## KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

Students can Download Chapter 8 Decimals Ex 8.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 6 Maths Chapter 8 Decimals Ex 8.1

Question 1.
Write the following as numbers in the given table. Solution:
It may be observed that Mixed number 2 1/32 to decimal.

Question 2.
Write the following decimals in the place value table
a) 19.4
b) 0.3
c) 10.6
d) 205.9
Solution:  How do you write 18 as a decimal? Algebra Conversion of Decimals, Fractions, and Percent.

Question 3.
Write each of the following as decimals:
Solution:
a) Seven – tenths b) Two tens and nine – tenths c) Fourteen point six
14.6

d) One hundred and two ones
100 + 2 = 102.0

e) Six hundred point eight
600.8

Question 4.
Write each of the following as decimals:-
Solution: Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form.
Solution:  Question 6.
Express the following as cm using decimals.
Solution:  Question 7.
Between Which two Whole numbers on the number line are the given numbers lie? Which of these Whole numbers is nearer the number? a) 0.8
b) 5.1
c) 2.6
d) 6.4
e) 9.1
f) 4.9
Solution:
a) 0.8 lies between 0 and 1, and is nearest to 1.
b) 5.1 lies between 5 and 6, and is nearest to 5.
c) 2.6 lies between 2 and 3, and is nears to 3.
d) 6.4 lies between 6 and 7, and is nears to 6.
e) 9.1 lies between 9 and 10, and is nears to 9.
f) 4.9 lies between 4 and 5, and is nears to 5.

Question 8.
Show the following numbers on the number line.
a) 0.2
b) 1.9
c) 1.1
d) 2.5
Solution:
a) 0.2 :- Represents a print between 0 and 1 on numbers line. Such that the space
between 0 and 1 is divided into 10 equal parts. Hence, each equal part will be equal to one – tenth.
Now 0.2 is the second print between 0 and 1. b) 1.9 :- Represents a print between 1 and 2 on number line. Such that the space between 1 and 2 is divided into 10 equal parts. Hence, each part will be equal to one – tenth
Now, 1.9 is the ninth print between 1 and 2. c) 1.1 Represents a point between 1 and 2 on number line, Such that the space between 1 and 2 is divided into 10 equal parts. Hence each equal part will be equal to one – tenth Now, 1.1 is the first point between 1 and 2. d) 2.5 Represents a point between 2 and 3 on number line, such that the space between 2
and 3 is divided into 10 equal parts. Hence each equal part will be equal to one – tenth Now, 2.5 is the fifth print between 2 and 3.  Question 9.
Write the decimal number represented by the points A,B,C,D on the given number line.
Solution: Prints A, B, C, D are represents 0.8, 1.3, 2.2, 2.9, respectively

Question 10.
a) The length of Ramesh’s note book is 9 cm 5 mm What will be its length in cm?
b) The length of a young gram plant is 65 mm Express it’s length in cm.
Solution:
a) The length of Ramesh’s note book is 9 cm 5 mm b) The length of a gram plant is 65 mm ## KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

Students can Download Chapter 3 Playing with Numbers Ex 3.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

Question 1.
Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.
Solution:
Weight of the two bags = 75 kg and 69 kg
Maximum weight = HCF (75, 69) 75 = 3 × 5 × 3
69 = 3 × 23
HCF = 3
Hence, the maximum value of weight, which can measure the weight of the fertilizer exact number of times, is 3 kg Question 2.
Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Solution:
Step measure of 1 Boy = 63 cm
Step measure of 2 Boy = 70 cm
Step measure of 3 Boy = 77 cm
LCM of 63, 70, 77 LCM = 2 × 3 × 3 × 5 × 7 × 11 = 6930
Hence, the minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm.

