Class 6

MCQ Questions for Class 6 Maths with Answers

Class 6 Maths MCQ Chapter Wise

Ganita Prakash Class 6 MCQ Questions

• Chapter 1 Patterns in Mathematics Class 6 MCQ
• Chapter 2 Lines and Angles Class 6 MCQ
• Chapter 3 Number Play Class 6 MCQ
• Chapter 4 Data Handling and Presentation Class 6 MCQ
• Chapter 5 Prime Time Class 6 MCQ
• Chapter 6 Perimeter and Area Class 6 MCQ
• Chapter 7 Fractions Class 6 MCQ
• Chapter 8 Playing with Constructions Class 6 MCQ
• Chapter 9 Symmetry Class 6 MCQ
• Chapter 10 The Other Side of Zero Class 6 MCQ

Also Read Ganita Prakash Class 6 Solutions

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 4 Data Handling and Presentation Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 4 Data Handling and Presentation Solutions

Ganita Prakash Class 6 Chapter 4 Solutions

Class 6 Maths Ganita Prakash Chapter 4 Solutions Data Handling and Presentation

Question 1.
Shri Nilesh is a teacher. He decided to bring sweets to the class to celebrate the new year. The sweets shop nearby has jalebi, gulab jamun, gujiya, barfi, and rasgulla. He wanted to know the choices of the children. He wrote the names of the sweets on the board and asked each child to tell him their preference. He put a tally mark ‘ | ’ for each student and when the count reached 5, he put a line through the previous four and marked it as

Complete the table to help Shri Nilesh to purchase the correct numbers of sweets.
(a) How many students chose jalebi? _______
(b) Barfi was chosen by _______ students?
(c) How many students chose gujiya? _______
(d) Rasgulla was chosen by _______ students?
(e) How many students chose gulab jamun? _______
Solution:

(a) It is clear from the table, Jalebi was chosen by 6 students.
(b) It is clear from the table, Barfi was chosen by 3 students.
(c) Gujiya was chosen by 13 students.
(d) Rasgulla was chosen by 7 students.
(e) Gulab jamun was chosen by 9 students.

Question 2.
Magan Bhai sells kites at Jamnagar. Six shopkeepers from nearby villages come to purchase kites from him. The number of kites he sold to these six shopkeepers are given below.

Shopkeeper	Number of Kites sold
Chaman	250
Rani	300
Rukhsana	100
Jasmeet	450
Jetha Lai	250
Poonam Ben	700

Prepare a pictograph using the symbol to represent 100 kites.
Answer the following questions:
(a) How many symbols represent the kites that Rani purchased?
(b) Who purchased the maximum number of kites?
(c) Who purchased more kites, Jasmeet or Chaman?
(d) Rukhsana says Poonam Ben purchased more than double the number of kites that Rani purchased. Is she correct? Why?
Solution:
Required pictograph is given below.

(a) Since Rani purchased 300 kites, it is clear from the pictograph, 3 symbols represent the kites that Rani purchased.

(b) The shopkeeper who purchased the maximum number of kites is the one with the most symbols in pictograph. It is clear from the pictograph that Poonam Ben purchased the maximum number of kites (7 × 100 = 700 kites).

(c) Number of kites purchased by Jasmeet = 450 Number of kites purchased by Chaman = 250 Jasmeet purchased more kites.

(d) Number of kites purchased by Poonam Ben = 700 Number of kites purchased by Rani = 300
Hence, Rukhsana is correct as 700 kites is more than double of 300 kites (300 × 2 = 600).

Question 3.
Samantha visited a tea garden and collected data of the insects and critters she saw there. Here is the data she collected —

Help her prepare a bar graph representing this data.
Solution:
To prepare a bar graph representing the given data, follow the point given below.
(i) Draw a horizontal line labeled, ‘Insects and Critters’ with each type of insect or critter evenly spaced.

(ii) Draw a vertical line labeled ‘Number of insects and critters seen’ with numbers starting from 0 up to the maximum number seen (in this case, 10) evenly spaced.

(iii) For each insect or critter, draw a bar that rises to the number seen. For example, the bar of caterpillars should reach up to 10.
Required bar graph is given alongside.

Question 4.
Chinu listed the various means of transport that passed across the road in front of his house from 9 a.m. to 10 a.m.

(a) Prepare a frequency distribution table for the data.
(b) Which means of transport was used the most?
(c) If you were there to collect this data, how could you do it? Write the steps or process.
Solution:
(a) Frequency distribution table for the given data is given below:

Means of transport	Bike	Car	Bus	Auto rickshaw	Bicycle	Bullock Cart	Scooter
Frequency	13	6	4	8	8	2	9

(b) The bike has the highest frequency in the table, indicating it was the most common means of transport observed.

(c) To collect this data, you could follow the steps given below. ;
Observation Timeframe: I will choose a specific timeframe, such as 9 a.m. to 10 a.m., to observe the road traffic.
Recording Data: 1 will use a tally chart or counting app to record the type of transport passing by during that hour.
Categorisation: Then I will organise the data into categories (e.g., bike, car, scooter, bus, etc.).
Final Count: After the observation period, I will get the total the number of occurrences for each category.
Analysis: Finally, I prepare a frequency distribution table based on the recorded data for analysis.

Question 5.
The number of girl students in each class of a school is depicted by the pictograph.

Observe this pictograph and answer the following questions.
(a) Which class has the least number of girl students?
(b) What is the difference between the number of girls in Class 5 and 6?
(c) If two more girls were admitted in Class 2, how would the graph change?
(d) How many girls are there in Class 7?
Solution:
(a) The pictograph shows the least number of symbols for Class 8.
Hence, Class 8 has the least number of girl students.

(b) Girl students in Class 5 = 2.5 × 4 = 10 Girl students in Class 6 = 4 × 4 = 16
Hence, there are 6 more girls in Class 6 than in Class 5.

(c) Here, adding 2 more girls would increase the count by 2 and requires a half additional symbol.
The pictograph would show half additional symbol for class 2, assuming each symbol represent 4 girls.

(d) Girl students in Class 7 = 3 × 4 = 12.

Data Handling and Presentation Class 6 Extra Questions

Data Handling and Presentation Class 6 Very Short Question Answer

Question 1.
Define the following terms:
(i) Observations
(ii) Data
Solution:
(i) Observations are the individual pieces of information we collect.
For example, if we ask five friends their ages, each friend’s age is one observation.

(ii) Any collection of facts, numbers, measures, observations or other descriptions of things that convey information about those things is called data.

Question 2.
Define the following terms:
(i) Tabulation of data
(ii) Tally marks
Solution:
(i) Tabulation of data is the process of systematically arranging data into rows and columns within a table to make it easier to understand, compare and analyse.

(ii) Tally marks are short vertical lines used to count items. Each time we count one item, we draw one line (|). After four lines (| | | |), we draw a oblique line (\) across them to show five.

Question 3.
Define the following terms:
(i) Raw data
(ii) Array
Solution:
(i) Raw data is the information as we first collect it (original form), before we sort or arrange it.
For example, writing ages on a sheet of paper as friends tell you, without putting them in any order.

(ii) Organising the data by arranging it in ascending or descending order is called an array.
For example, Meenal arranged the shoe sizes of the students in ascending order as follows:
3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7

Question 4.
Define the following terms:
(i) Frequency of observation
(ii) Frequency distribution
Solution:
(i) Frequency of observation tells us how many times a particular value appears in the data. For example, if the age ‘12’ appears three times in the list, then the frequency of age ‘12’ is 3.
(ii) Frequency distribution is a table that shows each value (or group of values) and how often it appears (frequency). For example:

Age	Frequency
10	2
11	1
12	3

Question 5.
A die was thrown 20 times and the following outcomes were noted:
1, 2, 5, 1, 6, 2, 3, 4, 2, 4, 2, 5, 6, 6, 6, 2, 3, 1, 1, 5
Represent the above data in the form of frequency distribution.
Solution:

Outcome	Frequency
1	4
2	5
3	2
4	2
5	3
6	4

Question 6.
Enlist 1 advantage and 1 disadvantage of presentation of data through pictograph.
Answer:
1 advantage and 1 disadvantage of pictograph are as follows:

Advantage	Disadvantage
1. Pictorial representation makes it easier to understand the data.	1. Drawing a pictograph is time consuming.

Data Handling and Presentation Class 6 Short Question Answer

Question 1.
Given below is the data showing the number of children in 15 families of a colony.
2, 3, 1, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 1, 3
Arrange the above data in ascending order and then form a tally table.
Solution:
Given data (number of children in 15 families) is as follows: 2, 3, 1, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 1, 3
(i) Data in ascending order: 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4.
(ii) Given data in tabular form is shown below:

Question 2.
The number of cricket bat and ball pairs, sold by a shopkeeper during a week are given below:

Day	Mon.	Tue.	Wed.	Thu.	Fri.	Sat.
Number of bat and ball pair sold	12	9	15	15	3	27

Decide the scaling and draw a pictograph to represent the given data.
Solution:

Question 3.
Total number of cows in five villages are as follows:

Village	A	B	C	D	E
Number of cows	40	60	20	100	120

Decide the scale and draw a pictograph to represent the given data.
Solution:

Question 4.
Following pictograph shows the data of the number of students who like a particular sport.

Answer the following questions:
(i) How many students like cricket?
(ii) How many more students like basketball than volleyball?
Solution:
(i) Number of symbols for cricket = 4
Students who like cricket = 4 × 3= 12

(ii) Number of symbols for basketball = 3
∴ Students who like basketball = 3 × 3 = 9
Number of symbols for volleyball = 2
∴ Students who like volleyball = 2 × 3 = 6
Difference = 9 – 6 = 3
Hence, 3 more students like basketball than volleyball.

Question 5.
The bar graph below shows the production of Kharif and Rabi crops (in tons) in 5 different states in India. Draw a table that represents the data in the bar graph.

Solution:
Table for the given bar graph can be created as follows:

States	Rabi crops (in tons)	Kharif crops (in tons)
West Bengal	30	60
Karnataka	45	40
Madhya Pradesh	20	54
Tamil Nadu	36	22
Uttar Pradesh	50	42

Data Handling and Presentation Class 6 Long Question Answer

Question 1.
The weights of newborn babies (in Kg) in a hospital on a particular day are as follows:
2.1, 2.5, 3.5, 2.7, 2.9, 3.1, 2.6, 2.5, 2.8, 2.3, 2.9, 3.3, 3.4, 2.7, 2.8
Answer the following questions:
(i) How many babies are born on that day?
(ii) Arrange the above data in descending order.
(iii) Determine the highest weight.
(iv) Determine the lowest weight.
(v) Determine the range of weight.
(vi) How many babies weigh below 2.5 kg?
Solution:
(i) There are 15 babies born on that day.

(ii) Writing the weights from highest to lowest:
3.5, 3.4, 3.3, 3.1, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.3, 2.1

(iii) The highest weight is the first weight in the descending order of weights.
Highest weight = 3.5 kg

(iv) The lowest weight is the last weight in the descending order of weights.
Lowest weight = 2.1 kg

(v) We know, range = highest value – lowest value
∴ Range of weight = highest weight – lowest weight
= 3.5 kg – 2.1 kg = 1.4 kg

(vi) The weights below 2.5 kg are 2.1 kg and 2.3 kg. So, the required number of babies is 2.

Question 2.
Study the following pictograph:

Answer the following questions:
(i) In which city was the rainfall maximum and how much?
(ii) In which city was the rainfall minimum and how much?
(iii) Which cities had rainfall of more than 50 cm?
Solution:
(i) Chennai has three full icons, which is maximum.
∴ Rainfall in Chennai = 3 × 50 cm = 150 cm

(ii) Delhi has one full icon, which is minimum.
∴ Rainfall in Delhi = 1 × 50 cm = 50 cm.

(iii) Mumbai has 2 full icons.
∴ Rainfall in Mumbai = 2 × 50 cm = 100 cm (which is greater than 50 cm)
Hyderabad has 1 full and 1 half icons.
∴ Rainfall in Hyderabad = 50 cm + 25 cm = 75 cm (which is greater than 50 cm) Chennai has 3 full icons.
∴ Rainfall in Chennai = 50 × 3 = 150 cm (which is greater than 50 cm)
Kolkata has 2 full and 1 half icons.
∴ Rainfall in Kolkata =100 cm + 25 cm =125 cm (which is greater than 50 cm)
Hence, Mumbai, Hyderabad, Chennai, and Kolkata had rainfall of more than 50 cm.

Question 3.
In a shop, there are 5 different types of toys. The number of each toy is given in the table below. Construct a bar graph representing this data. Use appropriate scaling.

Toys Name	Blocks	Cars	Dolls	Balls	Bikes
Number of Toys	40	48	60	35	30

Solution:
The required bar graph can be easily created using the following steps:
Step 1: On a graph sheet draw two mutually perpendicular lines, a horizontal and a vertical line.

Step 2: Label horizontal line as ‘Toys name’ & write names of toys from the table with equal gaps between them: Blocks, Cars, Dolls, Balls, Bikes. Label vertical line as ‘Number of Toys’.

Step 3: Scaling: Choose a suitable scale to show the number of toys.
Here, we take: 1 unit (1 big division) = 10 toys. Each big division is further divided into 10 sub-divisions, representing 1 toy.

Step 4: Calculate the height of each bar:
10 toys = 1 large divisions
∴ 1 toy = \(\frac{1}{10}\) large divisions
= 1 sub-division
Now, the height of the bar for Blocks
= 40 × (\(\frac{1}{10}\)) = 4 large divisions
The height of the bar for Cars
48 × (\(\frac{1}{10}\)) = 4 large divisions and 8 sub-divisions
The height of the bar for Dolls = 60 × (\(\frac{1}{10}\)) = 6 large divisions
The height of the bar for Balls = 35 × (\(\frac{1}{10}\)) = 3 large divisions and 5 sub-divisions
The height of the bar for Bikes = 30 × (\(\frac{1}{10}\))

Step 5: Now draw vertical bars for each toy using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 4.
The zookeeper in Delhi is preparing a presentation for higher officers on the number of animals in the zoo. He intends to create a bar graph for better visualisations. The number of each animal is given below in the table. Construct a bar graph for the table.

Animals	Number of Animals
Lion	8
Elephant	12
Gorilla	9
Zebra	15
Giraffe	7
Cheetah	6

Solution:
Step 1: Take a graph sheet and draw two mutually perpendicular lines, a horizontal line and a vertical line.

Step 2: Label the horizontal line as ‘Animals’ and write the names of the animals from the table with equal gaps between them: Lion, Elephant, Gorilla, Zebra, Giraffe, Cheetah Label the vertical lines ‘ Number of Animals’.

Step 3 (Scaling) : Choose a suitable scale to show the number of animals.
Here, we take: 1 unit (1 large division)
= 2 animals
∴ 1 animal = \(\frac{1}{2}\) unit = 1 sub-division
This scale helps us draw the graph clearly.

Step 4: Calculate the height of each bar using the scale:
Height of the bar for the Lion = 8 ÷ 2 = 4 large divisions
Height of the bar for the Elephant = 12 ÷ 2 = 6 large divisions
Height of the bar for the Gorilla = 9 ÷ 2 = 4 large divisions and 1 subdivision
Height of the bar for the Zebra = 15 ÷ 2 = 7 large divisions and 1 sub-division
Height of the bar for the Giraffe = 7 + 2 = 3 large divisions and 1 subdivision
Height of the bar for the Cheetah = 6 + 2 = 3 large divisions

Step 5: Now draw vertical bars for each animal using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 5.
The following bar graph shows the revenue of your country from exports of various items in 1 year. (1 unit = 10 crore rupees)

(a) Write the information given by the bar graph in a table.
(b) What is the difference between the maximum revenue and the minimum revenue?
(c) What is the total revenue from exports?
Solution:
(a) The required table is shown below:

Items	Revenue (Rupees in crores)
Electronics	30
Foodgrains	70
Livestock	25
Metals	50
Softwares	80

(b) From the table, it can be observed that the maximum revenue was generated by the export of software, i.e. 80 crore rupees.
Also, the minimum revenue was generated by the exports of livestock, i.e. 25 crore rupees.
∴ The difference between the revenue from software and livestock = 80 – 25 = 55 crore

(c) Total revenue from exports can be calculated by summing up the revenues from all the commodities.
∴ Total revenue from exports = 30 + 70 + 25 + 50 + 80 = 255 crore rupees

Question 6.
The table shows how much money Imran’s family spends each month on different items:

Items	Expenditure (in ₹)
House rent	3,000
Food	3,400
Education	800
Electricity	400
Transport	600
Miscellaneous	1,200

Answer the following questions:
(a) Represent the given data in the form of a bar graph
(b) On which item does the Imran’s family spend the most and the second most?
(c) Is the cost of electricity about one half of the cost of education?
Solution:
(a) To represent this data in the form of a bar graph, here are the steps:

Step 1: Draw two perpendicular lines, one horizontal and one vertical.
Step 2: Along the horizontal line, mark the ‘items’ with equal spacing between them and mark the corresponding expenditures along the vertical line.
Step 3: Take bars of the same width, keeping a uniform gap between them.
Step 4: Choose a suitable scale along the vertical line. Let, 1 unit length = ₹ 200, and then mark and write the corresponding values (₹ 200, ₹400, etc.) representing each unit length and calculate the heights of the bars for various items as shown below:

House rent	3000 ÷ 200	15 units
Food	3400 ÷ 200	17 units
Education	800 ÷ 200	4 units
Electricity	400 ÷ 200	2 units
Transport	600 ÷ 200	3 units
Miscellaneous	1200 ÷ 200	6 units

Here is the bar graph that we obtained based on the above steps.

(b) From the given data, we can say that on food Imran’s family spend the most and on house rent they spend the second most.

(c) Yes. As the cost of electricity is ₹ 400 and the cost of eduction is ₹ 800, we can say that the cost of electricity is about one-half the cost of education.

Question 7.
Your school is collecting data on various transportation options that students used to opt. The table below shows the data collected by you. Prepare a bar graph to be given to your school principal, also use appropriate scaling.

Transportation	Number of Students
Bus	200
Cycle	50
Walk	80
Auto	120
Car	40

Solution:
The required bar graph can be easily created using the following steps:

Step 1: Take a graph sheet and draw two mutually perpendicular lines, a horizontal line and a vertical line.

Step 2: Label the horizontal line as‘Transportation’ and write the names from the table with equal gaps between them: Bus, Cycle, Walk, Auto, Car.
Label the vertical line as ‘Number of Students’.

Step 3: (Scaling): Choose a suitable scale to show the number of students.
Here, we take:
1 unit (1 big division) = 20 students (This helps us fit large numbers like 200 easily on the graph.)

Step 4: Calculate the height of each bar:
= 200 × (\(\frac{1}{20}\)) = 10 large divisions
Height of the bar for the cycle
= 50 × (\(\frac{1}{20}\))= 2 large divisions and 5 sub-divisions
Height of the bar for the walk
= 80 × (\(\frac{1}{20}\)) = 4 large divisions
Height of the bar for the auto
= 120 × (\(\frac{1}{20}\)) = 6 large divisions
Height of the bar for the car
= 40 × (\(\frac{1}{20}\)) = 2 large divisions

Step 5: Now draw vertical bars for each mode of transportation using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 8.
The table below displays the number of bicycles produced at a factory from 1998 to 2002.

Year	1998	1999	2000	2001	2002
No. of Bicycles	800	600	900	1100	1200

Answer the following questions:
(a) Create a bar graph to represent this information. Select your preferred scale.
(b) Which year had the greatest number of bicycles produced?
(c) Which year had the least number of bicycles produced?
Solution:
(a) To represent this data in the form of a bar

Step 1: Draw two perpendicular lines, one horizontal and one vertical.
Step 2: Along the horizontal line, mark the ‘years’ with equal spacing between them and mark the corresponding ‘No. of bicycles’ along the vertical line.
Step 3: Take bars of the same width, keeping a uniform gap between them.
Step 4: Choose a suitable scale along the vertical line. Let, 1 unit length =100 bicycles, and then mark and write the corresponding values (100 bicycles, 200 bicycles, etc.) representing each unit length and calculate the heights of the bars for various years as shown below:

1998	800 ÷ 100	8 units
1999	600 ÷ 100	6 units
2000	900 ÷ 100	9 units
2001	1100 ÷ 100	11 units
2002	1200 ÷ 100	12 units

(b) The greatest number of bicycles produced in 2002.
(c) The least number of bicycles produced in 1999.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 3 Number Play Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 3 Number Play Solutions

Ganita Prakash Class 6 Chapter 3 Solutions

Class 6 Maths Ganita Prakash Chapter 3 Solutions Number Play

Question 1.
Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.

Solution:

5346

8643

1166

1258

1056

2012

8000

9635

9905

Question 2.
Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.

