Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 19 Mean And Median

Exercise 19(A)

Question 1. Find the mean of the numbers: 43, 51, 50, 57, and 54.
Answer: The given dataset contains the values: 43, 51, 50, 57, and 54.
To calculate the arithmetic mean, we divide their total sum by the count of numbers:
Mean = \( \frac{43 + 51 + 50 + 57 + 54}{5} \)
\( = \frac{255}{5} \)
\( = 51 \)
In simple words: Add all five numbers together to get 255, then divide by 5 because there are 5 numbers in total. This gives a mean of 51.

Exam Tip: Always double-check your addition of the terms to prevent simple calculation mistakes on basic mean questions.

Question 2. Find the mean of the first six natural numbers.
Answer: The initial six natural numbers are 1, 2, 3, 4, 5, and 6.
Their arithmetic mean is calculated as:
Mean = \( \frac{1 + 2 + 3 + 4 + 5 + 6}{6} \)
\( = \frac{21}{6} \)
\( = 3.5 \)
In simple words: The first six counting numbers are 1 through 6. Adding them up gives 21, and dividing this total by 6 gives 3.5.

Exam Tip: Remember that natural numbers start from 1, whereas whole numbers start from 0. Misunderstanding this distinction is a common trap.

Question 3. Find the mean of the first ten odd natural numbers.
Answer: The first ten odd numbers belonging to natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19.
We find the mean of these ten values as follows:
Mean = \( \frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10} \)
\( = \frac{100}{10} \)
\( = 10 \)
In simple words: List the first ten odd numbers starting from 1. Their sum is 100, and dividing by 10 gives a mean of 10.

Exam Tip: The sum of the first n odd natural numbers is always \( n^2 \). Thus, for 10 terms, the sum is \( 10^2 = 100 \), which can save you time in verifying your sum.

Question 4. Find the mean of all the factors of 10.
Answer: The complete set of factors for the number 10 is 1, 2, 5, and 10.
We can compute the mean of these factors like this:
Mean = \( \frac{1 + 2 + 5 + 10}{4} \)
\( = \frac{18}{4} \)
\( = 4.5 \)
In simple words: The numbers that divide 10 exactly are 1, 2, 5, and 10. Adding these four factors gives 18, and dividing by 4 gives 4.5.

Exam Tip: Be careful to list all factors of the given number. Missing a factor like 1 or the number itself will lead to an incorrect mean.

Question 5. Find the mean of the values: \( x + 3 \), \( x + 5 \), \( x + 7 \), \( x + 9 \), and \( x + 11 \).
Answer: The collection of provided algebraic terms is \( x + 3 \), \( x + 5 \), \( x + 7 \), \( x + 9 \), and \( x + 11 \).
Their average value is found using:
Mean = \( \frac{(x + 3) + (x + 5) + (x + 7) + (x + 9) + (x + 11)}{5} \)
\( = \frac{5x + 35}{5} \)
\( = \frac{5(x + 7)}{5} \)
\( = x + 7 \)
In simple words: Add the five algebraic expressions together to get \( 5x + 35 \). Dividing this total by 5 gives the mean, which is \( x + 7 \).

Exam Tip: When dealing with algebraic terms, group all the \( x \) variables together first, and then sum the constant numbers separately before simplifying.

Question 6. For the given numbers: 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5, 11.1.
(i) Find their mean.
(ii) Find the value of \( \sum_{i=1}^{10} (x_i - \bar{x}) \).
Answer:
(i) Let us find the average of the given set of ten numbers:
Mean (\( \bar{x} \)) = \( \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}}{10} \)
\( = \frac{9.8 + 5.4 + 3.7 + 1.7 + 1.8 + 2.6 + 2.8 + 8.6 + 10.5 + 11.1}{10} \)
\( = \frac{58}{10} \)
\( = 5.8 \)

(ii) We need to evaluate the summation \( \sum_{i=1}^{10} (x_i - \bar{x}) \). It is a standard statistical property that the sum of deviations of all observations from their arithmetic mean is always equal to zero:
\( \sum_{i=1}^{n} (x_i - \bar{x}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + \dots + (x_n - \bar{x}) = 0 \)
Using our calculated mean of \( \bar{x} = 5.8 \):
\( \sum_{i=1}^{10} (x_i - \bar{x}) = (9.8 - 5.8) + (5.4 - 5.8) + (3.7 - 5.8) + (1.7 - 5.8) + (1.8 - 5.8) + (2.6 - 5.8) + (2.8 - 5.8) + (8.6 - 5.8) + (10.5 - 5.8) + (11.1 - 5.8) \)
\( = 4.0 + (-0.4) + (-2.1) + (-4.1) + (-4.0) + (-3.2) + (-3.0) + 2.8 + 4.7 + 5.3 \)
\( = 0 \)
In simple words: For part (i), add all ten numbers to get 58, then divide by 10 to find the mean, which is 5.8. For part (ii), subtracting the mean from each number and adding those differences up will always equal 0.

