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 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

   

Students can Download Chapter 2 Fractions and Decimals Ex 2.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Multiplying and Dividing Rational Expressions Calculator.

Question 1.
Solve:
i)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 1
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 2

ii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 3
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 4

iii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 5
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 6
L.C.M = 5 × 7 × 1 × 1 = 35
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 7
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 8

iv)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 9
Solution:

Online Subtracting Fractions Calculator subtracts the fractions 9/11 and 4/5 i.e. 1/55

The given fractions are 9/11 and 4/5

Firstly the L.C.M should be done for the denominators of the two fractions 9/11 and 4/5

9/114/5

The LCM of 11 and 5 (denominators of the fractions) is 55

Given numbers has no common factors except 1. So, there LCM is their product i.e 55

 

= 9 x 5 – 4 x 11/55

= 45 – 44/55

= 1/55

Result: 1/55

v)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 12
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 13
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 14

vi)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 15
⇒ convert mixed fractions to improper fraction
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 16

vii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 17
convert mixed fractions to improper fraction
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 18

Using a fraction and whole number calculator in situations where you struggle to solve your arithmetic assignments can be useful.

Question 2.
Arrange the following in descending order:
i)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 19
= We need to arrange these in descending order,
To find which number is greater or smaller, we make their denominators equal.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 20

ii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 21
⇒ We make their denominators equal, to find the descending order.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 22

Free quadratic equation completing the square calculator – Solve quadratic equations using completing the square step-by-step.

Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square ?
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 23
Solution:
For Row,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 24
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 25
For diagonals,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 26
Since Sum of air rows, columns and diagonals are equal.

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 4.
A rectangular sheet of paper is 12\(\frac{1}{2}\) cm long and 10\(\frac{2}{3}\) cm wide. Find its perimeter.
Solution:
Length of rectangular sheet of paper = 12\(\frac{1}{2}\) cm
(breadth) width of rectangular sheet paper = 10\(\frac{2}{3}\) cm
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 27
Perimeter of rectangle = 2 (length + breadth)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 28
converting the above fraction to mixed fraction,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 29

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

2 2/3 as an improper fraction in its simplest form.

Question 5.
Find the perimeters of
(i) ∆ ABE
(ii) the rectangle BCDE in this figure. Whose perimeter is greater ?
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 30
Solution:
(i) ∆ ABE
perimeter of ∆ ABE
perimeter = AB + AE + BE
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 31
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 311

ii) The rectangle BCDE in this figure
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 32
Perimeter of rectangle BCDE,
As it is a rectangle, opposite sides are equal
BC = DE CD = BE
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 33
perimeter of rectangle = 2(l + b)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 34
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 35
Also,
We have to find which perimeter is greater
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 36
To find which fraction is greater, we make its denominator equal.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 37
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 38
Perimeter of ∆ ABE > Perimeter of BCDE
(Thus, Perimeter of ∆ ABE is greater)

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 6.
Salil wants to put a picture in a frame. The picture is 7\(\frac{3}{10}\) cm wide. How much should the picture be trimmed ?
Solution:
There are two things here – picture, and frame
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 39
so, picture has to be trimmed
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 40
\(=\frac{3}{10}\)
∴ Picture has to trimmed by = \(=\frac{3}{10}\) cm

Question 7.
Ritu ate \(\frac{3}{5}\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much ?
Solution:
Total part = 1
Part eaten by Ritu = \(\frac{3}{5}\)
Part eaten by Somu = Total part – part eaten by Ritu
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 41
Now,
We have to tell who ate the larger share so, we have to compare,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 42
∴ Ritu ate the larger share.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 43
Ritu ate large share by \(\frac{1}{5}\)

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 8.
Michael finished colouring in \(\frac{7}{12}\) hour. Vaibhav finished colouring the same picture in \(\frac{3}{4}\) hour. Who worked longer ? By what fraction was it longer ?
Solution:
Michael finished work in = \(\frac{7}{12}\) hour
Vaibhav finished work in = \(\frac{3}{4}\)
We need to find who worked longer.
i.e., we need to find greater of = \(\frac{7}{12}\) & \(\frac{3}{4}\)
We make their denominators equal
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 44
∴ Vaibhav worked longer.
We also need to find by how much
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 45
Vaibhav worked longer by \(\frac{1}{6}\) hours.