Solution:

999

102

909

110

918

210

928

350

958

Question 3.
Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Solution:
No, we cannot fill a supercell table without repeating numbers such that there are no supercells because at least one number will always be larger than its adjacent cell unless repetition is allowed.
For example,

3

6

9

12

15

18

21

20 or 22

Here, if the last number is greater than 21, it is a supercell and if less than 21, then 21 itself is a supercell number.

Question 4.
Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not?
Solution:
Yes, the cell having the largest number in a table will always be a supercell, because largest number will always be greater than its neighbouring numbers.
But, the cell having smallest number in a table can never be a supercell, because it will always be smaller than its neighbouring numbers.

Question 5.
Fill a table such that the cell having the second largest number is not a supercell but the second smallest number is a supercell. Is it possible?
Solution:
Yes, it is possible.

245

147

368

313

696

758

532

590

485

In this table, the second largest number is 696, it is not a supercell number and second smallest number is 245, which is a supercell number.

Question 6.
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution:
Smallest 5-digit palindrome = 10001
Largest 5-digit palindrome = 99999
Sum = 10001 + 99999 = 110000
Difference = 99999 – 10001 = 89998

Question 7.
Write an example for each of the below scenarios whenever possible.

Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates.
Solution:
(i) 45000 + 50000 = 95000 > 90250
(ii) 99999 + 900 = 100899
(iii) 4-digit 4- 4-digit to give a 6-digit sum. Since, the maximum sum for two 4-digit numbers (9999 + 9999) is 19998. Thus, it is not possible to get a 6 -digit number by adding two 4-digit numbers.
(iv) 50000 + 61000 = 111000
(v) The minimum sum of two 5-digit numbers (10000 + 10000) is 20000, which is greater than 18500. Therefore, it is not possible to get a sum of 18500 with two 5-digit numbers.
(vi) 70000 – 15000 = 55000 < 56503
(vii) 10000 – 999 = 9001
(viii) 12000 – 8000 = 4000
(ix) 20000 – 19500 = 500
(x) The difference between the largest 5-digit number and smallest 5-digit number (99999 – 10000) is 89999 which is less than 91500.
Therefore, it not possible to get the difference of 91500 between two 5-digit numbers.

Question 8.
There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells.

Figure out which digits to swap.
Solution:
Swap the digits 6 and 1 in the number 62,871. The number becomes 12,876 which will be the smallest among all the numbers.

16,200	39,344	29,765
23,609	12,876	45,306
19,381	50,319	38,408

Now, there are four supercells.

Question 9.
We are the group of 5-digit numbers between 35000 and 75000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000?
Solution:
The possible odd digits are: 1, 3, 5, 7 and 9.
Largest odd number is 73,999. (between 35000 and 75000)
Smallest odd number is 35,111. (between 35000 and 75000)
Closest odd number to 50,000 is 51,111.

Question 10.
Recall the sequence of powers of 2 from chapter 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?
Solution:
Powers of 2 → 2, 4, 8, 16,
For 2 (even) → 2 divide by 2 = 1
For 4 (even) → 4 divide by 2 = 2 (even) → 2 divide by 2 = 1
For 8 (even) → 8 divide by 2 = 4 (even) → 4 divide by 2 = 2 (even) → 2 divide by 2 = 1
Thus, for starting numbers that are powers of 2, the Collatz conjecture holds true because the sequence of operations simply involves a series of divisions by 2, which eventually leads to 1.

InText Questions

[Instruction for Q1 to Q7]
Some Children in a park are standing in a line. Each one says a number.

• A child says ‘1’ if there is only one taller child standing next to them.
• A child says ‘2’ if both the children standing next to them are taller.
• A child says ‘O’, if neither of the children standing next to them are taller.

That is each person says the number of taller neighbours they have.

Question 1.
Can the children rearrange themselves so that the children standing at the ends say ‘2’?
Solution:
No, the children cannot line up in a way that the ones at the ends say ‘2′ because a child only says ‘2’ when both of their neighbours are taller and children at the ends have only one neighbour, not two.

Question 2.
Can we arrange the children in a line so that all would say only 0s?
Solution:
No, it’s not possible to arrange the children in a line so that they all say only ‘0’, because a child says ‘0’ only when neither of their neighbours is taller. This would only happen if all the children are the same height.

Question 3.
Can two children standing next to each other say the same number?
Solution:
Yes, two children standing next to each other can say the same number i.e. 1 because there can be one taller and one smaller child standing next to them.

Question 4.
There are 5 children in a group, all of different heights. Can they stand such that four of them say 1 and the last one says ‘0’? Why or why not?
Solution:
Yes, because a child can say ‘0’ only if neither of the children standing next to them are taller. Arranging them in increasing or decreasing order of heights will eventually put tallest child at one of the end. So it is obvious that last child will say ‘0’.

Question 5.
For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
Solution:
No, the sequence 1, 1, 1, 1, 1 is not possible because there are 5 children and first four will say 1 only if they are arranged in increasing order of height. So, the last child will be tallest and will say 0 (not 1).

Question 6.
Is the sequence 0, 1,2, 1,0 possible? Why or why not?
Solution:
Yes, the sequence 0, 1, 2, 1, 0 is possible. It represents a situation where the middle child is the smallest (has two taller neighbours). The two children at the end are taller than the second last child from both the ends.

Question 7.
How would you rearrange the five children so that the maximum number of children say ‘2’?
Solution:
We can rearrange the five children so that the maximum number of children say ‘2’ as 0, 2, 0, 2, 0.
Thus, maximum number of children who can say ‘2’ is 2 . ,

Question 8.
Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours.

Once you have filled the table above, put commas appropriately after the thousands digit.
The biggest number in the table is _______.
The smallest even number in the table is ______.
The smallest number greater than 50,000 in the table is _______.
Solution:

96,310	96,301	36,109	36,190′
93,610.	13,609	60,319	19,306
93,106	10,639	60,193	30,196
10,369	10,963	10,396	31,906

The biggest number in the table is 96,310.
The smallest even number in the table is 10,396.
The smallest number greater than 50,000 in the table is 60,193.

Question 9.
Among the numbers 1-100, how many times will the digit 7’ occur? Among the numbers 1-1000, how many times will the digit ‘7’ occur?
Solution:
Among the numbers 1-100, we can divide the range into two parts: When ‘7’ appears in the ones place and when ‘7’ appears in the tens place.

In the ones place: The numbers that have ‘7’ in the ones place are: 7, 17, 27, 37, 47, 57, 67, 77, 87, and 97. So, there are 10 numbers where ‘7’ appears in the ones place.

In the tens place: The numbers 70 to 79 contain the digit ‘7’ in the tens place. So, there are 10 numbers where ‘7’ appears in the tens place.

Total occurrences of ‘7’ between 1 and 100:
Ones place: 10 times and Tens place: 10 times
Total = 10 + 10 = 20
So, digit ‘7’ appears 20 times among the numbers 1-100.

Among the numbers 1-1000, we can divide the range into three parts:
hundreds place, tens place and ones place
In the hundreds place: The numbers 700 to 799 contain the digit ‘7’ in the hundreds place. There are 100 numbers front 700 to 799.

In the tens place: We already know that in each set of 100 numbers (i.e. 0-99, 100-199, …, 900-999),
‘7’ will appear in the tens place 10 times (just like in the 1-100 range). Since we have 10 sets of 100 numbers, ‘7’ will appear in the tens place, 10 × 10 = 100 times.

In the ones place: Similarly, for each set of 100 numbers (i.e. 0-99, 100-199, …. 900-999), 7’ will appear in the ones place 10 times (like in the 1-100 range). So, across 10 sets of 100 numbers, 7’ will appear in the ones place 10 × 10 = 100 times.

Total occurrences of 7’ between 1 and 1000:
Hundreds place: 100 times, Tens place: 100 times, Ones place: 100 times
Total = 100 + 100 + 100 = 300
So, digit ‘7’ appears 300 times among the numbers 1-1000.

Question 10.
Write all possible 3-digit palindromes using 1, 2, 3.
Solution:
All the possible 3-digit palindromes using the digits ‘1’, ‘2’, and ‘3’ are: 111, 121, 131, 212, 222, 232, 313, 323 and 333.
There are 9 possible palindromes using the digits ‘1’, ‘2’, and ‘3’.

Question 11.
Puzzle time:

I am a 5-digit palindrome.
I am an odd number.
My ‘t’ digit is double of my ‘M’ digit.
My ‘h’ digit is double of my ‘t’ digit.
Who am I? _______
Solution:
Since palindrome is an odd number, u digit would be 1, 3, 5, 7 or 9 and tth digit would be same as u digit.
Since double of 5, 7 and 9 is not a digit, u digit would be either 1 or 3.
Double of 1 is 2 and double of 3 is 6. So, t digit would be either 2 or 6.
Since double of 6 is not a digit, u digit is 1 and t digit is 2.
Then, h digit would be 4.
So, the required 5-digit odd palindrome number is 12421.
In words: Twelve thousand four hundred twenty one

Number Play Class 6 Extra Questions

Number Play Class 6 Very Short Question Answer

Question 1.
What is the largest 4-digit number with non-zero digits such that its digits add up to 14?
Solution:
The largest 4-digit number must have largest digit ‘9’ at thousands place and the remaining digits in descending order such that the sum of all digits is 14.
Hence, the required number is 9311.

Question 2.
Write all 2-digit palindromic numbers.
Solution:
The 2-digit palindromic numbers are: I 1, 22, 33, 44, 55, 66, 77, 88 and 99.

Question 3.
Write all possible 3-digit palindromes using the digits 7, 8, 9.
Solution:
Using digits 7, 8 and 9, we get the following palindromes:
777, 787, 797, 878, 888, 898, 979, 989, 999

Question 4.
Write two 4-digit numbers such that their difference is a 2-digit number.
Solution:
Let us take 5500 and 5460 as two 4-digit numbers.
Clearly, 5500 – 5460 = 40 which is a 2-digit number.

Question 5.
Write two 5-digit numbers whose sum is 32500.
Solution:
Let us take 12500 and 20000 as two 5-digit numbers.
Clearly, 12500 + 20000 = 32500

Question 6.
Write two 4-digit numbers whose difference is 4680.
Solution:
Let us take 7500 and 2820 as two 4-digit numbers.
Clearly, 7500 – 2820 = 4680

Question 7.
Write the smallest and largest 4-digit palindromes. Find their sum and difference.
Solution:
Smallest 4-digit palindrome = 1001 and largest 4-digit palindrome = 9999
Required sum = 1001 + 9999 = 11000
Required difference = 9999 – 1001 = 8998

Question 8.
Write the second smallest and second largest 5-digit palindromes. Find their sum.
Solution:
Second smallest 5-digit palindrome = 10101
Second largest 5-digit palindrome = 99899
Required sum = 10101 + 99899 = 110000

Question 9.
Write a 6-digit number and a 4-digit number such that their difference is a 5-digit number.
Solution:
Let us take 103200 as a 6-digit number and 8200 as a 4-digit number.
Clearly, 103200 – 8200 = 95000 which is a 5-digit number.

Question 10.
Write a 6-digit number and a 5-digit number such that their difference is a 5-digit number.
Solution:
Let us take 120500 as a 6-digit number and 40500 as a 5-digit number.
Clearly, 120500 – 40500 = 80000, which is a 5-digit number.

Question 11.
Rearrange the digits of 48900125 to get the largest 8-digit number?
Solution:
In order to get the largest 8-digit number using the digits of 48900125, digits should be in descending order on moving from left to right.
Therefore, the required largest number is 98542100.

Question 12.
What is the smallest number whose digit sum is 16?
Solution:
There is no single digit number to give digit sum equal to 16.
The 2-digit numbers whose digits add up to 16 are: 79, 88 and 97.
We see that 79 is the smallest number among these numbers.
Therefore, 79 is the smallest number whose digit sum is 16.

Question 13.
Rearrange the digits of 4500731 to get the smallest 7-digit number?
Solution:
In order to get the smallest 7-digit number using the digits of 4500731, the first digit from left to right should be the smallest digit except 0. So, in this case, it is 1. Then, on moving from left to right further, digits should be in ascending order.
Therefore, the required smallest number is 1003457.

Question 14.
What is the largest 4-digit number whose digits add up to 15?
Solution:
The largest 4-digit number should have the largest digit at thousands place. Since 9 is the largest digit, it takes thousands place. Then, 6 takes the hundreds place because 9 + 6 = 15. As the sum is exhausted by thousands and hundreds place digits, rest places will take digit ‘0’.
Therefore, the required largest 4-digit number is 9600.

Number Play Class 6 Short Question Answer

Question 1.
Fill the given table such that the cell having the 2nd largest number is not a supercell but the cell having 2nd smallest number is a supercell. Use numbers between 10 and 100.

Solution:
To make sure the cell with the second largest number is not a supercell, it has to be next to the cell with the largest number. This is because the cell with the largest number is always a supercell, and two supercells cannot be next to each other.

To make the cell with the second smallest number a supercell, it should be one of the extreme end cells and the smallest number should be in the adjacent cell.

53

88

72

64

22

32

Here, 72 is the second largest number and 32 is the second smallest number
Activity: Think of another numbers by yourself.

Question 2.
Mark the supercells in the table below.

6828	670	9435	2180
3780	3708	7308	9225
8000	5583	52	5001

Solution:
6828 is a supercell because it is larger than its neighbours 670 and 3780.
9435 is a supercell because it is larger than its neighbours 670, 2180 and 7308.
9225 is a supercell because ii is larger than its neighbours 2180, 7308 and 5001.
8000 is a supercell because it is larger than its neighbours 3780 and 5583.

6828	670	9435	2180
3780	3708	7308	9225
8000	5583	52	5001

Question 3.
Fill in the table using only 3-digit numbers, making sure that the supercells line up exactly with the coloured cells.

Solution
Since 2^nd cell is a supercell, it must contain a number greater than 576. Let it he 832.

Since 3^rd and 5^th cells are not supercells and 4^th cell is a supercell, the numbers in 3^rd and 5^th cells should be smaller than the number in 4^th cell i.e., 188. Let them he 112 and 176 respectively.

576

832

112

188

176

912

The last cell is a supercell, and its neighbouring cell contains 912. So, the number in the last cell should be greater than 912. Let it be 950.

Since 7^th cell is not a supercell, it must contain number smaller than 912. Let it be 492.

Also, 6^th cell is not a supercell, it must contain a number smaller than the number present in 7^th cell, i.e. 492. Let it be 350.

Now, the complete table is as follows:

576

832

112

188

176

350

492

912

950

Activity: Think of another numbers by yourself.

Question 4.
Fill in the table using only 3-digit numbers, making sure that the supercells line up exactly with the coloured cells.

Solution:
Since 2^nd cell is a supercell, it must contain a number greater than 576. Let it be 832. Then, number in 1^st cell should be less than 432. Let it be 100.

100

432

120

312

731

512

Since 4^th cell is not supercell, it may Contain any 3-digit number less than 312. Let it be 230.

Also. 7^th cell is not supercell and second last cell is a supercell. it must contain number smaller than 731 and 512. I et it be 219.

The second last cell is a supercell, and it contains 512. So, the number in the last cell should be smaller than 512. Let it he 194.

Now, the complete table is as follows:

100

432

120

230

312

731

219

512

494

Question 5.
Calculate the digit sums of 2-digit numbers whose digits are consecutive. Do you observe a pattern?
Solution:
2-digit numbers having consecutive digits are: 12, 23, 34, 45, 56, 67, 78 and 89 .
The sums of digits of these 8 numbers arc:
1 + 2 = 3
2 + 3 = 5
3 + 4 = 7
4 + 5 = 9
5 + 6 = 11
6 + 7 = 13
7 + 8 = 15
8 + 9 = 17
We find that each sum is 2 more than the preceding sum.

Question 6.
What is the largest 5-digit number with non-zero digits whose digits add up to 17?
Solution:
Non-zero digits are 1, 2, 3, 4, 5, 6, 7, 8, 9.
The largest 5-digit number must have largest digit ‘9’ at ten thousands place (tth) and the remaining digits in descending order such that the sum of all digits is 17.

tth

th

h

t

u

Question 7.
Calculate the digit sums of 4-digit numbers whose digits are consecutive. Do you observe any pattern?
Solution:
The 4-digit numbers having consecutive digits are 1234,2345, 3456, 4567, 5678 and 6789.
The sum of digits of these 6 numbers are 10, 14, 18, 22, 26 and 30 respectively.
We find that each sum is 4 more than the preceding sum.

Question 8.
What is the smallest 5-digit number whose digits add up to 18?
Solution:
In order to write smallest 5-digit number, ten thousands place must be occupied by digit ‘1’. Then, sum of the rest of the digits should be 18 – 1 = 17.

The smallest 5-digit number should have the largest digit at units place. Since 9 is the largest digit, it takes units place. Then, 8 takes the tens place because 9 + 8 = 17. As the sum is exhausted by ten thousands, tens and units place digits, rest places will take digit ‘0’.

Therefore, the required smallest 5-digit number is 10089.

Question 9.
Convert the following times from 12-hour format to 24-hour format.
(i) 02:00 PM
(ii) 10:00 AM
(iii) 07:30 PM
(iv) 12:20 AM
(v) 12:15 PM
Solution:

	12-hour Format	24-hour Format
(i)	02:00 PM	14:00 hours
(ii)	10:00 AM	10:00 hours
(iii)	07:30 PM	19:30 hours
(iv)	12:20 AM	00:20 hours
(v)	12:15 PM	12:15 hours

Question 10.
What can be the previous number in the Collatz sequence to 40?
Solution:
Let x be the required number.
If x is an even number, then
\(\frac{x}{2}\) = 40 ⇒ x = 2 × 40 ⇒ x = 80
If x is an odd number, then
3x + 1 = 40 ⇒ 3x = 40 – 1
⇒ 3x = 39 ⇒ x = \(\frac{39}{3}\) = 13
Thus, either 80 or 13 can be the number preceding 40 in a Collatz sequence.

Question 11.
Observe the following numbers written in a pattern. Find their sum.

Solution:
We find that there are 3 rows each consisting of 5 boxes written with 30 and 2 rows each consisting of 6 boxes written with 70.
∴ Total sum = 3(5 × 30) + 2(6 × 70)
= 3 × 150 + 2 × 420
= 450 + 840 = 1290

Question 12.
Apply Kaprekar’s routine on the number 3562 to get Kaprekar constant.
Solution:
The number is 3562.
Round 1: A = Largest number using digits of the number 3562 = 6532
B = Smallest number using digits of the number 3562 = 2356
C = A – B = 6532 – 2356 = 4176

Round 2: A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174

Question 13.
Apply Kaprekar’s routine on the number 8825 to get Kaprekar constant.
Solution:
The number is 8825.
Round 1:
A = Largest number using digits of the number 8825 = 8852
B = Smallest number using digits of the number 8825 = 2588
C = A – B = 8852 – 2588 = 6264

Round 2:
A = Largest number using digits of the number 6264 = 6642
R = Smallest number using digits of the number 6264 = 2466
C = A – B = 6642 – 2466 = 4176

Round 3:
A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174

Question 14.
Convert the following times from 24-hour format to 12-hour format.
(i) 18:00 hours
(ii) 06:00 hours
(iii) 12:30 hours
(iv) 00:45 hours
(v) 23:15 hours
Solution:

	24-hour Format	12-hour Format
(i)	18:00 hours	06:00 PM
(ii)	06:00 hours	06:00 AM
(iii)	12:30 hours	12:30 PM
(iv)	00:45 hours	12:45 AM
(v)	23:15 hours	11:15 PM

Question 15.
Count all the suns in the following pattern.

Solution:
There are 26 boxes with 1 sun each and 26 boxes with 4 suns each.
∴ Total number of suns = (26 × 1) + (26 × 4)
= 26 + 104 = 130

Question 16.
Find the sum of the numbers in the number pattern shown in the figure.