Question 3.
The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Solution:
Length = 825 cm = 3 × 5 × 5 × 11
Breadth = 675 cm = 3 × 3 × 3 × 5 × 5
Height = 450 cm = 2 × 3 × 3 × 5 × 5
Longest tape = HCF of 825, 675, and 450 = 3 × 5 × 5 = 75 cm
Therefore, the longest tape is 75 cm. The LCM of 15 and 20 is 60.

Question 4.
Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12
Solution:
Smallest number = LCM of 6, 8, 12 LCM = 2 × 2 × 2 × 3 = 24
the smallest 3-digit multiple of 24.
It can be seen that 24 × 4 = 96 and 24 × 5 = 120.
Hence, the smallest 3-digit number which is exactly divisible by 6, 8, and 12 is 120

Question 5.
Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.
Solution:
LCM of 8, 10, and 12
2, 8,10,12 LCM = 2 × 2 × 2 × 3 × 5 = 120
We have to and the greatest 3-digit multiple of 120.
It can be seen that 120 × 8 = 960 and 120 × 9 = 1080.
Hence, the greatest 3-digit number exactly divisible by 8, 10, and 12 is 960. Question 6.
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?
Solution:
Time period after which these lights will change = LCM of 48, 72, 108 LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432
They will change together after every 432 seconds i.e., 7 min 12 seconds.
Hence, they will change simultaneously at 7 : 07 : 12 am.

Question 7.
Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
Solution:
Maximum capacity of the required tanker = HCF of 403, 434, 465
403 = 13 × 31
434 = 2 × 7 × 31
465 = 3 × 5 × 31
HCF = 31
A container of capacity 31 l can measure the diesel of 3 containers exact number of times

Question 8.
Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Solution:
LCM of 6, 15, 18 LCM = 2 × 3 × 3 × 5 = 90
Required number = 90 + 5 = 95

Question 9.
Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Solution:
LCM of 18, 24, and 32 LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
We have to and the smallest 4-digit multiple of 288.
It can be observed that 288 × 3 = 864 and 288 × 4 = 1152.
Therefore, the smallest 4-digit number which is divisible by 18, 24, and 32 is 1152. Question 10.
Find the LCM of the following numbers:
(a) 9 and 4
(b) 12 and 5
(c) 6 and 5
(d) 15 and 4
Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?
Solution:
(a) 9 and 4 LCM = 2 × 2 × 3 × 3 = 36 (b) 12 and 5 LCM = 2 × 2 × 3 × 5 = 60

c) 6 and 5 LCM = 2 × 3 × 5 = 30

d) 15 and 4 LCM = 2 × 2 × 3 × 5 = 60
Yes, it can be observed that in each case, the LCM of the given numbers is the product of these numbers.
When two numbers are co-prime, their LCM is the product of those numbers. Also, in each case, LCM is a multiple of 3. Question 11.
Find the LCM of the following numbers in which one number is the factor of the other.
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45
Solution: LCM = 2 × 2 × 5 = 20 LCM = 2 × 3 × 3 = 18 LCM = 2 × 2 × 2 × 2 × 3 = 48 LCM = 3 × 3 × 5 = 45
Yes, it can be observed that in each case, the LCM of the given numbers is the larger number. When one number is a factor of the other number, their LCM will be the larger number.

## KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1

Students can Download Chapter 12 Ratio and Proportion Ex 12.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1

Least common multiple (LCM) of 12 and 16 is 48.

Question 1.
There are 20 girls and 15 boys in a class.
a) What ¡s the ratio of number of girls to the number of boys?
b) What is the ratio of number of girls to the total number of students in the class?
Solution:
Numbers of girls in a class = 20
Numbers of boys in a class = 15
Total numbers of students in a class = 20 + 15 = 35
a) Ratio of number of girls to number of boys b) Ratio of numbers of girls to total number of students  Question 2.
Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of
a) Number of students liking football to numbers of students liking tennis.
b) Numbers of students liking cricket to total number of students. Solution:
Number of students Who like football = 6
number of students Who like cricket = 12
Number of students who like tennis = 30 – 6 – 12 = 12
a) Ratio of the number of students liking football to the number of students liking tennis = b) Ratio of the number of students liking cricket to the total number of students = 2 : 5

