

Lowest age in combined data = 7

Range for combined data = $12 - 7 = 5$

Range for the combined data is 5.

Given:

A bar graph showing the production of motor bikes over a period of 6 months (January to June).

From the bar graph, we can read the production for each month:

• January: 600 motor bikes
• February: 800 motor bikes
• March: 700 motor bikes
• April: 1000 motor bikes
• May: 900 motor bikes
• June: 1200 motor bikes

(a) What information does the bar graph depict?

The bar graph depicts the number of motor bikes produced in a factory during the months from January to June.

(b) How many motor bikes were produced in the first three months?

The first three months are January, February, and March.

Production in January = 600

Production in February = 800

Production in March = 700

Total production in the first three months = Production (Jan) + Production (Feb) + Production (Mar)

Total production = $600 + 800 + 700$

Total production = $1400 + 700$

Total production = 2100

2100 motor bikes were produced in the first three months.

(c) Calculate the increase in production in May over the production in January.

Production in May = 900

Production in January = 600

Increase in production = Production (May) - Production (January)

Increase = $900 - 600$

Increase = 300

The increase in production in May over the production in January is 300 motor bikes.

(d) In which month the production was minimum and what was it?

Look at the heights of the bars to find the lowest one:

• January: 600
• February: 800
• March: 700
• April: 1000
• May: 900
• June: 1200

The minimum production is 600, which occurred in January.

The production was minimum in January, and it was 600 motor bikes.

(e) Calculate the average (mean) production of bikes in 6 months.

The average (mean) production is the total production over 6 months divided by the number of months (6).

Total production over 6 months = Production (Jan) + Feb + Mar + Apr + May + Jun

Total production = $600 + 800 + 700 + 1000 + 900 + 1200$

Total production = $1400 + 700 + 1000 + 900 + 1200$

Total production = $2100 + 1000 + 900 + 1200$

Total production = $3100 + 900 + 1200$

Total production = $4000 + 1200$

Total production = 5200

Number of months = 6

Average (Mean) production = $\frac{\text{Total production}}{\text{Number of months}}$

Average production = $\frac{5200}{6}$

Average production = $\frac{2600}{3}$

Average production $\approx 866.67$

The average production of bikes in 6 months is $\frac{2600}{3}$ or approximately 866.67 motor bikes.

Reading the given bar graph, we find the number of students in different mark ranges are as follows:

Marks 0-19: 2 students

Marks 20-39: 3 students

Marks 40-49: 4 students

Marks 50-59: 3 students

Marks 60-69: 3 students

Marks 70-79: 5 students

Marks 80-89: 2 students

Marks 90-100: 2 students

(a) If 40 is the pass mark, then how many students have failed?

Students who scored less than 40 marks have failed.

These are the students in the mark ranges 0-19 and 20-39.

Number of failed students = (Number of students in 0-19 range) + (Number of students in 20-39 range)

Number of failed students = 2 + 3 = 5

(b) How many students got marks from 50 to 69?

These are the students in the mark ranges 50-59 and 60-69.

Number of students with marks from 50 to 69 = (Number of students in 50-59 range) + (Number of students in 60-69 range)

Number of students with marks from 50 to 69 = 3 + 3 = 6

(c) How many students scored 90 marks and above?

These are the students in the mark range 90-100.

Number of students who scored 90 marks and above = 2

(d) If students who scored marks above 80 are given merits then how many merit holders are there?

Students who scored marks above 80 are those in the mark ranges 80-89 and 90-100.

Number of merit holders = (Number of students in 80-89 range) + (Number of students in 90-100 range)

Number of merit holders = 2 + 2 = 4

(e) What is the strength of the class?

The strength of the class is the total number of students in all ranges.

Total strength = (Students in 0-19) + (Students in 20-39) + (Students in 40-49) + (Students in 50-59) + (Students in 60-69) + (Students in 70-79) + (Students in 80-89) + (Students in 90-100)

Total strength = 2 + 3 + 4 + 3 + 3 + 5 + 2 + 2 = 24

Reading the given bar graph, we find the production of rice in tonnes for each year:

Year 2005: 40 tonnes

Year 2006: 50 tonnes

Year 2007: 70 tonnes

Year 2008: 50 tonnes

Year 2009: 80 tonnes

(a) What information does the above bar graph represent?

The bar graph represents the production of rice in tonnes during the years 2005 to 2009.

(b) In which year was production the least?

Looking at the heights of the bars, the shortest bar corresponds to the least production.

The production is lowest in the year 2005 (40 tonnes).

(c) After which year was the maximum rise in the production?

Let's calculate the rise in production between consecutive years:

Rise from 2005 to 2006 = $50 - 40 = 10$ tonnes

Rise from 2006 to 2007 = $70 - 50 = 20$ tonnes

Rise from 2007 to 2008 = $50 - 70 = -20$ tonnes (This is a fall)

Rise from 2008 to 2009 = $80 - 50 = 30$ tonnes

The maximum rise is 30 tonnes, which occurred from 2008 to 2009.

The question asks "After which year was the maximum rise". The maximum rise occurred after the year 2008.

So, the maximum rise in production was after the year 2008.

(d) Find the average production of rice during the 5 years.

Average production = $\frac{\text{Sum of production for 5 years}}{\text{Number of years}}$

Sum of production = (Production in 2005) + (Production in 2006) + (Production in 2007) + (Production in 2008) + (Production in 2009)

Sum of production = $40 + 50 + 70 + 50 + 80 = 290$ tonnes

Number of years = 5

Average production = $\frac{290}{5} = 58$ tonnes

The average production of rice during the 5 years is 58 tonnes.

(e) Find difference of rice production between years 2006 and 2008.

Production in 2006 = 50 tonnes

Production in 2008 = 50 tonnes

Difference in production = (Production in 2008) - (Production in 2006)

Difference in production = $50 - 50 = 0$ tonnes

The difference of rice production between years 2006 and 2008 is 0 tonnes.

Reading the given bar graph, we find the marks obtained by the student in each subject:

Hindi: 60 marks

English: 50 marks

Mathematics: 75 marks

Science: 80 marks

Social Studies: 70 marks

(a) What information is depicted from the bar graph?

The bar graph depicts the marks obtained by a student in different subjects.

(b) In which subject is the student very good?

A student is considered very good in the subject where they scored the highest marks.

Comparing the marks: Hindi (60), English (50), Mathematics (75), Science (80), Social Studies (70).

The highest marks obtained are 80 in Science.

So, the student is very good in Science.

(c) Calculate the average marks of the student.

Average marks = $\frac{\text{Sum of marks in all subjects}}{\text{Number of subjects}}$

Sum of marks = $60 + 50 + 75 + 80 + 70$

Sum of marks = $335$

Number of subjects = 5

Average marks = $\frac{335}{5}$

Average marks = $67$

The average marks of the student are 67.

(d) If 75 and above marks denote a distinction, then name the subjects in which the student got distinction.

We need to identify subjects where the student scored 75 or more marks.

