OP Malhotra Class 10 Maths Solutions Chapter 18 Arithmetic Mean Median Mode and Quartiles Exercise 18 (D)

Question 1. Find the median of the following sets of data:
(i) 2, 3, 5, 7, 9
(ii) 4, 8, 12, 16, 20, 23, 28, 32
(iii) 60, 33, 63, 61, 44, 48, 51
(iv) 13, 22, 25, 8, 11, 19, 17, 31, 16, 10
(v) First ten prime numbers
(vi) Prime numbers between 51 and 80.
Answer:
(i) Given data: 2, 3, 5, 7, 9
The numbers are already in ascending order.
Count of numbers \( n = 5 \). This is an odd number.
The median is found using the formula: \( \frac{n+1}{2} \)th term.
Median \( = \frac{5+1}{2} \)th term
\( \implies = \frac{6}{2} \)th term
\( \implies = 3 \)rd term
The 3rd term in the data set is 5.
Therefore, the median is 5.

(ii) Given data: 4, 8, 12, 16, 20, 23, 28, 32
The numbers are already in ascending order.
Count of numbers \( n = 8 \). This is an even number.
For an even number of data points, the median is the average (mean) of the two middle terms: \( \frac{n}{2} \)th term and \( (\frac{n}{2}+1) \)th term.
Here, \( \frac{n}{2} \)th term \( = \frac{8}{2} \)th term \( = 4 \)th term (which is 16).
And \( (\frac{n}{2}+1) \)th term \( = (4+1) \)th term \( = 5 \)th term (which is 20).
Median \( = \frac{1}{2} \) (4th term + 5th term)
\( \implies = \frac{1}{2} (16 + 20) \)
\( \implies = \frac{1}{2} (36) \)
\( \implies = 18 \)
Therefore, the median is 18.

(iii) Given data: 60, 33, 63, 61, 44, 48, 51
First, arrange the numbers in ascending order:
33, 44, 48, 51, 60, 61, 63
Count of numbers \( n = 7 \). This is an odd number.
The median is found using the formula: \( \frac{n+1}{2} \)th term.
Median \( = \frac{7+1}{2} \)th term
\( \implies = \frac{8}{2} \)th term
\( \implies = 4 \)th term
The 4th term in the ordered data set is 51.
Therefore, the median is 51.

(iv) Given data: 13, 22, 25, 8, 11, 19, 17, 31, 16, 10
First, arrange the numbers in ascending order:
8, 10, 11, 13, 16, 17, 19, 22, 25, 31
Count of numbers \( n = 10 \). This is an even number.
For an even number of data points, the median is the average (mean) of the two middle terms: \( \frac{n}{2} \)th term and \( (\frac{n}{2}+1) \)th term.
Here, \( \frac{n}{2} \)th term \( = \frac{10}{2} \)th term \( = 5 \)th term (which is 16).
And \( (\frac{n}{2}+1) \)th term \( = (5+1) \)th term \( = 6 \)th term (which is 17).
Median \( = \frac{1}{2} \) (5th term + 6th term)
\( \implies = \frac{1}{2} (16 + 17) \)
\( \implies = \frac{1}{2} (33) \)
\( \implies = 16.5 \)
Therefore, the median is 16.5.

(v) First ten prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
The numbers are already in ascending order.
Count of numbers \( n = 10 \). This is an even number.
For an even number of data points, the median is the average (mean) of the two middle terms: \( \frac{n}{2} \)th term and \( (\frac{n}{2}+1) \)th term.
Here, \( \frac{n}{2} \)th term \( = \frac{10}{2} \)th term \( = 5 \)th term (which is 11).
And \( (\frac{n}{2}+1) \)th term \( = (5+1) \)th term \( = 6 \)th term (which is 13).
Median \( = \frac{1}{2} \) (5th term + 6th term)
\( \implies = \frac{1}{2} (11 + 13) \)
\( \implies = \frac{1}{2} (24) \)
\( \implies = 12 \)
Therefore, the median is 12. Prime numbers are whole numbers greater than 1 that have only two factors: 1 and themselves.