Solution:
We note that the number 75 occurs 20 times around the outer edge, the number 40 appears 12 times along the second layer, the number 50 is present 4 times on the edge of the inner layer, and the number 1500 is located at the centre.
∴ Required sum
= (20 × 75) + (12 × 40) + (4 × 50) + 1500
= 1500 + 480 + 200 + 1500 = 3680

Question 17.
Consider the numbers in the following boxes:

Using operations addition/subtraction on these numbers, we can get
13100 = 32000 – 20000 + 700 + 200 + 200
Similarly, obtain the following numbers by using the numbers in the boxes.
(i) 25800
(ii) 28000
(iii) 7500
(iv) 14400
(v) 53000
Solution:
(i) 25800 = 20000 + 6000 – 200
(ii) 28000 = 32000 – 3500 – 700 + 200
(iii) 7500 = 3500 + 3500 + 700 – 200
(iv) 14400 = 20000 – 6000 + 200 + 200
(v) 53000 = 32000 + 20000 – 6000 + 3500 + 3500

Question 18.
Consider the numbers in the following boxes:

Using operations addition/subtraction on these numbers, we can get
23300 = 16000 + 16000 – 9000 + 300
Similarly, obtain the following numbers by using the numbers in the boxes.
(i) 60200
(ii) 2500
(iii) 7700
(iv) 64600
(v) 17300
Solution:
(i) 60200 = 45000 4 16000-800
(ii) 2500 = 9000 + 9000 – 16000 + 800 – 300
(iii) 7700 = 9000 – 800 – 800 + 300
(iv) 64600 = 45000 + 9000 + 9000 + 800 + 800
(v) 17300 = 16000 + 800 + 800 – 300

Number Play Class 6 Long Question Answer

Question 1.
Colour or mark the supercells in the given table:

6235

970

8145

4780

4708

7084

9000

160

Solution:
Clearly, 6235 is greater than the number 970 in the neighbouring cell.
Hence, the cell containing 6235 is the supercell.
970 is smaller than 6235 and 8145, so the cell containing 970 is not a supercell.
As 8145 is greater than the numbers 970 and 4780 in the neighbouring cells, the cell containing 8145 is a supercell.
As 4780 is smaller than 8145, the cell containing 4780 is not a supercell.
As 4708 is smaller than 4780 and 7084, the cell containing 4708 is not a supercell.
As 7084 is smaller than 9000, the cell containing 7084 is not a supercell.
As 9000 is greater than the numbers 7084 and 160 in the neighbouring cells, the cell containing 9000 is a supercell.
As 160 is smaller than 9000, the cell containing 160 is not a supercell.

6235

970

8145

4780

4708

7084

9000

160

Question 2.
Mark supercells in the following grid:

3462	2198	6757	5678
1001	5982	4723	2345
2723	5600	7210	2316
3298	1465	3467	4621

Solution:
The cell containing 3462 is a supercell as the numbers 2198 and 1001 in neighbouring cells are smaller than 3462.
The cell containing 6757 is a supercell as the numbers 2198, 4723 and 5678 in neighbouring cells are smaller than 6757.
The cell containing 5982 is a supercell as the numbers 2198, 4723, 5600 and 1001 in neighbouring cells are smaller than 5982.
The cell containing 7210 is a supercell as the numbers 4723, 2316, 3467 and 5600 in neighbouring cells are smaller than 7210.
The cell containing 3298 is a supercell as the numbers 2723 and 1465 in neighbouring cells are smaller than 3298.
The cell containing 4621 is a supercell as the numbers 2316 and 3467 in neighbouring cells are smaller than 4621.

3462	2198	6757	5678
1001	5982	4723	2345
2723	5600	7210	2316
3298	1465	3467	4621

Question 3.
Identify the numbers marked on the number lines below, and label the remaining positions.

Put a circle around the smallest number and a box around the largest number in each of the sequences above.
Solution:
(i) There are 3 sub-divisions between 729 and 732.
Therefore, each sub-division represents \(\frac{732-729}{3}\) = \(\frac{3}{3}\) = 1 number.

(ii) There are 7 sub-divisions between 6236 and 6250.
Therefore, each sub-division represents \(\frac{6250-6250}{7}\) = \(\frac{14}{7}\) = 2 numbers.

(iii) There are 6 sub-divisions between 12100 and 12160.
Therefore, each sub-division represents \(\frac{12100-12160}{6}\) = \(\frac{60}{6}\) = 10 numbers.

(iv) There is 1 sub-division between 53312 and 54312.
Therefore, each sub-division represents \(\frac{53312-54312}{1}\) = 1000 numbers

Question 4.
Among the numbers 100 – 1000, how many times will the digit ‘3’ occur?
Solution:
There are 901 numbers from 100 to 1000.

When digit ‘3’ is at hundreds place:
Numbers with ‘3’ in the hundreds place look like: 300 – 399
So, total count of digit ‘3’ at hundreds place = 100

When digit ‘3’ is at tens place:
To count how often ‘3’ appears in the tens place, fix the hundreds and units digit, and loop through all possibilities:
There are:

• 9 choices for hundreds digit (1 to 9)
• 1 choice for tens digit (3)
• 10 choices for units digit (0 to 9)

So, total count of digit ‘3’ at tens place = 9 × 1 × 10 = 90

When digit ‘3’ is at units place:
To count how often ‘3’ appears in the units place, fix the hundreds and tens digit, and loop through all possibilities:
There are:

• 9 choices for hundreds digit (1 to 9)
• 10 choices for tens digit (0 to 9)
• 1 choice for units digit (3)

So, total count of digit ‘3’ at units place = 9 × 1 × 10 = 90
Total count of digit ‘3’ = 100 + 90 + 90 = 280
Thus, digit ‘3’ occurs 280 times.

Question 5.
How big a number can you form having the digit sum 16? Can you make an even bigger number?
Solution:
2-digit numbers having the digit sums 16 are: 79, 88 and 97.
3-digit numbers having the digit sums 16 are: 169, 178, 187, 196, 259, 268, 277, 286, 295, 349, 358, 367, 376, 385, 394, 439, 448, 457, 466, 475, 484, 493, 529, 538, 547, 556, 565, 574, 583, 592, 619, 628, 637, 646, 655, 664, 673, 682, 691, 709, 718, 727, 736, 745, 754, 763, 772, 781, 790, 808, 817, 826, 835, 844, 853, 862, 871, 880, 907, 916, 925, 934, 943, 952, 961 and 970.
To form a 4-digit number having the digit sum 16, put one 0 in any of the above 3-digit numbers anywhere after first digit from left or two 0 in any of the above 2-digit numbers anywhere after first digit from left.

Similarly, we can form w-digit numbers by putting an appropriate number of 0 in any of the above numbers.

Thus, there will be infinitely many numbers with digit sum 16.

Yes, we can make even much bigger numbers.

Question 6.
Solve the puzzle:
My father’s salary is a 6-digit palindrome.
It is an even number.
Its tens digit is double of units digit and hundreds digit is double of tens digit.
What is my father’s salary?
Solution:
Units digit of an even number is 0 or 2 or 4 or 6 or 8. Since it is a palindrome, 0 cannot be the units digit. It is given that tens digit is double of units digit. So, 6 and 8 cannot be the units digit as 6 × 2 = 12 and 8 × 2 = 16 but 12 and 16 are not digits. If 4 is the units digit, then 8 would be the tens digit. But it is given that hundreds digit is double of tens digit, then hundreds digit would be 16, which is not possible.

So, units digit must be 2 only. Then, tens digit would be 4 and hundreds digit would be 8.

2

4

8

8

4

2

Thus, the palindromic number is 248842.
Hence, father’s salary is 248842.

Question 7.
List all 3-digit palindromic numbers and highlight those whose digit sum is greater than 15.
Solution:
3-digit palindromic numbers are given in the below table:

101	111	121	131	141	151	161	171	181	191
202	212	222	232	242	252	262	272	282	292
303	313	323	333	343	353	363	373	383	393
404	414	424	434	444	454	464	474	484	494
505	515	525	535	545	555	565	575	585	595
606	616	626	636	646	656	666	676	686	696
707	717	727	737	747	757	767	777	787	797
808	818	828	838	848	858	868	878	888	898
909	919	929	939	949	959	969	979	989	999

Highlighted (bold) palindromes have their digit sum greater than 15.

Question 8.
Write a 5-digit number, a 4-digit number and a 3-digit number such that their sum is 24680.
Solution:
Let us take 20000 as a 5-digit number. Then,
20000 + Sum of a 4-digit number and a 3-digit number = 24680
⇒ Sum of a 4-digit number and a 3-digit number
= 24680 – 20000 = 4680
Let us take 4500 as a 4-digit number. Then,
4500 + a 3-digit number = 4680
⇒ 3-digit number = 4680 – 4500 = 180
Hence, 20000 is a 5-digit number, 4500 is a 4-digit number and 180 is a 3-digit number such that their sum is 24680.
There can also be other numbers like
19000 + 5100 + 580 = 24680,
17500 + 7000 4 180 = 24680 etc.
Think of another numbers by yourself.

Question 9.
State whether the following statements are True/ False.
(i) The sum of two 5-digit numbers is always a 5-digit number.
(ii) The sum of a 5-digit number and a 3-digit number is always a 5-digit number.
(iii) 4’be sum of a 5-digit number and a 3-digit number is always a 6-digit number.
(iv) The difference of two 4-digit numbers is always a 4-digit number.
(ii) The difference between a 5-digit number and a 3-digit number is always a 4-digit number.
Solution:
(i) False; 26000 + 42000 = 68000 → a 5-digit number
76000 + 84000 = 160000 + not → a 5-digit number
Thus, the sum of two 5-digit numbers may or may not be a 5-digit number.

(ii) False; 10000 + 100 = 10100 → a 5-digit number
99900 + 900 = 100800 → not a 5-digit number
Thus, the sum of a 5-digit number and a 3-digit number may or may not be a 5-digit number.

(iii) False; 10000 + 100 = 10100 → not a 6-digit number
99900 + 900 = 100800 → a 6-digit number
Thus, the sum of a 5-digit number and a 3-digit number may or may not be a 6-digit number.

(iv) False; 8040 – 6080 = 1960 → a 4-digit number
1980 – 1200 = 780 → not a 4-digit number
Thus, the difference of two 4-digit numbers may or may not be a 4-digit number.

(v) False; 10250 – 750 = 9500 → a 4-digit number
15500 – 900 = 14600 → not a 4-digit number
Thus, the difference between a 5-digit number and a 3-digit number may or may not be a 4-digit number.

Question 10.
Apply Kaprekar’s routine on a 3-digit number. What number will start repeating?
Solution:
Let the number be 729.
Round 1:
A = Largest number using digits of the number 729 = 972
B = Smallest number using digits of the number 729 = 279
C = A – B = 972 – 279 = 693

Round 2:
A = Largest number using digits of the number 693 = 963
B = Smallest number using digits of the number 693 = 369
C = A – B = 963 – 369 = 594

Round 3:
A = Largest number using digits of the number 594 = 954
B = Smallest number using digits of the number 594 = 459
C = A – B = 954 – 459 = 495

Round 4:
A = Largest number using digits of the number 495 = 954
B = Smallest number using digits of the number 495 = 459
C = A – B = 954 – 459 = 495
Clearly, 495 repeats after 3 iterations.

Question 11.
How many rounds does the number 7443 take to reach the Kaprekar constant.
Solution:
The number is 7443.
Round 1:
A = Largest number using digits of the number 7443 = 7443
B = Smallest number using digits of the number 7443 = 3447
C = A – B = 7443 – 3447 = 3996

Round 2:
A = Largest number using digits of the number 3996 = 9963
B = Smallest number using digits of the number 3996 = 3699
C = A – B = 9963 – 3699 = 6264

Round 3:
A = Largest number using digits of the number 6264 = 6642
B = Smallest number using digits of the number 6264 = 2466
C = A – B = 6642 – 2466 = 4176

Round 4:
A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174, which is the Kaprekar constant
Thus, we reach at the Kaprekar constant in four rounds.

Question 12.
Make the Collatz sequence by starting with number 24.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till 1 is obtained.
First term: 24
Second term: \(\frac{24}{2}\) = 12 [As 24 is an even number.]
Third term: \(\frac{12}{2}\) = 6 [As 12 is an even number.] 0
Fourth term: \(\frac{6}{2}\) = 3 [As 6 is an even number.]
Fifth term: 3 × 3 + 1 = 10 [As 3 is an odd number.]
Sixth term: \(\frac{10}{2}\) = 5 [As 10 is an even number.]
Seventh term: 3 × 5 + 1 = 16 [As 5 is an odd number.]
Eighth term: \(\frac{16}{2}\) = 8 [As 16 is an even number.]
Ninth term: \(\frac{8}{2}\) = 4 [As 8 is an even number.]
Tenth term: \(\frac{4}{2}\) = 2 [As 4 is an even number.]
Eleventh term: \(\frac{2}{2}\) = 1 [As 2 is an even number.]
Hence, the Collatz sequence starting with 24 is:
24, 12, 6, 3, 10, 5, 16, 8, 4, 2, 1

Question 13.
Make the Collatz sequence by starting with number 17.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till I is obtained.
1^st term : 17
2^nd term : 3 × 17 + 1 = 52 [As 17 is an odd number]
3^rd term : \(\frac{52}{2}\) = 26 [As 52 is an even number]
4^th term : \(\frac{26}{2}\) = 13 [As 26 is an even number]
5^th term : 3 × 13 + 1 = 40 [As 13 is an odd number]
6^th term : \(\frac{40}{2}\) = 20 [As 40 is an even number]
7^th term : \(\frac{20}{2}\) = 10 [As 20 is an even number]
8^th term : \(\frac{10}{2}\) = 5 [As 10 is an even number]
9^th term : 3 × 5 + 1 = 16 [As 5 is an odd number]
10^th term : \(\frac{16}{2}\) = 8 [As 16 is an even number]
11^th term : \(\frac{8}{2}\) = 4 [As 8 is an even number]
12^th term : \(\frac{4}{2}\) = 2 [As 4 is an even number]
13^th term : \(\frac{2}{2}\) = 1 [As 2 is an even number]
Hence, the Collatz sequency starting with 17 is: 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

Question 14.
Find the Collatz sequence by starting with number 11.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till 1 is obtained.
1^st term : 11
2^nd term : 3 × 11 + 1 = 34 [As 11 is an odd number]
3^rd term : \(\frac{34}{2}\) = 17 [As 34 is an even number]
4^th term : 3 × 17 + 1 = 52 [As 1 7 is an odd number]
5^th term : \(\frac{52}{2}\) = 26 [As 52 is an even number]
6^th term : \(\frac{26}{2}\) = 13 [As 26 is an even number]
7^th term : 3 × 13 + 1 = 40 [As 13 is an odd number]
8^th term : \(\frac{40}{2}\) = 20 [As 40 is an even number]
9^th term : \(\frac{20}{2}\) = 10 [As 20 is an even number]
10^th term : \(\frac{10}{2}\) = 5 [As 10 is an even number]
11^th term : 3 × 5 + 1 = 16 [As 5 is an odd number]
12^th term : \(\frac{16}{2}\) = 8 [As 16 is an even number]
13^th term : \(\frac{8}{2}\) = 4 [As 8 is an even number]
14^th term : \(\frac{4}{2}\) = 2 [As 4 is an even number]
15^th term : \(\frac{2}{2}\) = 1 [As 2 is an even number]

Question 15.
Make a quick estimate. Take about 30 seconds. Then compare your answer with your friends.
(i) Number of tiles on your classroom floor:
(a) More than 200
(b) Less than 200
(Hint: Count how many tiles there are*in one row and one column.)

(ii) Time taken to write your full name:
(a) More than 10 seconds
(b) Less than 10 seconds

(iii) Estimate the number of books in your school library.
(a) More than 2000
(b) Less than 2000

(iv) Weight of a fully packed school bag:
(a) More than 4 kg
(b) Less than 4 kg

(v) Estimate the number of leaves on a big tree near your school.
(a) Around 1000
(b) Around 10,000
(c) More than 50,000
Solution:
(i) Estimate the number of tiles in one row and one column. Multiply the two to get the total. If it is 20 rows and 15 columns, then 20 × 15 = 300 tiles.
(ii) Say your name aloud and imagine writing each letter. Most students take 5-12 seconds depending on length.
(iii) Think of how many shelves there are and how many books per shelf. For example, 50 shelves × 50 books = 2500 books.
(iv) Lift your bag and guess based on how heavy it feels. Most packed school bags weigh around 3-5 kg.
(v) A large tree has thousands of branches and leaves. It’s very hard to count, but an estimate can be made.

Number Play Class 6 Case Based Questions

Question 1.
During a school trip, Class 6 students noticed a unique design featuring a 9-box row on a wall. Inspired, they decided to fill the boxes with numbers between 300 and 700, using each number only once.
They were excited to experiment with different number arrangements and see what interesting patterns might emerge.

Based on the above information, answer the following:
(ii) Fill the table such that we get maximum number of supercells.
(ii) How many supercells are there in the table.
Solution:
(i) If there are n cells in a row, then maximum number of supercells = \(\left\{\begin{array}{c}
\frac{n}{2}, \text { if } n \text { is even } \\
\frac{n+1}{2}, \text { if } n \text { is odd }
\end{array}\right.\)
Since total number of cells is 9, maximum 9 +1
number of supercell = \(\frac{9+1}{2}\) = 5
We know that two adjacent cells can never be supercells. So, the maximum number of supercells would be obtained when every alternate cell is a supercell.

In order to get the maximum number of supercells, always consider the 1st cell to be the supercell.
One such table with numbers between 300 and 700 is as follows:

699

301

698

302

697

303

696

304

695

(ii) Clearly, there are 5 supercells.

Question 2.
One day, Raghu bought a packet of pencils and found a lottery scratch coupon tucked inside. Curious, he scratched it off and revealed six two-digit numbers. The first number was 34, and to his surprise, the rest of the numbers followed the mysterious pattern of the Collatz sequence.

Based on the above information, answer the following questions:
(i) What number is written in second box?
(ii) What number is written in last box?
(iii) How many boxes are supercells?
Solution:
(i) As the numbers on the lottery ticket follow the Collatz sequence, and the first number is 34, which is even, the number in second box = \(\frac{34}{2}\) = 17

(ii) In the given Collatz sequence
First term = 34
Second term = \(\frac{34}{2}\) = 17 [As 34 is an even number]
Third term = 3 × 17 + 1 = 52 [As 17 is an odd number]
Fourth term = \(\frac{52}{2}\) = 26 [As 52 is an even number]
Fifth term = \(\frac{26}{2}\) = 13 [As 26 is an even number]
Sixth term = 3 × 13 + 1 = 40 [As 13 is an odd number]
Thus, the number in the last i.e. sixth box is 40.
The list of numbers obtained on the lottery ticket is given below.

(iii) The list of numbers obtained on the lottery ticket is given below.

34

17

52

26

13

40

Clearly, 3 boxes are supercells.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 2 Lines and Angles Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 2 Lines and Angles Solutions

Ganita Prakash Class 6 Chapter 2 Solutions

Class 6 Maths Ganita Prakash Chapter 2 Solutions Lines and Angles

Question 1.
Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle.

Solution:
Yes,

Question 2.
Mark any four points on your paper so that no three of them are on one line. Label them A, B, C and D. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C and D? Write them all down and mark each of them with a curve.
Solution:
A, B, C and D are four points.

The six possible lines are \(\overleftrightarrow{A B}\), \(\overleftrightarrow{B C}\), \(\overleftrightarrow{C D}\), \(\overleftrightarrow{A D}\), \(\overleftrightarrow{A C}\) and \(\overleftrightarrow{B D}\).
And, 12 possible angles arc ∠BAC, ∠CAD, ∠BAD, ∠ADB, ∠BDC, ∠ADC, ∠DCA. ∠ACB, ∠DCB, ∠ABD, ∠DBC, and ∠ABC.

Question 3.
Find out the number of acute angles in each of the figures below.

What will be the next figure and how many acute angles will it have? Do you notice any pattern in the numbers?
Solution:
First figure: There are three acute angles in the first figure.
Second figure: There are 12 acute angles in second figure (each of the 4 smaller triangles has 3 acute angles).
Third figure: There are 21 acute angles in the third figure (each of the 7 smaller triangles has 3 acute angles).

The next figure is given alongside. And, it has 30 acute angles.
The number of acute angles: 3, 12, 21, 30, …
Pattern: number of acute angles increase by 9 in each step.

Question 4.
Angles in a clock
(a) The hands of a clock make different angles at different times. At 1 o’clock, the angle between the hands is 30°. Why?
(b) What will be the angle at 2 o’clock? And at 4 o’clock? 6 o’clock?
(c) Explore other angles made by the hands of a clock.

Solution:
(a) The clock is divided into 12 hours, so each hour mark is 30° apart, (\(\frac{360^{\circ}}{12}\) = 30°)
Therefore, at 1 o’clock the hour hand is at 1 and the minute hand is at 12, forming a 30° angle.

(b) At 2 o’clock, it is 60° (i.e. 30° × 2 = 60°),
At 4 o’clock, it is 120° (i.e. 30° × 4 = 120°)
At 6 o’clock, it is 180° (i.e. 30° × 6 = 180°)

(c) The angle increases by 30° for each hour. Other angle includes 90° at 3 o’clock, 150° at 5 o’clock and so on. ‘ Thus, on multiplying the hour by 30°, we can find the angle at any hour.

Question 5.
Vidya is enjoying her time on the swing. She notices that the greater the angle with which she starts the swinging, the greater is the speed she achieves on her swing. But where is the angle? Are you able to see any angle?

Solution:
Yes, the angle is visible, and it is formed between the rope and the tree branch.

