OP Malhotra Class 9 Maths Solutions Chapter 15 Mean Median and Frequency Polygon Exercise 15 (A)

 

Question 1. Find the mean of each of the following set of numbers:
(a) First 5 natural numbers
(b) First 7 whole numbers
(c) First 4 prime numbers
(d) 1.3 cm, 5.7 cm, 9.8 cm, 6.4 cm, 6.9 cm
(e) Rs. 7, Rs. 19, Rs. 31, Rs. 43, Rs. 70
Answer:
(a) The first 5 natural numbers are 1, 2, 3, 4, 5. To find their mean, we add these numbers and divide by how many there are. \[ \text{Mean} = \frac{1+2+3+4+5}{5} = \frac{15}{5} = 3 \] The mean of these numbers is 3. Finding the mean helps us understand the average value of a set.
(b) The first 7 whole numbers are 0, 1, 2, 3, 4, 5, 6. We sum these numbers and then divide by 7 to get the mean. \[ \text{Mean} = \frac{0+1+2+3+4+5+6}{7} = \frac{21}{7} = 3 \] So, the mean of the first 7 whole numbers is 3. Whole numbers include zero, which is important for calculations.
(c) The first 4 prime numbers are 2, 3, 5, and 7. To calculate their mean, we add them up and divide by 4. \[ \text{Mean} = \frac{2+3+5+7}{4} = \frac{17}{4} = 4.25 \] The mean of these prime numbers is 4.25. Prime numbers are special because they can only be divided by 1 and themselves.
(d) The given measurements are 1.3 cm, 5.7 cm, 9.8 cm, 6.4 cm, and 6.9 cm. We find the mean by adding all these lengths and dividing by 5. \[ \text{Mean} = \frac{1.3+5.7+9.8+6.4+6.9}{5} = \frac{30.1}{5} = 6.02 \text{ cm} \] The average length is 6.02 cm. Understanding mean helps to find a central value for a set of data.
(e) The amounts are Rs. 7, Rs. 19, Rs. 31, Rs. 43, and Rs. 70. To find the mean, we sum these amounts and divide by 5. \[ \text{Mean} = \frac{7+19+31+43+70}{5} = \frac{170}{5} = \text{Rs. } 34 \] The average amount is Rs. 34. This calculation is useful for budgeting or finding average costs.
In simple words: For each part, add all the given numbers together. Then, divide the sum by how many numbers there are. This gives you the mean, which is the average value. Remember to use "Rs." for money and "cm" for lengths.

🎯 Exam Tip: Always double-check that you've included all numbers in the sum and divided by the correct count of items to avoid simple calculation errors.

 

Question 2. Seema obtained the following scores (out of 100) on a set of spelling tests: What is her mean score?
Answer: Seema's scores on the spelling tests were 80, 85, 90, 71, 60, and 100. To find her mean score, we add all these marks and then divide by the number of tests, which is 6. \[ \text{Mean} = \frac{80+85+90+71+60+100}{6} = \frac{486}{6} = 81 \] Her average score across these tests is 81. The mean gives us a good overall idea of her performance.
In simple words: We add up all of Seema's scores: 80, 85, 90, 71, 60, and 100. This sum is 486. Then we divide 486 by 6 (because there are 6 scores), which gives 81.

🎯 Exam Tip: When calculating a mean from a list of scores, ensure every score is included in the sum and that the count used for division is accurate.

 

Question 3. Shaleen's last six batting scores were 138, 144, 155, 142, 167, 172. What was his mean score?
Answer: Shaleen's last six batting scores were 138, 144, 155, 142, 167, and 172. To find his mean score, we add all these runs and divide by 6, which is the number of innings. \[ \text{Mean score} = \frac{138+144+155+142+167+172}{6} = \frac{918}{6} = 153 \] His average batting score is 153 runs. A higher mean score usually indicates a more consistent and better player.
In simple words: Add up Shaleen's six scores: 138, 144, 155, 142, 167, and 172. The total is 918. Divide 918 by 6 to get 153.

🎯 Exam Tip: In statistics questions, clearly list the data points first before summing them to avoid missing any values, especially with longer lists.

