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

 

Question 1. Find the mode of the following data :
(i) 8, 5, 6, 8, 8, 4, 6, 10, 8, 2;
(ii) 1, 2, 3, 3, 3, 5, 6, 8, 8, 8, 9;
(iii) 2, 5, 6, 6, 5, 3, 5, 3, 6, 5, 3, 5, 7, 6, 5, 7, 5;
(iv) 4, 5, 6, 4, 1, 4, 2, 4, 4.
Answer:
(i) To find the mode, we count how often each number appears. In the given data (8, 5, 6, 8, 8, 4, 6, 10, 8, 2):

The number 8 appears 4 times, which is more than any other number. Thus, the mode is 8.
(ii) For the data (1, 2, 3, 3, 3, 5, 6, 8, 8, 8, 9):Both 3 and 8 appear 3 times, which is the highest frequency. So, the data has two modes: 3 and 8.
(iii) For the data (2, 5, 6, 6, 5, 3, 5, 3, 6, 5, 3, 5, 7, 6, 5, 7, 5):The number 5 appears 7 times, which is the highest frequency. Therefore, the mode is 5.
(iv) For the data (4, 5, 6, 4, 1, 4, 2, 4, 4):The number 4 appears 7 times, which is the highest frequency. Thus, the mode is 4. The mode helps us understand the most common value in a group of data.
In simple words: To find the mode, just count which number shows up most often in the list. That number is the mode. If two numbers appear the same highest number of times, both are modes.

🎯 Exam Tip: When finding the mode, it's often helpful to first arrange the data in ascending order or create a frequency table to easily spot the most frequent value(s).

 

Question 2. Find the median and mode for the set of numbers: 2, 2, 3, 5, 5, 5, 6, 8, 9
Answer: First, arrange the numbers in ascending order: 2, 2, 3, 5, 5, 5, 6, 8, 9. The total count of numbers, \( n \), is 9, which is an odd number. To find the median, we use the formula for an odd number of data points: Median \( = \frac { (n+1) }{ 2 } \)th term
\( \implies \) Median \( = \frac { (9+1) }{ 2 } \)th term
\( \implies \) Median \( = \frac { 10 }{ 2 } \)th term
\( \implies \) Median \( = 5 \)th term The 5th term in the ordered list is 5. So, the median is 5. For the mode, we look for the number that appears most frequently. In the list (2, 2, 3, 5, 5, 5, 6, 8, 9), the number 5 appears 3 times, which is more than any other number. Therefore, the mode is 5. The median is a good measure of central tendency because it is not affected by extremely high or low values.
In simple words: First, put all the numbers in order from smallest to largest. The middle number is the median. For the mode, just find the number that shows up the most times in the list. In this case, both the middle number and the most common number are 5.

🎯 Exam Tip: Always arrange the data in order before finding the median. If there are an odd number of data points, the median is exactly the middle one. If there are an even number, it's the average of the two middle ones.

 

Question 3. A boy scored the following marks in various class tests during a term, each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16
(i) What are his modal marks ?
(ii) What are his median marks ?
(iii) What are his mean marks ?
Answer: First, arrange the marks in ascending order: 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19.
(i) **Modal Marks:** The mode is the mark that appears most frequently. In the ordered list, the mark 16 appears 3 times, which is more than any other mark. So, the modal mark is 16.
(ii) **Median Marks:** The total number of marks, \( n \), is 11, which is an odd number. Median \( = \frac { (n+1) }{ 2 } \)th term
\( \implies \) Median \( = \frac { (11+1) }{ 2 } \)th term
\( \implies \) Median \( = \frac { 12 }{ 2 } \)th term
\( \implies \) Median \( = 6 \)th term The 6th term in the ordered list is 15. So, the median mark is 15.
(iii) **Mean Marks:** The mean is the sum of all marks divided by the total number of marks. Mean \( = \frac { \sum x_i }{ n } \)
\( \implies \) Mean \( = \frac { 7+10+12+12+14+15+16+16+16+17+19 }{ 11 } \)
\( \implies \) Mean \( = \frac { 154 }{ 11 } \)
\( \implies \) Mean \( = 14 \) The mean mark is 14. This shows the average performance of the student in the tests.
In simple words: First, put all the boy's scores in order. The modal mark is the score he got most often. The median mark is the score exactly in the middle of the ordered list. The mean mark is found by adding all his scores together and then dividing by how many tests he took.

🎯 Exam Tip: When calculating mean, median, and mode for a single set of data, always start by arranging the data in ascending order. This helps prevent errors, especially with median and mode.