Marks in Mathematics = 75 (Distinction)

Marks in Science = 80 (Distinction)

Marks in Hindi = 60 (No distinction)

Marks in English = 50 (No distinction)

Marks in Social Studies = 70 (No distinction)

The subjects in which the student got distinction are Mathematics and Science.

(e) Calculate the percentage of marks the student got out of 500.

Total marks obtained = 335 (from part c)

Maximum possible marks = 500

Percentage of marks = $\left(\frac{\text{Total marks obtained}}{\text{Maximum possible marks}}\right) \times 100\%$

Percentage of marks = $\left(\frac{335}{500}\right) \times 100\%$

Percentage of marks = $\left(\frac{335}{\cancel{500}_{\cancel{5}}}\right) \times \cancel{100}^1 \%$

Percentage of marks = $\frac{335}{5} \%$

Percentage of marks = $67 \%$

The percentage of marks the student got out of 500 is 67%.

Reading the given bar graph, and noting that the figures are in hundreds, we find the circulation of newspapers in each language:

Hindi: 80 hundreds = 8000

English: 30 hundreds = 3000

Punjabi: 40 hundreds = 4000

Urdu: 10 hundreds = 1000

Marathi: 60 hundreds = 6000

Tamil: 20 hundreds = 2000

(a) Find the total number of newspapers read in Hindi, Punjabi, Urdu, Marathi and Tamil.

Total newspapers = (Hindi circulation) + (Punjabi circulation) + (Urdu circulation) + (Marathi circulation) + (Tamil circulation)

Total newspapers = $8000 + 4000 + 1000 + 6000 + 2000$

Total newspapers = $12000 + 1000 + 6000 + 2000$

Total newspapers = $13000 + 6000 + 2000$

Total newspapers = $19000 + 2000$

Total newspapers = 21000

(b) Find the excess number of newspapers read in Hindi than those in English.

Excess newspapers = (Hindi circulation) - (English circulation)

Excess newspapers = $8000 - 3000$

Excess newspapers = 5000

(c) Name the language in which the least number of newspapers are read.

Comparing the circulation figures for all languages: Hindi (8000), English (3000), Punjabi (4000), Urdu (1000), Marathi (6000), Tamil (2000).

The least circulation is 1000.

This corresponds to the language Urdu.

(d) Write the total circulation of newspapers in the town.

Total circulation = Sum of circulation in all languages

Total circulation = (Hindi) + (English) + (Punjabi) + (Urdu) + (Marathi) + (Tamil)

Total circulation = $8000 + 3000 + 4000 + 1000 + 6000 + 2000$

Total circulation = $11000 + 4000 + 1000 + 6000 + 2000$

Total circulation = $15000 + 1000 + 6000 + 2000$

Total circulation = $16000 + 6000 + 2000$

Total circulation = $22000 + 2000$

Total circulation = 24000

Reading the given double bar graph, we can find the number of students in Class VII and Class VIII who like different sports:

Class VII:

Cricket: 12 students

Football: 8 students

Hockey: 6 students

Tennis: 4 students

Badminton: 10 students

Class VIII:

Cricket: 10 students

Football: 10 students

Hockey: 6 students

Tennis: 5 students

Badminton: 8 students

(a) Which sport is liked the most by Class VIII students?

Comparing the number of Class VIII students for each sport: Cricket (10), Football (10), Hockey (6), Tennis (5), Badminton (8).

The highest number of students in Class VIII (10 students) like Cricket and Football.

So, the sports liked the most by Class VIII students are Cricket and Football.

(b) How many students of Class VII like Hockey and Tennis in all?

Number of Class VII students who like Hockey = 6

Number of Class VII students who like Tennis = 4

Total students liking Hockey and Tennis in Class VII = $6 + 4 = 10$

Total students of Class VII who like Hockey and Tennis are 10.

(c) How many students are there in Class VII?

The total number of students in Class VII is the sum of students who like each sport in Class VII.

Total students in Class VII = (Cricket) + (Football) + (Hockey) + (Tennis) + (Badminton)

Total students in Class VII = $12 + 8 + 6 + 4 + 10 = 40$

The strength of Class VII is 40.

(d) For which sport is the number of students of Class VII less than that of Class VIII?

Comparing the number of students in Class VII and Class VIII for each sport:

Cricket: Class VII (12) $>$ Class VIII (10)

Football: Class VII (8) $<$ Class VIII (10)

Hockey: Class VII (6) $=$ Class VIII (6)

Tennis: Class VII (4) $<$ Class VIII (5)

Badminton: Class VII (10) $>$ Class VIII (8)

The number of students in Class VII is less than that of Class VIII for Football and Tennis.

(e) For how many sports students of Class VIII are less than Class VII?

Comparing the number of students in Class VIII and Class VII for each sport:

Cricket: Class VIII (10) $<$ Class VII (12)

Football: Class VIII (10) $>$ Class VII (8)

Hockey: Class VIII (6) $=$ Class VII (6)

Tennis: Class VIII (5) $>$ Class VII (4)

Badminton: Class VIII (8) $<$ Class VII (10)

Students of Class VIII are less than Class VII for Cricket and Badminton.

This occurs for 2 sports.

(f) Find the ratio of students who like Badminton in Class VII to students who like Tennis in Class VIII.

Number of students who like Badminton in Class VII = 10

Number of students who like Tennis in Class VIII = 5

Ratio = (Students liking Badminton in Class VII) : (Students liking Tennis in Class VIII)

Ratio = $10 : 5$

Simplifying the ratio by dividing both numbers by 5:

Ratio = $\frac{\cancel{10}^2}{\cancel{5}_1} : \frac{\cancel{5}^1}{\cancel{5}_1} = 2 : 1$

The ratio is $2 : 1$.

Reading the given double bar graph, we can find the sales figures for Brand A and Brand B for each month:

January: Brand A = 100, Brand B = 80

February: Brand A = 120, Brand B = 100

March: Brand A = 130, Brand B = 120

April: Brand A = 140, Brand B = 110

May: Brand A = 110, Brand B = 130

June: Brand A = 150, Brand B = 140

(a) What information is represented by the above double bar graph?

The double bar graph represents the sales of two brands, Brand A and Brand B, over a period of six months (from January to June).

(b) In which month sales of Brand A decreased as compared to the previous month?

Let's examine the sales of Brand A month by month:

January to February: $120 - 100 = 20$ (Increase)

February to March: $130 - 120 = 10$ (Increase)

March to April: $140 - 130 = 10$ (Increase)

April to May: $110 - 140 = -30$ (Decrease)

May to June: $150 - 110 = 40$ (Increase)

Sales of Brand A decreased in the month of May as compared to April.

(c) What is the difference in sales of both the Brands for the month of June?

Sales of Brand A in June = 150

Sales of Brand B in June = 140

Difference in sales = Sales of Brand A - Sales of Brand B

Difference in sales = $150 - 140 = 10$

The difference in sales of both the Brands for the month of June is 10.