 

Question 4. Madhu worked \( 2\frac { 1 }{ 2 } \) hours on Monday, \( 3\frac { 1 }{ 4 } \) hrs. on Tuesday, and \( 2\frac { 3 }{ 4 } \) hrs. on Wednesday. What is the mean number of hours she worked on these three days?
Answer: Madhu worked \( 2\frac{1}{2} \) hours on Monday, \( 3\frac{1}{4} \) hours on Tuesday, and \( 2\frac{3}{4} \) hours on Wednesday. To find the mean number of hours she worked, we add the hours for all three days and divide by 3. First, convert mixed fractions to improper fractions: \( 2\frac{1}{2} = \frac{5}{2} \) \( 3\frac{1}{4} = \frac{13}{4} \) \( 2\frac{3}{4} = \frac{11}{4} \) Now, sum the hours by finding a common denominator (4): \[ \text{Total hours} = \frac{5}{2} + \frac{13}{4} + \frac{11}{4} = \frac{10}{4} + \frac{13}{4} + \frac{11}{4} = \frac{10+13+11}{4} = \frac{34}{4} = \frac{17}{2} \] Next, calculate the mean: \[ \text{Mean hours} = \frac{\text{Total hours}}{3} = \frac{\frac{17}{2}}{3} = \frac{17}{2 \times 3} = \frac{17}{6} \] Convert the improper fraction back to a mixed number: \[ \frac{17}{6} = 2\frac{5}{6} \text{ hours} \] So, the mean number of hours Madhu worked per day is \( 2\frac{5}{6} \) hours. Working with fractions requires converting them to a common denominator before adding.
In simple words: Add up the hours Madhu worked each day: \( 2\frac{1}{2} \), \( 3\frac{1}{4} \), and \( 2\frac{3}{4} \). First, change them to fractions with the same bottom number (denominator), then add them to get \( \frac{17}{2} \) hours. Divide this total by 3 to get the average, which is \( 2\frac{5}{6} \) hours.

🎯 Exam Tip: Always convert mixed numbers to improper fractions and find a common denominator before adding or subtracting them to ensure accuracy.

 

Question 5. Ayushree sat for six tests and Ananya sat for seven tests. Their percentage scores were: Ayushree: 68 75 70 45 57 77 Ananya: 52 87 64 53 74 81 86 Who has the higher mean score?
Answer: We need to find the mean score for both Ayushree and Ananya to see who has the higher mean. For Ayushree, her scores are 68, 75, 70, 45, 57, 77. There are 6 scores. \[ \text{Ayushree's Mean} = \frac{68+75+70+45+57+77}{6} = \frac{392}{6} = 65\frac{1}{3}\% \] For Ananya, her scores are 52, 87, 64, 53, 74, 81, 86. There are 7 scores. \[ \text{Ananya's Mean} = \frac{52+87+64+53+74+81+86}{7} = \frac{497}{7} = 71\% \] Comparing their mean scores, Ayushree's mean is \( 65\frac{1}{3}\% \) and Ananya's mean is \( 71\% \). Since \( 71\% > 65\frac{1}{3}\% \), Ananya has the higher mean score.
In simple words: We find the average score for Ayushree by adding her 6 scores and dividing by 6. This gives \( 65\frac{1}{3}\% \). We do the same for Ananya by adding her 7 scores and dividing by 7. This gives \( 71\% \). Ananya's average score is higher.

🎯 Exam Tip: Clearly show the calculation for each individual's mean, then explicitly state the comparison and conclusion to earn full marks.

 

Question 6. Find the mean of first ten odd natural numbers.
Answer: The first ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19. To find their mean, we add all these numbers and divide by 10. \[ \text{Mean} = \frac{1+3+5+7+9+11+13+15+17+19}{10} = \frac{100}{10} = 10 \] The mean of the first ten odd natural numbers is 10. Interestingly, the mean of the first 'n' odd natural numbers is always 'n'.
In simple words: Add the first ten odd numbers (1, 3, 5, 7, 9, 11, 13, 15, 17, 19). The total is 100. Divide 100 by 10, which gives 10.

🎯 Exam Tip: Remember that natural numbers start from 1, and odd numbers are not divisible by 2. This rule can save time on similar problems.

 

Question 7. If the mean of 16, 14, x, 23, 20 is 18 find the value of x.
Answer: We are given that the mean of five numbers (16, 14, \( x \), 23, 20) is 18. First, calculate the total sum these numbers should have based on the mean. Total sum = Mean \( \times \) Number of terms Total sum = \( 18 \times 5 = 90 \) Next, find the sum of the given numbers, including \( x \): Sum of numbers = \( 16 + 14 + x + 23 + 20 = 73 + x \) Now, we set this sum equal to the total sum we calculated: \( 73 + x = 90 \)
\( \implies x = 90 - 73 \)
\( \implies x = 17 \) So, the value of \( x \) is 17. The mean helps us find missing values when other data points and the average are known.
In simple words: The average of 5 numbers is 18, so their total sum should be \( 18 \times 5 = 90 \). When we add the known numbers (16, 14, 23, 20), we get 73. So, 73 plus \( x \) must equal 90. To find \( x \), subtract 73 from 90, which gives 17.