 

Question 4. Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks :
0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8
Answer: The marks are already arranged in ascending order: 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8. The total number of students, \( n \), is 16.
(i) **Mean Marks:** Mean \( = \frac { \sum x_i }{ n } \)
\( \implies \) Mean \( = \frac { 0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8 }{ 16 } \)
\( \implies \) Mean \( = \frac { 64 }{ 16 } \)
\( \implies \) Mean \( = 4 \) The mean mark is 4.
(ii) **Median Marks:** Since \( n = 16 \) (an even number), the median is the average of the two middle terms. Median \( = \frac { 1 }{ 2 } [ \frac { n }{ 2 } \text{th term} + ( \frac { n }{ 2 } + 1 ) \text{th term} ] \)
\( \implies \) Median \( = \frac { 1 }{ 2 } [ \frac { 16 }{ 2 } \text{th term} + ( \frac { 16 }{ 2 } + 1 ) \text{th term} ] \)
\( \implies \) Median \( = \frac { 1 }{ 2 } [ 8\text{th term} + 9\text{th term} ] \) From the ordered list, the 8th term is 4 and the 9th term is 5.
\( \implies \) Median \( = \frac { 1 }{ 2 } [ 4+5 ] \)
\( \implies \) Median \( = \frac { 9 }{ 2 } \)
\( \implies \) Median \( = 4.5 \) The median mark is 4.5.
(iii) **Modal Marks:** The mode is the mark that appears most frequently. In the given data, the mark 5 appears 4 times, which is the highest frequency. So, the mode is 5. Knowing these measures helps understand the overall performance of the students.
In simple words: The numbers are already in order. For the mean, add them all up and divide by 16. For the median, since there are 16 numbers, take the 8th and 9th numbers, add them, and divide by 2. For the mode, see which number appears most often.

🎯 Exam Tip: When the number of data points (n) is even, remember to find the average of the two middle terms for the median. Carefully count frequencies for the mode.

 

Question 5. Find the mode from the following distributions :
Answer: We are given the following distribution of marks and the number of students who obtained them:

In this distribution, the frequency corresponding to a mark of 25 is 20. This is the highest frequency shown in the table for a single mark. Therefore, the mode is 25 marks. The mode tells us which score was most common among the students.
In simple words: Look at the table and find the "No. of students" row. The biggest number in that row is the highest frequency. Now look at the "Marks" row directly above it to find which mark has that highest frequency. That mark is the mode.

🎯 Exam Tip: For discrete data presented in a frequency table, the mode is simply the data value (x) that corresponds to the highest frequency (f).

 

Question 6. At a shooting competition the scores of a competitor were as given below:
(i) What was his modal score ?
(ii) median score ?
(iii) What was his total score ?
(iv) What was his mean score ?
Answer: We can create a frequency table with cumulative frequency (c.f.) and the product of score and frequency (f x x) to solve this:

(i) **Modal Score:** From the frequency table, the score 4 has the highest frequency of 7. So, the modal score is 4.
(ii) **Median Score:** The total number of shots \( n = \sum f = 25 \), which is an odd number. Median \( = \frac { (n+1) }{ 2 } \)th term
\( \implies \) Median \( = \frac { (25+1) }{ 2 } \)th term
\( \implies \) Median \( = \frac { 26 }{ 2 } \)th term
\( \implies \) Median \( = 13 \)th term Looking at the cumulative frequency (c.f.) column, the 13th term falls within the score range where the c.f. becomes 13 (which is for score 3). So, the median score is 3.
(iii) **Total Score:** The total score is the sum of all (f x x) values, which is \( \sum (f \times x) = 80 \).
(iv) **Mean Score:** Mean \( = \frac { \sum (f \times x) }{ \sum f } \)
\( \implies \) Mean \( = \frac { 80 }{ 25 } \)
\( \implies \) Mean \( = \frac { 16 }{ 5 } \)
\( \implies \) Mean \( = 3.2 \) The mean score is 3.2. These measures help understand the competitor's performance, from their most frequent score to their average.
In simple words: To solve this, make a table to count each score, keep a running total of shots, and multiply score by shots. The modal score is the one with the most shots. The median is the middle score from the cumulative shots count. The total score is all (score times shots) added up. The mean score is the total score divided by the total number of shots.

🎯 Exam Tip: For grouped data, creating a full frequency distribution table with cumulative frequency and (f x x) columns is crucial for accurately calculating the mode, median, and mean.