(d) Find the average sales of Brand B for the six months.

Sum of sales of Brand B for six months = Sales in Jan + Feb + Mar + Apr + May + Jun

Sum of sales of Brand B = $80 + 100 + 120 + 110 + 130 + 140$

Sum of sales of Brand B = $180 + 120 + 110 + 130 + 140$

Sum of sales of Brand B = $300 + 110 + 130 + 140$

Sum of sales of Brand B = $410 + 130 + 140$

Sum of sales of Brand B = $540 + 140 = 680$

Number of months = 6

Average sales of Brand B = $\frac{\text{Sum of sales}}{\text{Number of months}}$

Average sales of Brand B = $\frac{680}{6} = \frac{340}{3}$

Average sales of Brand B = $113.\overline{33}$ (approximately)

The average sales of Brand B for the six months is $\frac{340}{3}$ or approximately 113.33.

(e) List all months for which the sales of Brand B was less than that of Brand A.

We compare sales of Brand B with Brand A for each month:

January: Brand B (80) $<$ Brand A (100) - Yes

February: Brand B (100) $<$ Brand A (120) - Yes

March: Brand B (120) $<$ Brand A (130) - Yes

April: Brand B (110) $<$ Brand A (140) - Yes

May: Brand B (130) $>$ Brand A (110) - No

June: Brand B (140) $<$ Brand A (150) - Yes

The months for which the sales of Brand B was less than that of Brand A are January, February, March, April, and June.

(f) Find the ratio of sales of Brand A as compared to Brand B for the month of January.

Sales of Brand A in January = 100

Sales of Brand B in January = 80

Ratio of sales of Brand A to Brand B = $\frac{\text{Sales of Brand A}}{\text{Sales of Brand B}}$

Ratio = $\frac{100}{80} = \frac{\cancel{100}^{10}}{\cancel{80}_{8}} = \frac{\cancel{10}^5}{\cancel{8}_4} = \frac{5}{4}$

The ratio of sales of Brand A as compared to Brand B for the month of January is $5 : 4$.

Reading the given double bar graph, we can find the minimum temperatures ($^\circ$C) for the years 2008 and 2009 for four months:

November 2008: $10^\circ\text{C}$

November 2009: $15^\circ\text{C}$

December 2008: $8^\circ\text{C}$

December 2009: $12^\circ\text{C}$

January 2008: $6^\circ\text{C}$

January 2009: $10^\circ\text{C}$

February 2008: $9^\circ\text{C}$

February 2009: $8^\circ\text{C}$

(a) What information is compared in the above given double bar graph?

The double bar graph compares the minimum temperatures ($^\circ$C) recorded during four months (November, December, January, and February) in two different years, 2008 and 2009.

(b) Calculate the ratio of minimum temperatures in the year 2008 to the year 2009 for the month of November.

Minimum temperature in November 2008 = $10^\circ\text{C}$

Minimum temperature in November 2009 = $15^\circ\text{C}$

Ratio of minimum temperature in 2008 to 2009 for November = $\frac{\text{Min Temp Nov 2008}}{\text{Min Temp Nov 2009}}$

Ratio = $\frac{10}{15}$

Simplifying the ratio:

Ratio = $\frac{\cancel{10}^2}{\cancel{15}_3} = \frac{2}{3}$

The ratio is $2:3$.

(c) For how many months was the minimum temperature in the year 2008 greater than that of year 2009? Name those months.

Comparing minimum temperatures for each month:

November: $10^\circ\text{C}$ (2008) vs $15^\circ\text{C}$ (2009) $\implies 10 < 15$

December: $8^\circ\text{C}$ (2008) vs $12^\circ\text{C}$ (2009) $\implies 8 < 12$

January: $6^\circ\text{C}$ (2008) vs $10^\circ\text{C}$ (2009) $\implies 6 < 10$

February: $9^\circ\text{C}$ (2008) vs $8^\circ\text{C}$ (2009) $\implies 9 > 8$

The minimum temperature in the year 2008 was greater than that of year 2009 only in February.

This occurred for 1 month, which is February.

(d) Find the average minimum temperature for the year 2008 for the four months.

Sum of minimum temperatures in 2008 = (Nov) + (Dec) + (Jan) + (Feb)

Sum of minimum temperatures in 2008 = $10 + 8 + 6 + 9 = 33^\circ\text{C}$

Number of months = 4

Average minimum temperature = $\frac{\text{Sum of temperatures}}{\text{Number of months}}$

Average minimum temperature = $\frac{33}{4}^\circ\text{C}$

Average minimum temperature = $8.25^\circ\text{C}$

The average minimum temperature for the year 2008 for the four months is $8.25^\circ\text{C}$.

(e) In which month is the variation in the two temperatures maximum?

We calculate the absolute difference (variation) between the temperatures in 2009 and 2008 for each month:

November Variation: $|15 - 10| = 5^\circ\text{C}$

December Variation: $|12 - 8| = 4^\circ\text{C}$

January Variation: $|10 - 6| = 4^\circ\text{C}$

February Variation: $|8 - 9| = |-1| = 1^\circ\text{C}$

Comparing the variations ($5, 4, 4, 1$), the maximum variation is $5^\circ\text{C}$.

This maximum variation occurs in the month of November.

To draw a double bar graph comparing the average intake of nutrients in calories by rural and urban groups, follow these steps:

Step 1: Draw the axes.

Draw two perpendicular lines on a graph paper. The horizontal axis (X-axis) will represent the Foodstuff categories, and the vertical axis (Y-axis) will represent the Average Intake of Nutrients in Calories.

Step 2: Choose a suitable scale for the Y-axis.

The highest value in the table is 250 (Urban Milk). A suitable scale would be $1$ unit (or $1$ cm) on the Y-axis representing $25$ or $50$ calories. Let's use $1$ unit = $25$ calories for better resolution.

Mark points on the Y-axis at equal intervals corresponding to the chosen scale (e.g., 0, 25, 50, 75, ..., 250, 275).

Step 3: Label the axes.

Label the X-axis as 'Foodstuff' and the Y-axis as 'Average Intake of Nutrients (Calories)'.

Step 4: Draw the bars for each foodstuff.

For each Foodstuff category (Pulses, Leafy vegetables, etc.), draw two adjacent bars:

One bar representing the average intake for the Rural group.

Another bar representing the average intake for the Urban group.

The height of each bar should correspond to the value given in the table according to the chosen scale on the Y-axis.

Leave some space between each pair of bars representing different Foodstuff categories.

Step 5: Use different colours or patterns for the bars.

Use one colour/pattern for all bars representing the Rural group and a different colour/pattern for all bars representing the Urban group.

Step 6: Add a legend.

Create a legend to indicate which colour/pattern corresponds to the Rural group and which corresponds to the Urban group.

Step 7: Give the graph a title.

Give the bar graph a suitable title, such as "Average Nutrient Intake by Rural and Urban Groups".