Question 3.
See that figure and find the ratio of
a) Numbers of triangles to the number of circles inside the rectangle
b) Number of squares to all the figures inside the rectangle
c) Number of circles to all the figures inside the rectangle. Solution:
Number of triangle in a rectangles = 3
Number of circles in a rectangles = 2
Numbers of squares in a rectangles = 2
Total number of figures in a rectangles = 7
a) Ratio of the number of triangles to the number of circles = $$\frac{3}{2}$$
b) Ratio of the number of squares to all the figures in rectangle = $$\frac{2}{7}$$
c) Ratio of the number of circles to tall the figures in the rectangle = $$\frac{2}{3}$$

Question 4.
Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of speed of Hamid to the speed of Akhsta.
Solution:
The distance travelled in an hour by a certain object is caused the speed of the object. Distance travelled by Hamid in one Hour = 9 km / hr
& Distance travelled by Akhtar in one hour = 12 km / hr
Ratio of speed of Hamid to the speed of Akhtar  Question 5.
Fill in the following Blanks : Solution: Therefore, 5,12, 25 be the answer True (Yes) all these are equivalent ratios

Question 6.
Find the ratio of the following:
a) 81 to 108
b) 98 to 63
c)33 km to 121 km
d) 30 minutes to 45minutes
Solution:  Question 7.
Find the ratio of the following
a) 30 minutes to 1.5 hours
b) 40 cm to 1.5 m
c) 55 paise to Rs 1
d) 500 mi to litres
Solution:
a) 1 hour = 60 min b) 40 cm to 1.5 m
1.5 m = 100 cm
1.5 m = 150 cm
Requires Ratio c) 55 paise to Rs 1
Rs 1 = 100 paise
Requires Ratio d) 500 ml to 2l
1l= 1000 ml
2l = 2 × 1000 ml = 2000 ml Question 8.
In a year, Seema earns Rs 1,50,000 and saves Rs 50,000. Find the ratio of
a) Money that Seema earns to the money she saves.
b) Money that she saves to the money she spends.
Solution:
Money earned by seema = Rs 1,50,000
Money Saved by Seemed = Rs 50,000
Money spent = Rs 1,50,000 – Rs 50,000 = Rs 1,00,000  Question 9.
There are 102 teachers in a school of 3300 Students, Find the ratio of the number of teachers to the number of students.
Solution: Question 10.
In a college, out of 4320 Students, 2300 are girls. Find the ratio of
a) Number of girls to the total number of students.
b) Number of boys to the number of girls.
c) Number of boys to the total number of students
Solution:
Total number of students = 4320
Number of girls in a college = 2300
Number of boys in a college = 4320 – 2300 = 2020
a) Required ratio n for number of girls to the  b) Required ratio for number of boys to the number of girls  c) Required ratio for number of boys to the total number of students  Question 11.
Out of 1800 students in a school 750 opted basket ball, 80 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of
a) Number of students who opted basket ball to the number of students Who opted table tennis.
b) Number of students Who opted cricket to the number of students opting basketball
c) Number of students Who opted basket ball to the total number of students.
Solution:
Total number of students = 1800
Number of students opted basketball = 750
Number of students opted cricket = 800
Number of students opted table tennis = 1800 – 750 – 800 = 250  Question 12.
Cost of dozen pens is Rs 180 and cost of 8 ball pens is Rs 56. Find the ratio of the cost of pen to the cost of a ball pen.
Solution:
Cost of dozen pens = Rs 180
Cost of 1 per = $$\frac{180}{12}$$ = Rs 15
Cost of 8 ball pens = Rs 56
Cost of ball pens=$$\frac{56}{8}$$ = Rs 7
Required ratio of the cost of a pen to the cost of a ball pen = $$\frac{15}{7}$$ = 15 : 7

Question 13.
Consider the statement: Ratio of breadth and length of a hall is 2 : 5. Complete the following table that shows some possible breadths and lengths of the hall. Solution:
i) Length = 50 m 5 × Breadth = 50 × 2 (by cross multiplication )  2 × Length = 5 × 40 (By cross – Multiplication) Length = 100 m.