🎯 Exam Tip: To find a missing value in a mean problem, always calculate the required total sum first, then set up an equation to solve for the unknown term.

 

Question 8. If the mean of x, x + 2, x + 4, x + 6, x + 8 is 24, find x.
Answer: We are given that the mean of five terms (\( x \), \( x+2 \), \( x+4 \), \( x+6 \), \( x+8 \)) is 24. First, calculate the total sum of these terms using the mean. Total sum = Mean \( \times \) Number of terms Total sum = \( 24 \times 5 = 120 \) Next, find the sum of the given terms algebraically: Sum of terms = \( x + (x+2) + (x+4) + (x+6) + (x+8) \) Sum of terms = \( x+x+x+x+x + 2+4+6+8 \) Sum of terms = \( 5x + 20 \) Now, set the algebraic sum equal to the total sum: \( 5x + 20 = 120 \) To solve for \( x \), subtract 20 from both sides:
\( \implies 5x = 120 - 20 \)
\( \implies 5x = 100 \) Then, divide by 5:
\( \implies x = \frac{100}{5} \)
\( \implies x = 20 \) So, the value of \( x \) is 20. This problem shows how to find an unknown value when terms are related and the mean is given.
In simple words: The average of 5 terms is 24, so their total sum is \( 24 \times 5 = 120 \). When we add the terms (\( x \), \( x+2 \), \( x+4 \), \( x+6 \), \( x+8 \)), we get \( 5x + 20 \). So, \( 5x + 20 \) must equal 120. Solving this gives \( 5x = 100 \), which means \( x = 20 \).

🎯 Exam Tip: When terms are expressed with variables, group the variables and the constant numbers separately before forming the total sum equation.

 

Question 9. Madhu practised on her sitar 45 minutes, 30 minutes, 60 minutes, 50 minutes and 20 minutes. What was her mean practice time?
Answer: Madhu practiced on her sitar for 45, 30, 60, 50, and 20 minutes. To find her mean practice time, we sum all these times and divide by the number of practice sessions, which is 5. \[ \text{Mean practice time} = \frac{45+30+60+50+20}{5} = \frac{205}{5} = 41 \text{ minutes} \] Her average practice time is 41 minutes per session. Consistent practice over time helps to improve skills.
In simple words: Add all of Madhu's practice times: 45, 30, 60, 50, and 20 minutes. The total is 205 minutes. Divide 205 by 5 to get the average practice time of 41 minutes.

🎯 Exam Tip: Ensure all data points are in the same units (e.g., minutes) before calculating the mean. If different units were given, a conversion would be necessary.

 

Question 10. Nisha secured 73, 86, 78 and 75 marks in four tests. What is the least number of marks she can secure in her next test, if she has to have a mean score of 80 marks in five tests?
Answer: Nisha scored 73, 86, 78, and 75 marks in four tests. She wants to have a mean score of 80 marks after five tests. Let \( x \) be the marks she needs to score in the fifth test. If the mean of five tests is 80, the total sum of marks in five tests should be: Total marks = Mean \( \times \) Number of tests Total marks = \( 80 \times 5 = 400 \) Now, sum the marks from her first four tests and add \( x \) for the fifth test: Sum of 5 test marks = \( 73 + 86 + 78 + 75 + x \) Sum of 5 test marks = \( 312 + x \) Set this sum equal to the desired total marks: \( 312 + x = 400 \) To find \( x \), subtract 312 from both sides:
\( \implies x = 400 - 312 \)
\( \implies x = 88 \) So, Nisha must secure at least 88 marks in her fifth test to achieve a mean score of 80. This problem shows how to work backwards from a desired mean to find a missing data point.
In simple words: Nisha needs an average of 80 marks over 5 tests, meaning a total of \( 80 \times 5 = 400 \) marks. Her current 4 test scores add up to \( 73+86+78+75 = 312 \). To reach 400 total, she needs \( 400 - 312 = 88 \) marks in the fifth test.

🎯 Exam Tip: When solving for a minimum score to achieve a target mean, ensure your final answer is a positive whole number, as marks are typically counted this way.

