RS Aggarwal Class 7 Mathematics Solutions Chapter 21 Collection and Organisation of Data

 

Question 1. Define the following terms: (i) Data (ii) Raw data (iii) Array (iv) Tabulation of data (v) Observations (vi) Frequency of an observation (vii) Statistics
Answer: (i) Data refers to information presented as numerical values. (ii) Raw data is information obtained in its original, unorganized form. (iii) Array is the arrangement of raw data in order from smallest to largest or largest to smallest. (iv) Tabulation of data involves organizing information systematically in table form. (v) Observations are individual numerical values that make up a dataset. (vi) Frequency of an observation refers to how many times a particular value shows up in the data. (vii) Statistics is the branch of mathematics dealing with gathering, displaying, analyzing, and interpreting numerical information.
In simple words: Data is numbers. Raw data is messy numbers. Array means arranging them in order. Tabulation means putting them in a table. An observation is one number. Frequency means how often it appears. Statistics means studying all this information.

Exam Tip: Learn each definition word-for-word and be ready to state it clearly in one sentence per term - examiners value precision in terminology.

 

Question 2. Arrange the following data in ascending order and prepare a frequency table: 1, 1, 2, 2, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6
Answer: Data in ascending order: 1, 1, 2, 2, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6

In simple words: Write the data from smallest to biggest. Then count how many times each number appears and show it in a table.

Exam Tip: Always verify the frequency table by adding all frequencies - the total should match the number of data points in the original list.

 

Question 3. Construct a frequency table for the daily wages (in Rs.) of workers: 130, 150, 180, 150, 200, 130, 150, 180, 180, 180, 200, 200
Answer: Daily wages in ascending order: 130, 130, 150, 150, 150, 180, 180, 180, 180, 200, 200, 200

In simple words: Organize the wage data from smallest to largest. Count how many workers earn each wage amount and present in table format.

Exam Tip: Include all unique values that appear in the data, even if some wages don't repeat - the frequency table should capture the complete distribution.

 

Question 4. Arrange the following data in ascending order and prepare a frequency table: 5, 7, 6, 6, 7, 8, 8, 8, 8, 9, 10, 10
Answer: Data in ascending order: 5, 6, 6, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10

In simple words: Organize the values from smallest to largest. Count how many times each value repeats and record in a frequency table.

Exam Tip: Check that every data value appears exactly once when you're arranging - missing or extra values indicate a counting error.

 

Question 5. Write the steps to find the mean. List them in order: (i) numerical (ii) original (iii) array (iv) frequency (v) tabulation
Answer: The sequence of steps to find the mean is: (i) numerical - data shown as numbers; (ii) original - the starting form of the information; (iii) array - arranging information in order; (iv) frequency - counting occurrences; (v) tabulation - organizing into table form.
In simple words: Collect numbers, arrange them in order, count how often each appears, make a table, then calculate the mean from that table.

Exam Tip: Remember that these steps create a logical flow from raw data collection to final calculation - skip any step and your result may be incorrect.

 

Question 6. Find the mean of the first five natural numbers.
Answer: The first five natural numbers are 1, 2, 3, 4, and 5.

Mean = \( \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3 \)

The mean of the first five natural numbers is 3.
In simple words: Add up all the numbers: 1 + 2 + 3 + 4 + 5 = 15. Divide by how many numbers there are: 15 ÷ 5 = 3.

Exam Tip: Always divide the total sum by the count of values - forgetting the division step is the most common error when calculating means.

 

Question 7. Find the mean of the first six odd natural numbers.
Answer: The first six odd natural numbers are 1, 3, 5, 7, 9, and 11.

Mean = \( \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{1 + 3 + 5 + 7 + 9 + 11}{6} = \frac{36}{6} = 6 \)

The mean of the first six odd natural numbers is 6.
In simple words: Add the first six odd numbers: 1 + 3 + 5 + 7 + 9 + 11 = 36. Divide by 6 to get 36 ÷ 6 = 6.

Exam Tip: Odd numbers skip every other value - double-check your list to make sure you've included exactly the right number of odd values.

 

Question 8. Find the mean of the first seven even natural numbers.
Answer: The first seven even natural numbers are 2, 4, 6, 8, 10, 12, and 14.

Mean = \( \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{2 + 4 + 6 + 8 + 10 + 12 + 14}{7} = \frac{56}{7} = 8 \)

The mean of the first seven even natural numbers is 8.
In simple words: Add the first seven even numbers: 2 + 4 + 6 + 8 + 10 + 12 + 14 = 56. Divide by 7 to get 56 ÷ 7 = 8.