Exam Tip: The algebraic sum of deviations from the mean is always zero. This is a fundamental property of the arithmetic mean that you can use to check your calculations.

Question 7. The mean of 15 observations is 32. Find the new mean if each observation is:
(i) increased by 3
(ii) decreased by 7
(iii) multiplied by 2
(iv) divided by 0.5
(v) increased by 60%
(vi) decreased by 20%
Answer: We are given that the average of 15 observations is 32. Any arithmetic operation applied to every observation changes the mean by that same operation:
(i) Since each observation is increased by 3, the new mean increases by 3:
New Mean = \( 32 + 3 = 35 \)

(ii) When each observation is decreased by 7, the new mean drops by 7:
New Mean = \( 32 - 7 = 25 \)

(iii) Multiplying each observation by 2 scales the mean by 2:
New Mean = \( 32 \times 2 = 64 \)

(iv) Dividing each observation by 0.5 divides the mean by 0.5:
New Mean = \( \frac{32}{0.5} = 64 \)

(v) Increasing each observation by 60% means the new mean increases by 60% of its original value:
New Mean = \( 32 + \frac{60}{100} \times 32 \)
\( = 32 + 19.2 \)
\( = 51.2 \)

(vi) Decreasing each observation by 20% means the new mean decreases by 20% of its original value:
New Mean = \( 32 - \frac{20}{100} \times 32 \)
\( = 32 - 6.4 \)
\( = 25.6 \)
In simple words: When you change every number in a list by a certain rule (like adding, subtracting, or multiplying), the average changes by that exact same rule.

Exam Tip: Instead of recalculating with individual values, apply the change directly to the old mean to solve these types of questions instantly.

Question 8. The mean of 5 numbers is 18. If one number is excluded, the mean of the remaining numbers becomes 16. Find the excluded number.
Answer: The mean of a group of 5 numbers is 18. Therefore, the combined sum of these 5 numbers is:
Sum of 5 numbers = \( 18 \times 5 = 90 \)
When one number is removed, the remaining 4 numbers have a mean of 16. The sum of these 4 numbers is:
Sum of remaining 4 numbers = \( 16 \times 4 = 64 \)
To find the value of the excluded number, we subtract the sum of the 4 numbers from the original sum of the 5 numbers:
Excluded number = Total sum of 5 numbers - Total sum of 4 numbers
\( = 90 - 64 \)
\( = 26 \)
In simple words: The total sum of all 5 numbers is 90. When one number is left out, the sum of the remaining 4 numbers is 64. The missing number must be the difference, which is 26.

Exam Tip: Remember the basic relation: \( \text{Sum of observations} = \text{Mean} \times \text{Number of observations} \). This is key to solving missing value problems.

Question 9. The mean of five observations \( x, x + 2, x + 4, x + 6, x + 8 \) is 11.
(i) Find the value of \( x \).
(ii) Find the mean of the first three observations.
Answer:
(i) We are given that the arithmetic mean of five algebraic quantities, \( x, x+2, x+4, x+6 \), and \( x+8 \), is 11. Using the mean formula:
Mean = \( \frac{\text{Sum of observations}}{n} \)
\( 11 = \frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} \)
\( 11 = \frac{5x + 20}{5} \)

\( \implies 55 = 5x + 20 \)

\( \implies 5x = 35 \)

\( \implies x = 7 \)

(ii) We need to find the mean of the first three terms, which are \( x \), \( x+2 \), and \( x+4 \). Substituting the value \( x = 7 \):
Mean of first three observations = \( \frac{x + (x + 2) + (x + 4)}{3} \)
\( = \frac{3x + 6}{3} \)
Since \( x = 7 \):
Mean = \( \frac{3(7) + 6}{3} \)
\( = \frac{21 + 6}{3} \)
\( = \frac{27}{3} \)
\( = 9 \)
In simple words: For part (i), set up the average equation and solve for \( x \), which gives 7. For part (ii), substitute 7 back into the first three terms to find their average, which is 9.