Question 14.
Divided 20 pens between sheela and sangeeta in the ratio of 3 : 2.
solution:
Terms of 3 : 2 are 3 and 2
Sum of these two terms = 3 + 2 = 5
Sheela will get $$\frac{3}{5}$$ of total pens and sangeeta will get $$\frac{2}{5}$$ of total pens. Question 15.
Mother wants to divide Rs 36 between her daughters shreya and bhoomika in the ratio of their ages. If age of shreya is 15 years and age of bhoomika is 12 years, find how much shreya and bhoomika will get.
Solution: Therefore, mother wants to divide Rs 36 in a ratio of 5 : 4 sum of these terms = 5 + 4 = 9
Shreya will get $$\frac{5}{9}$$ of the total money and bhoomika will get $$\frac{4}{9}$$ of it Therefore, Shreya and bhoomika will get Rs 20 and Rs 16 respectively. Question 16.
Present age of father is 42 years and that of his son is 14 years. Find the ratio of
a) Present age of father to the present age of son
b) Age of the father to the age of son, When son was 12 years old.
c) Age of father after 10 years to the age of son after 10 years
d) Age of father to the age of son when father was 30 years old
Solution:
a) Present age of father = 42 years
Present age of son = 14 years b) Two years ago, the age of the son was 12 years and the age of the father was 42 – 2 = 40 years c) After 10 years, the age of the father and son will be 52 years and 24 years respectively d) 12 years ago, the father was 30 years old, At that time age of son = 14 – 12 = 2 years ## KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

Students can Download Chapter 3 Playing with Numbers Ex 3.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

Question 1.
Find the HCF of the following numbers:
(a) 18, 48
(b) 30,42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75
Solution:
a) 18, 48
18 = 1, 2, 3, 6, 9, 18
48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Common factors = 1, 2, 3, 6
H.C.F of 18 & 48 = $$\boxed { 6 }$$

b) 30, 42
30 = 1, 2, 3, 5, 6, 10, 15, 30
42 = 1, 2, 3, 4, 6, 7, 14, 21, 42
Common factors = 1, 2, 3, 6
∴ H.C.F. = $$\boxed { 6 }$$
∴ H.C.F of 30 & 42 is $$\boxed { 6 }$$ c) 18 & 60
18 = 1, 2, 3, 6, 9, 18
60 = 1, 2, 3, 4, 6, 10, 12, 15, 30, 60
Common factors = 1, 2, 3, 6
∴ H.C.F. of 18 & 60 is $$\boxed { 6 }$$

d) 27 & 63
27 = 1, 3, 9, 27
63 = 1, 3, 7, 9, 21, 63
∴ Common factors = 1, 3, 9
∴ H.C.F. of 27 & 63 is = 9.

e) 36 & 84
36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
∴ Common factors = 1, 2, 3, 4, 6, 12
∴ H.C.F. of 36 & 84 is 12.

f) 34 & 102
34 = 1, 2, 17, 34
102 = 1, 2, 3, 6, 17, 34, 51, 102
∴ Common factors = 1, 2, 17, 34
∴ H.C.F. of 34 & 102 is 34.

g) 70, 105, 175
70 = 1, 2, 5, 7, 10, 14, 35, 70
105 = 1, 3, 5, 7, 10, 14, 35, 105
175 = 1, 5, 7, 25, 35, 175
∴ Common factors 1, 5, 7, 35
∴ H.C.F. of 70, 105, 175 is 35

h) 91, 112, 49
91 = 1, 7, 13, 91
112 = 1, 2, 4, 7, 8, 14, 16, 28, 56, 112
49 = 1, 7, 49
∴ Common factors = 1, 7
∴ H.C.F. of 91, 112, 49 is 7 i) 12, 45, 75
12 = 1, 2, 3, 4, 6, 12
45 = 1, 3, 5, 9, 15, 45
75 = 1, 3, 5, 15, 26, 75
∴ Common factors 1, 3
∴ H.C.F of 12, 45, 75 is 3.