Exam Tip: Even numbers always increment by 2 - verify this pattern in your list to avoid accidentally including odd values.

 

Question 9. Find the mean of the first five prime numbers.
Answer: The first five prime numbers are 2, 3, 5, 7, and 11.

Mean = \( \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{2 + 3 + 5 + 7 + 11}{5} = \frac{28}{5} = 5.6 \)

The mean of the first five prime numbers is 5.6.
In simple words: Add the first five primes: 2 + 3 + 5 + 7 + 11 = 28. Divide by 5 to get 28 ÷ 5 = 5.6.

Exam Tip: Prime numbers have no factors other than 1 and themselves - 2 is the only even prime, so don't forget to include it.

 

Question 10. Find the mean of the first six multiples of 5.
Answer: The first six multiples of 5 are 5, 10, 15, 20, 25, and 30.

Mean = \( \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{5 + 10 + 15 + 20 + 25 + 30}{6} = \frac{105}{6} = 17.5 \)

The mean of the first six multiples of 5 is 17.5.
In simple words: Add the first six multiples: 5 + 10 + 15 + 20 + 25 + 30 = 105. Divide by 6 to get 105 ÷ 6 = 17.5.

Exam Tip: Multiples of a number follow a consistent pattern - each is the previous one plus 5, which makes them simple to list correctly.

 

Question 11. Find the mean weight of workers from the following data.

Mean weight = \( \frac{\sum(fi × xi)}{\sum fi} = \frac{975}{15} = 65 \) kg.
In simple words: Multiply each weight by how many workers have that weight. Add all these products. Divide by the total number of workers.

Exam Tip: When finding the mean from a frequency table, always multiply (value × frequency) for each row before summing - multiplying the totals separately will give the wrong answer.

 

Question 12. Find the mean daily wages of workers from the following frequency table.

Mean daily wages = \( \frac{\sum(fi × xi)}{\sum fi} = \frac{9540}{60} = Rs 159 \).
In simple words: For each wage, multiply it by the count of workers earning that wage. Add these products together. Divide by the total worker count.

Exam Tip: Frequency tables compress data, but the calculation stays the same - don't confuse the frequency table format with the simpler direct-data formula.

 

Question 13. Find the mean height of plants from the following data.

Mean height = \( \frac{\sum(fi × xi)}{\sum fi} = \frac{5634}{90} = 62.6 \) cm.
In simple words: Multiply each height value by how many plants have that height. Sum all products. Divide by the total plant count to find the mean height.

Exam Tip: Check that your sum of all frequencies equals the total number of observations stated in the problem - this is a quick way to catch arithmetic errors.

 

Question 14. Find the mean age of players from the following frequency table.

Mean age = \( \frac{\sum(fi × xi)}{\sum fi} = \frac{770}{50} = 15.4 \) years.
In simple words: For each age, multiply it by the number of players at that age. Add all these results. Divide by the total number of players.

Exam Tip: Ensure your multiplication (age × number of players) is accurate for each row - single arithmetic errors here will throw off your final answer.

 

Question 15. Find the mean height of boys from the following frequency table.

Mean height = \( \frac{\sum(fi × xi)}{\sum fi} = \frac{6930}{40} = 173.25 \) cm.
In simple words: Multiply each height by the number of boys at that height. Add all these products. Divide by the total number of boys.

Exam Tip: When the total number of observations is even, the mean does not need to be a whole number - express your answer as a decimal if required by the calculation.

 

Question 1. Find the median of the following data: (i) 3, 11, 7, 2, 5, 9, 9, 2 and 10
Answer: Arranging the data in ascending order: 2, 2, 3, 5, 7, 9, 9, 10, 11

Number of terms, N = 9
Since N is an odd number,

Median = \( \left( \frac{N+1}{2} \right) \)th observation

Median = \( \left( \frac{9+1}{2} \right) \)th observation

Median = 5th observation = 7
In simple words: Sort the numbers from smallest to largest. Since there are 9 numbers (odd), the middle one is the median. Counting to the 5th position gives 7.

Exam Tip: For odd-numbered datasets, the median is always one of the values in your list - for even-numbered datasets, it's typically between two values.

 

Question 1. Find the median of the following data: (ii) 9, 25, 18, 15, 6, 16, 8, 22, 21
Answer: Arranging the data in ascending order: 6, 8, 9, 15, 16, 18, 21, 22, 25

Number of terms, N = 9
Since N is an odd number,

Median = \( \left( \frac{N+1}{2} \right) \)th observation

Median = \( \left( \frac{9+1}{2} \right) \)th observation

Median = 5th observation = 16
In simple words: Arrange all numbers from lowest to highest. With 9 values, the 5th value in the middle is your median, which is 16.