Exam Tip: Simplify algebraic expressions in the numerator before multiplying with the denominator to keep your steps clear and easy to read.

Question 10. The mean of 100 observations was found to be 40. Later, it was discovered that an observation of 53 was misread as 83. Find the correct mean.
Answer: The initial calculation for 100 observations gave a mean of 40. The sum of these observations was assumed to be:
Incorrect sum = \( 40 \times 100 = 4000 \)
Since the observation 53 was incorrectly recorded as 83, this sum is incorrect. We can find the correct sum by subtracting the wrong observation and adding the correct one:
Correct sum = Incorrect sum - Incorrect observation + Correct observation
\( = 4000 - 83 + 53 \)
\( = 3970 \)
Using this correct sum, we find the accurate mean:
Correct mean = \( \frac{\text{Correct sum}}{10} \)
\( = \frac{3970}{100} \)
\( = 39.7 \)
In simple words: The original total sum of 4000 was wrong because 83 was recorded instead of 53. Subtracting the extra 30 gives the correct sum of 3970, which leads to a correct mean of 39.7.

Exam Tip: Be very careful with which number is subtracted and which is added. Subtract the incorrect value, and add the correct one.

Question 11. The mean of 200 items was 50. Later, it was found that two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.
Answer: For 200 items, the recorded mean was 50, which gives an incorrect total sum of:
Incorrect sum = \( 50 \times 200 = 10000 \)
We are told that two items were misread as 92 and 8 instead of 192 and 88. To find the correct sum, we remove the two wrong values and add the two correct ones:
Correct sum = \( 10000 - (92 + 8) + (192 + 88) \)
\( = 10000 - 100 + 280 \)
\( = 10180 \)
Using this corrected sum, we compute the accurate mean:
Correct mean = \( \frac{\text{Correct sum}}{n} \)
\( = \frac{10180}{200} \)
\( = 50.9 \)
In simple words: The starting total was 10,000. We correct this by subtracting the wrong numbers (92 and 8) and adding the real numbers (192 and 88), giving 10,180. Dividing by 200 gives the correct mean of 50.9.

Exam Tip: For multiple error corrections, group the incorrect values together and the correct values together to simplify your calculation steps.

Question 12. The mean of 45 numbers is 18, and the mean of the remaining 30 numbers is 13. Find the mean of all 75 numbers together.
Answer: We are given that 45 of the numbers have a mean of 18. Their combined total is:
Sum of 45 numbers = \( 18 \times 45 = 810 \)
The remaining 30 numbers have a mean of 13, making their combined total:
Sum of remaining 30 numbers = \( 13 \times 30 = 390 \)
The total sum of all 75 numbers is the sum of these two groups:
Total sum of 75 numbers = \( 810 + 390 = 1200 \)
Thus, the overall mean is:
Mean of all 75 numbers = \( \frac{1200}{75} \)
\( = 16 \)
In simple words: Find the sum of the first group (810) and the second group (390). Add these sums together to get 1200, then divide by the total count of 75 numbers to get 16.

Exam Tip: Never just find the average of the two means (i.e., \( \frac{18 + 13}{2} \)) because the two groups have a different number of items. You must use the weighted average method.

Question 13. The mean weight of 120 students in a class is 52.75 kg. If the mean weight of 50 of them is 51 kg, find the mean weight of the remaining students.
Answer: The overall average weight of 120 students is 52.75 kg. The total mass of all 120 students is:
Total weight of 120 students = \( 120 \times 52.75 = 6330 \text{ kg} \)
For a subgroup of 50 students, the average weight is 51 kg, which gives a total mass of:
Total weight of 50 students = \( 50 \times 51 = 2550 \text{ kg} \)
The total weight of the rest of the class (70 students) is found by subtracting the subgroup's weight from the whole class's weight:
Total weight of remaining 70 students = \( 6330 - 2550 = 3780 \text{ kg} \)
The mean weight of these remaining 70 students is:
Mean weight of remaining 70 students = \( \frac{3780}{70} = 54 \text{ kg} \)
In simple words: The total weight of all students is 6330 kg, and 50 of them weigh 2550 kg in total. This leaves 3780 kg for the other 70 students, giving them an average weight of 54 kg.

Exam Tip: Keep your units (kg) consistent throughout your calculations to maintain accuracy and earn full marks.