Based on the data provided:

Pulses: Rural = 35, Urban = 49

Leafy vegetables: Rural = 14, Urban = 21

Other vegetables: Rural = 51, Urban = 89

Fruits: Rural = 35, Urban = 66

Milk: Rural = 70, Urban = 250

Fish and flesh foods: Rural = 10, Urban = 22

Fats and Oils: Rural = 9, Urban = 35

Sugar/Jaggery: Rural = 19, Urban = 31

Draw pairs of bars for each foodstuff category with the heights corresponding to these values on the chosen scale. For example, for Milk, the Rural bar will go up to the mark for 70, and the Urban bar will go up to the mark for 250 on the Y-axis.

Reading the given double bar graph, we find the number of boys and girls in each section of Class VII:

Section A: Boys = 45, Girls = 25

Section B: Boys = 40, Girls = 30

Section C: Boys = 50, Girls = 35

Section D: Boys = 35, Girls = 40

(a) What information does the double bar graph represent?

The double bar graph represents the number of boys and girls in different sections (A, B, C, D) of Class VII.

(b) Find the total number of boys in all sections of Class VII.

Total number of boys = (Boys in A) + (Boys in B) + (Boys in C) + (Boys in D)

Total number of boys = $45 + 40 + 50 + 35$

Total number of boys = $85 + 50 + 35$

Total number of boys = $135 + 35$

Total number of boys = 170

(c) In which sections, the number of girls is greater than the number of boys?

Comparing the number of girls and boys for each section:

Section A: Girls (25) vs Boys (45) $\implies 25 < 45$

Section B: Girls (30) vs Boys (40) $\implies 30 < 40$

Section C: Girls (35) vs Boys (50) $\implies 35 < 50$

Section D: Girls (40) vs Boys (35) $\implies 40 > 35$

The number of girls is greater than the number of boys in Section D.

(d) In which section, the number of boys is the maximum?

Comparing the number of boys in each section: Section A (45), Section B (40), Section C (50), Section D (35).

The maximum number of boys is 50, which is in Section C.

The number of boys is maximum in Section C.

(e) In which section, the number of girls is the least?

Comparing the number of girls in each section: Section A (25), Section B (30), Section C (35), Section D (40).

The least number of girls is 25, which is in Section A.

The number of girls is the least in Section A.

The given table shows the number of Newspaper Readers and Magazine Readers in a public library on six different days of a week.

(a) Draw a double bar graph choosing an appropriate scale.

To draw a double bar graph, follow these steps:

Draw the horizontal and vertical axes. Label the horizontal axis (X-axis) as 'Days' and the vertical axis (Y-axis) as 'Number of Readers'.

Choose a suitable scale for the Y-axis. The maximum number of readers for a single category on any day is 600. An appropriate scale could be $1$ unit on the Y-axis representing $50$ or $100$ readers. Let's choose $1$ unit = $50$ readers.

Mark points on the Y-axis at intervals of 50 (0, 50, 100, 150, 200, ..., 600, ...).

For each day (Mon, Tue, Wed, Thurs, Fri, Sat), draw two adjacent bars. One bar will represent the number of Newspaper Readers, and the other will represent the number of Magazine Readers.

The height of the bar for Newspaper Readers on Monday should be 400 according to the scale ($400/50 = 8$ units). The height of the bar for Magazine Readers on Monday should be 150 ($150/50 = 3$ units).

Repeat this for all six days, drawing pairs of bars with heights corresponding to the values in the table.

Use different colours or shading patterns for the Newspaper Reader bars and the Magazine Reader bars.

Include a legend to show which bar represents Newspaper Readers and which represents Magazine Readers.

Give the graph a title, such as "Number of Library Readers in a Week".

(b) On which day, the number of readers in the library was maximum?

To find the day with the maximum number of readers, we need to find the total number of readers (Newspaper + Magazine) for each day:

Monday: $400 + 150 = 550$ readers

Tuesday: $600 + 100 = 700$ readers

Wednesday: $350 + 200 = 550$ readers

Thursday: $550 + 300 = 850$ readers

Friday: $500 + 250 = 750$ readers

Saturday: $350 + 200 = 550$ readers

Comparing the total numbers for each day ($550, 700, 550, 850, 750, 550$), the maximum number is 850.

This occurred on Thursday.

(c) What is the mean number of magazine readers?

The mean number of magazine readers is the sum of magazine readers over the six days divided by the number of days.

Sum of magazine readers = $150 + 100 + 200 + 300 + 250 + 200 = 1200$

Number of days = 6

Mean number of magazine readers = $\frac{\text{Sum of magazine readers}}{\text{Number of days}}$

Mean number of magazine readers = $\frac{1200}{6}$

Mean number of magazine readers = 200

The mean number of magazine readers is 200.

The given table shows the total number of students and the number of students present on a particular day (15.4.2009) for classes VI to X in a government school.

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

To draw the double bar graph:

Draw horizontal (X-axis) and vertical (Y-axis) axes.

Label the X-axis as 'Class' and the Y-axis as 'Number of Students'.

Choose a suitable scale for the Y-axis. Since the maximum number of students is 95, a scale of $1$ unit = $10$ students or $1$ unit = $5$ students would be appropriate. Let's choose $1$ unit = $5$ students, marking the axis at 0, 5, 10, ..., 100.

For each class (VI, VII, VIII, IX, X), draw two adjacent bars. One bar represents the 'Total Students' and the other represents the 'Number of Students Present'.

The height of the 'Total Students' bar for Class VI will be 90 units on the Y-axis (if 1 unit = 1 student) or $90/5 = 18$ units (if 1 unit = 5 students). Similarly, the height of the 'Students Present' bar for Class VI will be 81 units or $81/5 = 16.2$ units.

Use different colours or patterns to distinguish between 'Total Students' bars and 'Students Present' bars.

Include a legend specifying which bar represents 'Total Students' and which represents 'Students Present'.

Give the graph a title, e.g., "Class-wise Total Students and Attendance on 15.4.2009".

Inference from the bar graph:

By observing the graph, we can infer the total strength of each class and the attendance percentage. Generally, the number of students present is very close to the total number of students in all classes, indicating a high attendance rate. Comparing the heights of the bars for 'Total Students' across classes gives the class strength. Comparing the difference in heights for each class (the gap between the two bars) shows the number of absent students.

(b) Which class has the maximum number of students?

Looking at the 'Total Students' column:

Class VI: 90

Class VII: 82

Class VIII: 95

Class IX: 70

Class X: 63

The maximum number among these is 95.

Therefore, Class VIII has the maximum number of students.

(c) In which class, the difference of total students and number of students present is minimum?

The difference between total students and students present is the number of absent students.

Absent students in Class VI = $90 - 81 = 9$

Absent students in Class VII = $82 - 76 = 6$

Absent students in Class VIII = $95 - 91 = 4$

Absent students in Class IX = $70 - 65 = 5$

Absent students in Class X = $63 - 62 = 1$

Comparing the differences ($9, 6, 4, 5, 1$), the minimum difference is 1.