Exam Tip: Always arrange data in order first - the median's position matters much more than the values themselves.

 

Question 1. Find the median of the following data: (iii) 21, 15, 6, 25, 18, 13, 20, 9, 16, 8, 22
Answer: Arranging the data in ascending order: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25

Number of terms, N = 11
Since N is an odd number,

Median = \( \left( \frac{N+1}{2} \right) \)th observation

Median = \( \left( \frac{11+1}{2} \right) \)th observation

Median = 6th observation = 16
In simple words: List all numbers from smallest to largest. Since there are 11 values, the 6th value in the center is the median, which is 16.

Exam Tip: For any odd number of values, use the formula (N + 1) ÷ 2 to find which position holds the median.

 

Question 2. Find the median of the following data: (i) 10, 32, 17, 19, 21, 22, 9, 35
Answer: Arranging the data in ascending order: 9, 10, 17, 19, 21, 22, 32, 35

Number of terms, N = 8

Median = \( \frac{1}{2} \left \{ \left ( \frac{N}{2} \right ) \text{th observation} + \left( \frac{N}{2} + 1 \right ) \text{th observation} \right \} \)

Median = \( \frac{1}{2} \) (4th observation + 5th observation)

Median = \( \frac{1}{2} \) (19 + 21) = 20

Median = 20
In simple words: Arrange numbers from smallest to largest. With 8 values (even), take the average of the two middle numbers: (19 + 21) ÷ 2 = 20.

Exam Tip: For even-numbered datasets, the median falls between two middle values - you must average them rather than just picking one.

 

Question 2. Find the median of the following data: (ii) 55, 60, 35, 51, 29, 63, 72, 91, 85, 82
Answer: Arranging the data in ascending order: 29, 35, 51, 55, 60, 63, 72, 82, 85, 91

Number of terms, N = 10
Since N is an even number,

Median = \( \frac{1}{2} \left \{ \left ( \frac{N}{2} \right ) \text{th observation} + \left( \frac{N}{2} + 1 \right ) \text{th observation} \right \} \)

Median = \( \frac{1}{2} \) (5th observation + 6th observation)

Median = \( \frac{1}{2} \) (60 + 63)

Median = 61.5
In simple words: Order all values from smallest to largest. With 10 numbers, find the two middle values (5th and 6th positions). Average them: (60 + 63) ÷ 2 = 61.5.

Exam Tip: Double-check your count of terms - miscounting odd vs. even will lead you to use the wrong median formula.

 

Question 3. Find the median of the first 15 odd numbers.
Answer: The first 15 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, and 29.

Number of terms, N = 15
Since N is an odd number,

Median = \( \left( \frac{N+1}{2} \right) \)th observation

Median = \( \left( \frac{15+1}{2} \right) \)th observation

Median = 8th observation = 15
In simple words: List the first 15 odd numbers in order. Count to the 8th position, which is right in the middle. That position holds 15.

Exam Tip: Odd numbers follow the pattern 2n - 1 (or start at 1 and add 2 each time) - use this to quickly build your list and verify it's complete.

 

Question 4. Find the median of the first 10 even numbers.
Answer: The first 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20.

Number of terms, N = 10
Since N is an even number,

Median = \( \frac{1}{2} \left \{ \left ( \frac{N}{2} \right ) \text{th observation} + \left( \frac{N}{2} + 1 \right ) \text{th observation} \right \} \)

Median = \( \frac{1}{2} \) (5th observation + 6th observation)

Median = \( \frac{1}{2} \) (10 + 12) = 11
In simple words: Write the first 10 even numbers. Take the two middle values: 10 and 12. Their average is (10 + 12) ÷ 2 = 11.

Exam Tip: The middle values for a 10-item list are at positions 5 and 6 - knowing this position formula prevents errors.

 

Question 5. Find the median of the first 50 whole numbers (0, 1, 2, 3, ..., 49).
Answer: The first 50 whole numbers are 0, 1, 2, 3, ..., and 49.

Number of terms, N = 50
Since N is an even number,

Median = \( \frac{1}{2} \left \{ \left ( \frac{N}{2} \right ) \text{th observation} + \left( \frac{N}{2} + 1 \right ) \text{th observation} \right \} \)

Median = \( \frac{1}{2} \) {(25th observation) + (26th observation)}

Median = \( \frac{1}{2} \) {24 + 25}

Median = 24.5
In simple words: The first 50 whole numbers go from 0 to 49. The two middle values are at positions 25 and 26, which are 24 and 25. Their average is 24.5.