Question 14. The mean marks of boys in an examination is 70 and that of girls is 73. If the mean marks of all the students is 71, find the ratio of the number of boys to the number of girls.
Answer: Let \( x \) represent the count of boys and \( y \) represent the count of girls.
The total marks scored by the boys is:
Sum of boys' marks = \( 70x \)
The total marks scored by the girls is:
Sum of girls' marks = \( 73y \)
The combined marks for all \( x + y \) students is:
Total marks of all students = \( 71(x + y) \)
Setting these equal to solve for the ratio of boys to girls:
\( 71(x + y) = 70x + 73y \)
\( \implies 71x + 71y = 70x + 73y \)

\( \implies 71x - 70x = 73y - 71y \)

\( \implies x = 2y \)

\( \implies \frac{x}{y} = \frac{2}{1} \)
Thus, the ratio of the number of boys to the number of girls is 2:1.
In simple words: Let there be \( x \) boys and \( y \) girls. The boys' total marks is \( 70x \) and the girls' is \( 73y \). Together, their total is \( 71(x + y) \). Solving this gives a ratio of 2 boys for every 1 girl.

Exam Tip: Be careful with the ratio order requested in the question. "Ratio of boys to girls" means finding \( \frac{\text{boys}}{\text{girls}} \) or \( x : y \).

Exercise 19(B)

Question 1. Find the median of the following observations:
(i) 25, 16, 26, 16, 35, 31, 28, 32, 19
(ii) 243, 258, 257, 241, 261, 292, 299, 271, 350, 327, 347
(iii) 21, 14, 17, 9, 25, 50, 34, 43, 63, 50
(iv) 185, 173, 189, 194, 194, 200, 204, 220, 208, 223
Answer: To find the median, we first arrange each dataset in ascending order and determine whether the number of terms \( n \) is odd or even.

(i) Sorting the given data in increasing order:
16, 16, 19, 25, 26, 28, 31, 32, 35
Since \( n = 9 \) is odd, the median is the middle term:
Median = \( \left( \frac{n + 1}{2} \right)^{\text{th}} \text{ term} \)
\( = \left( \frac{9 + 1}{2} \right)^{\text{th}} \text{ term} \)
\( = 5^{\text{th}} \text{ term} \)
\( = 26 \)
Thus, the median is 26.

(ii) Arranging the observations in ascending order:
241, 243, 257, 258, 261, 271, 292, 299, 327, 347, 350
Here, \( n = 11 \) is odd, so the median is the middle term:
Median = \( \left( \frac{n + 1}{2} \right)^{\text{th}} \text{ term} \)
\( = \left( \frac{11 + 1}{2} \right)^{\text{th}} \text{ term} \)
\( = 6^{\text{th}} \text{ term} \)
\( = 271 \)
Thus, the median is 271.

(iii) Arranging the observations in ascending order:
9, 14, 17, 21, 25, 34, 43, 50, 50, 63
Since \( n = 10 \) is an even number, the median is the average of the two middle terms, which are the 5th and 6th terms:
Median = \( \frac{1}{2} \left[ \left( \frac{n}{2} \right)^{\text{th}} \text{ term} + \left( \frac{n}{2} + 1 \right)^{\text{th}} \text{ term} \right] \)
\( = \frac{1}{2} \left[ \left( \frac{10}{2} \right)^{\text{th}} \text{ term} + \left( \frac{10}{2} + 1 \right)^{\text{th}} \text{ term} \right] \)
\( = \frac{1}{2} \left[ 5^{\text{th}} \text{ term} + 6^{\text{th}} \text{ term} \right] \)
\( = \frac{1}{2} [ 25 + 34 ] \)
\( = \frac{1}{2} [ 59 ] \)
\( = 29.5 \)
Thus, the median is 29.5.

(iv) Arranging the observations in ascending order:
173, 185, 189, 194, 194, 200, 204, 208, 220, 223
Since \( n = 10 \) is even, the median is the average of the 5th and 6th terms:
Median = \( \frac{1}{2} \left[ \left( \frac{n}{2} \right)^{\text{th}} \text{ term} + \left( \frac{n}{2} + 1 \right)^{\text{th}} \text{ term} \right] \)
\( = \frac{1}{2} \left[ 5^{\text{th}} \text{ term} + 6^{\text{th}} \text{ term} \right] \)
\( = \frac{1}{2} [ 194 + 200 ] \)
\( = \frac{1}{2} [ 394 ] \)
\( = 197 \)
Thus, the median is 197.
In simple words: To find the median, write the numbers in order from smallest to largest. If the count of numbers is odd, the median is the exact middle number. If the count is even, add the two middle numbers and divide by 2.