Question 2.
What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?
Find the HCF of the following:
(i) 24 and 36
(ii) 15, 25 and 30
(iii) 8 and 12
(iv) 12, 16 and 28
Solution:
(i) 1 e.g., HCF of 2 and 3 is 1.
(ii) 2 e.g., HCF of 2 and 4 is 2.
(iii) 1 e.g., HCF of 3 and 5 is 1. To do the prime factorization of 84, successively divide 84 by prime numbers.

Question 3.
HCF of co-prime numbers 4 and 15 was found as follows by factorisation:
4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?
Solution:
No. The answer is not correct. 1 is the correct HCF.

## KSEEB Solutions for Class 6 Science Chapter 3 Fibre to Fabric

Students can Download Chapter 3 Fibre to Fabric Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Science, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 6 Science Chapter 3 Fibre to Fabric

### Class 6 Science Fibre to Fabric NCERT Textbook Questions and Answers

Question 1.
Classify the following fibres as natural or synthetic, nylon, wool, cotton, silk, polyester, jute.
Natural fibres : Cotton, jute, silk, wool.
Synthetic fibres: Nylon, polyester.

Question 2.
State the following statement are true or false
a) Yarn is made from fibres
b) Spinning is a process of making fibres
c) Jute is the outer covering of coconut
d) The process of removing seed from cotton is called ginning.
e) Weaving of yam makes a piece of fabric.
f) Silk fibre is obtained from the stem of a plant
g) Polyester is a natural fibre.
a) – True,
b) – False,
c) – False,
d) – True
e) – True,
f) – False,
g) – False

Question 3.
Fill in the blanks :
a) Plant fibres are obtained from ___ and ___
b) Animal fibres are ___ and ___
a) Jute and cotton
b) Silk and wool

Question 4.
From which parts of the plant cot¬ton and jute are obtained?
Jute is obtained from stem and cotton is obtained from cotton bolls which are directly from surface of cotton seeds.

Question 5.
Name any two things got from coconut fibre
Ropes and Mats.

Question 6.
Explain the process of making yarn from fibre.
The thin strands are called as yarns which are made from fibre. Spinning is the process of making yarn. The process where the cotton fibers are drawn out and being twisted. This process brings all the fibres together to from a yarn.

### Class 6 Science Fibre to Fabric Additional Important Questions and Answers

Question 1.
The process of twisting fibres to make yarn is called ___
a. India
b. America
c. Spinning
d. Knotting
c. Spinning

Question 2.
Weaving is done on devices called ___
a. Takli
b. Charka
c. Looms
d. Needles
c. Looms

Question 3.
The process of removing seeds from ____
a. Camel
b. Sheep
c. Knitting
d. Shearing
a. Camel

Question 4.
Where was silk first produced in 3000B.C
a. India
b. America
c. Australia
d. China
d. China

Question 5.
The process of growing silkworms to obtain silk from is_____
a. Agriculture
b. Sericulture
c. Pisciculture
d. Horticulture
b. Sericulture

Question 6.
‘Pashmina’wool is obtained from___
a. Camel
b. Sheep
c. Rabbit
d. Goat
d. Goat

Question 7.
What do you mean by shearing.
a. The process of shearing things
b. The process of making woollen fabric from fibre.
c. The process of removing hair from the body of animals
d. The process of removing seeds from cotton
c. The process of removing hair from the body of animals