Exam Tip: Whole numbers include zero - don't accidentally start from 1, or your median will be off by a full value.

 

Question 6. Find the median of students' marks (out of 50) in an examination.
Answer: Marks of the students in ascending order: 17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40

Number of terms, N = 15
Since N is an odd number,

Median = \( \left( \frac{N+1}{2} \right) \)th observation

Median = \( \left( \frac{15+1}{2} \right) \)th observation

Median = 8th observation

Median = 23

The median marks are 23.
In simple words: Sort the marks from lowest to highest. With 15 values, the 8th position gives the middle value, which is 23.

Exam Tip: Always arrange marks or scores in numerical order first - the position matters more than remembering the raw numbers.

 

Question 7. Find the median age of 10 teachers in a school.
Answer: Ages (in years) of 10 teachers in a school are given as: 34, 37, 53, 46, 43, 31, 36, 40, 50

Arranging them in ascending order: 31, 34, 36, 37, 40, 43, 46, 50, 52, 53

Number of terms, N = 10
Since N is an even number,

Median = \( \frac{1}{2} \left \{ \left ( \frac{N}{2} \right ) \text{th observation} + \left( \frac{N}{2} + 1 \right ) \text{th observation} \right \} \)

Median = \( \frac{1}{2} \) {5th observation + 6th observation}

Median = \( \frac{1}{2} \) {40 + 43}

Median = 41.5

The median age is 41.5 years.
In simple words: Order the teachers' ages from youngest to oldest. Take the 5th and 6th values (40 and 43). Average them: (40 + 43) ÷ 2 = 41.5 years.

Exam Tip: When finding the average of two middle values, be careful with the arithmetic - forgetting to divide by 2 is a common mistake.

 

Question 8. Find the median weight from a cumulative frequency table of boys' weights.
Answer:

Number of terms, N = 41
Since N is an odd number,

Median = \( \left \{ \left( \frac{N+1}{2} \right) \text{th observation} \right \} = \left \{ \left( \frac{41+1}{2} \right) \text{th observation} \right \} = \{21\text{st observation}\} = 50 \) kg

The median weight is 50 kg.
In simple words: Build a cumulative frequency table by adding frequencies as you go down. Since there are 41 boys total, the middle boy is the 21st. Find which weight group contains the 21st cumulative count - that's 50 kg.

Exam Tip: The cumulative frequency shows you how many observations fall at or below each value - use this to locate the median position quickly.

 

Question 9. Find the median from the following marks and cumulative frequency data.
Answer: First, arrange the raw marks in ascending order: 15, 17, 20, 22, 25, 30

Frequency table:

Number of terms, N = 37
Since N is an odd number,

Median = \( \left \{ \left ( \frac{N+1}{2} \right) \text{th observation} \right \} = \left \{ \left ( \frac{37+1}{2} \right) \text{th observation} \right \} = \{19\text{th observation}\} = 22 \)

The median is 22.
In simple words: Build a frequency table, then add a cumulative frequency column. Find the 19th observation (since N = 37, position is (37+1) ÷ 2 = 19). The cumulative frequency table shows 19 falls in the marks = 22 row.

Exam Tip: When reading from a cumulative frequency table, find the first cumulative frequency value that is greater than or equal to your target position - that row gives you the median.

 

Question 10. Find the median height of students from the following data.
Answer: Arranging the terms in ascending order, we have:

Cumulative frequency table:

Number of terms, N = 50

Median = \( \frac{1}{2} \left \{ \left ( \frac{N}{2} \right ) \text{th observation} + \left( \frac{N}{2} + 1 \right ) \text{th observation} \right \} = \frac{1}{2} \{25\text{th observation} + 26\text{th observation}\} = \frac{1}{2} \{154 + 155\} \)

Median = 154.5
In simple words: Build a cumulative frequency table for the heights. With 50 students, find the 25th and 26th observations. Both are in the height range that shows cumulative frequency reaching 25 and 26, which corresponds to heights 154 and 155. Average them: (154 + 155) ÷ 2 = 154.5 cm.

Exam Tip: Always verify your cumulative frequency column adds up correctly and matches the total number of observations - mismatches indicate errors in your table.

 

Exercise 21C

 

Question 1. Find the mode of the given data: (i) 10, 8, 4, 7, 8, 11, 15, 8, 6, 8 and (ii) 27, 23, 39, 18, 27, 21, 27, 40, 36, 27
Answer: We need to find the mode of the given data.

Mode is that value of the variables that occurs most frequently.

(i) 10, 8, 4, 7, 8, 11, 15, 8, 6, 8

Looking at the data, 8 shows up most often. The mode is 8.