Exam Tip: Always make sure to write the numbers in ascending (or descending) order first. Forgetting to sort the numbers before finding the middle value is the most common mistake in median problems.

Question 2. The following observations are arranged in ascending order: 34, 37, 53, 55, x, x + 2, 77, 83, 89, 100. If the median of the data is 63, find the value of x.
Answer: The total number of observations is \( n = 10 \), which is an even number. The formula for the median when \( n \) is even is: \[ \text{Median} = \frac{1}{2} \left[ \text{value of } \left(\frac{n}{2}\right)^{\text{th}} \text{ term} + \text{value of } \left(\frac{n}{2} + 1\right)^{\text{th}} \text{ term} \right] \] Substituting \( n = 10 \) into the formula:
\( \implies \text{Median} = \frac{1}{2} \left[ \text{value of } 5^{\text{th}} \text{ term} + \text{value of } 6^{\text{th}} \text{ term} \right] \) From the given dataset, the 5th term is \( x \) and the 6th term is \( x + 2 \). Given that the median value is 63:
\( \implies 63 = \frac{1}{2} [x + (x + 2)] \)
\( \implies 63 = \frac{2x + 2}{2} \)
\( \implies 63 = x + 1 \)
\( \implies x = 62 \)
In simple words: Since there is an even number of values, the median is the average of the two middle numbers. Solving this equation gives us the unknown value.

Exam Tip: When dealing with an even number of terms, always average the two middle terms. Remember to simplify the fraction to avoid calculation errors.

Question 3. For the data given in Question 2, find the effect on the median if the 7th number is diminished by 8.
Answer: Let the 10 terms from the dataset be: 34, 37, 53, 55, \( x \), \( x + 2 \), 77, 83, 89, 100. Since \( x = 62 \), the terms are: 34, 37, 53, 55, 62, 64, 77, 83, 89, 100. The 7th term in this dataset is 77. If the 7th term is decreased by 8, its new value becomes:
\( \implies 77 - 8 = 69 \) The updated sequence in ascending order becomes: 34, 37, 53, 55, 62, 64, 69, 83, 89, 100. Since the terms are still arranged in ascending order, the 5th term remains 62 and the 6th term remains 64. The median depends only on these two middle terms: \[ \text{Median} = \frac{5^{\text{th}} \text{ term} + 6^{\text{th}} \text{ term}}{2} = \frac{62 + 64}{2} = 63 \] Therefore, decreasing the 7th term by 8 does not alter the median.
In simple words: The median only depends on the two middle terms (the 5th and 6th terms). Changing the 7th term does not shift their position or values, so the median stays the same.

Exam Tip: If a value changes but its new value doesn't change its position relative to the middle terms, the median of the dataset remains unaffected.

Question 4. Find the median score of a group of 10 students whose marks, when arranged in ascending order, are such that three students scored less than 30 marks, and three students scored more than 75 marks. The remaining four students scored 35, 40, 48, and 66 marks respectively.
Answer: The total number of observations is \( n = 10 \) (even). The median is calculated as: \[ \text{Median} = \frac{1}{2} \left[ \text{value of } \left(\frac{10}{2}\right)^{\text{th}} \text{ term} + \text{value of } \left(\frac{10}{2} + 1\right)^{\text{th}} \text{ term} \right] \]
\( \implies \text{Median} = \frac{1}{2} \left[ \text{value of } 5^{\text{th}} \text{ term} + \text{value of } 6^{\text{th}} \text{ term} \right] \) Let's tabulate the given marks in ascending order:

From the table, the 5th term is 40 and the 6th term is 48.
\( \implies \text{Median} = \frac{1}{2} (40 + 48) = \frac{88}{2} = 44 \) Thus, the median score of the group is 44.
In simple words: Since there are 10 terms, we find the middle two terms (the 5th and 6th terms), which are 40 and 48, and take their average to find the median.

Exam Tip: When data is given in ranges, identify the specific positions of the middle terms first. Only their exact values are needed to compute the median.

Question 5. The median of 9 observations arranged in ascending order is 18. If the fifth observation is x + 5, find the value of x.
Answer: The total number of observations is \( n = 9 \), which is odd. For an odd number of observations: \[ \text{Median} = \text{value of } \left(\frac{n + 1}{2}\right)^{\text{th}} \text{ term} \] Substituting \( n = 9 \):
\( \implies \text{Median} = \text{value of } \left(\frac{9 + 1}{2}\right)^{\text{th}} \text{ term} \)
\( \implies \text{Median} = \text{value of } 5^{\text{th}} \text{ term} \) The 5th term is given as \( x + 5 \). Since the median is 18:
\( \implies x + 5 = 18 \)
\( \implies x = 13 \)
In simple words: For 9 observations, the 5th term is the exact middle. Since the median is 18, the 5th term must be 18, which gives us x = 13.