Question 8.
Which fibre is obtained from flax seeds
a. Cotton
b. Jute
c. Linen
d. Nylon.
c. Linen

Question 9.
Which of these fibres is non – biodegradable
a. Cotton
b. Jute
c. Wool
d. Nylon
d. Nylon

Question 10.
The fibres obtained from animals are made up of
a. Proteins
b. Carbohydrates
c. Cellulose
d. Vitamins.
a. Proteins

II. Fill in the blanks :

1. ___ and ___ are animals fibres.
2. ___ is used for packing material, sacks, bags and ropes
3. ___ is an example of dietry fibre.
4. Jute is obtained from a plant called ___
5. Give two example of unstitched garments ___ and ___
1. Silk and wool
2. Jute
3. Cellulose
4. Patsun
5. Shawl and Dhoti

Question 1.
What is fibre?
Fibre is a thread like a strand from which yam is made

Question 2.
What is yarn?
Yarn is a long continuous length of interlocked fibre which is woven knitted for making garments.

Question 3.
Which kind of soil is needed for growing cotton plant?
Black Soil.

Question 4.
Why cotton Clothes are preferred over synthetic fibre?
Because they are natural fibre and are biodegradable . they do not cause any pollution and are also good for our skin as they have natural ventilation.

Question 5.
Name some items that are made from coir, jute, hemp, silk and cotton
coir : Ropes, floor covering , stuffing in mattresses and pillows.
jute : Jute bags, hat, slipper, rope, matand sack
hemp : Ropes , carpets, nets, clothes and paper
Flax : Used to make a fabric called linen, rope and quality paper

Question 6.
Make a flow chart of cotton ball to fabric Question 7.
What are looms?
The devices on which weaving of fabrics take place are called looms The looms are either hand operated or power operated

Question 8.
When we burn nylon, why we do not get the smell of burning paper or burning hair?
Nylon is a synthetic fibre made from chemicals. On burning nylon these chemicals don’t produce the smell of burning paper or hair which are natural substances.

Question 9.
When we burn wool, why do we get the smell of hair burning?
Wool is obtained from the fleece of sheep, goat, yak etc. This is the reason why burning of wool smells like the burning of hair.

Question 10.
What were the materials used by people in ancient times in place of clothes?
It appears that in those days people used the bark and big leaves of trees or animal skin and furs in place of clothes.

Question 11.
What are natural fibres ? Explain with example
The fibres obtained from plants and animals are called natural fibres For Ex: Cotton from cotton balls, jute from jute plant, silk from cocoon of silkworm and from hair of animals like sheep or goat.

Question 12.
Are all fibres produced by plants?
No

Question 13.
Name the states where jute plants are mainly grown in India.
West Bengal, Bihar and Assam.

Question 14.
Define ginning of cotton
The process of separating fibers from the seeds of cotton is called ginning of cotton.

Question 15.
Describe the process of spinning and weaving?
Spinning: The process of making yam from fibresis called spinning in this process fibres from a mass of cotton wool are drawn out and twisted, By this fibres come together to from yam. Spinning can be done by hand,by takli and charka. On a large scale, spinning is done with the help of machines.
Weaving: The process of arranging two sets of yarn together to make a fabric is called weaving. The process of weaving can be done on looms. The looms are either-hand operated or power operated.

## KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.2

Students can Download Chapter 3 Playing with Numbers Ex 3.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 6 Maths Chapter 3 Playing with Numbers Ex 3.2

Question 1.
What is the sum of any two
a) odd numbers
b) Even numbers?
Solution:
a) The sum of two odd numbers is even
eg:- 1 + 5 = 6; 16 + 18 = 34
b) The sum of two even numbers is even
eg:- 4 + 2 = 6; 10 + 18 = 28 In this quick guide we’ll describe what the factor pairs of 96 are, how you find them and list them out for you to prove the calculation works.