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

   

Students can Download Chapter 2 Fractions and Decimals Ex 2.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 1.
Which is greater ?
i) 0.5 or 0.05
Solution:
0.5 > 0.05 (∵ 0.5 = 0.50)

ii) 0.7 or 0.5
Solution:
0.7 > 0.5

iii) 7 or 0.7
Solution:
7 > 0.7 (∵ 7 = 7.0)

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

iv) 1.37 or 1.49
Solution:
1.49 > 1.37

v) 2.03 or 2.30
Solution:
2.30 > 2.03

vi) 0.8 or 0.88
Solution:
0.88 > 0.8 (∵ 0.8 = 80)

1 1/2 as a decimal is equal to 1.5.

Question 2.
Express as rupees using decimals :
i) 7 paise
Solution:
7 paise = Rs. 0.07

ii) 7 rupees 7 paise
Solution:
7 rupees 7 paise Rs. 7.07

iii) 77 rupees 77paise
Solution:
77 rupees 77 paise = ₹ 77.77

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

iv) 50 paise
Solution:
50 paise = 0.50

v) 235 paise
Solution:
235 paise = 2.35

Question 3.
i) Express 5cm in metre and kilometre
Solution:
5 cm = 0.05m = 0.00005km.

ii) Express 35mm in cm, m and km
Solution:
35mm = 3.5cm = 0.000035km

Question 4.
Express in kg :
i) 200g
Solution:
200g = 0.200kg = 0.2 kg

ii) 3470g
Solution:
3470g = 3.470kg

iii) 4 kg 8 g
Solution:
4kg 8g = 4.008 kg

Question 5.
Write the following decimal numbers in the expanded form :
i) 20.03
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 1

ii) 2.03
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 21

iii) 200.03
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 22

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

iv) 2.034
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 23

Question 6.
Write the place value of 2 in the following decimal numbers :

i) 2.56
Solution:
2.56 place value of 2 in the decimal number 2.56 = 2 ones place)

ii) 21.37
Solution:
21.37 place value of 2 in the decimal number 21.37 = 2 × 10 = 20. (2 is in tens place)

iii) 10.25
Solution:
10.25 = palce value of 2 in the decimal
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 24

iv) 9.42
Solution:
9.42 = place value of 2 in the decimal number
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 25
(2 is in hundredths place)

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

v) 63.352
Solution:
63.352 = place value of 2 in the decimal number
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 26

Question 7.
Dinesh went from place A to place B and from 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelles more and by how much ?
Solution:
Distance travelled from A to B = 7.5kms
Distance travelled from B to C = 12.7kms
Total distance travelled from A to B & B to C = 20.2 kms.
Ayub
Distance travelled from A to D = 9.3 kms
Distance travelled from D to C = 11.8 kms
Total distance travelled from A to D & D to C = 21.1 kms
∴ Ayub travelled more distance than Dinesh by = 21.1 – 20.2 = 0.9 kms.

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 8.
Shyama bought 5 kg 300g apples and 3 kg 250 g mangoes. Sarala bought 4 kg of 800 g oranges and 4 kg 150g bananas. Who bought more fruits?
Solution:
Shyama:
Weight of Apples bought = 5.300kgs
Weight of Mangoes bought = 3.250kgs
Total weight of fruits = 8.550kgs

Sarala:
Weight of oranges bought = 4.800kgs
Weight of Bananas bought = 4.150kgs
Total weight of fruits = 8.950kgs
8.950 >8.550
∴ Sarala bought more fruits than Shyama.

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 9.
How much less is 28 km than 42.6 km?
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 30
∴ 28 km is less than 42.6 km by 14.6 kms.