Exam Tip: For odd numbers of observations, the median is a single actual data point from the sorted list, making it straightforward to solve for the variable.

Exercise 19(C)

Question 1. The mean of the numbers 8, 12, 16, 22, 10, and 4 is 12. Find the new mean if each number is:
(i) Multiplied by 3
(ii) Divided by 2
(iii) Multiplied by 3 and then divided by 2
(iv) Increased by 25%
(v) Decreased by 40%
Answer: The mean of the original data set is: \[ \text{Mean} = \frac{8 + 12 + 16 + 22 + 10 + 4}{6} = \frac{72}{6} = 12 \]
(i) **Multiplied by 3**: If each observation \( x_i \) is multiplied by a constant \( a \), the new mean is also multiplied by \( a \).
\( \implies \text{New Mean} = 12 \times 3 = 36 \)
(ii) **Divided by 2**: If each observation is divided by a constant \( a \), the new mean is divided by \( a \).
\( \implies \text{New Mean} = \frac{12}{2} = 6 \)
(iii) **Multiplied by 3 and then divided by 2**: If each observation is multiplied by 3 and divided by 2, the mean undergoes the same operations.
\( \implies \text{New Mean} = 12 \times \frac{3}{2} = 18 \)
(iv) **Increased by 25%**: If each observation is increased by 25%, the mean will also increase by 25%.
\( \implies \text{New Mean} = 12 + (25\% \text{ of } 12) \)
\( \implies \text{New Mean} = 12 + \left(\frac{25}{100} \times 12\right) = 12 + 3 = 15 \)
(v) **Decreased by 40%**: If each observation is decreased by 40%, the mean will also decrease by 40%.
\( \implies \text{New Mean} = 12 - (40\% \text{ of } 12) \)
\( \implies \text{New Mean} = 12 - \left(\frac{40}{100} \times 12\right) = 12 - 4.8 = 7.2 \)
In simple words: Any math operation done to every single number in a group will change the mean in the exact same way.

Exam Tip: Instead of performing operations on individual data points, apply the change directly to the original mean to save time and prevent errors.

Question 2. The mean of five observations 18, 24, 15, 2x + 1, and 12 is 21. Find the value of x.
Answer: The sum of the five observations divided by 5 equals the mean: \[ \text{Mean} = \frac{18 + 24 + 15 + (2x + 1) + 12}{5} \] Given that the mean is 21:
\( \implies 21 = \frac{70 + 2x}{5} \) Multiplying both sides by 5:
\( \implies 105 = 70 + 2x \) Subtracting 70 from both sides:
\( \implies 2x = 105 - 70 \)
\( \implies 2x = 35 \)
\( \implies x = 17.5 \)
In simple words: Add up all five values and divide by 5 to equal the mean of 21. Then, rearrange the equation to find x.

Exam Tip: When solving for a variable in a mean equation, ensure you combine all the constants in the numerator correctly before multiplying.

Question 3. The mean of 6 numbers is 42. If the mean of first 5 of these numbers is 45, find the sixth number.
Answer: Let the six observations be \( x_1, x_2, x_3, x_4, x_5, \) and \( x_6 \). The mean of these 6 numbers is 42: \[ \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6}{6} = 42 \]
\( \implies x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 42 \times 6 = 252 \) - (Equation 1) The mean of the first 5 numbers is 45: \[ \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5} = 45 \]
\( \implies x_1 + x_2 + x_3 + x_4 + x_5 = 45 \times 5 = 225 \) - (Equation 2) Substitute Equation 2 into Equation 1:
\( \implies 225 + x_6 = 252 \)
\( \implies x_6 = 252 - 225 \)
\( \implies x_6 = 27 \) So, the sixth number is 27.
In simple words: Find the total sum of all 6 numbers, then subtract the sum of the first 5 numbers to get the 6th number.

Exam Tip: This type of question relies on the relation: \( \text{Sum} = \text{Mean} \times \text{Number of observations} \). Calculating the two sums makes finding the missing value easy.