Question 6.
Observe the images below where there is an insect and its rotated version. Can angles be used to describe the amount of rotation?

How? What will be the arms of the angle and the vertex?
Solution:
Yes, angles can he used to show the amount of rotation.

One arm of the angle is the horizontal line that the insects are on, and the other arm is the line that meets it at a corner point. The point where the two lines meet is the vertex of the angle.

Question 7.
In this figure, ∠TER = 80°. What is the measure of ∠BET? What is the measure of ∠SET?

Solution:
Given, ∠TER = 80° and ∠SER = 90°
Now, ∠SET = ∠SER – ∠TER = 90° – 80° = 10°; ∠BET = ∠BES + ∠SET = 90° + 10° = 100°

Question 8.
Draw the letter ‘M’ such that the angles on the sides are 40° each and the angle in the middle is 80°.
Solution:

Note: In NCERT, the middle angle is given as 60°, which is not possible with side angles as 40°.

Question 9.
Draw the letter Y such that the three angles formed are 150°, 60° and 150°.
Solution:

Question 10.
The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between two spokes next to each other? What is the largest acute angle formed between two spokes?

Solution:
The angle between two spokes is 15° (i.e. \(\frac{360^{\circ}}{24}\)) because the Ashoka Chakra is a circle and dividing 360° by the number of spokes 24 gives the angle between two adjacent spokes.
Now, each pair of spokes can form different angles depending on how many spokes apart they are:
1 spoke apart → 15°;
2 spokes apart → 30°;
3 spokes apart → 45° ;
4 spokes apart → 60°;
5 spokes apart → 75°;
6 spokes apart → 90° → Not acute
7 spokes apart → 105° → Not acute (but obtuse)
So, the largest acute angle formed between two spokes = 75°.

InText Questions

Question 1.
In given figure, we have ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠EOF = ∠FOG = ∠GOH = ∠HOI = ______. Why?

Solution:
In the figure, it is shown that a straight angle is divided into 8 equal angles:
∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOG, ∠GOH, and ∠HOI.
A straight angle around a point measures 180°.
Since all 8 angles are equal, we divide 180° equally among them:
∠AOB = ∠BOC = … = ∠HOI = \(\frac{180^{\circ}}{8}\) = 22.5°

Lines and Angles Class 6 Extra Questions

Lines and Angles Class 6 Very Short Question Answer

Question 1.
In the figure, 8 points are given. Join the points A to E, E to F, F to B, B to G, G to C, C to H, H to D and D to C. How many line segments are formed?

Solution:
While joining the points A to E, E to F, F to B, B to G, G to C, C to H, H to D and D to C, we get the figure given alongside.

From the figure, the line segments are \(\overline{A E}, \overline{E F}, \overline{F B}, \overline{B G}, \overline{G C}, \overline{C H}, \overline{H D}\) and \(\overline{D C}\).
So, 8 line segments are formed.

Question 2.
In the given figure, write concurrent lines and their points of concurrence.

Solution:
We know that three or more lines that pass through the same point are called concurrent lines.

Concurrent Lines	Point of Concurrence
line n, line q, line r	A
line l, line p, line q	B
line l, line m, line r	D

Question 3.
Complete the statements given in Column I using the appropriate statements from Column II.

	Column I		Column II
(i)	Three or more points are collinear	(a)	part of line having two endpoints.
(ii)	Through a point	(b)	can be either parallel or intersecting.
(iii)	Line segment is a	(c)	more than one line can pass.
(iv)	Through two points	(d)	if they lie on the same line.
(v)	Two lines in a plane	(e)	only one line can pass.

Solution:
We know that,
Three or more points are collinear if they lie on the same line.
Through a point more than one line can pass.
Line segment is a part of line having two endpoints. Through two points only one line can pass.
Two lines in a plane can be either parallel or intersecting.
∴ (i) – (d), (ii) – (c), (iii) – (a), (iv) – (e), (v) – (b)

Question 4.
Match the expressions in Column I with their correct descriptions from Column II.

	Column I		Column II
(i)	\(\overrightarrow{P Q}\)	(a)	Line PQ
(ii)	\(\overrightarrow{Q P}\)	(b)	Line segment PQ
(iii)	\(\overleftrightarrow{P Q}\)	(c)	A ray with starting point P
(iv)	\(\overline{P Q}\)	(d)	A ray with starting point Q

Solution:
Here,
(i) \(\overrightarrow{P Q}\) represents a ray with starting point P.
(ii) \(\overrightarrow{Q P}\) represents a ray with starting point Q.
(iii) \(\overleftrightarrow{P Q}\) represents a line PQ.
(iv) \(\overline{P Q}\) represents a line segment PQ.
∴ (i) – (c), (ii) – (d), (iii) – (a), (iv) – (b)

Question 5.
Write the name of the angles in the given figure.

Solution:
In the given figure, the possible angles are as follows:

Vertex	Arms	Name of angle
O	\(\overrightarrow{O P}\) and \(\overrightarrow{O Q}\)	∠POQ or ∠QOP
O	\(\overrightarrow{O P}\) and \(\overrightarrow{O R}\)	∠POR or ∠ROP
O	\(\overrightarrow{O Q}\) and \(\overrightarrow{O R}\)	∠ROQ or ∠QOR

Question 6.
How many acute angles are present in the given figure? Also, name them.

Solution:
We know that acute angles measure less than 90°.
In the given figure,
Acute angles: ∠ABC, ∠BCD, ∠DEF, ∠EFG and ∠GHI
So, 5 acute angles are present in the given figure.

Question 7.
In the figure, find ∠AOC.

Solution:
Given, ∠AOB = 51° and ∠BOC = 46°
Now, ∠AOC = ∠AOB + ∠BOC = 51° + 46° = 97°

Question 8.
In the figure, find the missing angle.

Solution:
From figure, ∠ACB is a straight angle.
∴ ∠ACB = 180°
⇒ ∠ACD + ∠BCD = 180° ,
⇒ 123° + ∠BCD = 180°
⇒ ∠BCD = 180°- 123° = 57°

Question 9.
Find the measure of the angle POQ given below using a protractor.

Solution:
Using protractor, we get ∠POQ = 137°.

Question 10.
Find the degree measures of ∠POR, ∠QOR and ∠QOS in the below figure.

Solution:
Using protractor, we get
∠POR = 98°, ∠QOR = 50°, ∠QOS = 76°

Question 11.
In the figure, find the missing angle.

Solution:
From the figure, ∠POQ is a straight angle.
∴ ∠POQ = 180°
⇒ ∠QOR + ∠ROS + ∠SOP = 180°
⇒ 26° + ∠ROS + 32° = 180°
⇒ ∠ROS + 58° = 180°
⇒ ∠ROS = 180°-58° = 122°

Question 12.
In the figure given below, find the value of x°.

Solution:
We know that the complete angle at a point measures 360°
∴ ∠AOB + ∠BOC + ∠COD + ∠AOD = 360°
⇒ 98° + 23° + 76° + x° = 360°
⇒ 197° + x° = 360°
⇒ x° = 360°- 197° = 163°

Lines and Angles Class 6 Short Question Answer

Question 1.
What is the maximum and minimum number of points of intersection of four lines in a plane?

Solution:
When four lines intersect each other in such a way that each pair of lines meet at a different point, we get the maximum number of intersecting points.

Let the four lines be p, q, r and s. Then, the lines can intersect as shown in the figure to give the maximum number of points of intersection.

So, the maximum number of points of intersection of four lines in a plane is six.

Formula to confirm the answer:
Maximum number of points of intersection of n lines = \(\frac{n(n-1)}{2}\)
If all four lines are parallel, then they will not intersect at any point.
So, the minimum number of points of intersection of four lines in a plane is zero.

Question 2.
Lines l, m and n are concurrent. Also, lines p, m and n are concurrent. State whether lines p, l, m and n are concurrent or not.
Solution:
Three or more lines that pass through one common point is called concurrent lines. We are told that:

Lines l, m, and n are concurrent. So, they all meet at a point, let’s call it point A. Also, lines p, m, and n are concurrent. That means these lines also meet at point A. Now, let’s look at all the lines:
Line l goes through point A.
Line p goes through point A.
Line m goes through point A.
Line n goes through point A.
So, all four lines (l, m, n, and p) go through the same point A.
That means they are all concurrent.

Question 3.
Name all the line segments in each of the following figures.

Solution:
All the line segments in figure (i) are \(\overline{A B}\), \(\overline{A C}\) and \(\overline{A D}\).
All the line segments in figure (ii) are \(\overline{P Q}\), \(\overline{P T}\), \(\overline{P S}\), \(\overline{P R}\), \(\overline{Q T}\), \(\overline{S R}\) and \(\overline{S T}\).
All the line segments in figure (iii) are \(\overline{L M}\), \(\overline{M N}\), \(\overline{N O}\), \(\overline{O P}\), \(\overline{P Q}\) and \(\overline{Q L}\).

Question 4.
There are four non-collinear points as shown in the figure. Draw lines through these points taking two at a time. Name the lines. How many such different lines can be drawn?

Solution:
The lines through the given points taking two at a time are \(\overleftrightarrow{P Q}, \overleftrightarrow{P R}, \overleftrightarrow{P S}, \overleftrightarrow{Q R}, \overleftrightarrow{Q S}\) and \(\overleftrightarrow{R S}\)

So, 6 different lines can be drawn from the given four non-collinear points.

Question 5.
In the given figure, name

(i) lines containing the point P.
(ii) lines passing through the point Q.
(iii) line on which H lies.
(iv) three pairs of intersecting lines.
Solution:
(i) Lines containing the Point P are \(\overleftrightarrow{P Q}\), \(\overleftrightarrow{A X}\) and \(\overleftrightarrow{P G}\).
(ii) Lines through the point Q are \(\overleftrightarrow{A Y}\) and \(\overleftrightarrow{P Q}\).
(iii) Line on which lines is \(\overleftrightarrow{X Y}\).
(iv) Three pairs of intersecting hues are \(\overleftrightarrow{P Q}\) and \(\overleftrightarrow{P G}\); \(\overleftrightarrow{A X}\) and \(\overleftrightarrow{X Y}\); \(\overleftrightarrow{A Y}\) and \(\overleftrightarrow{X Y}\).

Question 6.
Identify and name the line segments and rays in each of the following figures.

Solution:
We know that the shortest path from point A to point B (including A and B) is called the line segment AB. It is denoted by either \(\overline{A B}\) or \(\overline{B A}\).
And a ray is a portion of a line that starts at one point (called the starting point or initial point of the ray) and goes on endlessly in a direction.

Figure Number	Line Segment	Ray
(i)	\(\overline{A C}\), \(\overline{D E}\), \(\overline{A B}\) and \(\overline{C D}\)	\(\overrightarrow{A B}\) and \(\overrightarrow{D E}\)
(ii)	\(\overline{R T}\), \(\overline{R P}\), \(\overline{T Q}\), \(\overline{R S}\) and \(\overline{T S}\)	\(\overrightarrow{R P}\), \(\overrightarrow{T Q}\), \(\overrightarrow{R S}\) and \(\overrightarrow{T S}\)
(iii)	\(\overline{O N}\), \(\overline{Q L}\), \(\overline{Q P}\), \(\overline{L P}\), \(\overline{M N}\), \(\overline{N O}\) and \(\overline{M O}\)	\(\overrightarrow{Q L}\), \(\overrightarrow{N M}\), \(\overrightarrow{N O}\) and \(\overrightarrow{Q P}\)

Question 7.
In the given figure, name the point(s):

(i) in the interior of ∠POQ.
(ii) in the interior of ∠POR.
(iii) in the exterior of ∠QOR.
(iv) in the exterior of ∠QOS. ..
(v) on ∠POS.
(vi) on ∠ROS.
Solution:
(i) Point A is in the interior of ∠POQ.
(ii) Points A, B, Q and C are in the interior of ∠POR.
(iii) Points A, P, D. E and S are in the exterior of ∠QOR.
(iv) Points A and P arc in the exterior of ∠QOS.
(v) Points P, O, E and S are on ∠POS.
(vi) Points R, O, E and S are on ∠ROS.

Question 8.
Compare two angles given below using superimposition and find which angle is smaller?

Solution:
Suppose we have a transparent circular paper which can be moved and placed from one figure to another figure.

Let us place the circular paper on the ∠PQR

The circular paper is placed in such a way that its centre is on the vertex of the angle i.e. Q. Mark the points A and B on the edge of circular paper at the points where the arms of the angle PQR pass through the circular paper.

After that put it on the other given angle ∠LMN, such that B lies on the arm MN and check which is smaller.

Hence, ∠LMN is smaller than ∠PQR.

Question 9.
In the figure, write another name for:

(i) ∠1
(ii) ∠2
(iii) ∠3
(iv) ∠4
(v) ∠5
Solution:
(i) Arms of’ the ∠1 are BA and BC and its vertex is B.
So. another name for ∠1 is ∠ABG or ∠GBA.

(ii) Aims of tIme ∠2 are GB amid GC and its vertex is G.
So, another name for ∠2 is ∠BGC or ∠CGB.

(iii) Arms of the ∠3 are CG and CE and its vertex is C.
So, another name for ∠3 is ∠GCE or ∠ECG.

(iv) Arms of the ∠4 arc ED and EC and its vertex is E.
So. another name for ∠4 is ∠CED or ∠DEC.

(v) Arms of the ∠5 are FE and FG and its vertex is F.
So, another name for ∠5 is ∠EFG or ∠GFE.

Question 10.
In each case, determine which angle is greater and why?
(i) ∠PQR or ∠XYZ
(ii) ∠XYZ or ∠LMN
(iii) ∠PQR or ∠LMN

Solution:
On comparing by superimposition, we get

(i) ∠PQR is greater than ∠XYZ because the amount of rotation of ∠PQR is greater than the amount of rotation of ∠XYZ.
(ii) ∠XYZ is greater than ∠LMN because the amount of rotation of ∠XYZ is greater than the amount of rotation of ∠LMN.
(iii) ∠PQR is greater than ∠LMN because the amount of rotation of ∠PQR is greater than the amount of rotation of ∠LMN.

Question 11.
How many acute and obtuse angles are present in the given figure? Also, name them.

Solution:
We know that acute angles measure less than 90° and obtuse angles measure more than 90° but less than 180°.
In the given figure,
Acute angles: ∠OLP, ∠MLP, ∠LMQ, ∠NMQ, ∠MNR, ∠ONR, ∠NOS and ∠LOS
Obtuse angles: ∠LPS, ∠LPQ, ∠MQP, ∠MQR, ∠NRQ, ∠NRS, ∠OSR and ∠OSP
So, 8 acute angles and 8 obtuse angles are present in the given figure.

Question 12.
Classify the following angles (acute, obtuse, right, straight, complete):
(i) 80°
(ii) 90°
(iii) 180°
(iv) 175°
(v) 360°
Solution:
The given angles can be classified as follows:
(i) 80°: Acute angle
(ii) 90°: Right angle
(iii) 180°: Straight angle
(iv) 175°: Obtuse angle
(v) 360°: Complete angle

Question 13.
In the figure, if ∠POQ = 23° and ∠POR = 62°, then find ∠QOR.

Solution:
Given,
∠POQ = 23° and ∠POR = 62°
From the figure, ∠POR = ∠POQ + ∠QOR
⇒ 62° = 23° + ∠QOR
Subtracting 23° from both sides, we get
62° – 23° = 23° + ∠QOR – 23°
⇒ ∠QOR = 39°

Question 14.
Construct an angle of 115° using a protractor.
Solution:

To draw an angle of 115° using a protractor, we follow the steps laid down below:
Step 1: Draw a ray \(\overrightarrow{O A}\).
Step 2: Place the protractor in such a way that its centre is exactly on the point 0 and the base line lies along \(\overrightarrow{O A}\).
Step 3: Starting from 0° on the right, move and look for the mark 115° mark on the protractor.
Step 4: Mark a point B against this 115° mark.
Step 5: Remove the protractor and draw the ray \(\overrightarrow{O B}\).
Thus, ∠AOB is the required angle whose measure is 115°.

Question 15.
How many degrees are there in:
(i) \(\frac{1}{2}\) of a straight angle?
(ii) \(\frac{3}{4}\) of a right angle?
Solution:
(i) \(\frac{1}{2}\) of a straight angle = \(\frac{1}{2}\) × 180° = 90°
(ii) \(\frac{3}{4}\) of right angle = \(\frac{3}{4}\) × 90°
= 3 × 22.5° = 67.5°

Question 16.
Measure the following angles with the help of a protractor and write the measure in degrees.

Solution:
(i) Using protractor, we get ∠ABC = 68°.
(ii) Using protractor, we get ∠PQR = 109°.
(iii) Using protractor, we get ∠XYZ = 125°.

Question 17.
Name the different angles and write their degree measures.

Solution:
All the possible angles are as follows: ;
∠AOB, ∠AOC, ∠AOD, ∠AOE, ∠BOC, ∠BOD, ∠BOE, ∠COD, ∠COE, ∠DOE
Now, ∠AOB = 40°; ∠AOC = 65°; ∠AOD = 125°; ∠AOE = 180°
∠BOC = ∠AOC – ∠AOB = 65° – 40° = 25°
∠BOD = ∠AOD – ∠AOB = 125°- 40° = 85°
∠BOE = ∠AOE – ∠AOB = 180° – 40° = 140°
∠COD = ∠AOD – ∠AOC = 125° – 65° = 60°
∠COE = ∠AOE – ∠AOC = 180° – 65°= 115°
∠DOE = ∠AOE – ∠AOD = 180°- 125°= 55°

Question 18.
Draw a line segment PQ of length 8 cm. Take a point R on PQ such that PR = 5 cm. At point R, draw SR ⊥ PQ.

Solution:
To draw a required perpendicular line using a protractor, we follow the steps laid down below:
Step 1: Draw a line segment PQ of length 8 cm.
Step 2: Mark a point R on the line segment PQ such that PR = 5 cm.
Step 3: Draw SR perpendicular to PQ at R using protractor.
Thus, SR ⊥ PQ at point R.

Question 19.
State the type (obtuse, reflex, complete) of each of the following angles with their measure.

Solution:
Using protractor, we find
∠HOG = 125° (Obtuse angle)
∠EOE = 133°
⇒ Reflex of ∠EOE = 360°- 133° = 227° (Reflex angle)
∠NON = 360° (Complete angle)

Question 20.
Find the degree measures of ∠XOW, ∠XOY, ∠XOZ and ∠YOZ in the below figure.

Solution:
We need to find the measures of ∠XOW, ∠XOY, ∠XOZ and ∠YOZ.
So, we place the protractor in such-a way that its centre coincide with the vertex 0 of the given angles and the base line lies along \(\overrightarrow{O W}\).
Starting from 0° on the right on inner scale, \(\overrightarrow{O X}\) passes through the 26° mark, \(\overrightarrow{O Y}\) passes through the 65° mark and \(\overrightarrow{O Z}\) passes through the 106° mark.
So, ∠XOW = 26°, ∠WOY = 65° and ∠WOZ = 106°
Now, ∠XOW = 26°
∴ ∠XOY = ∠WOY – ∠WOX = 65° – 26° = 39° [∵ ∠XOW = ∠WOX]
Also, ∠XOZ =∠WOZ – ∠WOX = 106° – 26° = 80°
And, ∠YOZ = ∠WOZ – ∠WOY = 106° – 65° = 41°

Lines and Angles Class 6 Long Question Answer

Question 1.
In the given figure, name

(i) Five pairs of intersecting lines.
(ii) Four collinear points.
(iii) Three collinear points.
(iv) Three concurrent lines.
Solution:
(i) Five pairs of intersecting lines are \(\overleftrightarrow{A B}\) and \(\overleftrightarrow{I J}\) ; \(\overleftrightarrow{P Q}\) and \(\overleftrightarrow{I J}\) ; \(\overleftrightarrow{X Y}\) and \(\overleftrightarrow{I J}\) ; \(\overleftrightarrow{X Y}\) and \(\overleftrightarrow{L M}\) ; \(\overleftrightarrow{X Y}\) and \(\overleftrightarrow{A B}\).
(ii) Four collinear points are X, G, H and Y.
(iii) Three collinear points are P, K and Q. As X, G, H and Y are collinear points, we can select any three points from X, G, H and Y as well for three collinear points.
(iv) Three concurrent lines are \(\overleftrightarrow{A B}\), \(\overleftrightarrow{L M}\) and \(\overleftrightarrow{P Q}\).

Question 2.
Consider the line XV in the adjoining figure. Find whether the given statements are true or false.

