RS Aggarwal Solutions for Class 6 Chapter 21 Concept of Perimeter and Area

 

Question 1. Define the following terms:
(i) Observation
(ii) data
(iii) Frequency of an observation
(iv) Frequency distribution
Answer: (i) Observation refers to the active gathering of information from a primary source. (ii) A set of facts such as values and measurements is referred to as data. (iii) The number of times an observation takes place in a given set of data. (iv) A frequency distribution is a tabular arrangement showing how a variable distributes itself across its possible values. It presents a grouped collection of data organized into separate non-overlapping categories along with how many times each category happens.
In simple words: Observation is collecting information. Data is facts and numbers. Frequency is how many times something shows up. Frequency distribution organizes all this into a table showing each value and how often it appears.

 

Question 2. The final marks in mathematics of 30 students are as follows:
53, 61, 48, 60, 78, 68, 55, 100, 67, 95
75, 88, 77, 37, 84, 58, 60, 48, 62, 56
44, 58, 52, 64, 98, 59, 70, 39, 50, 60
(i) Arrange these marks in the ascending order. 30 to 39 one group, 40 to 49 second group, etc
(ii) What is the highest score?
(iii) What is the lowest score?
(iv) What is the range?
(v) If 40 is the pass mark how many have failed?
(vi) How many have scored 75 or more?
(vii) Which observations between 50 and 60 have not actually appeared?
(viii) How many have scored less than 50?
Answer:

(ii) From the given information we see that the highest score is 100. (iii) The information provided shows 37 as the lowest score. (iv) Range = highest score - lowest score = 100 - 37 = 63 (v) If 40 is the pass mark, learners who scored under 40 did not pass. So, the pupils who scored 37 and 39 did not succeed. Therefore number of pupils that did not pass in the test = 2 (vi) Pupils who scored 75, 77, 78, 84, 88, 90, 98 and 100 are the ones to score greater than 75. Therefore number of pupils who scored 75 or greater = 8 (vii) The observations 51, 54 and 57 have not shown up in the data set in the range 50 - 60. (viii) Pupils who scored 37, 39, 44, 48 and 48 are the ones to score under 50. Therefore number of pupils who received under 50 = 5
In simple words: The table organizes all scores into groups. The highest is 100, the lowest is 37. The difference (range) is 63. Two learners failed with scores below 40. Eight scored 75 or higher. The numbers 51, 54, and 57 never appear in the 50-60 range. Five students got below 50.

Exam Tip: Always organize data into clear groups before answering — it makes counting errors less likely and shows examiners you work systematically.

 

Question 3. The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9, 2.4
(i) Rearrange the weights in descending order.
(ii) Determine the highest weight.
(iii) Determine the lowest weight.
(iv) Determine the range.
(v) How many babies were born on that day?
(vi) How many babies weigh below 2.5 kg?
(vii) How many babies weigh more than 2.8 kg?
(viii) How many babies weigh 2.8 kg?
Answer: (i) Organizing the masses of the infants in decreasing order, we get: 3.1, 3.0, 2.9, 2.9, 2.8, 2.8, 2.7, 2.6, 2.5, 2.5, 2.4, 2.3, 2.2, 2.1, (ii) In a decreasing arrangement, the first value is always the greatest. Therefore, greatest mass = 3.1 kg. (iii) In a decreasing arrangement, the final value is always the smallest. Therefore, smallest mass = 2.1 kg (iv) Range = Greatest mass - smallest mass = 3.1 kg - 2.1 kg = 1.0 kg (v) We can calculate the number of newborns born on that specific day by tallying all the observations. Therefore, number of infants born on that day = 15. (vi) Infants weighing 2.1, 2.2, 2.3 and 2.4 kg are those weighing under 2.5 kg. (vii) Infants weighing 2.9, 2.9, 3.0 and 3.1 kg are those weighing greater than 2.8 kg. (viii) Number of infants weighing 2.8 kg = 2
In simple words: Put the weights from highest to lowest. The highest is 3.1 kg, the lowest is 2.1 kg. The difference is 1.0 kg. There were 15 babies born. Four weighed less than 2.5 kg, four weighed more than 2.8 kg, and two weighed exactly 2.8 kg.

Exam Tip: When finding "more than" or "less than," always exclude the boundary value unless the question says "or equal to."

 

Question 4. Following data gives the number of children in 40 families:
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 0, 4, 4, 3, 2
2, 0, 0, 1, 2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2
Represent it in the form of a frequency distribution.
Answer:

In simple words: Count how many families have 0 children, 1 child, 2 children, and so on. Use tally marks to track each count, then write the final frequency number for each group.

Exam Tip: Always ensure your tally mark groups sum to the total number of observations — a quick check prevents errors.