Question 4. The mean of 10 numbers is 24. If another number is included, the mean of the 11 numbers becomes 25. Find the included number.
Answer: Let the 10 numbers be \( x_1, x_2, \dots, x_{10} \). The mean of these 10 numbers is 24: \[ \frac{x_1 + x_2 + \dots + x_{10}}{10} = 24 \]
\( \implies x_1 + x_2 + \dots + x_{10} = 240 \) - (Equation 1) When the 11th number \( x_{11} \) is added, the mean becomes 25: \[ \frac{x_1 + x_2 + \dots + x_{10} + x_{11}}{11} = 25 \]
\( \implies x_1 + x_2 + \dots + x_{10} + x_{11} = 25 \times 11 = 275 \) - (Equation 2) Subtracting Equation 1 from Equation 2:
\( \implies 240 + x_{11} = 275 \)
\( \implies x_{11} = 275 - 240 = 35 \) Thus, the included number is 35.
In simple words: Find the sum of all 11 numbers and subtract the sum of the original 10 numbers to find the value of the added number.

Exam Tip: Always multiply the number of terms by their respective mean to find the total sum before calculating the difference.

Question 5. The observations 44, 47, 63, 65, x + 13, 87, 93, 99, 110 are arranged in ascending order. If the median of the data is 78, find the value of x.
Answer: The given dataset is: 44, 47, 63, 65, \( x + 13 \), 87, 93, 99, 110. The total number of observations is \( n = 9 \), which is odd. Since \( n \) is odd: \[ \text{Median} = \text{value of } \left(\frac{n + 1}{2}\right)^{\text{th}} \text{ observation} \]
\( \implies \text{Median} = \text{value of } \left(\frac{9 + 1}{2}\right)^{\text{th}} \text{ observation} \)
\( \implies \text{Median} = \text{value of } 5^{\text{th}} \text{ observation} \) From the sorted data, the 5th term is \( x + 13 \). Given that the median is 78:
\( \implies x + 13 = 78 \)
\( \implies x = 78 - 13 \)
\( \implies x = 65 \)
In simple words: With 9 sorted values, the middle term is the 5th one. Equating this term, which is x + 13, to the median of 78 gives x = 65.

Exam Tip: Ensure the data is confirmed to be in ascending order first before identifying the term at the median position.

Question 6. The observations 24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90 are arranged in ascending order. If their median is 58, find the value of x.
Answer: The given observations are: 24, 27, 43, 48, \( x - 1 \), \( x + 3 \), 68, 73, 80, 90. The total number of observations is \( n = 10 \), which is an even number. For even \( n \): \[ \text{Median} = \frac{1}{2} \left[ \text{value of } \left(\frac{n}{2}\right)^{\text{th}} \text{ term} + \text{value of } \left(\frac{n}{2} + 1\right)^{\text{th}} \text{ term} \right] \] Substituting \( n = 10 \):
\( \implies \text{Median} = \frac{1}{2} \left[ \text{value of } 5^{\text{th}} \text{ term} + \text{value of } 6^{\text{th}} \text{ term} \right] \) Here, the 5th term is \( x - 1 \) and the 6th term is \( x + 3 \). The given median is 58:
\( \implies 58 = \frac{1}{2} [(x - 1) + (x + 3)] \)
\( \implies 58 \times 2 = 2x + 2 \)
\( \implies 116 = 2x + 2 \)
\( \implies 2x = 114 \)
\( \implies x = 57 \)
In simple words: The median is the average of the two middle numbers, which are x - 1 and x + 3. Setting this average equal to 58 allows us to solve for x.

Exam Tip: Don't forget to multiply the median by 2 before trying to isolate the variable, which keeps the equation simple.

Question 7. For the following data: 30, 32, 24, 34, 26, 28, 30, 35, 33, 25:
(i) Find the mean and show that the sum of the deviations of the observations from their mean is zero.
(ii) Find the median of the data.
Answer:
(i) **Mean and deviations**: Let \( x \) represent the observations. The total number of observations is \( n = 10 \). The mean \( \bar{x} \) is computed as: \[ \bar{x} = \frac{30 + 32 + 24 + 34 + 26 + 28 + 30 + 35 + 33 + 25}{10} = \frac{297}{10} = 29.7 \] Now we construct a table of the deviations \( x_i - \bar{x} \):