(vi) Prime numbers between 51 and 80: 53, 59, 61, 67, 71, 73, 79
The numbers are already in ascending order.
Count of numbers \( n = 7 \). This is an odd number.
The median is found using the formula: \( \frac{n+1}{2} \)th term.
Median \( = \frac{7+1}{2} \)th term
\( \implies = \frac{8}{2} \)th term
\( \implies = 4 \)th term
The 4th term in the ordered data set is 67.
Therefore, the median is 67.
In simple words: To find the median, first put all the numbers in order from smallest to largest. If there's an odd number of items, the median is the middle one. If there's an even number of items, take the two middle numbers and find their average (add them up and divide by two).

🎯 Exam Tip: Always remember to arrange the data in ascending or descending order before calculating the median. Missing this step is a common mistake that leads to incorrect answers.

Question 2. The following table gives the monthly incomes of 12 families in a town:

Calculate the median of the above incomes.
Answer: The given monthly incomes are:
280, 180, 96, 98, 104, 75, 80, 94, 100, 75, 600, 200
First, arrange the incomes in ascending order:
75, 75, 80, 94, 96, 98, 100, 104, 180, 200, 280, 600
The total number of families (data points) \( n = 12 \). This is an even number.
For an even number of data points, the median is the average (mean) of the two middle terms: \( \frac{n}{2} \)th term and \( (\frac{n}{2}+1) \)th term.
Here, \( \frac{n}{2} \)th term \( = \frac{12}{2} \)th term \( = 6 \)th term (which is 98).
And \( (\frac{n}{2}+1) \)th term \( = (6+1) \)th term \( = 7 \)th term (which is 100).
Median \( = \frac{1}{2} \) (6th term + 7th term)
\( \implies = \frac{1}{2} (98 + 100) \)
\( \implies = \frac{1}{2} (198) \)
\( \implies = 99 \)
Therefore, the median monthly income is Rs. 99.
In simple words: First, list all the incomes from the smallest to the largest. Since there are 12 incomes (an even number), the median is the average of the 6th and 7th incomes in your ordered list.

🎯 Exam Tip: When working with data in tables, make sure to extract all values correctly and reorder them before applying the median formula. Double-check your ordering to avoid errors.

Question 3. Calculate the median from the following data:

Answer: To find the median for this frequency distribution, we first need to calculate the cumulative frequency (c.f.).

The total number of workers \( n = 120 \). This is an even number.
For an even number of data points, the median position is \( \frac{n}{2} \)th term.
Median position \( = \frac{120}{2} \)th term \( = 60 \)th term.
Now, we look for the cumulative frequency (c.f.) that is just greater than or equal to 60. The c.f. of 65 is the first one greater than 60.
The wage corresponding to this cumulative frequency of 65 is 24.
Therefore, the median wage is Rs. 24.
In simple words: For grouped data, first make a column for cumulative frequency. Then, find the position of the middle value by dividing the total number of workers by 2. Look in the cumulative frequency column for the number that is just bigger than or equal to this middle position. The wage amount next to that cumulative frequency is your median.

🎯 Exam Tip: When calculating the median for frequency distributions, always create a cumulative frequency column. The median value corresponds to the data point whose cumulative frequency is the first to be greater than or equal to \( N/2 \).

Question 4. Compute the median of the following distributions:
(i)
(ii)
Answer:
(i) Given distribution:

First, we arrange the data in ascending order and create a cumulative frequency (c.f.) table:

The total frequency \( n = 27 \). This is an odd number.
The median is found using the formula: \( \frac{n+1}{2} \)th term.
Median position \( = \frac{27+1}{2} \)th term
\( \implies = \frac{28}{2} \)th term
\( \implies = 14 \)th term
We look for the cumulative frequency (c.f.) that is just greater than or equal to 14. The c.f. of 14 corresponds to \( X = 12 \).
So, the 14th term is 12.
Therefore, the median is 12.

(ii) Given distribution:

First, we arrange the data in ascending order and create a cumulative frequency (c.f.) table:

The total frequency \( n = 83 \). This is an odd number.
The median is found using the formula: \( \frac{n+1}{2} \)th term.
Median position \( = \frac{83+1}{2} \)th term
\( \implies = \frac{84}{2} \)th term
\( \implies = 42 \)nd term
We look for the cumulative frequency (c.f.) that is just greater than or equal to 42. The c.f. of 48 is the first one greater than 42.
The value of \( x \) corresponding to this cumulative frequency of 48 is 5.
Therefore, the median is 5.
In simple words: When data comes with frequencies, first make a new column called 'cumulative frequency' by adding up the frequencies as you go down. Then, find the middle position based on the total frequency. The data value that matches this position in the cumulative frequency column is your median.