 

Question 11. A cricketer has a mean score of 60 runs in ten innings. Find out how many runs are to be scored in the eleventh innings to raise the mean score to 62.
Answer: A cricketer has a mean score of 60 runs in 10 innings. First, calculate the total runs scored in 10 innings: Total runs in 10 innings = Mean score \( \times \) Number of innings Total runs in 10 innings = \( 60 \times 10 = 600 \) runs Now, the cricketer wants to raise the mean score to 62 after 11 innings. Calculate the total runs needed for 11 innings to achieve a mean of 62: Total runs in 11 innings = Desired mean \( \times \) Number of innings Total runs in 11 innings = \( 62 \times 11 = 682 \) runs To find how many runs need to be scored in the 11th inning, subtract the total runs from 10 innings from the total runs needed for 11 innings: Runs needed in 11th inning = Total runs in 11 innings - Total runs in 10 innings Runs needed in 11th inning = \( 682 - 600 = 82 \) runs The cricketer needs to score 82 runs in the eleventh innings. This demonstrates how a single new data point can affect the overall mean significantly.
In simple words: The player scored \( 60 \times 10 = 600 \) runs in 10 games. To get an average of 62 runs over 11 games, they need a total of \( 62 \times 11 = 682 \) runs. So, in the 11th game, they must score \( 682 - 600 = 82 \) runs.

🎯 Exam Tip: Always clearly state the total sum for both the initial and final scenarios to correctly calculate the required additional score.

 

Question 12. Find the mean of the following frequency distributions: Weight: 30 31 32 33 34 No. of students: 8 10 15 8 9
Answer: To find the mean of the frequency distribution, we use the formula \( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \), where \( f_i \) is the frequency (number of students) and \( x_i \) is the weight. We need to create a table to calculate \( f_i x_i \) for each weight and then sum these values.

Using the formula for mean: \[ \text{Mean} (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{1600}{50} = 32 \text{ kg} \] The mean weight of the students is 32 kg. Frequency distributions are a way to organize data and easily calculate measures like the mean.
In simple words: Multiply each weight by its number of students, then add all these products together. This gives 1600. Add up the total number of students, which is 50. Divide 1600 by 50 to get the average weight of 32 kg.

🎯 Exam Tip: When creating the frequency table, clearly label your columns as \( x_i \), \( f_i \), and \( f_i x_i \) to make your calculations easy to follow and verify.

 

Question 13. x 2 5 7 8 f 2 4 6 3
Answer: To find the mean of this frequency distribution, we will use the formula \( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \). First, we create a frequency table to calculate the product of each \( x_i \) and its corresponding frequency \( f_i \).

Now, apply the mean formula: \[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{90}{15} = 6 \] The mean of this frequency distribution is 6. This method is effective for calculating the average when data points have different frequencies.
In simple words: Multiply each \( x \) value by its frequency \( f \), then add all these products (total 90). Add all the frequencies (total 15). Divide the sum of products by the sum of frequencies to get the mean, which is 6.

🎯 Exam Tip: Always double-check the multiplication for each \( f_i x_i \) product, as a small error there will affect the entire mean calculation.

 

Question 14. X 0.1 0.2 0.3 0.4 0.5 0.6 f 20 60 20 40 10 50
Answer: To find the mean of this frequency distribution with decimal values, we use the formula \( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \). First, we create a table to calculate the product of each \( x_i \) and its corresponding frequency \( f_i \).

Now, apply the mean formula: \[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{71}{200} = 0.355 \] The mean of this frequency distribution is 0.355. This calculation is a good way to find the average of data points that occur with different frequencies, even if they are decimals.
In simple words: Multiply each \( x \) value by its frequency \( f \), then add all these products together (total 71). Add all the frequencies (total 200). Divide the sum of products by the sum of frequencies to get the mean, which is 0.355.

🎯 Exam Tip: Be careful with decimal multiplications. Ensure you place the decimal point correctly in each \( f_i x_i \) product to get the accurate sum for the numerator.

 

Question 15. The marks scored in a test by a class of 25 boys are as follows: 24 25 23 20 20 19 22 20 24 22 18 23 23 18 20 16 25 24 17 18 23 22 23 20 24 Draw a frequency table and calculate the mean.
Answer: We are given the marks scored by 25 boys in a test. We need to create a frequency table and then calculate the mean. First, list all the unique scores and count how many times each score appears (frequency). The scores are: 24, 25, 23, 20, 20, 19, 22, 20, 24, 22, 18, 23, 23, 18, 20, 16, 25, 24, 17, 18, 23, 22, 23, 20, 24. Let's make a frequency table:

Now, we calculate the mean using the formula: \( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \) \[ \text{Mean} = \frac{533}{25} = 21.32 \] The mean marks scored by the boys is 21.32. Creating a frequency table first helps organize raw data, making calculations much easier.
In simple words: First, count how many times each mark appears and write it in a table. Then, multiply each mark by its count and add these products to get 533. Add all the counts (total 25 boys). Divide 533 by 25 to find the average mark, which is 21.32.

🎯 Exam Tip: Always verify that the sum of frequencies (\( \sum f_i \)) matches the total number of data points given in the question (e.g., 25 boys) to ensure no data was missed.