 

Question 5. Prepare a frequency table of the following scores obtained by 50 students in a test
42 51 21 42 37 37 42 49 38 52
7 33 17 44 39 7 14 27 39 42
42 62 37 39 67 51 53 53 59 41
29 38 27 31 54 19 53 51 22 61
42 39 59 47 33 34 16 37 57 43
Answer:

In simple words: Go through all 50 scores and tally each one. Then count the tally marks for every distinct score value and write the frequency. This shows how often each score appears.

Exam Tip: Count items twice to verify — once while making tally marks and once while adding up frequencies — so the total always reaches 50.

 

Question 6. A die was thrown 25 times and following scores were obtained
1 5 2 4 3
6 1 4 2 5
1 6 2 6 3
5 4 1 3 2
3 6 1 5 2
Answer:

In simple words: Count how many times each number (1 through 6) appears when the die is rolled 25 times. Record these counts in the frequency table.

Exam Tip: For die throws, always show all six possible values even if some appear zero times — it demonstrates complete understanding of the sample space.

 

Question 7. In a study of number of accidents per day, the observations for 30 days were obtained as follows:
6 3 5 6 4 3 2 5 4 2
4 2 1 2 2 0 5 4 6 1
6 0 5 3 6 1 5 5 2 6
Prepare a frequency distribution table
Answer:

In simple words: Count how many days had 0 accidents, 1 accident, 2 accidents, and so forth over the 30-day period. Organize this into a table with tally marks and frequencies.

Exam Tip: Always verify the total frequency adds to the total number of observations (30 in this case) — if not, recount the tally marks.

 

Question 8. Prepare a frequency table of the following ages (in years) of 30 students of class VIII in your school:
13, 14, 13, 12, 14, 13, 14, 15, 13, 14, 13, 14, 16, 12, 14
13, 14, 15, 16, 13, 14, 13, 12, 17, 13, 12, 13, 13, 13, 14
Answer:

In simple words: List all distinct ages of the 30 students. Use tally marks to count how many students are each age, then record the frequency for each age group.

Exam Tip: Organize ages in order from youngest to oldest for a clear presentation — examiners value systematic arrangement.

 

Question 9. Following figures relate the weekly wages (in Rs) of 15 workers in a factory:
300, 250, 200, 250, 200, 150, 350, 200, 250, 200, 150, 300, 150, 200, 250
(i) Prepare a frequency table
(ii) What is the range in wages (in Rs)?
(iii) How many Workers are getting Rs 350?
(iv) How many of workers are getting the minimum wages?
Answer:

(ii) Lowest wage = Rs. 150, Highest wage = Rs. 350. Therefore, Range = Maximum wage - Minimum wage = Rs. 350 - Rs. 150 = Rs. 200 (iii) Number of workers getting Rs. 350 = 1 worker (iv) Here, Lowest wages Rs. 150. Number of workers getting Rs. 150 = 3 workers. Therefore, number of workers getting lowest wages = 3 workers.
In simple words: The wage table shows five different amounts. The range from lowest to highest is Rs. 200. Only 1 worker earns Rs. 350. Three workers earn the minimum wage of Rs. 150.

Exam Tip: Range is always highest minus lowest — state both values clearly before subtracting to show complete reasoning.

 

Question 10. Construct a frequency distribution table for the following marks obtained by 25 students in a history test in class VI of a school:
9, 17, 12, 20, 9, 18, 25, 17, 19, 9, 12, 9, 18, 17, 19, 20, 25, 9, 12, 17, 19, 19, 20, 9
(i) What is the range of marks?
(ii) What is the highest mark?
(iii) Which mark is occurring more frequently?
Answer:

(i) Number of learners who have received marks equal to or more than 7 = Frequency of 7 + frequency of 8 + frequency of 9 = 5 + 4 + 3 = 12 (ii) Numbers of pupils who have received marks less than 4 = Frequency of 1 + frequency of 2 + frequency of 3 = 2 + 3 + 3 = 8
In simple words: The range is from 9 (lowest) to 25 (highest), giving a range of 16 marks. The highest mark is 25. The mark 9 appears most often with 6 learners, making it the most frequent value.

Exam Tip: The most frequently occurring value is called the mode — state it clearly when asked "which appears most often."

 

Question 11. Following is the choice of sweets of 30 students of class VI: Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Ladoo.
(i) Arrange the names of sweets in a table using tally marks
(ii) Which sweet is preferred by most of the students.
Answer:

(ii) The frequency of Ladoo is 12 i.e. maximum. Therefore, Ladoo is the sweet that is preferred by most of the pupils.
In simple words: Count how many times each sweet is mentioned. Ladoo appears 12 times, Rasgulla 9 times, Jalebi 6 times, and Barfi 3 times. Ladoo is clearly the favorite.

Exam Tip: When finding the mode (most preferred/frequent item), always show the frequency for each option — the one with the highest frequency is the answer.