🎯 Exam Tip: Always check if the cumulative frequency exactly matches the median position or if you need to find the next highest cumulative frequency. This determines the correct median value for discrete frequency distributions.

Question 5. Marks obtained by 38 students are given below.

Calculate the median marks:
Answer: To calculate the median marks, we first need to arrange the marks in ascending order and then find the cumulative frequency.

The total number of students \( n = 38 \). This is an even number.
For an even number of data points, the median position is \( \frac{n}{2} \)th term.
Median position \( = \frac{38}{2} \)th term \( = 19 \)th term.
We look for the cumulative frequency (c.f.) that is just greater than or equal to 19. The c.f. of 28 is the first one greater than 19.
The marks corresponding to this cumulative frequency of 28 is 70.
Therefore, the median marks are 70.
In simple words: Organize the marks and their counts in a table. Add a 'cumulative frequency' column. Find the total number of students and divide it by two to get the median position. Then, find the mark that corresponds to the cumulative frequency just above this position.

🎯 Exam Tip: When dealing with discrete frequency distributions for median, it's crucial to correctly identify the \( N/2 \)th term or the value corresponding to the cumulative frequency just greater than \( N/2 \). Always double check the ordering of X values and the calculation of cumulative frequencies.

Question 6. Find the arithmetic mean of the first 10 natural numbers and show that it is equal to their median.
Answer: The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Arithmetic Mean (\( \bar{x} \)):
Arithmetic Mean \( = \frac{\text{Sum of all observations}}{\text{Number of observations}} \)
\( \implies \bar{x} = \frac{1+2+3+4+5+6+7+8+9+10}{10} \)
\( \implies \bar{x} = \frac{55}{10} \)
\( \implies \bar{x} = 5.5 \)

Median:
The numbers are already in ascending order: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Count of numbers \( n = 10 \). This is an even number.
For an even number of data points, the median is the average (mean) of the two middle terms: \( \frac{n}{2} \)th term and \( (\frac{n}{2}+1) \)th term.
Here, \( \frac{n}{2} \)th term \( = \frac{10}{2} \)th term \( = 5 \)th term (which is 5).
And \( (\frac{n}{2}+1) \)th term \( = (5+1) \)th term \( = 6 \)th term (which is 6).
Median \( = \frac{1}{2} \) (5th term + 6th term)
\( \implies = \frac{1}{2} (5 + 6) \)
\( \implies = \frac{1}{2} (11) \)
\( \implies = 5.5 \)
Since the arithmetic mean is 5.5 and the median is 5.5, the arithmetic mean is equal to the median in this case. Natural numbers are whole numbers starting from 1 up to infinity.
In simple words: First, find the average of the first ten counting numbers by adding them up and dividing by ten. Then, arrange the same numbers in order and find the middle value (the median). You will see that both answers are the same.

🎯 Exam Tip: When proving equality between mean and median, ensure you clearly show the calculation steps for both measures. For natural numbers, the mean and median often coincide or are very close, especially for symmetric distributions.

Question 7. Find whether the following statement is true or false:
(i) The median of a frequency distribution is the most commonly occurring value.
(ii) The median of a discrete ungrouped frequency distribution containing a number of items is the value of the middle item, the data being arranged in ascending or descending order.
Answer:
(i) False: The most commonly occurring value in a frequency distribution is called the mode, not the median. The median is the middle term after arranging the data in order.
(ii) True: The definition of the median for ungrouped data is indeed the value of the middle item once the data is sorted in either ascending or descending order. This helps us find the central point of the data set.
In simple words: For the first statement, "most common value" is the definition of mode, not median, so it's false. For the second statement, the median is indeed the middle value when numbers are sorted, so it's true.

🎯 Exam Tip: Clearly understand the definitions of mean, median, and mode. They are distinct measures of central tendency, and confusing them is a common source of errors in true/false questions.