(i) B is a point on \(\overrightarrow{G Y}\).
(ii) A is a point on \(\overrightarrow{H Y}\).
(iii) G, B and H are points on the line segment GC.
(iv) \(\overrightarrow{H Y}\) is same as \(\overrightarrow{G Y}\).
(v) \(\overrightarrow{B X}\) is different from \(\overrightarrow{B Y}\).
(vi) A, B, C, G, H, X and Y are points on the line XY.
Solution:
(i) True
From the figure, the point B lies between the points G and Y.
Therefore, B is a point on \(\overrightarrow{G Y}\).

(ii) False
From the figure, the point A does not lie between the points H and Y.
Therefore, A is not a point on \(\overrightarrow{H Y}\).

(iii) True
From the figure, the point G is one endpoint of the line segment GC and the points B and H lie between points G and C.
Therefore, G, B and H are points on the line segment GC.

(iv) False
From the figure, the initial points of \(\overrightarrow{H Y}\) and \(\overrightarrow{G Y}\) are different.
Therefore, \(\overrightarrow{H Y}\) is not the same as \(\overrightarrow{G Y}\).

(v) True
From the figure, the initial points of the \(\overrightarrow{B X}\) and \(\overrightarrow{B Y}\) are same but they are moving in different directions.
Therefore, \(\overrightarrow{B X}\) is different from \(\overrightarrow{B Y}\).

(vi) True
From the figure, the points A, B, C, G, H, X and F lie on the line \(\overrightarrow{X Y}\).
Therefore, A, B, C, G, H, X and Y are points on the line \(\overrightarrow{X Y}\).

Question 3.
Draw rough diagrams of two angles for the condition which is possible.
(i) Vertex in common.
(ii) One arm in common.
(iii) Two arms in common.
(iv) Three points in common.
Solution:
Rough diagrams of two angles for the given condition are as follows:
(i) In the figure, ∠POQ and ∠XOY have vertex O in common.

(ii) In the figure, ∠POO and ∠QOR have arm \(\overrightarrow{O Q}\) in common.

(iii) In the figure, ∠POQ and ∠XOY have two arms in common.

(iv) In the figure, ∠POQ and ∠QOR have three points O, S and Q in common.

Question 4.
In each of the grids, join A to other grid points in the figure by a straight line to get:
(i) An acute angle

Solution:
(i) We know that an acute angle is greater than 0° and less than 90°.

Here, ∠BAX, ∠BAY, ∠DCX and ∠DCY are acute angles.

(ii) We know that an obtuse angle is greater than 90° and less than 180°.

Here, ∠BAX, ∠BAY, ∠DCX and ∠DCY are obtuse angles.

(iii) We know that an obtuse angle is greater than 180° and less than 360°.

Here, ∠BAX, ∠BAY, ∠DCX and ∠DCY are reflex angles.

(iv) We know that a right angle is equal to 90°

Here, ∠BAX, ∠BAY, ∠DCX and ∠DCY are right angles.

Question 5.
Find, the degree measures of ∠AOB, ∠BOE, ∠BOD, ∠COE and ∠COD.

Solution:
For ∠AOB, \(\overrightarrow{O A}\) is at 0° mark on the right and \(\overrightarrow{O B}\) passes through the 35° mark from right.
Therefore, ∠AOB = 35°
For ∠BOE, \(\overrightarrow{O E}\) is at 0° mark on the left and \(\overrightarrow{O B}\) passes through the 145° mark from left.
Therefore, ∠BOE = 145°
For ∠COE, \(\overrightarrow{O E}\) is at 0° mark on the left and \(\overrightarrow{O C}\) passes through the 105° mark from left.
Therefore, ∠COE = 105°
For ∠BOD, we have
∠BOD = ∠BOE – ∠DOE …….. (i)
For ∠DOE, \(\overrightarrow{O E}\) is at 0° mark on the left and \(\overrightarrow{O D}\) passes through the 65° mark from left.
Therefore, ∠DOE = 65°
From (i), ∠BOD = 145° – 65° = 80° [∵ ∠BOE = 145° and ∠DOE = 65°]
For ∠COD, we have
∠COD = ∠COE – ∠DOE = 105°- 65° = 40°

Question 6.
Identify the types of angles represented by shaded portion in each:

Solution:
(i) As the angle is less than quarter of a full turn, it is an acute angle.
(ii) As the angle is greater than quarter of a full turn but less than half of a full turn, it is an obtuse angle.
(iii) As the angle is greater than half of a full turn but less than a full turn, it is a reflex angle.
(iv) As the angle is equal to quarter of a full turn, it is a right angle.
(v) As the angle is equal to half of a full turn, it is a straight angle.
(vi) As the angle is equal to a full turn, it is a complete angle.

Lines and Angles Class 6 Case Based Questions

Question 1.
The Ashoka Chakra, found at the centre of the Indian national flag, is a symbol of righteousness and progress. It consists of 24 equally spaced spokes.
Based on the above information, answer the following questions:
(i) What is the angle between any two consecutive spokes of the Ashoka Chakra?
(ii) What is the largest acute angle that can be formed between any two spokes of the Ashoka Chakra?
(iii) What is the largest obtuse angle that can be formed between any two spokes of the Ashoka Chakra?
(iv) What is the smallest reflex angle that can be formed between any two spokes of the Ashoka Chakra?
Solution:
There are 24 spokes in the Ashoka Chakra.
(i) Since the spokes divide the circle into 24 equal parts, the angle between any two consecutive spokes is calculated by dividing the complete angle of a circle (360°) by 24. Thus, the angle between two consecutive spokes is \(\frac{360^{\circ}}{24}\) = 15°.

(ii) The largest acute angle that can be formed between any two spokes would be the greatest multiple of 15° that is still less than 90°.
The largest multiple of 15° and less than 90° is 75°.
Therefore, the largest acute angle between any two spokes is 75°.

(iii) The largest obtuse angle that can he formed between two spokes is the greatest multiple of 15°, which is greater than 90° but less than 1.80°.
The multiples of 15° (between 90° and 180°) are 105°, 120°, 135°, 150°, 165°.
Thus, the largest obtuse angle between any two spokes is 165°.

(iv) The smallest reflex angle that can be formed between two spokes is the smallest angle multiple of 15° greater than 180°, which would be 195° (since 180° + 15° = 195°).

Question 2.
Priya drawn a figure in her notebook as shown.

Based on the above information, answer the following questions:
(i) Write the collinear points.
(ii) Write the concurrent lines.
(iii) How many lines are concurrent at point N?
(iv) How many lines have point Q as the point of intersection?
Solution:
(i) Three or more points that lie on the same straight line are known as collinear points.
From the figure, N, P and O are collinear points.
And, M, S, R and Q are collinear points.

(ii) We know that three or more lines in a plane that pass through one point are known as concurrent lines.
From the figure, \(\overleftrightarrow{M N}\), \(\overleftrightarrow{S N}\) and \(\overleftrightarrow{Q N}\) are concurrent lines.

(iii) From the figure, we can see that three lines \(\overleftrightarrow{M N}\), \(\overleftrightarrow{S N}\) and \(\overleftrightarrow{Q N}\) are concurrent at point N.

(iv) From the figure, we can see that two lines. \(\overleftrightarrow{M Q}\) and \(\overleftrightarrow{N Q}\) intersect at point Q.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 1 Patterns in Mathematics Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 1 Patterns in Mathematics Solutions

Ganita Prakash Class 6 Chapter 1 Solutions

Class 6 Maths Ganita Prakash Chapter 1 Solutions Patterns in Mathematics

Question 1.
Why are 1, 3, 6, 10, 15, … called triangular numbers? Why are 1, 4, 9, 16, 25, … called square numbers or squares? Why are 1, 8, 27, 64, 125, … called cubes?
Solution:
As the dot representation of sequence 1,3, 6, 10, 15, … forms triangles, it is called triangular numbers sequence. As the dot representation of sequence 1,4, 9, 16, 25, … forms squares, it is called square numbers sequence.

As the dot representation of sequence 1,8, 27, 64, 125, … forms cubes, it is called cube numbers sequence.

Question 2.
You will have noticed that 36 is both a triangular number and a square number! That is, 36 dots can be arranged perfectly both in a triangle and in a square. Make pictures in your notebook illustrating this!
This shows that the same number can be represented differently and play different roles, depending on the context. Try representing some other numbers pictorially in different ways!
Solution:
Representation of 36 as a triangular number:

Representation of 36 as a square number:

Representation of 10 as even number:

Representation of 10 as triangular number:

Question 3.
Can you think of pictorial way to visualise the sequence of Powers of 2? Powers of 3?
Solution:
Powers of 2 can be visualised as:

Powers of 3 can be visualised as:

Question 4.
Can you find a similar pictorial explanation for why adding counting numbers up and down, i.e., 1, 1+2 + 1,1+2 + 3 + 2 + 1, … , gives square numbers?
Solution:
Yes,
As we can see, the dot representation of the addition of counting numbers up and down forms the dot representation of square numbers.

Question 5.
Which sequence do you get when you start to add the All 1’s sequence up? What sequence do you get when you add the All l’s sequence up and down?
Solution:
Adding all 1 ’s sequence up
1 = 1
1 + 1 = 2
1 + 1 + 1 = 3
1 + 1 + 1 + 1 = 4 and so on.
Here, we get a sequence of counting numbers i.e. 1, 2, 3, 4, … .

Adding all l’s sequence up and down 1 = I
1 + 1 + 1 = 3
1 + 1 + 1 + 1 + 1 = 5 and so on.
Here, we get a sequence of odd numbers.

Question 6.
What happens when you add up pairs of consecutive triangular numbers? That is, take 1 + 3, 3 + 6, 6 + 10, 10 + 15,… ? Which sequence do you get?
Solution:
Adding up pairs of consecutive triangular numbers, we get
1 + 3 = 4;
3 + 6 = 9;
6 + 10 = 16;
10 + 15 = 25 and so on.
Here, we get a sequence of square numbers.

Question 7.
What happens when you multiply the triangular numbers by 6 and add 1 ? Which sequence do you get?
Solution:
Triangular numbers are 1, 3, 6, 10, 15, …
On multiplying triangular numbers by 6 and add 1 to it, we get
1 × 6 + 1 = 7;
3 × 6 + 1 = 19;
6 × 6 + 1 = 37;
10 × 6 + 1 = 61;
15 × 6 + 1 = 91 and so on.
Hence, we get a sequence of hexagonal numbers.

InText Questions

Question 1.
Observe the pattern given below:
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
Why does this happen? Do you think it will happen forever?
Solution:
The sequence of odd numbers is: 1, 3, 5, 7, 9, 11,…
Sum of first 1 odd number = 1 = 1²
Sum of first 2 odd numbers = 1 + 3 = 4 = 2²
Sum of first 3 odd numbers = 1 + 3 + 5 = 9 = 3²
Sum of first 4 odd numbers =1 + 3 + 5 + 7 = 16 = 4²
Sum of first 5 odd numbers =1+3 + 5 + 7 + 9 = 25 = 5²
Sum of first 6 odd numbers =1 + 3 + S + 7 + 9 + 11 = 36 = 6²
Each time we add another odd number, the total becomes a perfect square.
Since the sequence of odd numbers keeps going forever, and the sum of the first n odd numbers is always n², this pattern will continue forever.

Question 2.
Pictorially represent and find the sum of the first 10 odd numbers.
Solution:
From figure, it is clear that
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100 = 10²

Patterns in Mathematics Class 6 Extra Questions

Patterns in Mathematics Class 6 Very Short Question Answer

Question 1.
Find the next term of the sequence 1, 4, 9, 16,… .
Solution:
Given sequence is 1, 4, 9, 16, …, i.e. 1², 2², 3², 4², …, which is a sequence of squares.
So, next term, i.e. fifth term = 5² = 25

Question 2.
What is the sum of first 10 terms of the sequence of counting numbers?
Solution:
We know, the sequence of counting numbers is 1, 2, 3, 4, ….
Now, sum of first 10 terms of the sequence of counting numbers
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55

Question 3.
Identify the rule in the following number pattern and write the missing entries:
111 × 11 = 1221
121 × 11 = 1331
131 × 11 = 1441
141 × _____ = ______
_____ × 11 = 1661
161 × 11 = ______
Solution:
We observe that the first number on the left side increases by 10 each time: 111, 121, 131, 141, 151, 161,…
The result also increases by 110 each time: 1221, 1331, 1441, 1551, 1661, 1771, …
The first number is multiplied by 11 each time.
Thus, the missing numbers are as follows:
141 × 11 = 1551;
151 × 11 = 1661;
161 × 11 = 1771

Question 4.
Find the next term of the sequence 2 + 1, 2 + 2, 2 + 3, 2 + 4,… .
Solution:
Given sequence is 2 + 1, 2 + 2, 2 + 3, 2 + 4, ….
First term = 2 + 1
Second term = 2 + 2
Third term = 2 + 3
Fourth term = 2 + 4
So, next term, i.e. fifth term = 2 + 5

Question 5.
Find the next term of the sequence 2, 6, 12, 20, …
Solution:
C liven sequence is 2, 6, 12, 20, ….
Here, 2 = 1 × 2
6 = 2 × 3
12 = 3 × 4
20 = 4 × 5
So, next term = 5 × 6 = 30

Question 6.
Find the next term of the sequence 1, 3, 6, 10, 15,
Solution:
Given sequence is 1, 3, 6, 10, 15
First term = 1
Second term = 3 = 1 + 2 (First term + 2)
Third term = 6 = 3 + 3 (Second term + 3)
Fourth term = 10 = 6 + 4 (Third term + 4)
Fifth term = 15 = 10 + 5 (Fourth term + 5)
So, next term = Fifth term + 6 = 15 + 6 = 21

Question 7.
What is the sum of first 12 terms of the sequence of all 1s?
Solution:
We know, the sequence of all 1 s is 1, 1, 1, 1, ….
Now, sum of first 12 terms of the sequence of all 1s
= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12

Question 8.
What is the sum of first 6 terms of the sequence of odd numbers?
Solution:
The sequence of odd numbers is 1, 3, 5, 7
Now, sum of first 6 terms of the sequence of odd numbers = 1 + 3 + 5 + 7 + 9 + 11 = 36 = 6²

Question 9.
Write the first 5 square numbers.
Solution:
We know, the sequence of square numbers is 1, 4, 9, 16, 25, 36, …….
So, the first 5 square numbers are 1, 4, 9, 16 and 25.

Question 10.
Write the first 4 triangular numbers.
Solution:
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, … .
So, the first 4 triangular numbers are 1, 3, 6 and 10.

Question 11.
Which sequence do you get on adding the counting numbers up?
Solution:
We know, the sequence of counting numbers is 1. 2, 3, 4, 5, … .
Now, on adding the counting numbers up, we get the following sequence:
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
So, the sequence is 1, 3, 6, 10, 15, …, which is the sequence of triangular numbers.

Patterns in Mathematics Class 6 Short Question Answer

Question 1.
Find the next term of the sequence 2, 16, 54, 128,
Solution:
Given sequence is 2, 16, 54, 128, ….
First term = 2 = 2 × 1 = 2 × 1³
Second term = 16 = 2 × 8 = 2 × 2³
Third term = 54 = 2 × 27 = 2 × 3³
Fourth term = 128 = 2 × 64 = 2 × 4³
So, next term, i.e. fifth term = 2 × 53 = 2 × 125 = 250

Question 2.
Can you identify numbers which are both triangular as well as square numbers? Find 2 such numbers.
Solution:
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, 36, 45, ….
And the sequence of squares is 1,4, 9, 16, 25, 36, 49, 64, 81, ….
So, 1 and 36 are both triangular as well as square numbers.

Question 3.
Observe the pattern shown below and write the next three steps:
1 × 9 + 2 = 11
12 × 9 + 3 = 111
123 × 9 + 4 = 1111
Solution:
We observe that

• On the left side, the first number starts at 1, then becomes 12, then 123, each time we add the next digit in order.
• We multiply by 9 each time.
• Then, we add next number (2, then 3, then 4)
• On the right side the answer is made of all 1 s and the number of Is is one more than the number of digits in starting number.

So, next three steps will be:
1234 × 9 + 5 = 11111
12345 × 9 + 6 = 111111
123456 × 9 + 7 = 1111111

Question 4.
Identify the pattern in the following number pattern and write the missing terms:
2178 × 4 = 8712
21978 × 4 = 87912
219978 × 4 = _______
_____ × 4 = 8799912
21999978 × 4 = ________
219999978 × ______ = 879999912
Solution:
In the first number, we start with 2178 and keep adding one more 9 before 78 in each step.
Then, we multiply the number by 4. The result is a number starting with 87, followed by the same number of 9s, and ending with 12.
The missing numbers are as follows:
2178 × 4 = 8712
21978 × 4 = 87912
219978 × 4 = 879912
2199978 × 4 = 8799912
21999978 × 4 = 87999912
219999978 × 4 = 879999912

Question 5.
Pairs of consecutive triangular numbers are added (i.e. 1 + 3, 3 + 6, …). Which sequence will you get on such addition? Write the 6^th term of the new obtained sequence.
Solution:
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, … .
Now, on adding pairs of consecutive triangular numbers we get the following sequence:
1 + 3 = 4
3 + 6 = 9
6 + 10 = 16
10 + 15 = 25
15 + 21 = 36
21 + 28 = 49
So, the sequence is 4, 9, 16, 25, 36, 49, …, which represents the square numbers starting with 4.
Now, 6^th term of the new obtained sequence is 49.

Question 6.
Which sequence do you get on adding the odd numbers up?
Solution:
We know, the sequence of odd numbers is 1, 3, 5, 7, 9, … .
Now, on adding the odd numbers up, we get the following sequence:
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
So, the sequence is 1,4, 9, 16, 25, …, which is the sequence of square numbers.

Patterns in Mathematics Class 6 Long Question Answer

Question 1.
What will happen if you multiply the triangular numbers by 6 and add 1? Which sequence do you get?
Solution:
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, 36, 45, … .
Now, if we multiply the triangular numbers by 6 and add 1, we get the following sequence:
1 × 6 + 1 = 6 + 1 = 7
3 × 6 + 1 = 18 + 1 = 19
6 × 6 + 1 = 36 + 1 = 37
10 × 6 + 1 = 60 + 1 = 61
15 × 6 + 1 = 90 + 1 = 91
So, the required sequence is 7, 19, 37, 61, 91, …, which represents hexagonal numbers starting with 7.

Question 2.
Find the number of line segments connecting any two distinct vertices of the polygon as shown in the figure.

Solution:
We know, in a complete graph, every pair of vertices is connected by a unique line segment. The number of line segments in a complete graph with n vertices is given by \(\frac{n(n-1)}{2}\).
For K₇ (Heptagon):
Number of vertices = 7
∴ Number of line segments
= \(\frac{n(n-1)}{2}\) = \(\frac{7(7-1)}{2}\) = 7 × 3 = 21

Question 3.
Find the number of sides of a Koch Snowflake obtained after 5 iterations.
Solution:
To get from one shape to the next shape in the Koch Snowflake sequence, each line segment ‘_______’ is replaced by a speed bump ‘‘

Number of iterations	No. of sides
0	3 = 3 × 40
1	12 = 3 × 41
2	48 = 3 × 42
3	192 = 3 × 43
4	768 = 3 × 44

Thus,
After 5 iterations, number of sides = 3 × 4⁵ = 3 × 1024 = 3072.

Patterns in Mathematicse Class 6 Case Based Questions

Question 1.
Players wear different jersey numbers to help fans and broadcasters identify them, especially in games like cricket and football where they look similar on the field.

One day, Ashish went to the stadium to watch a cricket match. There, he observed the jersey numbers of some cricketers and found them to follow a sequence.
Based on the above information, answer the following questions:
(i) Write the rule for the sequence followed by the given numbers in words.
(ii) What will be the number on Captain’s jersey?
(iii) What will be the number on Vice-captain’s jersey?
Solution:
(i) Numbers on the jersey of players are given as
2 = 1² + 1;
5 = 2² + 1;
10 = 3² + 1;
17 = 4² + 1;
26 = 5² + 1
Rule of the sequence: n² + 1; n = 1, 2, 3, …
(ii) Since captain is at 6^th position, the number on his jersey is 6² + 1 = 36 + 1 = 37.
(iii) Since vice-captain is at 7^th position, the number on his jersey is 7² + 1 = 49 + 1 = 50

Question 2.
In Ms. Rina’s class, each student is given a unique roll number. One day, while arranging the books in the library, she noticed that the roll numbers of the students returning books followed a specific number pattern. She found that the roll numbers were: 1, 2, 3, 5, 8, 13, …
Based on the above information, answer the following questions:
(i) Write the rule for the sequence followed by the given numbers.
(ii) What is the name of the number sequence followed by the given numbers?
(iii) What will be the roll number of the 10^th student?
Solution:
The given numbers are 1, 2, 3, 5, 8, 13, …
(i) First number = 1
Second number = 2
Third number = 3 = 1 + 2
(First number + Second Number)
Fourth number = 5 = 2 + 3
(Second number + Third Number)
Fifth number = 8 = 3 + 5
(Third number + Fourth Number)
Sixth number = 13 = 5 + 8
(Fourth number + Fifth Number)
Therefore, the rule for the sequence is,
First number = 1, Second number = 2, any other number = sum of previous two numbers

(ii) The given numbers are known as Virahanka numbers.