 

Question 7. Find the value of k, if k, the mode of the following data is 17?
15, 16, 17, 13, 17, 16, 14,, 17, 16, 15, 15
Answer: The given data set is: 15, 16, 17, 13, 17, 16, 14, 17, 16, 15, 15. The problem states that \( k \) is part of the data, and the mode of the data is 17. This means 17 must be the most frequent number in the full data set including \( k \). Let's count the frequencies of the numbers in the given list: 13: 1 14: 1 15: 3 16: 3 17: 3 If the mode is 17, it means 17 must appear more times than any other number. Currently, 15, 16, and 17 all appear 3 times. If \( k \) is added to the data, and the mode becomes 17, then \( k \) must be 17. This would make the frequency of 17 become 4, which would be greater than the frequency of 15 and 16. So, \( k = 17 \).
In simple words: We are told that the number 17 is the mode, meaning it appears the most times. When we count the numbers in the list, 15, 16, and 17 all appear 3 times. For 17 to be the clear mode, it needs to appear more than 3 times. This means the missing number \( k \) must be 17, making 17 appear 4 times.

🎯 Exam Tip: When a missing value (like k) is stated to be the mode, it must be the value that increases the frequency of the mode to be uniquely the highest, or at least one of the highest if the data is bimodal by intention.

 

Question 8. Find the value of k for which the mode of the following is 7?
3, 5, 5, 7, 3, 6, 7, 9, 6, 7, 3, 5, 7, 3, k
Answer: The given data set is: 3, 5, 5, 7, 3, 6, 7, 9, 6, 7, 3, 5, 7, 3, k. We are told that the mode of this data is 7. This means that 7 must be the number that appears most frequently in the entire data set, including \( k \). Let's count the frequencies of the numbers in the given list (excluding \( k \) for now): 3: 4 5: 3 6: 2 7: 4 9: 1 Currently, both 3 and 7 appear 4 times. For 7 to be the unique mode, its frequency must be greater than 4. This implies that \( k \) must be 7, which would increase the frequency of 7 to 5.
In simple words: The problem says that 7 is the mode. This means 7 must show up more than any other number. When we count the numbers, both 3 and 7 appear 4 times. For 7 to be the only mode, the letter \( k \) must be 7, so that 7 appears 5 times, making it the most frequent number.

🎯 Exam Tip: If the mode is given and a variable is present, calculate the frequencies of all existing numbers first. The variable must take a value that makes the stated mode have the highest frequency.

 

Question 9. Find the mode of the distributions given in problems 9-13 by drawing a histogram.
(i) State the upper boundary of the last class;
(ii) State the class size
(iii) State the modal class
(iv) Determine the class which contains the median of the distribution.
Answer: The given data contains class marks and frequencies for the lengths of nails. To work with this data effectively, we first convert the class marks into continuous class intervals. Given class marks are 2.5, 3, 3.5, 4, 4.5, 5. The difference between consecutive class marks is 0.5. So, the class size is 0.5. To find the class boundaries, we subtract and add half of the class size (0.5 / 2 = 0.25) from each class mark. For example, for class mark 2.5, the class interval is \( (2.5 - 0.25) \) to \( (2.5 + 0.25) \), which is 2.25-2.75. The converted frequency distribution with cumulative frequency is:

The cumulative frequency table is:
(i) **Upper boundary of the last class:** The last class is 4.75-5.25. Its upper boundary is 5.25.
(ii) **Class size:** The difference between the upper and lower boundary of any class (e.g., 2.75 - 2.25) is 0.5.
(iii) **Modal class:** The modal class is the class with the highest frequency. Here, the frequency 50 is the highest, corresponding to the class 3.25-3.75.
(iv) **Class which contains the median of the distribution:** The total number of observations \( n = \sum f = 200 \). The median position is \( \frac { n }{ 2 } \)th term \( = \frac { 200 }{ 2 } = 100 \)th term. Looking at the cumulative frequency, the 100th term falls into the class where the cumulative frequency first exceeds 100. This is the class 3.75-4.25, which has a cumulative frequency of 140. So, the median lies in the class 3.75-4.25. Working with class intervals helps group large datasets into manageable categories.
In simple words: First, change the given class marks into proper class groups (like 2.25-2.75). The highest number in the last group is the upper boundary. The size of any group (like 2.75 minus 2.25) is the class size. The modal class is the group with the most items. For the median class, find half of the total number of items, then see which group's cumulative count reaches or passes that half-total first.

🎯 Exam Tip: For continuous data, convert class marks to exclusive class intervals first. The median class is identified by finding \( \frac{N}{2} \) and locating it in the cumulative frequency column.