Question 8. In a certain school, for a particular examination it is decided that exactly half the pupils will pass. Name the measure of central tendency that is used.
Answer: When it's decided that exactly half the pupils will pass, this means the passing mark is set so that 50% of students score above it and 50% score below it. This value that divides the data into two equal halves is the median. The median is a useful measure for setting benchmarks that split a group into two halves based on performance or scores.
In simple words: If you want exactly half the students to pass, you are looking for the score that cuts the class in half. This measure is called the median.

🎯 Exam Tip: Recognize that "dividing a data set into two equal halves" is the key phrase pointing directly to the definition and application of the median.

Question 9. {1, 2, 3, 6, 8} is a set of five positive integers whose mean is 4 and median is 3. Write down two other sets of five positive integers, each having the same mean and median as this set.
Answer: The given set is {1, 2, 3, 6, 8}.
Number of integers \( n = 5 \). This is an odd number.
The median for this set is the \( \frac{5+1}{2} \)th term \( = 3 \)rd term, which is 3.
The sum of the integers is \( 1+2+3+6+8 = 20 \).
The mean is \( \frac{20}{5} = 4 \).

To find other sets with the same mean (4) and median (3), we need to ensure:
1. The middle term (3rd term) is 3.
2. The sum of the five integers is 20 (since mean \( = 4 \) and count \( = 5 \), so sum \( = 4 \times 5 = 20 \)).
3. The first two integers are less than or equal to 3, and the last two integers are greater than or equal to 3 (to maintain 3 as the median).

Let the five integers be \( a_1, a_2, 3, a_4, a_5 \), where \( a_1 \le a_2 \le 3 \le a_4 \le a_5 \).
We know \( a_1 + a_2 + 3 + a_4 + a_5 = 20 \).
So, \( a_1 + a_2 + a_4 + a_5 = 17 \).

Here are two possible sets:
Set 1: {1, 2, 3, 4, 10}
- Arranged: 1, 2, 3, 4, 10
- Median: 3rd term = 3 (Correct)
- Sum: \( 1+2+3+4+10 = 20 \)
- Mean: \( \frac{20}{5} = 4 \) (Correct)

Set 2: {1, 2, 3, 5, 9}
- Arranged: 1, 2, 3, 5, 9
- Median: 3rd term = 3 (Correct)
- Sum: \( 1+2+3+5+9 = 20 \)
- Mean: \( \frac{20}{5} = 4 \) (Correct)

There are many possibilities, as long as the sum is 20 and the middle number is 3 after ordering. The values of the first two and last two numbers can be changed, but their sum must be 17 and they must maintain their relative positions around the median. Positive integers are whole numbers starting from 1, and they don't include fractions or decimals.
In simple words: You need to create two new lists of five positive whole numbers. Each list must have 3 as its middle number when sorted, and all numbers in each list must add up to 20 (so their average is 4). You can pick different numbers for the first two and last two spots, as long as they fit these rules.

🎯 Exam Tip: To create new sets with the same mean and median, keep the median value fixed and adjust the other numbers. Remember that the sum of the new numbers must equal \( \text{mean} \times \text{count} \).

Question 10. Find the median for the following data. 46, 64, 87, 41, 58, 77, 35, 90, 55, 33, 92 If in the data, the observation 92 is replaced by 19 determine the new median.
Answer:
Part (i): Find the initial median
Given data: 46, 64, 87, 41, 58, 77, 35, 90, 55, 33, 92
First, arrange the data in ascending order:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Count of numbers \( n = 11 \). This is an odd number.
The median is found using the formula: \( \frac{n+1}{2} \)th term.
Median position \( = \frac{11+1}{2} \)th term
\( \implies = \frac{12}{2} \)th term
\( \implies = 6 \)th term
The 6th term in the ordered data set is 58.
Therefore, the initial median is 58.

Part (ii): Find the new median after replacement
If the observation 92 is replaced by 19, the new data set is:
46, 64, 87, 41, 58, 77, 35, 90, 55, 33, 19
First, arrange the new data in ascending order:
19, 33, 35, 41, 46, 55, 58, 64, 77, 87, 90
Count of numbers \( n = 11 \). This is still an odd number.
The median is found using the formula: \( \frac{n+1}{2} \)th term.
Median position \( = \frac{11+1}{2} \)th term
\( \implies = \frac{12}{2} \)th term
\( \implies = 6 \)th term
The 6th term in the new ordered data set is 55.
Therefore, the new median is 55.
In simple words: First, sort the original list of numbers from smallest to largest and find the middle number. Then, change the number 92 to 19 in the list. Sort the new list again and find its middle number.