Sum of deviations: \[ \sum (x_i - \bar{x}) = 0.3 + 2.3 - 5.7 + 4.3 - 3.7 - 1.7 + 0.3 + 5.3 + 3.3 - 4.7 = 0 \] This demonstrates that the sum of all deviations from the mean is equal to 0.
(ii) **Median**: To find the median, arrange the data in ascending order: 24, 25, 26, 28, 30, 30, 32, 33, 34, 35. Since \( n = 10 \) is even: \[ \text{Median} = \frac{1}{2} \left[ \text{value of } 5^{\text{th}} \text{ term} + \text{value of } 6^{\text{th}} \text{ term} \right] \] The 5th term is 30 and the 6th term is 30.
\( \implies \text{Median} = \frac{30 + 30}{2} = 30 \)
In simple words: The mean is 29.7. If you subtract 29.7 from each number, some results are positive and some are negative, but they all add up to exactly 0. To find the median, we sort the numbers and average the two middle ones, which are both 30.

Exam Tip: The sum of deviations of a dataset from its mean is mathematically always zero. Double-check your arithmetic if your sum does not equal zero.

Question 8. Find the mean and median of the values: 35, 48, 92, 76, 64, 52, 51, 63, 71. If 51 is replaced by 66, what will be the new median?
Answer: The given numbers are: 35, 48, 92, 76, 64, 52, 51, 63, 71. The total number of observations is \( n = 9 \). The mean is calculated as: \[ \text{Mean} = \frac{35 + 48 + 92 + 76 + 64 + 52 + 51 + 63 + 71}{9} = \frac{552}{9} \approx 61.33 \] To find the median, let's arrange the numbers in ascending order: 35, 48, 51, 52, 63, 64, 71, 76, 92. Since \( n = 9 \) is odd: \[ \text{Median} = \text{value of } \left(\frac{9 + 1}{2}\right)^{\text{th}} \text{ observation} = 5^{\text{th}} \text{ observation} \] The 5th term is 63. So, the median is 63. If 51 is replaced by 66, the new dataset in ascending order is: 35, 48, 52, 63, 64, 66, 71, 76, 92. The 5th term in this new list is 64. So, the new median is 64.
In simple words: First, add up the numbers and divide by 9 to get the mean of 61.33. Sort them to find the middle (5th) term, which is 63. When 51 is swapped for 66, the order changes slightly, making 64 the new middle number.

Exam Tip: When substituting a number, always re-sort the entire dataset to determine the new median, as the position of other numbers might shift.

Question 9. The mean of five observations x, x + 2, x + 4, x + 6, and x + 8 is 11. Find the value of x and the mean of the first three observations.
Answer: The five observations are \( x \), \( x + 2 \), \( x + 4 \), \( x + 6 \), and \( x + 8 \). The mean of these five values is 11: \[ \frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} = 11 \]
\( \implies \frac{5x + 20}{5} = 11 \)
\( \implies x + 4 = 11 \)
\( \implies x = 7 \) The first three observations are: \( x = 7 \) \( x + 2 = 9 \) \( x + 4 = 11 \) The mean of these first three observations is: \[ \text{Mean} = \frac{7 + 9 + 11}{3} = \frac{27}{3} = 9 \]
In simple words: We write an equation using the mean of the five terms to find that x is 7. Using this, the first three numbers are 7, 9, and 11, which have an average of 9.

Exam Tip: Factor out the common term in the numerator \( 5x+20 \) to simplify it directly to \( x+4 \), saving valuable time.

Question 10. Find the mean and median of all the factors of 72.
Answer: The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. The total number of factors is \( n = 12 \). **Mean**: \[ \text{Mean} = \frac{1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 + 72}{12} \]
\( \implies \text{Mean} = \frac{195}{12} = 16.25 \) **Median**: Since the number of observations \( n = 12 \) is even: \[ \text{Median} = \frac{1}{2} \left[ \text{value of } \left(\frac{12}{2}\right)^{\text{th}} \text{ term} + \text{value of } \left(\frac{12}{2} + 1\right)^{\text{th}} \text{ term} \right] \]
\( \implies \text{Median} = \frac{1}{2} \left[ \text{value of } 6^{\text{th}} \text{ term} + \text{value of } 7^{\text{th}} \text{ term} \right] \) The 6th factor is 8 and the 7th factor is 9.
\( \implies \text{Median} = \frac{8 + 9}{2} = \frac{17}{2} = 8.5 \)
In simple words: First, list all the numbers that divide 72 perfectly. Add them together and divide by 12 to find the mean, then average the two middle numbers (8 and 9) to find the median.

Exam Tip: When listing factors of a number, work in pairs (like 1 and 72, 2 and 36) to make sure you do not miss any terms.