This minimum difference is in Class X.

(d) Find the ratio of number of students present to the total number of students of Class IX.

Number of students present in Class IX = 65

Total number of students in Class IX = 70

Ratio = (Students Present) : (Total Students)

Ratio = $65 : 70$

To simplify the ratio, divide both numbers by their greatest common divisor, which is 5.

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

The ratio is $13:14$.

(e) What per cent of Class VI students were absent?

Total students in Class VI = 90

Students present in Class VI = 81

Number of absent students in Class VI = Total students - Students present

Number of absent students = $90 - 81 = 9$

Percentage of absent students = $\left(\frac{\text{Number of absent students}}{\text{Total students}}\right) \times 100\%$

Percentage of absent students = $\left(\frac{9}{90}\right) \times 100\%$

Percentage of absent students = $\left(\frac{\cancel{9}^1}{\cancel{90}_{10}}\right) \times 100\%$

Percentage of absent students = $\frac{1}{10} \times 100\%$

Percentage of absent students = $10\%$.

The given table shows the number of mobile phone sets sold on different days of the week.

(a) Draw a bar graph to represent the above given information.

To draw the bar graph:

Draw horizontal (X-axis) and vertical (Y-axis) axes.

Label the X-axis as 'Days of the Week' and mark the days (Mon, Tue, Wed, Thurs, Fri, Sat) at equal intervals.

Label the Y-axis as 'Number of Mobile Phone Sets Sold'.

Choose a suitable scale for the Y-axis. The maximum number of sets sold is 60. A scale of $1$ unit on the Y-axis representing $5$ or $10$ mobile phone sets would be appropriate. Let's choose $1$ unit = $5$ sets, marking the axis at 0, 5, 10, 15, ..., 60, ...

Draw bars of equal width above each day on the X-axis. The height of each bar should correspond to the number of mobile phone sets sold on that day according to the chosen scale on the Y-axis.

For Monday, the height of the bar should be 50 units ($50/5 = 10$ units). For Tuesday, it should be 45 units ($45/5 = 9$ units), and so on.

Give the bar graph a title, such as "Mobile Phone Sets Sold During the Week".

Inference from the bar graph: The bar graph visually represents the sales trend over the week. It allows for easy comparison of sales on different days, quickly showing the day with the highest and lowest sales and days with similar sales figures.

(b) On which day of the week was the sales maximum?

Looking at the number of mobile phone sets sold each day: Mon (50), Tue (45), Wed (30), Thurs (55), Fri (27), Sat (60).

The maximum number of sets sold is 60.

This occurred on Saturday.

(c) Find the total sales during the week.

Total sales = Sum of sales on all days

Total sales = $50 + 45 + 30 + 55 + 27 + 60$

Total sales = $95 + 30 + 55 + 27 + 60$

Total sales = $125 + 55 + 27 + 60$

Total sales = $180 + 27 + 60$

Total sales = $207 + 60 = 267$

The total sales during the week were 267 mobile phone sets.

(d) Find the ratio of the minimum sale to the maximum sale.

Minimum sale = 27 (on Friday)

Maximum sale = 60 (on Saturday)

Ratio = Minimum sale : Maximum sale

Ratio = $27 : 60$

To simplify the ratio, divide both numbers by their greatest common divisor, which is 3.

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

The ratio of the minimum sale to the maximum sale is $9:20$.

(e) Calculate the average sale during the week.

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

Total sales = 267 (from part c)

Number of days = 6

Average sale = $\frac{267}{6}$

Average sale = $\frac{89}{2} = 44.5$

The average sale during the week is 44.5 mobile phone sets.

(f) On how many days of the week was the sale above the average sales?

Average sale = 44.5

We compare the sales on each day with the average:

Monday: $50 > 44.5$ (Above average)

Tuesday: $45 > 44.5$ (Above average)

Wednesday: $30 < 44.5$ (Below average)

Thursday: $55 > 44.5$ (Above average)

Friday: $27 < 44.5$ (Below average)

Saturday: $60 > 44.5$ (Above average)

The sale was above the average on Monday, Tuesday, Thursday, and Saturday.

This occurred on 4 days of the week.

From the given table, we have the following data for the tallest buildings in India (as per information till 2009):

Planet (Mumbai): Height = 181 m, Floors = 51

UB Tower (Bengaluru): Height = 184 m, Floors = 20

Ashok Towers (Mumbai): Height = 193 m, Floors = 49

The Imperial I (Mumbai): Height = 249 m, Floors = 60

The Imperial II (Mumbai): Height = 249 m, Floors = 60

RNA Mirage (Mumbai): Height = 180 m, Floors = 40

Oberoi Woods Tower I (Mumbai): Height = 170 m, Floors = 40

Oberoi Woods Tower II (Mumbai): Height = 170 m, Floors = 40

Oberoi Woods Tower III (Mumbai): Height = 170 m, Floors = 40

MVRDC (Mumbai): Height = 156 m, Floors = 35

(a) Find the height of each storey of the three tallest buildings and write them in the following table:

The three tallest buildings based on height are The Imperial I (249 m), The Imperial II (249 m), and Ashok Towers (193 m).

Height of each storey = Total Height / Number of Storeys

For The Imperial I: Height per storey = $249 \text{ m} / 60 \text{ floors} = 4.15 \text{ m/storey}$.

For The Imperial II: Height per storey = $249 \text{ m} / 60 \text{ floors} = 4.15 \text{ m/storey}$.

For Ashok Towers: Height per storey = $193 \text{ m} / 49 \text{ floors} \approx 3.94 \text{ m/storey}$ (rounded to two decimal places).

(b) The average height of one storey for the buildings given in (a) is ______________.

Average height per storey = $\frac{\text{Sum of heights per storey for the three buildings}}{3}$

Average height per storey = $\frac{4.15 + 4.15 + (193/49)}{3}$

Average height per storey = $\frac{8.3 + 3.93877...}{3}$

Average height per storey = $\frac{12.23877...}{3} \approx 4.0796...$ m

Rounding to two decimal places, the average height of one storey for these three buildings is approximately 4.08 m.

(c) Which city in this list has the largest percentage of skyscrappers? What is the percentage ?

Total number of buildings in the list = 10

Number of buildings in Mumbai:

Planet, Ashok Towers, The Imperial I, The Imperial II, RNA Mirage, Oberoi Woods Tower I, Oberoi Woods Tower II, Oberoi Woods Tower III, MVRDC. There are 9 buildings in Mumbai.

Number of buildings in Bengaluru: UB Tower. There is 1 building in Bengaluru.

Percentage of buildings in Mumbai = $\left(\frac{\text{Number of buildings in Mumbai}}{\text{Total number of buildings}}\right) \times 100\%$

Percentage of buildings in Mumbai = $\left(\frac{9}{10}\right) \times 100\% = 90\%$

Percentage of buildings in Bengaluru = $\left(\frac{\text{Number of buildings in Bengaluru}}{\text{Total number of buildings}}\right) \times 100\%$

Percentage of buildings in Bengaluru = $\left(\frac{1}{10}\right) \times 100\% = 10\%$

The city with the largest percentage of skyscrapers in this list is Mumbai.