Question 2.
State whether the following statements are True or False:
Solution:
a) The sum of three odd numbers is even.
False 3 + 5 + 7 = 15. i,e. odd

b) The sum of two odd numbers and one even number is even.
True 3 + 5 + 6 = 14 i,e = even

c) The product of three odd numbers is odd.
True 3 × 5 × 7 = 105 i.e, odd.

d) If an even number is divided by 2, the quotient is always odd.
False 4 ÷ 2 = 2 i.e, even

e) All prime numbers are odd.
False; 2 is even number.

f) Prime numbers do not have any factors.
False; 1 and the numbers itself are factors of the number.

g) Sum of two prime numbers is always even.
False 2 + 3 = 5 i,e odd h) 2 is the only even prime number.
True

i) All even numbers are composite numbers.
False, 2 is a prime numbers.

j) The product of two even numbers is always even.
True; 2 × 4 = 8, i.e even

Question 3.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.
Solution:
a) 17, 71
b) 37, 73
c) 79, 97 Question 4.
Write down separately the prime and composite numbers less than 20.
Solution:
Prime numbers less than 20 are
2, 3, 5, 7, 11, 13, 17, 19
Composite numbers less than 20 are
4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Question 5.
What is the greatest prime number between 1 and 10.
Solution:
Prime numbers between 1 and 10 are 2, 3, 5, 7 among these numbers 7 is the greatest.

Question 6.
Express the following as the sum of two odd primes,
a) 44
b) 36
c) 24
d) 18
Solution:
a) 44 = 37 + 7 = 13 + 31
b) 36 = 31 + 5 = 17 + 19
c) 24 = 19 + 5 = 11 + 13
d) 18 = 5 + 13 = 7 + 11 Question 7.
Give three pairs of prime numbers whose difference is 2.
[Remark: Two prime numbers whose difference is 2 are called twin primes]
Solution:
i) 3 & 5
ii) 5 & 7
iii) 11 & 13
iv) 17 & 19
v) 29 & 31 etc. Question 8.
Which of the following numbers are prime?
a) 23
b) 51
c) 37
d) 26
Solution:
a) 23 = 23 × 1 = 23; 23 = 1 × 23
23 has only two factors, 1 and 23 Therefore, it is a prime numbers

b) 5 = 1 × 51 = 51; 51 = 17 × 3
51 has four factors, 1, 3, 17, 51. Therefore, it is not a prime number, it is a composite number.

c) 37 It has only two factors 1 and 37.
It is a prime number

d) 26 26 has four factors (1, 2, 13, 26)
Therefore, it is not a prime number It is a composite number Question 9.
Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Solution:
Between 89 and 97, both of which are prime numbers there are 7 composite numbers.
They are 90, 91, 92, 93, 94, 95, 96, numbers factors
90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
91 = 1, 7, 13, 91
92 = 1, 2, 4, 23, 46, 92
93 = 1, 3, 31, 93
94 = 1, 2, 47, 94
95 = 1, 5, 19, 95
96 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Question 10.
Express each of the following numbers as the sum of three odd primes:
a) 21
b) 31
c) 53
d) 61
Solution:
a) 21 = 3 + 7 + 11
b) 31 = 5 + 7 + 19
c) 53 = 3 + 19 + 31
d) 61 = 11 + 19 + 31 Question 11.
Write five pairs of prime numbers less than 20 Whose sum is divisible by 5. (Hint: 3 + 7 = 10)
Solution:
The five pair of prime numbers less than 20 divisible by 5 are 2 + 3 = 5
2 + 13 = 15
3 + 17 = 20
7 + 13 = 20
19 + 11 = 30

Question 12.
Fill in the blanks:
1. A number which has only two factors is called a prime number
2. A numbers Which has more than two factors is called a composite number
3. 1 is neither prime number nor composite number
4. The smallest prime number is 2
5. The smallest composite number is 4
6. The smallest even number is 2

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