(iii) Observing the pattern (from above)
Seventh number = Fifth number + Sixth number
= 8 + 13 = 21
Eighth number = Sixth number + Seventh number
= 13 + 21 = 34
Ninth number = Seventh number + Eighth number
= 21 + 34 = 55
Tenth number = Eighth number + Ninth number
= 34 + 55 = 89
Therefore, the roll number of 10^th student is 89.

BSE Odisha Guru Class 6 Maths Solutions ଗଣିତ ପ୍ରକାଶ Ganita Prakash, 6th Class Math Book Odia Medium Question Answer Pdf Download, Class 6 Math Book Solution Odia Medium.

BSE Odisha Class 6 Math Solution

BSE Class 6 Maths Solutions

BSE Odisha Guru Class 6 Maths Solutions Ganita Prakash

• Chapter 1 Patterns in Mathematics Class 6 Solutions
• Chapter 2 Lines and Angles Class 6 Solutions
• Chapter 3 Number Play Class 6 Solutions
• Chapter 4 Data Handling and Presentation Class 6 Solutions
• Chapter 5 Prime Time Class 6 Solutions
• Chapter 6 Perimeter and Area Class 6 Solutions
• Chapter 7 Fractions Class 6 Solutions
• Chapter 8 Playing with Constructions Class 6 Solutions
• Chapter 9 Symmetry Class 6 Solutions
• Chapter 10 The Other Side of Zero Class 6 Solutions

Class 6 Math Solution Odia Medium

Class 6 Odia Medium Math Solution

• Class 6 Maths Chapter 1 Odia Medium – ଗଣିତରେ ସଂରଚନା
• Class 6 Maths Chapter 2 Odia Medium – ରେଖା ଓ କୋଣ
• Class 6 Maths Chapter 3 Odia Medium – ସଂଖ୍ୟା ଖେଳ
• Class 6 Maths Chapter 4 Odia Medium – ତଥ୍ୟ ଉପସ୍ଥାପନା ଓ ପରିଚାଳନା
• Class 6 Maths Chapter 5 Odia Medium – ମୌଳିକ ସଂଖ୍ୟା
• Class 6 Maths Chapter 6 Odia Medium – ପରିସୀମା ଓ କ୍ଷେତ୍ରଫଳ
• Class 6 Maths Chapter 7 Odia Medium – ଭଗ୍ନାଂଶ
• Class 6 Maths Chapter 8 Odia Medium – ଜ୍ୟାମିତିକ ଅଙ୍କନ
• Class 6 Maths Chapter 9 Odia Medium – ପ୍ରତିସମତା
• Class 6 Maths Chapter 10 Odia Medium – ପୂର୍ଣ୍ଣ ସଂଖ୍ୟା

6th Class Math Odia Medium Solutions (Old Syllabus)

Chapter 1 ସଂଖ୍ୟାମାନଙ୍କୁ ଜାଣିବା

• Chapter 1 ସଂଖ୍ୟାମାନଙ୍କୁ ଜାଣିବା Ex 1.1
• Chapter 1 ସଂଖ୍ୟାମାନଙ୍କୁ ଜାଣିବା Ex 1.2
• Chapter 1 ସଂଖ୍ୟାମାନଙ୍କୁ ଜାଣିବା Ex 1.3
• Chapter 1 ସଂଖ୍ୟାମାନଙ୍କୁ ଜାଣିବା Ex 1.4
• Chapter 1 ସଂଖ୍ୟାମାନଙ୍କୁ ଜାଣିବା Ex 1.5
• Chapter 1 ସଂଖ୍ୟାମାନଙ୍କୁ ଜାଣିବା InText Questions

Chapter 2 ସଂଖ୍ୟା ସମ୍ବନ୍ଧୀୟ ଅଧ୍ବକ ଆଲୋଚନା

• Chapter 2 ସଂଖ୍ୟା ସମ୍ବନ୍ଧୀୟ ଅଧ୍ବକ ଆଲୋଚନା Ex 2.1
• Chapter 2 ସଂଖ୍ୟା ସମ୍ବନ୍ଧୀୟ ଅଧ୍ବକ ଆଲୋଚନା Ex 2.2
• Chapter 2 ସଂଖ୍ୟା ସମ୍ବନ୍ଧୀୟ ଅଧ୍ବକ ଆଲୋଚନା Ex 2.3
• Chapter 2 ସଂଖ୍ୟା ସମ୍ବନ୍ଧୀୟ ଅଧ୍ବକ ଆଲୋଚନା Ex 2.4
• Chapter 2 ସଂଖ୍ୟା ସମ୍ବନ୍ଧୀୟ ଅଧ୍ବକ ଆଲୋଚନା Ex 2.5
• Chapter 2 ସଂଖ୍ୟା ସମ୍ବନ୍ଧୀୟ ଅଧ୍ବକ ଆଲୋଚନା InText Questions

Chapter 3 ଜ୍ୟାମିତିରେ ମୌଳିକ ଧାରଣା

• Chapter 3 ଜ୍ୟାମିତିରେ ମୌଳିକ ଧାରଣା Ex 3.1
• Chapter 3 ଜ୍ୟାମିତିରେ ମୌଳିକ ଧାରଣା Ex 3.2
• Chapter 3 ଜ୍ୟାମିତିରେ ମୌଳିକ ଧାରଣା Ex 3.3
• Chapter 3 ଜ୍ୟାମିତିରେ ମୌଳିକ ଧାରଣା Ex 3.4
• Chapter 3 ଜ୍ୟାମିତିରେ ମୌଳିକ ଧାରଣା Ex 3.5
• Chapter 3 ଜ୍ୟାମିତିରେ ମୌଳିକ ଧାରଣା Ex 3.6
• Chapter 3 ଜ୍ୟାମିତିରେ ମୌଳିକ ଧାରଣା InText Questions

Chapter 4 ସ୍ଵାଭାବିକ ସଂଖ୍ୟା

• Chapter 4 ସ୍ଵାଭାବିକ ସଂଖ୍ୟା Ex 4.1
• Chapter 4 ସ୍ଵାଭାବିକ ସଂଖ୍ୟା Ex 4.2
• Chapter 4 ସ୍ଵାଭାବିକ ସଂଖ୍ୟା Ex 4.3
• Chapter 4 ସ୍ଵାଭାବିକ ସଂଖ୍ୟା Ex 4.4
• Chapter 4 ସ୍ଵାଭାବିକ ସଂଖ୍ୟା Ex 4.5
• Chapter 4 ସ୍ଵାଭାବିକ ସଂଖ୍ୟା InText Questions

Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା

• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା Ex 5.1
• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା Ex 5.2
• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା Ex 5.3
• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା Ex 5.4
• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା Ex 5.5
• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା Ex 5.6
• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା Ex 5.7
• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା Ex 5.8
• Chapter 5 ଭଗ୍ନ ସଂଖ୍ୟା InText Questions

6th Class Math Book Pdf Odia Medium Chapter 6 ଦଶମିକ ସଂଖ୍ୟା

• Chapter 6 ଦଶମିକ ସଂଖ୍ୟା Ex 6.1
• Chapter 6 ଦଶମିକ ସଂଖ୍ୟା Ex 6.2
• Chapter 6 ଦଶମିକ ସଂଖ୍ୟା Ex 6.3
• Chapter 6 ଦଶମିକ ସଂଖ୍ୟା Ex 6.4
• Chapter 6 ଦଶମିକ ସଂଖ୍ୟା Ex 6.5
• Chapter 6 ଦଶମିକ ସଂଖ୍ୟା InText Questions

Chapter 7 ବ୍ୟାବସାୟିକ ଗଣିତ

• Chapter 7 ବ୍ୟାବସାୟିକ ଗଣିତ Ex 7.1
• Chapter 7 ବ୍ୟାବସାୟିକ ଗଣିତ Ex 7.2
• Chapter 7 ବ୍ୟାବସାୟିକ ଗଣିତ Ex 7.3
• Chapter 7 ବ୍ୟାବସାୟିକ ଗଣିତ Ex 7.4
• Chapter 7 ବ୍ୟାବସାୟିକ ଗଣିତ Ex 7.5
• Chapter 7 ବ୍ୟାବସାୟିକ ଗଣିତ InText Questions

Class 6 Math Book Odia Medium Chapter 8 ପୂର୍ଣ୍ଣ ସଂଖ୍ୟା

• Chapter 8 ପୂର୍ଣ୍ଣ ସଂଖ୍ୟା Ex 8.1
• Chapter 8 ପୂର୍ଣ୍ଣ ସଂଖ୍ୟା Ex 8.2
• Chapter 8 ପୂର୍ଣ୍ଣ ସଂଖ୍ୟା Ex 8.3
• Chapter 8 ପୂର୍ଣ୍ଣ ସଂଖ୍ୟା InText Questions

Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି

• Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି Ex 9.1
• Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି Ex 9.2
• Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି Ex 9.3
• Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି Ex 9.4
• Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି Ex 9.5
• Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି Ex 9.6
• Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି Ex 9.7
• Chapter 9 ସମତଳ ଉପରିସ୍ଥ ଜ୍ୟାମିତିକ ଆକୃତି InText Questions

Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ

• Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ Ex 10.1
• Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ Ex 10.2
• Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ Ex 10.3
• Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ Ex 10.4
• Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ Ex 10.5
• Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ Ex 10.6
• Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ Ex 10.7
• Chapter 10 ବୀଜଗଣିତ ସହିତ ପରିଚିତ InText Questions

Class 6 Math Odia Medium Question Answer Chapter 11 ପରିମିତି

• Chapter 11 ପରିମିତି Ex 11.1
• Chapter 11 ପରିମିତି Ex 11.2
• Chapter 11 ପରିମିତି Ex 11.3
• Chapter 11 ପରିମିତି InText Questions

Chapter 12 ତଥ୍ୟ ପରିଚାଳନା ଓ ସଂରଚନା

• Chapter 12 ତଥ୍ୟ ପରିଚାଳନା ଓ ସଂରଚନା Ex 12.1
• Chapter 12 ତଥ୍ୟ ପରିଚାଳନା ଓ ସଂରଚନା InText Questions

6th Class Math Question Answer Odia Medium Chapter 13 ଜ୍ୟାମିତିକ ଅଙ୍କନ

• Chapter 13 ଜ୍ୟାମିତିକ ଅଙ୍କନ Ex 13.1
• Chapter 13 ଜ୍ୟାମିତିକ ଅଙ୍କନ Ex 13.2
• Chapter 13 ଜ୍ୟାମିତିକ ଅଙ୍କନ Ex 13.3
• Chapter 13 ଜ୍ୟାମିତିକ ଅଙ୍କନ Ex 13.4
• Chapter 13 ଜ୍ୟାମିତିକ ଅଙ୍କନ Ex 13.5
• Chapter 13 ଜ୍ୟାମିତିକ ଅଙ୍କନ Ex 13.6
• Chapter 13 ଜ୍ୟାମିତିକ ଅଙ୍କନ InText Questions

BSE Odisha 6th Class Text Book Solutions

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 4 The Three Questions Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 4 The Three Questions

BSE Odisha 6th Class English Follow-Up Lesson 4 The Three Questions Text Book Questions and Answers

Session – 1 ( ସୋପାନ – ୧)

I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):
Here is a similar story for you. Read the story and do the activities. Some activities are given. The rest of the activities will be designed by your teacher. S/he will write them on the blackboard and help you do the tasks.
(ଏଠାରେ ତୁମ ପାଇଁ ଗୋଟିଏ ଏକାପରି ଗପ ଅଛି । ଗପଟିକୁ ପଢ଼ ଏବଂ କାର୍ଯ୍ୟାବଳୀଟିକୁ କର । କେତେକ କାର୍ଯ୍ୟାବଳୀ ଦିଆଯାଇଛି । ଅବଶିଷ୍ଟ କାର୍ଯ୍ୟାବଳୀ ତୁମ ଶିକ୍ଷକଙ୍କଦ୍ୱାରା ପ୍ରସ୍ତୁତ କରାଯିବ । ସେ (ପୁ/ସ୍ତ୍ରୀ) ସେଗୁଡ଼ିକୁ କଳାପଟାରେ ଲେଖିଦେବେ ଏବଂ ଏହି ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର କରିବାରେ ତୁମକୁ ସାହାଯ୍ୟ କରିବେ ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ)
Text – (ପାଠ୍ୟବସ୍ତୁ)

Read paragraphs 1-3 silently and answer the questions that follow.
(ଅନୁଚ୍ଛେଦ ୧ – ୩ କୁ ନୀରବରେ ପଢ଼ା ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)
1. Once there was a great king in Germany. He liked soldiers very much because he was a good soldier himself. He had a special liking for one section of tall soldiers in his army. He kept that section under his own care and loved to watch it. He wanted to make it the best section of the army and tried hard to get the best men for it.

2. At the first meeting, the king used to ask every new soldier the following three questions.
“How old are you?
How long have you served me already?
Do you like the food or the quarters here ?”

3. Once a man from another country came to join the German army. He did not know the German language, “How can I answer the king’s questions ?” he said, “I can’t understand them.”

ଓଡ଼ିଆ ଅନୁବାଦ :
(୧) ଏକଦା (ଥରେ) ଜର୍ମାନୀ ଦେଶରେ ଜଣେ ବିଶିଷ୍ଟ (ମହାନ୍) ରାଜା ଥିଲେ । ସେ ସୈନିକମାନଙ୍କୁ ବହୁତ ଭଲ ପାଉଥିଲେ କାରଣ ସେ ନିଜେ ଜଣେ ଭଲ ସୈନିକ ଥିଲେ । ତାଙ୍କର ତାଙ୍କ ସୈନ୍ୟବାହିନୀର ଏକ ଡେଙ୍ଗା ସୈନ୍ୟ ବିଭାଗ ପାଇଁ ସ୍ବତନ୍ତ୍ର ପସନ୍ଦ|ରୁଚି ଥିଲା । ସେ ସେହି ସୈନ୍ୟ ବିଭାଗ (ଦଳକୁ) ତାଙ୍କ ନିଜ ଦାୟିତ୍ଵରେ ରଖିଥିଲେ ଏବଂ ଏହାକୁ ଦେଖିବାକୁ ଭଲ ପାଉଥିଲେ । ସେ ଏହାକୁ ସୈନ୍ୟବାହିନୀର ସବୁଠାରୁ ଭଲ ବିଭାଗ (ଦଳ) ରୂପେ ଗଢ଼ିବାକୁ ଚାହୁଁଥିଲେ ଏବଂ ଏଥ‌ିପାଇଁ ସର୍ବୋତ୍ତମ ଲୋକମାନଙ୍କୁ ପାଇବାକୁ ବହୁତ ଚେଷ୍ଟା କରୁଥିଲେ ।

(୨) ପ୍ରଥମ ସାକ୍ଷାତ୍‌ରେ ରାଜା ପ୍ରତ୍ୟେକ ନୂଆ ସୈନିକଙ୍କୁ ପ୍ରାୟତଃ ନିମ୍ନଲିଖ୍ ତିନିଗୋଟି ପ୍ରଶ୍ନ ପଚାରୁଥିଲେ ।
ତୁମେ କେତେ ସମୟଧରି ମୋତେ ସେବା ପ୍ରଦାନ କରିଛ ? ତୁମେ ଏଠାରେ ଖାଦ୍ୟ କିମ୍ବା ବାସଗୃହ ପସନ୍ଦ କରୁଛ ?

(୩) ଥରେ ଅନ୍ୟ ଦେଶରୁ ଜଣେ ଲୋକ ଜର୍ମାନ୍ ସୈନ୍ୟବାହିନୀରେ ଯୋଗଦେବାକୁ ଆସିଲେ । ସେ ଜର୍ମାନ୍ ଭାଷା ଜାଣି ନଥୁଲେ, ସେ କହିଲେ, ‘‘କିପରି ମୁଁ ରାଜାଙ୍କର ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦେଇପାରିବି ? ମୁଁ ତ ସେଗୁଡ଼ିକୁ ବୁଝିପାରିବି ନାହିଁ ।?

Comprehension Questions – (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ) :

Question 1.
Why did the king like soldiers ?
(କାହିଁକି ରାଜା ସୈନ୍ୟବାହିନୀକୁ ଭଲ ପାଉଥିଲେ ?)
Answer:
The king liked soldiers very much because he was a good soldier himself.

Question 2.
How many questions did the king ask a new soldier?
(ଜଣେ ନୂଆ ସୈନିକକୁ ରାଜା କେତେଗୋଟି ପ୍ରଶ୍ନ ପଚାରୁଥିଲେ ?)
Answer:
The king used to ask three questions to a new soldier.

Question 3.
What was the problem with the new soldier?
(ନୂଆ ସୈନିକର କ’ଣ ସମସ୍ୟା ଥିଲା ?)
Answer:
The problem with the new soldier was that he did not know the German language.

Session – 2 (ସୋପାନ – ୨)
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ)

Read paragraphs 4-6 silently and answer the questions that follow.
(ଅନୁଚ୍ଛେଦ ୪ – ୬ କୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)
4. His officer told him all three questions and the way of answering them. He said the king always asked the same questions in the same order. So the soldier decided to answer them in that order

5. One day the king came to visit the army. He saw the new soldier and began to ask him questions. But this time the questions were in a different order. The soldier did not know this because he did not. understand the king’s words at all.

6. King: How long have you been in my service?
Soldier: Thirty years, sir.
King: How is it? You look so young! What’s your age?
Soldier: Three weeks, sir.
King: What do you mean? Are you mad, or am I?
Soldier: Both sir.

ଓଡ଼ିଆ ଅନୁବାଦ :
(୪). ତାଙ୍କର ଉଚ୍ଚ/ପଦସ୍ଥ କର୍ମଚାରୀ ତାଙ୍କୁ ସମସ୍ତ ତିନୋଟି ପ୍ରଶ୍ନ ଏବଂ ସେଗୁଡ଼ିକୁ ଉତ୍ତରଦେବା ଶୈଳୀ କହିଦେଲେ । ସେ କହିଲେ ଯେ ରାଜା ସର୍ବଦା ସେହି ଏକା ପ୍ରଶ୍ନ ଏକାକ୍ରମରେ ପଚାରନ୍ତି । ତେଣୁ ସୈନିକଟି ସ୍ଥିରକଲା ସେଗୁଡ଼ିକୁ ସେହି କ୍ରମରେ ଉତ୍ତର ଦେବାକୁ ।
(୫) ଦିନେ ରାଜା ସୈନ୍ୟବାହିନୀକୁ ପରିଦର୍ଶନ କରିବାକୁ ଆସିଲେ । ସେ ନୂଆ ସୈନିକକୁ ଦେଖିଲେ ଏବଂ ତାଙ୍କୁ ପ୍ରଶ୍ନଗୁଡ଼ିକ ପଚାରିବାକୁ ଆରମ୍ଭ କଲେ । କିନ୍ତୁ ଏଥର ପ୍ରଶ୍ନଗୁଡ଼ିକ ଭିନ୍ନ କ୍ରମରେ ଥିଲା । ସୈନିକଟି ଏହା ଜାଣିନଥୁଲା କାରଣ ସେ ରାଜାଙ୍କର ଶବ୍ଦଗୁଡ଼ିକୁ ଆଦୌ ବୁଝିପାରୁ ନଥିଲା ।

(୬) ରାଜା: କେତେକାଳ ହେଲା ତୁମେ ମୋ’ର ସେବାରେ ନିୟୋଜିତ ଅଛ ?
ସନିକ: ତିରିଶ ବର୍ଷ, ମହାଶୟ ।
ରାଜା: ଏହା କିପରି ସମ୍ଭବ ? ତୁମେ ଏତେ ଯୁବକ/କମ୍ ବୟସର ଦେଖାଯାଉଛ ? ତୁମର ବୟସ କେତେ ?
ସନିକ: ତିନି ସପ୍ତାହ, ମହାଶୟ ।
ରାଜା: ତୁମେ କ’ଣ ଭାବୁଛ ? ତୁମେ ପାଗଳ ନା ମୁଁ ?
ସନିକ: ଉଭୟ, ମହାଶୟ ।

Comprehension Questions – (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ) :

Question 1.
Who helped the new soldier in preparing to answer the king’s questions?
(କିଏ ନୂଆ ସୈନିକକୁ ରାଜାଙ୍କ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦେବାପାଇଁ ପ୍ରସ୍ତୁତ କରିବାରେ ସାହାଯ୍ୟ କଲା ?)
Answer:
His officer helped the new soldier, in preparing to answer the king’s questions.