The percentage is 90%.

(d) What is the range of data?

Assuming the question asks for the range of heights of the buildings.

Maximum height in the list = 249 m (The Imperial I & II)

Minimum height in the list = 156 m (MVRDC)

Range = Maximum value - Minimum value

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

The range of the height data is 93 m.

(e) Find the median of the data.

Assuming the question asks for the median of the heights of the buildings.

First, arrange the heights in ascending order:

156, 170, 170, 170, 180, 181, 184, 193, 249, 249

There are 10 data points. Since the number of data points is even, the median is the average of the two middle values (the 5th and 6th values).

5th value = 180

6th value = 181

Median = $\frac{180 + 181}{2} = \frac{361}{2} = 180.5$

The median height of the buildings is 180.5 m.

(f) Draw a bar graph for given data.

We can draw a bar graph to represent the height of each building listed.

Draw the horizontal and vertical axes. Label the horizontal axis (X-axis) as 'Building Name' and the vertical axis (Y-axis) as 'Height (in meters)'.

Choose a suitable scale for the Y-axis. Since the heights range from 156 m to 249 m, a scale of $1$ unit on the Y-axis representing $20$ meters would be appropriate. Mark points on the Y-axis at intervals (e.g., 0, 20, 40, ..., 260).

Draw bars of equal width above the name of each building on the X-axis. The height of each bar should correspond to the height of the building according to the chosen scale on the Y-axis.

Give the bar graph a title, such as "Height of Tallest Buildings in India".

The bar graph would visually compare the heights of the 10 buildings.

The given table shows the marks obtained by Kunal and Soni in six subjects in the Half Yearly Examination, where the maximum marks for each subject are 100.

(a) Draw a double bar graph by choosing appropriate scale.

To draw the double bar graph:

Draw horizontal (X-axis) and vertical (Y-axis) axes. Label the X-axis as 'Subjects' and mark the subjects (English, Hindi, Maths, Science, S. Science, Sanskrit) at equal intervals. Label the Y-axis as 'Marks Obtained'.

Choose a suitable scale for the Y-axis. Since the maximum marks are 100, a scale of $1$ unit on the Y-axis representing $10$ marks would be appropriate. Mark points on the Y-axis at intervals of 10 (0, 10, 20, ..., 100).

For each subject, draw two adjacent bars of equal width. One bar represents Kunal's marks, and the other represents Soni's marks.

The height of the bar for Kunal's marks in English should be 72 according to the scale ($72/10 = 7.2$ units). The height of the bar for Soni's marks in English should be 86 units ($86/10 = 8.6$ units). Repeat this for all six subjects, drawing pairs of bars with heights corresponding to the marks obtained by Kunal and Soni.

Use different colours or shading patterns to distinguish between Kunal's bars and Soni's bars.

Include a legend specifying which bar represents Kunal and which represents Soni.

Give the graph a title, e.g., "Marks of Kunal and Soni in Half Yearly Examination".

(b) Calculate the total percentage of marks obtained by Soni.

Total marks obtained by Soni = Sum of marks in all subjects

Total marks obtained by Soni = $86 + 89 + 90 + 82 + 75 + 82$

Total marks obtained by Soni = $175 + 90 + 82 + 75 + 82$

Total marks obtained by Soni = $265 + 82 + 75 + 82$

Total marks obtained by Soni = $347 + 75 + 82$

Total marks obtained by Soni = $422 + 82 = 504$

Maximum possible marks = Number of subjects $\times$ Maximum marks per subject

Maximum possible marks = $6 \times 100 = 600$

Percentage of marks obtained by Soni = $\left(\frac{\text{Total marks obtained}}{\text{Maximum possible marks}}\right) \times 100\%$

Percentage of marks obtained by Soni = $\left(\frac{504}{600}\right) \times 100\%$

Percentage of marks obtained by Soni = $\left(\frac{504}{\cancel{600}_{\cancel{6}}}\right) \times \cancel{100}^1 \%$

Percentage of marks obtained by Soni = $\frac{504}{6} \%$

Percentage of marks obtained by Soni = $84 \%$

The total percentage of marks obtained by Soni is 84%.

(c) Calculate the total percentage of marks obtained by Kunal.

Total marks obtained by Kunal = Sum of marks in all subjects

Total marks obtained by Kunal = $72 + 81 + 92 + 96 + 64 + 85$

Total marks obtained by Kunal = $153 + 92 + 96 + 64 + 85$

Total marks obtained by Kunal = $245 + 96 + 64 + 85$

Total marks obtained by Kunal = $341 + 64 + 85$

Total marks obtained by Kunal = $405 + 85 = 490$

Maximum possible marks = 600

Percentage of marks obtained by Kunal = $\left(\frac{\text{Total marks obtained}}{\text{Maximum possible marks}}\right) \times 100\%$

Percentage of marks obtained by Kunal = $\left(\frac{490}{600}\right) \times 100\%$

Percentage of marks obtained by Kunal = $\left(\frac{490}{\cancel{600}_{\cancel{6}}}\right) \times \cancel{100}^1 \%$

Percentage of marks obtained by Kunal = $\frac{490}{6} \%$

Percentage of marks obtained by Kunal = $\frac{245}{3} \% \approx 81.67 \%$

The total percentage of marks obtained by Kunal is $\frac{245}{3}\%$ or approximately 81.67%.

(d) Compare the percentages of marks obtained by Kunal and Soni.

Soni's percentage = 84%

Kunal's percentage = $\frac{245}{3}\% \approx 81.67\%$

Comparing the two percentages, $84\% > 81.67\%$.

Thus, Soni obtained a higher total percentage of marks than Kunal.

(e) In how many subjects did Soni get more marks than Kunal? Which are those subjects?

Comparing marks subject-wise:

English: Soni (86) vs Kunal (72) $\implies 86 > 72$ (Soni got more)

Hindi: Soni (89) vs Kunal (81) $\implies 89 > 81$ (Soni got more)

Maths: Soni (90) vs Kunal (92) $\implies 90 < 92$ (Kunal got more)

Science: Soni (82) vs Kunal (96) $\implies 82 < 96$ (Kunal got more)

S. Science: Soni (75) vs Kunal (64) $\implies 75 > 64$ (Soni got more)

Sanskrit: Soni (82) vs Kunal (85) $\implies 82 < 85$ (Kunal got more)

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

This occurred in 3 subjects. The subjects are English, Hindi, and S. Science.

(f) Who got more marks in S. Science and what was the difference of marks?

Marks in S. Science: Kunal = 64, Soni = 75

Comparing the marks, $75 > 64$.

So, Soni got more marks in S. Science.

Difference of marks = Soni's marks - Kunal's marks

Difference of marks = $75 - 64 = 11$

The difference of marks was 11.

(g) In which subject the difference of marks was maximum and by how much?