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 5 Maths Chapter 3 Mental Arithmetic

   

Students can Download Maths Chapter 3 Mental Arithmetic Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 5 Maths Chapter 3 Mental Arithmetic

KSEEB Class 5 Maths Mental Arithmetic Revision Exercise

I. Round off each of the following numbers to nearest thousands place:

1. 7,547

7,547 lie between 7000 and 8000
Replace by zero
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 1

2. 3,469
3,469 lie between 2000 and 3000
Replace by zero
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 2

KSEEB Solutions

3. 15,238
15,038
Replace by zero
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 3

4. 32,658
32,658
Replace by zero
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 4

II. Round off each of the following numbers to nearest ten thousands place:

1. 26,674
26,674
Replace to zero
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 5

2. 32,464
32,464
Replace to zero
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 6

3. 46,379
46,379
Replace to zero
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 7

4. 53,668
53,668
Replace to zero
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 8

III. Estimate sum of the following by rounding off to nearest thousands place:

1. 42,125 + 35,637
42,125 is rounded off to nearest thousand – 42,000
35,637 is rounded off to nearest thousand – 36,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 9
Estimate sum is 78,000
Verification by actual addition – 42125
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 10

KSEEB Solutions

2. 54,837 + 41,354
54,837 is rounded off to nearest thousand – 55,000
41,354 is rounded offto nearest thousand – 41,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 11
Estimate sum is 96,000
Verification by actual addition
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 12

3. 33,231 + 20,097
33,321 is rounded off to nearest thousand – 33,000
20,097 is rounded off to nearest thousand – 20,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 13
Estimate sum is 53,000
Verification by actual addition
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 14

4. 47,463 + 41,541
47,463 is rounded off to nearest thousand – 47,000
41,541 is rounded off to nearest thousand – 42,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 15
Estimate sum is 89,000
Verification by actual addition
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 16

IV. Estimate the sum of the following by rounding off to nearest ten thousands place:

1. 56,256 + 24,872
56,256 is rounded off to nearest ten thousand – 55,000
24,872 is rounded off to nearest ten thousand – 25,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 17
Estimate sum is 80,000
Verification by actual addition
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 18

KSEEB Solutions

2. 47,671 + 28,745
47,671 is rounded off to nearest ten thousand – 50,000
28,745 is rounded off to nearest ten thousand – 30,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 19
Estimate sum is 80,000
Verification by actual addition
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 20

3. 32,184 + 45,138
32,184 is rounded off to nearest ten thousand – 30,000
45,138 is rounded off to nearest ten thousand – 50,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 21
Estimate, sum is 80,000
Verification by actual addition
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 22

4. 15,025 + 40,165
15,025 is rounded off to nearest ten thousand – 20,000
40,165 is rounded off to nearest ten thousand – 40,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 23
Estimate sum is 60,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 24

Detailed Steps to Round to Nearest Hundredth by Hand ·

V. Estimate the difference of the following by rounding off to nearest thousands place:

1. 65,487 – 46,502
65,487 is rounded off to nearest ten thousand – 64,000
46,502 is rounded off to nearest ten thousand – 46,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 25
Estimate sum is 18,000
Verification by actual substruction
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 26

KSEEB Solutions

2. 45,630 – 32,148
45,630 is rounded off to nearest ten thousand – 44,000
32,148 is rounded off to nearest ten thousand – 30,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 27
Estimate sum is 14,000
Verification by actual substruction
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 28

3. 57,146 – 25,472
57,146 is rounded off to nearest ten thousand – 56,000
25,472 is rounded off to nearest ten thousand – 24,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 29
Estimate sum is 32,000
Verification by actual substraction
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 30

4. 60,046 – 15,247
60,046 is rounded off to nearest ten thousand – 60,000
15,247 is rounded off to nearest ten thousand – 15,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 31
Estimate sum is 45,000
Verification by actual substraction
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 32

VI. Estimate the difference of thefolhming hy rounding off to nearest ten thousands place:

1. 51,689 – 34,685
51,689 is rounded off to nearest ten thousand – 50,000
34,685 is rounded off to nearest ten thousand – 30,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 33
Estimate sum is 20,000
Verification by actual substraction
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 34

