KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4

   

Students can Download Chapter 8 Decimals Ex 8.4 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 8.4

Question 1.
Express as rupees using decimals.
a) 5 paise
b) 75 paise
c) 20 paise
d) 50 rupees 90 paise
e) 725 paise.
Solution:
W.K.T. 1 Rupee = 100 paise
a) 5 paise = \(\frac{5}{100}\) rupees = 0.05 Re
b) 75 paise = \(\frac{75}{100}\)rupees = 0.75 Re
c) 20 paise = \(\frac{20}{100}\) rupees = 0.20 Re
d) 50 rupees 90 paise = \(\left(50+\frac{90}{100}\right)\) rupees
e) 725 paise = \(\frac{725}{100}\) rupees = 7.25

KSEEB Solutions

Click here to get an answer to your question ✍️ How do you write 9/20 as a decimal?

Question 2.
Express as metres using decimals.
a) 15 cm
b) 6 cm
c) 2 m 45 cm
d) 9 m 7 cm
e) 416 cm
Solution:
we know that 1 metre = 100 cm
a) 15 cm = \(\frac{15}{100}\) m = 0.15 m
b) 6 cm = \(\frac{6}{100}\) m = 0.06 m
c) 2 m 45 cm = \(\left(2+\frac{45}{100}\right)\) m = 2.45 m
d) 9m 7 cm = \(\left(9+\frac{7}{100}\right)\) m = 9.07 m
e) 416 cm = \(\frac{419}{100}\) m = 4.19 m

KSEEB Solutions

Question 3.
Express as cm using decimals.
a) 5 mm
b) 60 mm
c) 164 mm
d) 9 cm 8 mm
e) 93 mm
Solution:
a) 5mm = \(\frac{5}{10}\) cm = 0.5 cm
b) 60 mm = \(\frac{60}{10}\) cm = 6.0 cm
c) 164 mm = \(\frac{164}{10}\) cm = 16.4 cm
d) 9 cm 8 mm = \(\left(9+\frac{8}{10}\right)\) cm = 9.8 cm
e) 93 mm = \(\frac{93}{10}\) cm = 9.3 cm

KSEEB Solutions

Question 4.
Express as km using decimals.
a) 8 m
b) 88 m
c) 8888 m
d) 70 km 5 m
Solution:
a) 8 m = \(\frac{8}{1000}\) = 0.008 km
b) 88 m = \(\frac{88}{1000}\) = 8.888 km
c) 8888 m = \(\frac{8888}{1000}\) km = 8.888 km
d) 70 km 5 m = \(\left(70+\frac{5}{1000}\right)\) km = 70.005 km

KSEEB Solutions

Question 5.
Express as kg using decimals.
a) 2g
b) 100 g
c) 3750 g
d) 5 kg 8 g
e) 26 kg 50 g
Solution:
W.K.T. 1 kg = 1000 grams
a) 2g = \(\frac{2}{1000}\) kg = 0.002 kg
b) 100g = \(\frac{100}{1000}\) g = 0.10 kg
c) 3750 g = \(\frac{3750}{1000}\) kg = 3.750 kg
d) 5 kg 8 g = \(\left(5+\frac{8}{1000}\right)\) kg = 5.008 kg
e) 26 kg 50 g = \(\left(26+\frac{50}{1000}\right)\) kg = 26.050 kg

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.
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 1
Solution:
It may be observed that
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 2

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:
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 3

KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

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
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 4

b) Two tens and nine – tenths
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 5

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

Rounding to One Decimal Place Calculator will round the value of a number to 1 decimal place accurately and displays the work quickly.

Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form.
Solution:
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 7

KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

Question 6.
Express the following as cm using decimals.
Solution:
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 8

KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 9

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?
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 10
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.
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 11

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.
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 12

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.
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 13

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.
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 14

KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

Question 9.
Write the decimal number represented by the points A,B,C,D on the given number line.
Solution:
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 15
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
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 16

b) The length of a gram plant is 65 mm
KSEEB Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 17

KSEEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1

   

Students can Download Chapter 1 Algebra Ex 1.1 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 1 Algebra Ex 1.1

The fifth digit right of the decimal point is the hundred thousandth place where you have to round off.

Question 1.
Fill in the blanks:-
a. 1 lakh = 10 ten thousand
b. 1 million = 10 hundred thousand
c. 1 crore = 10 ten lakh
d. 1 crore = 10 million
e. 1 million = 10 lakh

KSEEB Solutions

Question 2.
Place common correctly and write the numerals :
a. Seventy three lakh seventy five thousand three hundred seven.
73,75,307

b. Nine crore five lakh forty one.
9,05,00,041

c. Seven crore fifty two lakh twenty one thousand three hundred two.
7,52,21,302

d. Fifty eight million four hundred twenty three thousand two hundred two.
58,423,202

e . Twenty three lakh thirty thousand ten.
23,30,010

KSEEB Solutions

Question 3.
Insert comas suitably and write the names according to indian system of numeration:
a. 87595762
Eight crore seventy five lakh ninty five thousand seven hundred sixty two.

b. 8546283
Eighty five lakh forty six thousand two hundred eighty three.

c. 99900046
Nine crore ninety nine lakh forty six.

d. 98432701
Nine crore eighty four lakh thirty two thousand seven hundred one

KSEEB Solutions

Question 4.
Insert commas suitably and write the names according to international system of numeration:

a. 78921092
Seventy eight million nine hundred twenty one thousand ninety two

b. 7452283
Seven million four hundred fifty two thousand two hundred eighty three

c. 99985102
Ninety nine million nine hundred eighty five thousand one hundred two

d. 48049831
Forty eight million forty nine thousand eight hundred thirty one.