Let's calculate the absolute difference in marks for each subject:

English: $|86 - 72| = 14$

Hindi: $|89 - 81| = 8$

Maths: $|90 - 92| = |-2| = 2$

Science: $|82 - 96| = |-14| = 14$

S. Science: $|75 - 64| = 11$

Sanskrit: $|82 - 85| = |-3| = 3$

Comparing the differences ($14, 8, 2, 14, 11, 3$), the maximum difference is 14.

This maximum difference occurs in English and Science.

The maximum difference in marks was 14.

The given table shows the number of girls and boys from Class VII who chose different clubs.

(a) Draw a double bar graph using appropriate scale to depict the above data.

To draw the double bar graph:

Draw horizontal (X-axis) and vertical (Y-axis) axes. Label the X-axis as 'Clubs' and mark the clubs (Music, Dance, Yoga, Dramatics, Fine Arts, Electronics) at equal intervals.

Label the Y-axis as 'Number of Students'.

Choose a suitable scale for the Y-axis. The maximum number of students in any category is 30 (Boys in Electronics) and 27 (Girls in Fine Arts). A scale of $1$ unit on the Y-axis representing $2$ or $4$ students would be appropriate. Let's choose $1$ unit = $2$ students, marking the axis at 0, 2, 4, ..., 30.

For each club, draw two adjacent bars of equal width. One bar represents the number of girls, and the other represents the number of boys.

The height of the bar for Girls in Music should be 15 units ($15/2 = 7.5$ units). The height of the bar for Boys in Music should be 12 units ($12/2 = 6$ units). Repeat this for all six clubs, drawing pairs of bars with heights corresponding to the number of girls and boys.

Use different colours or shading patterns to distinguish between the bars for girls and boys.

Include a legend specifying which bar represents Girls and which represents Boys.

Give the graph a title, e.g., "Club Preferences of Class VII Students".

(b) How many students are there in Class VII?

Total students in Class VII = (Total number of girls) + (Total number of boys)

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

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

Total students = $116 + 94 = 210$

There are 210 students in Class VII.

(c) Which is the most preferred club by boys?

Comparing the number of boys for each club: Music (12), Dance (16), Yoga (8), Dramatics (17), Fine Arts (11), Electronics (30).

The maximum number of boys is 30, which corresponds to the Electronics club.

The most preferred club by boys is Electronics.

(d) Which is the least preferred club by girls?

Comparing the number of girls for each club: Music (15), Dance (24), Yoga (10), Dramatics (19), Fine Arts (27), Electronics (21).

The minimum number of girls is 10, which corresponds to the Yoga club.

The least preferred club by girls is Yoga.

(e) For which club the difference between boys and girls is the least?

We calculate the absolute difference between the number of girls and boys for each club:

Music: $|15 - 12| = 3$

Dance: $|24 - 16| = 8$

Yoga: $|10 - 8| = 2$

Dramatics: $|19 - 17| = 2$

Fine Arts: $|27 - 11| = 16$

Electronics: $|21 - 30| = |-9| = 9$

Comparing the differences ($3, 8, 2, 2, 16, 9$), the least difference is 2.

This occurs for the Yoga club and the Dramatics club.

(f) For which club is the difference between boys and girls the maximum?

Comparing the differences calculated in part (e) ($3, 8, 2, 2, 16, 9$), the maximum difference is 16.

This occurs for the Fine Arts club.

The given table shows the production of motor bikes in a factory for certain months in the years 2007 and 2008.

(a) Draw a double bar graph using appropriate scale to depict the above information and compare them.

To draw the double bar graph:

Draw horizontal (X-axis) and vertical (Y-axis) axes. Label the X-axis as 'Months' and mark the months (Feb, May, August, October, December) at equal intervals. Label the Y-axis as 'Production of Motor Bikes'.

Choose a suitable scale for the Y-axis. The maximum production figure is 6000. An appropriate scale could be $1$ unit on the Y-axis representing $500$ or $1000$ motor bikes. Let's choose $1$ unit = $500$ motor bikes, marking the axis at 0, 500, 1000, ..., 6000.

For each month, draw two adjacent bars of equal width. One bar represents the production in 2008, and the other represents the production in 2007.

The height of the bar for February 2008 should be 2700 units according to the scale ($2700/500 = 5.4$ units). The height of the bar for February 2007 should be 2800 units ($2800/500 = 5.6$ units). Repeat this for all five months, drawing pairs of bars with heights corresponding to the production figures for 2008 and 2007.

Use different colours or shading patterns to distinguish between the bars for 2008 and 2007.

Include a legend specifying which bar represents the production in 2008 and which represents the production in 2007.

Give the graph a title, e.g., "Motor Bike Production in 2007 and 2008".

Comparison from the bar graph: The double bar graph allows for a visual comparison of production levels between the two years for the same months. It helps to quickly identify months where production was higher or lower in one year compared to the other and to see which year had generally higher production.

(b) In which year was the total output the maximum?

Calculate the total production for each year for the given months:

Total production in 2008 = Production in Feb + May + Aug + Oct + Dec

Total production in 2008 = $2700 + 3200 + 6000 + 5000 + 4200 = 21100$

Total production in 2007 = Production in Feb + May + Aug + Oct + Dec

Total production in 2007 = $2800 + 4500 + 4800 + 4800 + 5200 = 22100$

Comparing the total production figures, $22100 > 21100$.

The total output was maximum in the year 2007.

(c) Find the mean production for the year 2007.

Mean production for 2007 = $\frac{\text{Total production in 2007}}{\text{Number of months}}$

Total production in 2007 = 22100 (from part b)

Number of months = 5

Mean production for 2007 = $\frac{22100}{5} = 4420$

The mean production for the year 2007 for these months was 4420 motor bikes.

(d) For which month was the difference between the production for the two years the maximum?

Calculate the absolute difference in production between the two years for each month:

February: $|2700 - 2800| = |-100| = 100$

May: $|3200 - 4500| = |-1300| = 1300$

August: $|6000 - 4800| = 1200$

October: $|5000 - 4800| = 200$

December: $|4200 - 5200| = |-1000| = 1000$

Comparing the differences ($100, 1300, 1200, 200, 1000$), the maximum difference is 1300.

This occurred in the month of May.

(e) In which month for the year 2008, the production was the maximum?

Looking at the production figures for 2008: Feb (2700), May (3200), August (6000), October (5000), December (4200).

The maximum production in 2008 is 6000.

This was in the month of August.

(f) In which month for the year 2007, the production was the least?

Looking at the production figures for 2007: Feb (2800), May (4500), August (4800), October (4800), December (5200).

The least production in 2007 is 2800.

This was in the month of February.

The given table shows the population (in hundreds) of four towns (A, B, C, D) in two different years, 2007 and 2009.

(a) Draw a double bar graph using appropriate scale to depict the above information.

To draw the double bar graph:

Draw horizontal (X-axis) and vertical (Y-axis) axes. Label the X-axis as 'Towns' and mark the towns (A, B, C, D) at equal intervals.