Question 2.
What did the king ask first?
(ରାଜା ପ୍ରଥମେ କ’ଣ ପଚାରିଲେ ?)
Answer:
The king first asked, “How long have you been in my service?

Question 3.
What did the soldier answer?
(ସୈନିକଟି କ’ଣ ଉତ୍ତର ଦେଲା ?)
Answer:
The soldier answered, “Thirty years, sir.”

Session – 3 ( ସୋପାନ – ୩)
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ)

1. Writing- (ଲେଖୁବା) :
(a) Answer the following questions
(କବିତାଟି କେଉଁ ବିଷୟରେ ଆଧାରିତ ?)

(i) Why did the king like soldiers?
(କାହିଁକି ରାଜା ସୈନିକମାନଙ୍କୁ ଭଲ ପାଉଥିଲେ ?)
Answer:
The king liked the soldiers very much because he was a good soldier himself.

(ii) How many questions did the king ask?
(ରାଜା କେତୋଟି ପ୍ରଶ୍ନ ପଚାରିଲେ ?)
Answer:
The king asked three questions.

(iii) What was the problem with the new soldier ?
(ନୂଆ ସୈନିକର କ’ଣ ସମସ୍ୟା ଥିଲା ?)
The solder did not know.
Answer:
The soldier did not know the German language. So the problem with the new soldier was how he could answer the king’s questions.

(iv) Who helped him?
(କିଏ ତାକୁ ସାହାଯ୍ୟ କଲା ?)
Answer:
His officer helped him to answer the king’s questions.

(v) Did the help work ?
(ସାହାଯ୍ୟଟି କାମରେ ଲାଗିଲା କି ?)
Answer:
No, the help did not work at all.

(vi) Why? (କାହିଁକି ?)
Because the king did
Answer:
Because the king did not ask the same questions in the same order. As the questions were in a different order, the soldier did not answer the king’s questions correctly.

b. Match the questions under ‘A’ with the answers under ‘B’.
(‘A” ତଳେ ଥ‌ିବା ପ୍ରଶ୍ନଗୁଡ଼ିକୁ ‘B’ ତଳେ ଥିବା ଉତ୍ତରଗୁଡ଼ିକ ସହ ମିଳାଅ ।)

Answer:

Answer:

Now write on what really happened based on your matching.
(ବର୍ତ୍ତମାନ ତୁମର ଯୋଡ଼ିବାକୁ ଭିତ୍ତିକରି ଯାହା ପ୍ରକୃତରେ ଘଟିଗଲା ଲେଖ ।)
When the king asked, “________ old _______?” The soldier replied, “__________” . Next , when he “How long _________ ?” The soldier replied, “___________”Finally, When____________________________ The soldier replied,___________________.
Answer:
When the king asked, “How old are you ?” The soldier replied, “Three weeks, sir.” Next, when he asked, “How long have you served me already? The soldier replied, “Thirty years, sir.” Finally, when the king asked. “Do you like the food or the quarters here ?’’ The soldier replied, “Both sir.’’

Word Note – (ଶବ୍ଦାର୍ଥ)
(The words / phrases have been defined mostly on contextual meanings.)
(ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧିକାଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି ।)

great – ବହୁତ ଭଲ
soldier – ସନିକ
liking – ପସନ୍ଦ
section – ବିଭାଗ
tall – ଲମ୍ବା
army – ସେନା
own – ନିଜର
care – ଯତ୍ନ
hard – କଠିନ
used to – ଅଭ୍ୟସ୍ତ
served – ପରିବେଷିତ
quarters – କ୍ୱାର୍ଟର୍ସ
German language – ଜର୍ମାନ ଭାଷା
order – କ୍ରମ
understand – ମୋତେ ବୁ
mad – ପାଗଳ
developed – ବିକଶିତ
satisfaction – ସନ୍ତୁଷ୍ଟ
anger – କ୍ରୋଧ
deaf – ବଧିର
seriously ill – ଗୁରୁତର ଅସୁସ୍ଥ
special liking- ବିଶେଷ ପସନ୍ଦ
same order – ସମାନ କ୍ରମ
decided – ସ୍ଥିର କଲା
visit – ପରିଦର୍ଶନ କରନ୍ତୁ
how long – କେତେ ଦିନ
began – ଆରମ୍ଭ ହେଲା
at all – ସବୁ ସମୟରେ
service – ସେବା
so young – ଏତେ ଯୁବକ
age – ବୟସ
weeks – ସପ୍ତାହଗୁଡିକ
mean – ଅର୍ଥ
both – ଉଭୟ
classmate – ସହପାଠୀ
problems – ସମସ୍ୟାଗୁଡିକ
finally – ଶେଷରେ

Odisha State Board BSE Odisha 6th Class English Solutions Test-2(A) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-2(A)

BSE Odisha 6th Class English Test-2(A) Text Book Questions and Answers

□ The figures in the right-hand margin indicate the marks for each question.
1. Write the following names of persons in English.
(Teacher will provide names of six persons in Odia.)

ଡାକ୍ତର ରାଜେନ୍ଦ୍ର ପ୍ରସାଦ
ରଣଜିତ ସିଂହ
ଅରବିନ୍ଦ ଘୋଷ
ବାଘା ଜତିନ
ସମ୍ରାଟ ଅଶୋକ |
ରାଜା ଦଶରଥ

Answer:
Doctor Rajendra Prasad
Ranjit Singh
Aurobindo Ghose
Bagha Jatin
Emperor Ashok
King Dasaratha

2. Write the following names of places in English.
(Teacher will provide names of six places in Odia.)

ପୁଡୁଚେରୀ |
ଶ୍ରୀ ଲଙ୍କା
ବିଜୟ ନାଗର
ମଥୁରା |
ମୁମ୍ବାଇ |
କପିଳାସ

Answer:
Puducherry
Srilanka
Vijaya Nagar
Mathura
Mumbai
Kapilas

3. Your teacher will give a dictation of twelve words. Listen to him/her and write.

Answer:
village
classmate
friend
decided
deaf
Good Morning
fever
Germany
soldier
quarters
country
understand

4. Given below are some words. Your teacher will read aloud seven of them. Tick those s/he reads aloud.
language, young, neither, retire, gunny bag, religious, sight, greedy, weather, straightened, beautiful, special, elephant, bicycle
[Listen to your teacher carefully and tick those words as he reads aloud.]

5. Your teacher will read aloud a paragraph. Listen to him/her and fill in the gaps. (Question with Answer)
Once there lived a poor man in a village. He had a rich classmate. He lived in a town. They did not meet for a long time. In the meantime, the rich friend had problems with his ear and became deaf. His friend in the village could not know this.

6. Match the words which sound alike at the end. (Question with Answer)

Answer:

7. Read The poem and answer the questions incomplete sentences.

There was a dog and there was a cat.
One very thin and the other is fat.
Neither of them was a pet.
But the cat always sat on a mat.
and claimed she was a loving pet.

Question (i)
Who were there in the poem?
Answer:
In the poem, there were a dog and a cat.

Question (ii)
How were they?
Answer:
One was very thin and the other was fat.

Question (iii)
Were they pets?
Answer:
No, neither of them was a pet.

Question (iv)
Who sat on a mat?
Answer:
The cat always sat on a mat.

Question (v)
What did the cat claim?
Answer:
The cat claimed that she was a loving pet.

(vi)
Who was very thin?
Answer:
The dog was very thin.

8. Read the following paragraph and answer the questions in complete sentences.
“There is a special school in Karagudi. It is special because it is not for children. Can you guess for whom it is ? It is for baby elephants. Who teaches them? The elephant trainers teach them. Like our schools, they have a timetable. They learn, play and eat according to this timetable.

Question (i)
What is there in Karagudi?
Answer:
There is a special school in Karagudi.

Question (ii)
Why is this school special?
Answer:
This school is special because it is not for children.

Question (iii)
Who are the students?
Answer:
Baby elephants are the students.

Question (iv)
Who are the teachers?
Answer:
The elephant trainers are the teachers.

Question (v)
What do they have like our schools?
Answer:
Like our schools, they have a timetable.

Question (vi)
How do they follow it?
Answer:
According to this timetable, they learn, play, and eat in the special school.

9. Read the following poem and answer the questions in complete sentences.

It was a very cool night And there
was no crab in sight.
The fox looked for one
But there was none.
“Where did they go ?”
Not even one in sight!
They must be in their holes If
I’m right.”

Question (i)
How was the night?
Answer:
It was a very cool night.

Question (ii)
What was not in sight?
Answer:
There was no crab in sight.

Question (iii)
Who looked for the crab?
Answer:
The fox looked for the crab.

Question (iv)
Did he find one?
Answer:
No, he did not find any crab, because there was none.

Question (v)
Where did they go?
Answer:
They must have been in their holes.

Question (vi)
Who were there in the poem?
Answer:
There were the fox and the crab in the poem.

10. Read the following paragraph and answer the questions in complete sentences.
Mahagiri was a big elephant. He was trained at a special school. He was bought by a merchant. The merchant made a lot of money by putting Mahagiri to work. The elephant was often sent to the forest to carry heavy logs of wood. Sometimes, he carried people from one place to another. Once, he even carried a bridegroom to the bride’s house! At times he was sent to a famous temple in a village nearby to lead the festival procession.

Question (i)
What is this paragraph about?
Answer:
This paragraph is about an elephant.

Question (ii)
Who was Mahagiri?
Answer:
Mahagiri was a big elephant.

Question (iii)
Where was Mahagiri trained?
Answer:
Mahagiri was trained at a special school.

Question (iv)
Who bought it?
Answer:
A merchant bought it.

Question (v)
How did the merchant make a lot of money?
Answer:
The merchant made a lot of money by putting Mahagiri to work.

Question (vi)
Why was the elephant often sent to the forest?
Answer:
The elephant was often sent to the forest to carry heavy logs of wood.

Question (vii)
What did he sometimes carry?
Answer:
Sometimes, he carried people from one place to another.

Question (viii)
What did he once even carry?
Answer:
Once, he even carried a bridegroom to the bride’s house.

Question (ix)
Where was he sent to at times?
Answer:
At times he was sent to a famous temple in a village nearby.

Question (x)
Why was he sent to a famous temple?
Answer:
He was sent to a famous temple to lead the festival procession.

Odisha State Board BSE Odisha 6th Class English Solutions Test-2(B) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-2(B)

BSE Odisha 6th Class English Test-2(B) Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.
1. Write the following names of persons in English.
(Teacher will provide names of six persons ¡n Odia.)

ରାଜା ଦିବ୍ୟାସିଂହ ଦେବ
ରାଜା ରାମମୋହନ ରୟ |
ରାଜା ହରିଚନ୍ଦ୍ର
ରାମଚନ୍ଦ୍ର
ଚାଖି ଖୁଣ୍ଟିଆ
ଲକ୍ଷ୍ମଣ କୁମାର

Answer:
Raja Dibvasingh Deb
Raja Rammohan Roy
Raja Harischandra
Ramchandra
Chakhi Khuntia
Lakshmana Kumar

2. Write the following names of places ¡n English.
(Teacher will provide names of six places in Odia.)

ଦୌଲତାବାଦ
ସଂସଦ ଭବନ
ଜୁମ୍ମା ମସଜିଦ୍
ଦ୍ୱାରିକାପୁର
ଅୟୋଧ୍ୟା
ହସ୍ତିନା

Answer:
(i) Doulatabad
(ii) Parliament House
(iii) Jumma Mosque
(iv) Dwarikapura
(v) Ayodhya
(vi) Hasina

3. Your teacher will give a dictation of twelve words. Listen to him/her and write.

Answer:
Tamilnadu
children
elephant
instructions
trainer
circus
special
guess
naughty
raises
praised
popular

4. Given below are some words. Your teacher will read aloud seven of them. Tick those s/he reads aloud.
perhaps, sailor, zoo, trumpet, musical, private, driver, moon, doctor, pilot, cousin, farmer, builder, nurse, painter.
[Listen to your teacher carefully and tick those words as he reads aloud.]

5. Your teacher will read aloud a paragraph. You listen to him/her and fill in the gaps. (Question with Answer)
On Makar holidays Raghunath would come to his village Dandbose, a few kilometers away from Rairangpur town in the district of Mavurbhanj. He was then working at Baripada. In those days he was the only educated man in his area.

6. Match the words which sound alike at the end. (Question with Answer)

Answer:

7. Read the poem and answer the questions in complete sentences.

When I grow up
I want to be
A detective
With a master key.
I could be a soldier
Perhaps a sailor too.
Or become a keeper
At Nandan Kanan Zoo.
I’d like to own a trumpet
And play a musical tune.
Or buy a private space-ship
To fly to the moon.

Question (i)
Who is ‘I’ in the poem?
Answer:
The poet is T in the poem.

Question (ii)
What does the child want to be in the 1st stanza?
Answer:
In the first stanza, the child wants to be a detective.

Question (iii)
What does a detective have with him?
Answer:
A detective has a master key with him.

Question (iv)
In the second stanza, the child likes three professions. What are they?
Answer:
In the second stanza, the child likes three professions. They are soldiers, sailors, and keepers at Nandan Kanan zoo.

Question (v)
In which stanza does the poet describe a child’s interest in music?
Answer:
In the third stanza of the poem, the poet describes a child’s interest in music.

Question (vi)
How does he want to fly to the moon?
Answer:
He wants to buy a private spaceship in order to (960) fly him to the moon.

8. Read the following paragraph and answer the questions in complete sentences.
On Makar holidays Raghunath would come to his village Dandbose, a few kilometers away from Rairangpur town in the district of Mayurbhanj. He was then working at Baripada.

Question (i)
What is this paragraph about?
Answer:
This paragraph is about Raghunath.

Question (ii)
What is the name of his village?
Answer:
The name of his village is Dandbose.

Question (iii)
When would he come there?
Answer:
He would come there on the Makar holidays.

Question (iv)
How far is it from Rairangpur town?
Answer:
It is a few kilometers away from Rairangpur town.

Question (v)
Where is Rairangpur?
Answer:
Rairangpur is in the district of Mayurbhanj.

Question (vi)
Where was he working then?
Answer:
He was then working at Baripada.

9. Read the following poem and answer the questions in complete sentences.

“My father is a doctor.
My sister’s a doctor too.
My cousin works with animals.
He’s a keeper at the zoo.
What can I be?
What do I want to do?
I don’t want to be a farmer,
A builder or a nurse.
I don’t want to be a pilot, that is even worse.

Question (i)
What is the name of the poem?
Answer:
The name of the poem is “What can I be ?”

Question (ii)
Who is ‘I’ in the poem?
Answer:
The poet/child is T in the poem.

Question (iii)
What are the child’s father and sister?
Answer:
Both the child’s father and sister are doctors.

Question (iv)
Who is a keeper at the zoo?
Answer:
His cousin is a keeper at the zoo?

Question (v)
What doesn’t he want to be?
Answer:
He doesn’t want to be a farmer, a builder, a nurse, and a pilot.

Question (vi)
Whose job is worse?
Answer:
A pilot’s job is even worse.

10. Read the following paragraph and answer the questions in complete ententes.
“Raghunath Murmu was also a great writer. He had written many plays, novels, and poems in Santali. His most important play is ‘Kherwar Bir”. Martin Orans, a foreign scholar and writer, called this the Santali Mahabharata. Raghunath was awarded by the Odisha Sahitya Academy for his contribution to the Santali language and literature. The Government of Odisha has named the Medical College at Baripada after his name. What Fakir Mohan Senapati is to Odia language and literature, Raghunath Murmu is to Santali language and literature.

Question (i)
What else was Raghunath Murmu?
Answer:
Raghunath Murmu was also a great writer.

Question (ii)
What did he write in Santali?
Answer:
He had written many plays, novels and poems in Santali.

Question (iii)
Which book is Raghunath’s most important play?
Answer:
Raghunath’s most important play is “Kherwar Bir”.

Question (iv)
Who was Martin Orans?
Answer:
Martin Orans was a foreign scholar and writer.

Question (v)
Was he in high praise of Raghunath’s writings?
Answer:
Surely, he was in high praise of Raghunath’s writings.

Question (vi)
Which book is called (he Santal Mahabharat)?
Answer:
The book “Kherwar Bir” is called the Santal Mahabharata.

Question (vii)
What did the Odisha Sahitya Academy award him for?
Answer:
Raghunath Murmu was awarded by the Odisha Sahitya Academy for his contribution to Santali language and literature.

Question (viii)
What has the Government of Odisha done in his honor?
Answer:
In his honor, the Government of Odisha has named the Medical College at Baripada after his name.

Question (ix)
Who is Raghunath Murmu compared to?
Answer:
Raghunath Murmu is compared to Fakir Mohan Senapati.

Question (x)
How are they equal?
Answer:
What Fakir Mohan Senapati is to Odia language and literature, Raghunath Murmu is to Santali language and literature.

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 7 What Can I Be? Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 7 What Can I Be?

BSE Odisha 6th Class English Follow-Up Lesson 7 What Can I Be? Text Book Questions and Answers

Session – 1

I – Pre-Reading
□ Pre-reading questions

1. What is your father’s job? What is your mother’s job?
(ତୁମ ବାପାଙ୍କର ବୃତ୍ତି କ’ଣ ? ତୁମ ମାଆଙ୍କର ବୃତ୍ତି କ’ଣ ?)
2. What would you like to be in the future? What do you see in the picture?
(ତୁମେ ଭବିଷ୍ୟତରେ କ’ଣ ହେବାକୁ ପସନ୍ଦ କର ? ତୁମେ ଛବିରେ କ’ଣ ଦେଖୁଛ ?)

II. While-Reading

Text

Read the poem silently and answer the questions that follow.

My father is a doctor.
My sister’s a doctor too.
My cousin works with animals.
He’s- a keeper at the zoo. 4
What can I be?
What do I want to do?
I don’t want to be a farmer,
A builder or a nurse.
I don’t want to be a pilot,
that is even worse. 10
I don’t want to be a painter
But a keeper at the zoo.
That’s what I’ll be.
That’s what I want to do. 14

କବିତାର ଓଡ଼ିଆ ଉଚ୍ଚାରଣ :
ମାଇ ଫାଦର୍‌, ଇଜ୍ ଏ ଡକ୍ଟର ।
ମାଇ ସିଷ୍ଟର୍’ଜ୍ ଏ ଡକ୍ଟର ଠୁ।
ମାଇ କଜିନ୍ ୱାକ୍‌ସ୍‌ ଉଇଥ୍ ଆନିମାଲ୍‌ ।
ହି’ଜ୍ ଏ କିପର୍ ଆଟ୍ ଦ’ ଜୁ ।
ହ୍ମାଟ୍ କ୍ୟାନ୍ ଆଇ ବି ?
ଦ୍ଵାଟ୍ ଡୁ ଆଇ ୱାଣ୍ଟ୍ ଟୁ ଡୁ ?
ଆଇ ଡୋ’ଣ୍ଟ ୱାଣ୍ଟ ଟୁ ବି ଏ ଫାର୍ମର,
ଏ ବିଲ୍‌ଡ଼ର୍ ଅର୍ ଏ ନର୍ସ ।
ଆଇ ଡୋ’ଣ୍ଟ୍‌ ୱାଣ୍ଟ୍ ଟୁ ବି ଏ ପାଇଲଟ୍
ଦ୍ୟାଟ୍ ଇଜ୍ ଇଭେନ୍ ଓର୍ସ୍ ।
ଆଇ ଡୋ’ଣ୍ଟ୍ ୱାଣ୍ଟ୍ ଟୁ ବି ଏ ପେଣ୍ଟର୍,
ବଟ୍ ଏ କିପର୍ ଆଟ୍ ଦ’ ଜୁ ।
ବ୍ୟାଟ୍’ଜ୍ ହ୍ୱାଟ୍ ଆଇ’ଲ ବି ।
ବ୍ୟାଟ୍’ଜ୍ ହ୍ୱାଟ୍ ଆଇ ୱାଣ୍ଟ୍ ଟୁ ଡୁ ।

କବିତାର ସାରକଥା :
ମୋ ବାପା ଜଣେ ଡାକ୍ତର ।
ମୋ ଭଉଣୀ ମଧ୍ୟ ଜଣେ ଡାକ୍ତର ।
ମୋ ସମ୍ପର୍କୀୟ ଭାଇ ପଶୁମାନଙ୍କର ସହ କାର୍ଯ୍ୟ କରେ ।
ସେ ଜଣେ ଚିଡ଼ିଆଖାନାର କର୍ମଚାରୀ ଅଟେ ।
ମୁଁ କ’ଣ ହୋଇପାରିବି ?
କ’ଣ କରିବାକୁ ଚାହେଁ ?
ଚାହେଁନା ଜଣେ କୃଷକ ହେବାକୁ,
ଜଣେ ନିର୍ମାଣକାରୀ କିମ୍ବା ଧାଈ ।
ମୁଁ ଚାହେଁନା ଜଣେ ଉଡ଼ାଜାହାଜ ଚାଳକ ହେବାକୁ,
ତାହା ମଧ୍ୟ ଏପରିକି ବହୁତ ଖରାପ ।
ମୁଁ ଜଣେ ଚିତ୍ରକର ହେବାକୁ ଚାହେଁନା,
କିନ୍ତୁ ଜଣେ ଚିଡ଼ିଆଖାନାର କର୍ମଚାରୀ ।
ତାହା ଅଟେ ଯାହା ମୁଁ ହେବି ।
ତାହା ଅଟେ ଯାହା ମୁଁ କରିବାକୁ ଚାହେଁ ।

• Your teacher reads the poem aloud. You listen to him/her without opening the book. Your teacher asks you: Who are there in this poem?
  (ତୁମ ଶିକ୍ଷକ କବିତାଟିକୁ ବଡ଼ପାଟିରେ ପଢ଼ିବେ । ବହି ନ ଖୋଲି, ତୁମେ ତାଙ୍କୁ (ପୁ/ସ୍ତ୍ରୀ) ମନଦେଇ ଶୁଣ । ତୁମ ଶିକ୍ଷକ ତୁମକୁ ପଚାରିବେ : ‘‘ଏହି କବିତାରେ କେଉଁମାନେ ଅଛନ୍ତି ?’’)
• Your teacher reads the poem aloud a second time. You listen to him/her and at the same time see the poem.
  (ତୁମ ଶିକ୍ଷକ କବିତାଟିକୁ ବଡ଼ପାଟିରେ ଦ୍ଵିତୀୟଥର ପଢ଼ିବେ । ତୁମେ ତାଙ୍କୁ (ପୁ/ସ୍ତ୍ରୀ) ମନଦେଇ ଶୁଣିବ ଏବଂ ସେହି ସମୟରେ କବିତାଟିକୁ ଦେଖୁବ ।)

Comprehension Questions

Question 1.
How many stanzas are there in the poem?
(କବିତାରେ କେତୋଟି ପଦ ଅଛି ?)
Answer:
There are three stanzas in the poem.