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)
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 1
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

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

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
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 2
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.

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

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
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 50
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
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 200
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.

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

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
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 201
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
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 30
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
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 35
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.

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

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
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 60
LCM = 2 × 2 × 3 × 3 = 36

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

(b) 12 and 5
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 61
LCM = 2 × 2 × 3 × 5 = 60

c) 6 and 5
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 62
LCM = 2 × 3 × 5 = 30

d) 15 and 4
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 63
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.

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

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:
KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 70
LCM = 2 × 2 × 5 = 20

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 71
LCM = 2 × 3 × 3 = 18

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 72
LCM = 2 × 2 × 2 × 2 × 3 = 48

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 73
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
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 1
b) Ratio of numbers of girls to total number of students
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 2

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

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.
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 3
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 =
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 352
b) Ratio of the number of students liking cricket to the total number of students
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 4
= 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.
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 5
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
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 50

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

Question 5.
Fill in the following Blanks :
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 51
Solution:
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 52
Therefore, 5,12, 25 be the answer
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 521
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:
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 53

KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 59

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
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 55

b) 40 cm to 1.5 m
1.5 m = 100 cm
1.5 m = 150 cm
Requires Ratio
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 56

c) 55 paise to Rs 1
Rs 1 = 100 paise
Requires Ratio
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 57

d) 500 ml to 2l
1l= 1000 ml
2l = 2 × 1000 ml = 2000 ml
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 58

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
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 60

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

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:
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 61

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
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 62
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 63

b) Required ratio for number of boys to the number of girls
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 64

KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 65

c) Required ratio for number of boys to the total number of students
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 66
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 661

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
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 67

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

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.
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 261
Solution:
i) Length = 50 m
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 69
5 × Breadth = 50 × 2 (by cross multiplication )
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 70

ii) Breadth = 40 m
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 71
2 × Length = 5 × 40 (By cross – Multiplication)
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 712
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.
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 72

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:
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 73
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
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 731
Therefore, Shreya and bhoomika will get Rs 20 and Rs 16 respectively.

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

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
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 74

b) Two years ago, the age of the son was 12 years and the age of the father was 42 – 2 = 40 years
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 75

c) After 10 years, the age of the father and son will be 52 years and 24 years respectively
KSEEB Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.1 76

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 12 Ratio and Proportion Ex 12.1 761

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 \(\boxed { 6 } \)

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

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

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

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.

KSEEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

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 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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena

   

Students can Download Kannada Lesson 2 Gandharvasena Questions and Answers, Summary, Notes Pdf, Siri Kannada Text Book Class 6 Solutions, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Siri Kannada Text Book Class 6 Solutions Gadya Bhaga Chapter 2 Gandharvasena

Gandharvasena Questions and Answers, Summary, Notes

Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 1

Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 2
Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 3

Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 4
Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 5
Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 6

Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 7
Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 8
Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 9
Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 10

Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 11
Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 12
Siri Kannada Text Book Class 6 Solutions Gadya Chapter 2 Gandharvasena 13

Gandharvasena Summary in Kannada

Gandharvasena Summary in Kannada 1

Gandharvasena Summary in Kannada 2
Gandharvasena Summary in Kannada 3

Gandharvasena Summary in Kannada 4
Gandharvasena Summary in Kannada 5
Gandharvasena Summary in Kannada 6
Gandharvasena Summary in Kannada 7

KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला

   

Students can Download Hindi Lesson 2 वर्णमाला Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 6 Hindi helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Hindi Chapter 2 वर्णमाला

वर्णमाला Questions and Answers, Summary, Notes

स्वराक्षर:

KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 1

KSEEB Solutions

व्यंजन:

KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 2

KSEEB Solutions

वर्णमाला – अभ्यास

स्वर पर गोले लगाओ:

KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 3

KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 4

KSEEB Solutions

खली जगह भरो:

KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 5

KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 6
KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 7

KSEEB Solutions

रंग भरो:

KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 8
KSEEB Solutions for Class 6 Hindi Chapter 2 वर्णमाला 9

KSEEB Solutions

KSEEB Solutions for Class 6 Hindi वल्लरी Karnataka State Syllabus

   

Expert Teachers at KSEEBSolutions.com has created KSEEB Solutions for Class 6 Hindi वल्लरी Pdf Free Download in English Medium and Kannada Medium of 6th Standard Karnataka Hindi Textbook Solutions Answers Guide, Textbook Questions and Answers, Notes Pdf, Model Question Papers with Answers, Study Material, are part of KSEEB Solutions for Class 6. Here we have given 3rd Language KTBS Karnataka State Board Syllabus for Class 6 Hindi Textbook Solutions Vallari.

Karnataka State Board Syllabus for Class 6 Hindi Solutions वल्लरी

KSEEB Solutions for Class 6 Hindi 3rd Language

6th Class Hindi Textbook Solutions Karnataka State Syllabus

We hope the given KSEEB Solutions for Class 6 Hindi वल्लरी Pdf Free Download in English Medium and Kannada Medium of 6th Std Karnataka Hindi Textbook Answers Solutions Guide, Textbook Questions and Answers, Notes Pdf, Model Question Papers with Answers, Study Material will help you. If you have any queries regarding Third Language KTBS Karnataka State Board Syllabus for Class 6 Hindi Textbooks Solutions Vallari, drop a comment below and we will get back to you at the earliest.

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