Label the Y-axis as 'Population (in hundreds)'.

Choose a suitable scale for the Y-axis. The maximum population figure is 9200 hundreds. An appropriate scale could be $1$ unit on the Y-axis representing $500$ or $1000$ hundreds. Let's choose $1$ unit = $500$ hundreds, marking the axis at 0, 500, 1000, 1500, ..., 9500.

For each town, draw two adjacent bars of equal width. One bar represents the population in 2007, and the other represents the population in 2009.

The height of the bar for Town A in 2007 should be 2900 units according to the scale ($2900/500 = 5.8$ units). The height of the bar for Town A in 2009 should be 3200 units ($3200/500 = 6.4$ units). Repeat this for all four towns, drawing pairs of bars with heights corresponding to the population figures for 2007 and 2009.

Use different colours or shading patterns to distinguish between the bars for 2007 and 2009.

Include a legend specifying which bar represents the population in 2007 and which represents the population in 2009.

Give the graph a title, e.g., "Population of Towns A, B, C, D in 2007 and 2009".

Inference from the bar graph: The double bar graph visually compares the population of each town in 2007 and 2009. It clearly shows the population growth in each town (by the difference in heights of the paired bars) and allows for easy comparison of the population size and growth rate across different towns.

(b) In which town was the population growth maximum?

Population growth = Population in 2009 - Population in 2007

Growth in Town A = $3200 - 2900 = 300$ (hundreds)

Growth in Town B = $7500 - 6400 = 1100$ (hundreds)

Growth in Town C = $9200 - 8300 = 900$ (hundreds)

Growth in Town D = $6300 - 4600 = 1700$ (hundreds)

Comparing the growth values (300, 1100, 900, 1700), the maximum growth is 1700 hundreds.

This occurred in Town D.

(c) In which town was the population growth least?

Comparing the growth values from part (b) (300, 1100, 900, 1700), the least growth is 300 hundreds.

This occurred in Town A.

The given table shows the population (in hundreds) of four towns (A, B, C, D) in two different years, 2007 and 2009.

(a) Draw a double bar graph using appropriate scale to depict the above information.

To draw the double bar graph:

Draw horizontal (X-axis) and vertical (Y-axis) axes. Label the X-axis as 'Towns' and mark the towns (A, B, C, D) at equal intervals.

Label the Y-axis as 'Population (in hundreds)'.

Choose a suitable scale for the Y-axis. The maximum population figure is 9200 hundreds. An appropriate scale could be $1$ unit on the Y-axis representing $500$ or $1000$ hundreds. Let's choose $1$ unit = $500$ hundreds, marking the axis at 0, 500, 1000, 1500, ..., 9500.

For each town, draw two adjacent bars of equal width. One bar represents the population in 2007, and the other represents the population in 2009.

The height of the bar for Town A in 2007 should correspond to 2900 units on the Y-axis ($2900/500 = 5.8$ units). The height of the bar for Town A in 2009 should correspond to 3200 units ($3200/500 = 6.4$ units). Repeat this for all four towns, drawing pairs of bars with heights corresponding to the population figures for 2007 and 2009.

Use different colours or shading patterns to distinguish between the bars for 2007 and 2009.

Include a legend specifying which bar represents the population in 2007 and which represents the population in 2009.

Give the graph a title, e.g., "Population of Towns A, B, C, D in 2007 and 2009".

Inference from the bar graph: The double bar graph visually compares the population of each town in 2007 and 2009. It clearly shows the population growth in each town (by the difference in heights of the paired bars) and allows for easy comparison of the population size and growth rate across different towns.

(b) In which town was the population growth maximum?

Population growth = Population in 2009 - Population in 2007

Growth in Town A = $3200 - 2900 = 300$ (hundreds)

Growth in Town B = $7500 - 6400 = 1100$ (hundreds)

Growth in Town C = $9200 - 8300 = 900$ (hundreds)

Growth in Town D = $6300 - 4600 = 1700$ (hundreds)

Comparing the growth values (300, 1100, 900, 1700), the maximum growth is 1700 hundreds.

This occurred in Town D.

(c) In which town was the population growth least?

Comparing the growth values from part (b) (300, 1100, 900, 1700), the least growth is 300 hundreds.

This occurred in Town A.

The given table shows the number of boys and girls who like different flavours of ice cream.

(a) Draw a double bar graph using appropriate scale to represent the above information.

To draw the double bar graph:

Draw horizontal (X-axis) and vertical (Y-axis) axes. Label the X-axis as 'Flavours' and mark the flavours (Vanilla, Chocolate, Strawberry, Mango, Butterscotch) at equal intervals.

Label the Y-axis as 'Number of Children'.

Choose a suitable scale for the Y-axis. The maximum number of children for any category is 13 (Boys for Butterscotch) and 12 (Girls for Chocolate). A scale of $1$ unit on the Y-axis representing $1$ or $2$ children would be appropriate. Let's choose $1$ unit = $1$ child, marking the axis at 0, 1, 2, ..., 14.

For each flavour, draw two adjacent bars of equal width. One bar represents the number of boys, and the other represents the number of girls.

The height of the bar for Boys liking Vanilla should be 4 units on the Y-axis. The height of the bar for Girls liking Vanilla should be 8 units. Repeat this for all five flavours, drawing pairs of bars with heights corresponding to the number of boys and girls.

Use different colours or shading patterns to distinguish between the bars for boys and girls.

Include a legend specifying which bar represents Boys and which represents Girls.

Give the graph a title, e.g., "Ice Cream Flavours Liked by Children".

(b) Which flavour is liked the most by the boys?

Comparing the number of boys for each flavour: Vanilla (4), Chocolate (9), Strawberry (3), Mango (8), Butterscotch (13).

The maximum number of boys is 13, which corresponds to the Butterscotch flavour.

The flavour liked the most by the boys is Butterscotch.

(c) How many girls are there in all?

Total number of girls = Sum of girls who like each flavour

Total girls = $8 + 12 + 7 + 9 + 10 = 46$

There are 46 girls in all.

(d) How many children like chocolate flavour of ice cream?

Total children who like chocolate = (Number of boys liking Chocolate) + (Number of girls liking Chocolate)

Total children liking chocolate = $9 + 12 = 21$

21 children like chocolate flavour of ice cream.

(e) Find the ratio of children who like strawberry flavour to vanilla flavour of ice cream.

Total children who like strawberry = (Boys liking Strawberry) + (Girls liking Strawberry) = $3 + 7 = 10$

Total children who like vanilla = (Boys liking Vanilla) + (Girls liking Vanilla) = $4 + 8 = 12$

Ratio = (Children liking Strawberry) : (Children liking Vanilla)

Ratio = $10 : 12$

To simplify the ratio, divide both numbers by their greatest common divisor, which is 2.

Ratio = $\frac{\cancel{10}^5}{\cancel{12}_6} = \frac{5}{6}$

The ratio is $5:6$.