(ii) 27, 23, 39, 18, 27, 21, 27, 40, 36, 27

Looking at the data, 27 shows up most often. The mode is 27.
In simple words: Mode means the number that appears the most times. In part (i), 8 appears four times while all others appear once - so 8 is the mode. In part (ii), 27 appears four times - so 27 is the mode.

Exam Tip: Count the frequency of each value carefully - the mode is whichever appears most often, and you may have multiple modes if two values tie for highest frequency.

 

Question 2. Find the mode of the following data: Following are the ages (in years) of 11 cricket players: 28, 34, 32, 41, 36, 32, 32, 38, 32, 40, 31
Answer: The data shows the ages of 11 cricket players: 28, 34, 32, 41, 36, 32, 32, 38, 32, 40, 31

Mode is the value of the variable that occurs most frequently.

Looking at the ages, 32 shows up a total of 4 times, while all other ages show up only once.

Therefore, 32 is the mode of the ages.
In simple words: Count how many times each age appears. The age 32 comes up four times while each other age comes up only once. So 32 is the most frequent and is the mode.

Exam Tip: In a list of ages or measurements, quickly tally how often each value appears - the one with the highest count is your mode.

 

Question 3. Find the median, mean, and mode for the given frequency distribution of daily wages.
Answer:

Here, N = 45, which is odd.

Median = \( \left \{ \left( \frac{N+1}{2} \right) \text{th observation} \right \} = \left \{ \left( \frac{45+1}{2} \right) \text{th observation} \right \} = 23 \)th observation

Median = 150

Mean = \( \frac{\sum(f × x)}{\sum f} = \frac{7050}{45} = 156.67 \)

Mode = \( 3 \times \text{Median} - 2 \times \text{Mean} = 3(150) - 2(156.67) = 450 - 313.34 = 136.6 \)

The median is 150, the mean is 156.67, and the mode is 136.6.
In simple words: Organize the wages in the table. Count to find the 23rd observation for median (it's 150). Multiply each wage by its frequency, add these products, and divide by 45 for the mean (156.67). Use the relationship Mode = 3(Median) - 2(Mean) to calculate mode (136.6).

Exam Tip: When you have all three measures (mean, median, mode), verify they're reasonably close to each other - a mode far from the other two may signal an error in your calculation or table.

 

Question 4. Find the mean, median and mode of the data given in the frequency distribution table.

Answer: The total number of terms, \( N \), is 41, which represents an odd number. To find the median, we use the formula \( \left( \frac{N+1}{2} \right) \)th observation, which equals the 21st observation, giving a value of 22. The median is therefore 22. For the mean, we apply the formula \( \frac{\sum(f_i \cdot x_i)}{\sum f_i} \), which becomes \( \frac{899}{41} = 21.92 \). Using the empirical relationship Mode \( = 3(\text{Median}) - 2(\text{Mean}) \), we get Mode \( = 3(22) - 2(21.92) = 66 - 43.84 = 22.16 \). Therefore, the median is 22, the mean is 21.92, and the mode is 22.16.
In simple words: Add up all the marks times their frequencies and divide by the total number of students to find the mean. The median is the middle value when the data is arranged in order. The mode can be found using a special formula that connects all three measures together.

Exam Tip: Always arrange calculations in a table with columns for f, x, and f × x to avoid arithmetic errors. Remember the empirical formula Mode = 3(Median) - 2(Mean) only applies to certain types of distributions.

 

Question 5. Find the mean, median and mode of the data given in the frequency distribution table.

Answer: The total number of terms, \( N \), is 12, which represents an even number. To locate the median, we take the average of the \( \left( \frac{N}{2} \right) \)th and \( \left( \frac{N}{2} + 1 \right) \)th observations, which are the 6th and 7th observations. These correspond to the values 50 and 50 respectively. Taking their average: \( \frac{1}{2}(50 + 50) = 50 \). The median is therefore 50. For the mean, we calculate \( \frac{\sum(f_i \cdot x_i)}{\sum f_i} = \frac{612}{12} = 51 \). Using the empirical formula Mode \( = 3(\text{Median}) - 2(\text{Mean}) \), we obtain Mode \( = 3(50) - 2(51) = 150 - 102 = 48 \). Therefore, the median is 50, the mean is 51, and the mode is 48.
In simple words: When there is an even number of observations, the median is found by taking the average of the two middle values. Multiply each weight by the number of players at that weight, sum them all, and divide by the total number of players to get the mean.

Exam Tip: For an even number of data points, always identify both middle positions correctly using N/2 and (N/2 + 1). Double-check your cumulative frequency column to locate the correct observations quickly.