🎯 Exam Tip: When a data point is replaced, remember to re-sort the entire dataset and recalculate the median. A single change can alter the order and thus the middle value.

Question 11. The following data have been arranged in ascending order:
24, 27, 28, 31, 34, x, 37, 40, 42, 45 If the median of the data is 35, find x. In the above data, if 45 changed to 33, find the new median.
Answer:
Part (i): Find the value of x
Given data in ascending order: 24, 27, 28, 31, 34, x, 37, 40, 42, 45
Count of numbers \( n = 10 \). This is an even number.
The median for an even number of data points is the average (mean) of the two middle terms: \( \frac{n}{2} \)th term and \( (\frac{n}{2}+1) \)th term.
Here, \( \frac{n}{2} \)th term \( = \frac{10}{2} \)th term \( = 5 \)th term (which is 34).
And \( (\frac{n}{2}+1) \)th term \( = (5+1) \)th term \( = 6 \)th term (which is x).
The median is given as 35.
Median \( = \frac{1}{2} \) (5th term + 6th term)
\( \implies 35 = \frac{1}{2} (34 + x) \)
\( \implies 35 \times 2 = 34 + x \)
\( \implies 70 = 34 + x \)
\( \implies x = 70 - 34 \)
\( \implies x = 36 \)
So, the value of x is 36. This value fits the ascending order: 34, 36, 37.

Part (ii): Find the new median if 45 is replaced by 33
The original ordered data (with \( x=36 \)) is: 24, 27, 28, 31, 34, 36, 37, 40, 42, 45
If 45 is replaced by 33, the new data set is: 24, 27, 28, 31, 34, 36, 37, 40, 42, 33
Now, arrange the new data set in ascending order:
24, 27, 28, 31, 33, 34, 36, 37, 40, 42
Count of numbers \( n = 10 \). This is still an even number.
The median will be the average of the 5th and 6th terms.
5th term is 33.
6th term is 34.
New Median \( = \frac{1}{2} \) (5th term + 6th term)
\( \implies = \frac{1}{2} (33 + 34) \)
\( \implies = \frac{1}{2} (67) \)
\( \implies = 33.5 \)
Therefore, the new median is 33.5.
In simple words: First, use the given median (35) and the formula for an even number of items to find the missing value 'x'. Then, change the number 45 to 33 in the original list. Sort this new list and find its median by averaging the two middle numbers.

🎯 Exam Tip: When finding a missing value using the median, ensure that the calculated value maintains the ascending (or descending) order of the data. Also, remember to re-sort the data completely after any replacements before calculating a new median.

Question 12. The median of the following observation arranged in ascending order is 24. Find x.
11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41.
Answer: Given observations arranged in ascending order:
11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41
Count of numbers \( n = 10 \). This is an even number.
The median for an even number of data points is the average (mean) of the two middle terms: \( \frac{n}{2} \)th term and \( (\frac{n}{2}+1) \)th term.
Here, \( \frac{n}{2} \)th term \( = \frac{10}{2} \)th term \( = 5 \)th term (which is \( x+2 \)).
And \( (\frac{n}{2}+1) \)th term \( = (5+1) \)th term \( = 6 \)th term (which is \( x+4 \)).
The median is given as 24.
Median \( = \frac{1}{2} \) (5th term + 6th term)
\( \implies 24 = \frac{1}{2} ( (x+2) + (x+4) ) \)
\( \implies 24 = \frac{1}{2} (2x+6) \)
\( \implies 24 = x+3 \)
\( \implies x = 24-3 \)
\( \implies x = 21 \)
Therefore, the value of x is 21.
In simple words: The list of numbers is already in order. Since there are 10 numbers, the median is the average of the 5th and 6th numbers. Set this average equal to 24 and solve for 'x'.

🎯 Exam Tip: When the terms involve variables like 'x', ensure you correctly identify the middle terms and set up the equation for the median. Always simplify the algebraic expression carefully to find 'x'.