How many lines are there in each stanza?
(ପ୍ରତ୍ୟେକ ପଦରେ କେତୋଟି ଧାଡ଼ି ରହିଛି ?)
Answer:
There are four lines each in 1st and last stanzas. But there are six lines in the second stanza.

Question 2.
Who is ‘F in the poem?
(କବିତାରେ ‘I? କିଏ ?)
Answer:
In the poem, T is the poet or the child.

Question 3.
What is the poem about?
(କବିତାଟି କେଉଁ ବିଷୟରେ ଲେଖାଯାଇଛି ?)
Answer:
The poem is about what the child/poet can be!

Question 4.
How many questions are there in the poem?
(କବିତାରେ କେତୋଟି ପ୍ରଶ୍ନ ରହିଛି ?)
Answer:
There are two questions in the poem.

What are they?
(ସେଗୁଡ଼ିକ କ’ଣ ?)
Answer:
They are – “What can I do? What do I want to do ?”

Question 5.
What are the child’s father and sister?
(ପିଲାଟିର ବାପା ଓ ଭଉଣୀ କ’ଣ ଅଟନ୍ତି ? )
Answer:
The child’s father and sister both are doctors.

Question 6.
Who is a zoo-keeper ?
(କିଏ ଚିଡ଼ିଆଖାନାର କର୍ମଚାରୀ ଅଟେ ?)
Answer:
His cousin is a zoo keeper.

Question 7.
Whose job is worse?
(କାହାର କାମ ଅଧିକ ଖରାପ ?)
Answer:
A pilot’s job is worse.

Question 8.
Which stanza tells you that the poet wants to be a keeper at the zoo?
(କେଉଁ ପଦଟି ତୁମକୁ କହେ ଯେ କବି ଚିଡ଼ିଆଖାନାର ଜଣେ କର୍ମଚାରୀ ହେବାକୁ ଚାହାଁନ୍ତି ?)
Answer:
The third stanza tells us that the poet wants to be a keeper at the zoo.

Question 9.
Is the poet happy? Why? Why not?
(କବି ସୁଖୀ କି ? କାହିଁକି ବା କାହିଁକି ନୁହେଁ ?)
Answer:
The poet is really happy. Because he wants to be a zoo keeper.

Question 10.
How many times the following words are repeated ? (Question with Answer)

Answer:
a. my 3 times
b. keeper 2 times
c. doctor 2 times
d. I 7 times

Session – 3

III. Post-Reading

1. Writing

(a) Answer the following questions.
(ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

Question (i)
What does the child / poet want to be ?
(ଶିଶୁ | କବି କ’ଣ ହେବାକୁ ଚାହାନ୍ତି ?)
Answer:
The child / poet wants to be a zoo-keeper.

Question (ii)
What is his father?
(ତାଙ୍କ ବାପା କ’ଣ ଅଟନ୍ତି ?)
Answer:
His father is a doctor.

Question (iii)
What is his sister?
(ତାଙ୍କ ଭଉଣୀ କ’ଣ ଅଟନ୍ତି ?)
Answer:
His sister is a doctor.

Question (iv)
What does the child not want to be?
(ପିଲାଟି କ’ଣ ହେବାକୁ ଚାହୁଁ ନାହିଁ ?)
Answer:
The child does not want to be a farmer, a builder, a nurse, a pilot, and a painter too.

(b) Write your own poem (the last word of the second and the last lines which rhyme are given. Rest you can choose).
(ତୁମ ନିଜର ଗୋଟିଏ କବିତା ଲେଖ । (ଦ୍ଵିତୀୟ ଓ ଶେଷ ଧାଡ଼ିର ଶେଷ ଶବ୍ଦ ଯାହା ଯତିପାତ ପଡ଼ୁଛି ଦିଆଯାଇଛି । ଅବଶିଷ୍ଟ ତୁମେ ପସନ୍ଦ କରିପାରିବ ।)
(Question with Answer)
I don’t want to be a builder or a nurse.
I don’t want to be a farmer.
I don’t want to be a pilot
I don’t want to be a teacher.

Word Note
(The words / phrases have been defined mostly on contextual meanings.)
(ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧିକାଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି ।)

cousin – ବ୍ୟତୀତ ଅନ୍ୟ ଭାଇ
detective – ଗୁଇନ୍ଦା ପୋଲିସ
keeper at zoo – ଚିଡିଆଖାନା
sailor – ନାବିକ
soldier – ସୈନିକ
spaceship – ମହାକାଶଯାନ
taking turn – ପାଳିକରି କୌଣସି କାମ କରିବା
trumpet – ତୂରୀ, ବିଗୁଲ୍
master key – ମୁଖ୍ୟ ଚାବି
perhaps – ବୋଧହୁଏ, ପ୍ରାୟ
own – ନିଜର
that’s what – ସେଇଟା ଯାହା
musical tune – ସଙ୍ଗୀତ ସ୍ବର
buy – କିଣିବା
private – ବ୍ୟକ୍ତିଗତ
moon – ଚନ୍ଦ୍ର
I’d like (I would like) – ମୁଁ ପସନ୍ଦ କରେ ।
express – ପ୍ରକାଶ କରିବା
light-house – ଆଲୋକସ୍ତମ୍ଭ
builder – ନିର୍ମାଣକାରୀ
nurse – ସେବିକା
pilot – ବିମାନ ଚାଳକ
even – ଏପରିକି
Worse – ଅଧ୍ଵ ଖରାପ
painter – ଚିତ୍ରକାର

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 1 I Like Bats Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 1 I Like Bats

BSE Odisha 6th Class English Follow-Up Lesson 1 I Like Bats Text Book Questions and Answers

Session – 1 (ପ୍ରଥମ ପର୍ଯ୍ୟାୟ):
I. Pre-Reading (ପ୍ରାକ୍-ପଠନ):

→ Socialisation (ସାମାଜିକୀକରଣ):
→ The teacher thinks of a pre-reading activity.
(ଶିକ୍ଷକ ଏକ ପଢ଼ିବା ପୂର୍ବବର୍ତ୍ତୀ କାର୍ଯ୍ୟ ବିଷୟରେ ଚିନ୍ତା କରନ୍ତୁ ।)

→ You may use pictures. You may also link the poem with the poem of the main lesson.
(ତୁମେ ଛବିଗୁଡ଼ିକୁ ବ୍ୟବହାର କରିପାର । ତୁମେ କବିତାଟିକୁ ମୁଖ୍ୟପାଠର କବିତା ସହ ଯୋଡ଼ିପାର ।)

→ What are there in the picture? What do they look like? That’s how bats sleep and rest hanging upside down. What an interesting way of resting and relaxing! Do you like to rest like bats hanging upside down?
(ଛବିରେ କ’ଣ ଅଛି ? ସେମାନେ କିପରି ଦେଖାଯାଉଛନ୍ତି ? ସେହିପରି ଭାବରେ ବାଦୁଡ଼ିମାନେ ଉପରୁ ତଳକୁ ଝୁଲିରହି ଶୁଅନ୍ତି ଏବଂ ବିଶ୍ରାମ ନିଅନ୍ତି । ବିଶ୍ରାମ ନେବା ଓ ନିଦ୍ରାଯିବାର କି କୌତୂହଳଜନକ ଉପାୟ ! ତୁମେ ବାଦୁଡ଼ିମାନଙ୍କ ପରି ଉପରୁ ତଳକୁ ଝୁଲିରହି ବିଶ୍ରାମ କରିବାକୁ ଭଲ ପାଅ କି ?)

→ In the poem ‘Mice’ the poet likes mice. Let’s read this poem to see if the poet likes bats.
(‘Mice” କବିତାରେ, କବି ମୂଷାମାନଙ୍କୁ ଭଲ ପାଇଛନ୍ତି । ଆସ ଆମେ ଏହି କବିତାଟି ପଢ଼ିବା ଏବଂ କବି ବାଦୁଡ଼ିମାନଙ୍କୁ ଭଲ ପାଆନ୍ତି କି ନାହିଁ ଦେଖୁବା ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ ):
Follow the steps of the main lesson.
(ମୁଖ୍ୟ ବିଷୟର ସୋପାନଗୁଡ଼ିକୁ ଅନୁସରଣ କର ।)
TEXT (ବିଷୟବସ୍ତୁ):
Read the poem silently and answer the questions that follow.
(କବିତାଟିକୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

I like bats
Hanging upside down
Like rats. Like silk-cotton fruits
Swinging in wind
What a way to relax and rewind.
I wish I could
Do that
Like a bat
A way to find
After a day’s work
To relax and rewind
Upside down
Hang like bats.
I like bats
Hanging upside down
Like rats.

ଓଡ଼ିଆ ଉଚ୍ଚାରଣ :
ଆଇ ଲାଇକ୍ ବ୍ୟାଟ୍‌ସ୍
ହ୍ୟାଙ୍ଗିଙ୍ଗ୍ ଅପସାଇଡ୍ ଡାଉନ୍
ଲାଇକ୍ ମ୍ୟାଟ୍‌ସ୍ । ଲାଇକ୍ ସିଲ୍‌କ୍-କଟନ୍ ଫୁସ୍
ସୁଇଙ୍ଗିଙ୍ଗ୍ ଇନ୍ ଉଇଣ୍ଡ୍
ଦ୍ଵାଟ୍ ଏ ୱେ ଟୁ ରିଲାକ୍ସ ଆଣ୍ଡ ରିୱାଇଣ୍ଡ୍ ।
ଆଇ ଉଇସ୍ ଆଇ କୁଡ଼୍
ଡୁ ଦ୍ଯାଟ୍
ଲାଇକ୍ ଏ ବ୍ୟାଟ୍
ଏ ୱେ ଟୁ ଫାଇଣ୍ଡ୍
ଆଫ୍‌ଟର୍ ଏ ଡେ’ଜ୍ ୱାର୍କ
ଟୁ ରିଲାକ୍ସ ଆଣ୍ଡ ରିୱାଇଣ୍ଡ୍
ଅପ୍‌ସାଇଡ୍ ଡାଉନ୍
ଆଇ ଲାଇକ୍ ବ୍ୟାଟ୍‌ସ୍ ।
ହ୍ୟାଙ୍ଗିଙ୍ଗ୍ ଅଫ୍‌ସାଇଡ୍ ଡାଉନ୍
ଲାଇକ୍ ଗ୍ୟାସ୍ ।

Knowing The Key Words (ମୁଖ୍ୟ ଶବ୍ଦଗୁଡ଼ିକୁ ଜାଣିବା):

like – ସଦୃଶ
bats – ବାଦୁଡି
hanging – ଫାଶୀ
upside – ଓଲଟା |
down – ତଳକୁ
like – ପରି
rats – ମୂଷା
silk – cotton – ଶିମିଳି-ତୁଳା
fruits – ଫଳ
swinging – ସୁଇଙ୍ଗ୍
in wind — ପବନରେ
could — କରିପାରନ୍ତି |
What a way — କି ଉପାୟ |
find — ଖୋଜିବା
relax — ବିଶ୍ରାମ କରିବା
day’s work — ଦିନର କାମ
rewind – ରିଭାଇଣ୍ଡ୍ |
wish — ଇଚ୍ଛା

ସାରକଥା | ଓଡ଼ିଆ ଅନୁବାଦ:
ମୁଁ ବାଦୁଡ଼ିମାନଙ୍କୁ ଭଲପାଏ
ସେମାନେ ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ମୂଷାମାନଙ୍କ ସଦୃଶ । | ସେମାନେ ପବନରେ ଦୋଳି ଖେଳୁଥା’ନ୍ତି ଶିମିଳି ତୁଳା ଫଳଗୁଡ଼ିକ ପରି । | କି ସୁନ୍ଦର ଉପାୟ ବିଶ୍ରାମ କରିବାର ଏବଂ ପବନରେ ଦୋହଲିବାର । | ମୁଁ ଭାବୁଛି ମୁଁ ସେହିପରି କରି ପାରିଥା’ନ୍ତି ଏକ ବାଦୁଡ଼ି ପରି । | ଗୋଟିଏ ଦିନକର କାମ ପରେ ଏକ ଉପାୟ ଖୋଜି ପାଇବାକୁ ବିଶ୍ରାମ କରିବାକୁ ଏବଂ ପବନରେ ଦୋହଲିବାକୁ । | ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ବାଦୁଡ଼ିମାନଙ୍କ ସଦୃଶ । ମୁଁ ଭଲପାଏ ବାଦୁଡ଼ିମାନଙ୍କୁ ଯେଉଁମାନେ ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ମୂଷାମାନଙ୍କ ଭଳି ।

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

The teacher is to try to frame his/her own questions. Here are some for him/her.
(ଶିକ୍ଷକ ତାଙ୍କର ନିଜର ପ୍ରଶ୍ନଗୁଡ଼ିକ ତିଆରି କରିବାକୁ ଚେଷ୍ଟା କରିବେ । ଏଠାରେ କେତେକ ତାଙ୍କ (ପୁ/ସ୍ତ୍ରୀ) ପାଇଁ ଅଛି ।)

Question 1.
How do bats hang?
(ବାଦୁଡ଼ିମାନେ କିପରି ଝୁଲୁଥା’ନ୍ତି ?)
Answer:
Bats hang upside down.

Question 2.
What are bats compared to ?
(ବାଦୁଡ଼ିମାନଙ୍କୁ କାହା ସହିତ ତୁଳନା କରାଯାଇଛି ?)
Answer:
Bats are compared to rats.

Question 3.
Have you seen bats hanging upside down on trees in great numbers?
(ଗଛଗୁଡ଼ିକରେ ବାଦୁଡ଼ିମାନେ ବହୁ ସଂଖ୍ୟାରେ ଝୁଲି ରହିଥ‌ିବାର ତୁମେ ଦେଖୁଛ କି ?)
Answer:
Yes, we have seen bats hanging upside down on trees in great numbers.

Question 4.
Have you seen silk-cotton fruit hanging in great numbers? Do they look alike?
(ଶିମିଳି-ତୁଳା ଫଳ ବହୁ ସଂଖ୍ୟାରେ ଝୁଲୁଥ‌ିବା ତୁମେ ଦେଖୁଛ କି ? ସେଗୁଡ଼ିକ ଏକାପରି ଦେଖାଯାଆନ୍ତି କି ?)
Answer:
Yes, we have seen silk-cotton fruit hanging in great numbers. Really, they look alike.

Question 5.
What is the meaning of the word ‘rewind’? See the dictionary at the end of this lesson.
(‘ରିୱାଇଣ୍ଡ୍’ ଶବ୍ଦର ଅର୍ଥ କ’ଣ ? ଏହି ଅଧ୍ୟାୟର ଶେଷରେ ଥ‌ିବା ଅଭିଧାନ ବା ଶବ୍ଦାର୍ଥ ଦେଖ ।)
Answer:
The meaning of the word ‘rewind’ is taking a rest with occasional backward movement.
Teacher is to frame some questions from the second stanza.
(ଶିକ୍ଷକ ଦ୍ଵିତୀୟ ପଦରୁ କିଛି ପ୍ରଶ୍ନ ତିଆରି କରିବେ ।)

Session – 2 (ସୋପାନ – ୨)
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):
6. Writing (ଲେଖିବା ):

(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

(i) How do bats hang ?
(ବାଦୁଡ଼ିମାନେ କିପରି ଝୁଲନ୍ତି ?)
Bats hang ______________________________________.
Answer:
Bats hang upside down.

(ii) What are bats compared to ?
(ବାଦୁଡ଼ିମାନଙ୍କୁ କାହା ସହ ତୁଳନା କରାଯାଇଛି ?)
Bats are ______________________________________.
Answer:
Bats are compared to rats.

(iii) What swings in the wind?
(ପବନରେ କ’ଣ ଝୁଲିଥାଏ ?)
The silk – ______________________________________.
Answer:
The silk-cotton fruits swing in the wind.

(iv) Why does the poet like to hang like bats upside down?
(କବି କାହିଁକି ବାଦୁଡ଼ିମାନଙ୍କ ପରି ଉପରପାଖ ତଳକୁ କରି ଝୁଲିବାକୁ ଭଲପାଆନ୍ତି ?)
______________________________________to relax and ____________________________.
Answer:
The poet likes to hang like a bats upside down because it is a nice way to relax and rewind.

(b) Let us summarise the poem. Fill in the gaps.
(ଆସ ଆମେ କବିତାର ସାରାଂଶ ବାହାର କରିବା । ଶୂନ୍ୟସ୍ଥାନଗୁଡ଼ିକୁ ପୂରଣ କର ।) (Question with Answer)

The poet _____________________to see bats. _____________________. Bats hang like _____________. This is a good way to relax and ____________. The poet wants to_____________ _____________bats to relax ________________ ________________after the day’s ________________.
Answer:
The poet likes to see bats. They are hanging upside down. Bats hang like rats and like silk-cotton fruits swinging in wind. This is a good way to relax and rewind. The poet wants to hang upside down like a bat to relax and rewind after the day’s work.

(c) Think of writing a poem. Start with replacing ‘bats’ with some fruits and make minimum changes in the poem. Change the title accordingly.
(ଗୋଟିଏ କବିତା ଲେଖିବା କଥା ଭାବ । ‘ବାଦୁଡ଼ିମାନଙ୍କ’’ ବଦଳରେ କେତେକ ଫଳକୁ ନେଇ ଏବଂ କବିତାରେ ସ୍ଵଳ୍ପ ପରିବର୍ତ୍ତନ କରି ଆରମ୍ଭ କର । ସେହି ଅନୁସାରେ କବିତାର ଶିରୋନାମା ପରିବର୍ତ୍ତନ କର ।)

APPLES
Answer:
I like apples
Hanging upside down
Like rats. Like silk-cotton fruits
Swinging in wind
What a way to relax and rewind.
I wish I could
Do that
Like an apple
A way to find
After a day’s work
To relax and rewind
Upside down Hang like apples.
I like apples
Hanging upside down
Like rats.

Knowing The Key Words (ମୁଖ୍ୟ ଶବ୍ଦଗୁଡ଼ିକୁ ଜାଣିବା):
(The words/phrases have been defined mostly on contextual meanings.)

Nibble – gentle and playful bite of a mouse. ମୂଷାର କୁଟ୍ କୁଟ୍ କରି କାଟି ଖାଇବା
Pink – (colour) pale red, ଫିକା ନାଲି |
Swinging – hanging and moving (bats have)
relax and rewind – taking rest (with occassional backward movement) ଆରାମରେ ବିଶ୍ରାମ ନେବା
upside down – legs upward and head downward. ଗୋଡ଼ ଉପରକୁ ଓ ମୁଣ୍ଡ ତଳକୁ କରି ଓଲଟା ରହିବା ।