2. 86,853 – 47,829
86,853 is rounded off to nearest ten thousand – 85,000
47,829 is rounded off to nearest ten thousand – 45,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 35
Estimate difference is 40,000
Verification by actual substraction
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 36

KSEEB Solutions

3. 80,808 – 55,055
80,808 is rounded off to nearest ten thousand – 80,000
55,055 is rounded off to nearest ten thousand – 50,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 37
Estimate difference is 30,000
Verification by actual substractiori
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 38

4. 77,777 – 44,444
77,777 is rounded off to nearest ten thousand – 70,000
44,444 is rounded off to nearest ten thousand – 40,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 39
Estimate difference is 30,000
Verification by actual substraction
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 40

VII. Estimate the product of each of the following hy rounding off to its highest place:

1. 428 × 54
428 is rounded off to nearest hundred as 400
54 is rounded off to nearest ten as 50
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 41
Estimate product is 20,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 42

2. 878 × 46
876 is rounded off to nearest hundred as 900
46 is rounded off to nearest ten as 50
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 43
Verification by actual multiplication

Verification by actual multiplication
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 44

3. 5,476 × 11
5,476 is rounded off to nearest hundred as 5000
11 is rounded off to nearest ten as 10
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 45
Verification by actual multiplication

Verification by actual multiplication
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 46

KSEEB Solutions

4. 2,645 × 18
2,645 is rounded off to nearest hundred as 3000
18 is rounded off to nearest ten as 20
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 47
Verification by actual multiplication

Verification by actual multiplication
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 48

VIII. Estimate the quotient of each of the following hy rounding off to its highest place.

1. 398 ÷ 82
398 is rounded off to nearest hundred aS 400
82 is rounded off to nearest ten as ÷ 80
Estimate quotient \(\frac{800}{40}\) = 5

2. 786 ÷ 22
786 is rounded off to nearest hundred as 800
22 is rounded off to nearest ten as ÷ 20
Estimate quotient \(\frac{800}{20}\) = 40

3. 3,265 ÷ 58
3,265 is rounded off to nearest hundred as 3000
58 is rounded off to nearest ten as ÷ 60
Estimate quotient \(\frac{3000}{60}\) = 50

4. 7,687 ÷ 43
7,687 is rounded off to nearest hundred as 8000
43 is rounded off to nearest ten as ÷ 40
Estimate quotient \(\frac{8000}{40}\) = 200

IX. Solve the following problems.

Question 1.
A garment company stitched 16,783 shirts and 12,438 pants in a month. Estimate the total number of dresses stitched to the nearest ten thousands place.
Answer:
A garment company stitched shirts – 16,783
A garment company stitched pants – 12,438.
(To the nearest ten thousand place) – 20,000
(To the nearest ten thousand place) – 10,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 49
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 50

Question 2.
A news paper agent sells 36,721 papers in first month and 24,172 papers in sec¬ond month. Estimate the decrease in sale of the newspaper in second month to the nearest ten thousands place.
Answer:
In first month a news paper agents sells – 36,721
In Second month a news paper agents sells – 24,172
(To the nearest ten thousand place)
first month – 40,000
Second month – 20,000
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 51
The estimated substraction of sale of the newspaper 20,000

KSEEB Solutions

Question 3.
A train can cover 225 km in one hour. Estimate the distance covered in a day to the highest place.
Answer:
A train can cover 225 Km in 1 hour
The distance covered in a day = 24 hours
(To the highest place) 200
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 52

Question 4.
A carpenter earned Rs. 18,634 during the month of November and Rs. 32,645 in December. Estimate how much more he earned in December to the nearest ten thousands place.
Answer:
A carpet earned in november – 18,634
A carpet earned in december – 32,645
(To the nearest ten thousand place)
KSEEB Solutions for Class 5 Maths Chapter 3 Mental Arithmetic 53

KSEEB Solutions

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 5 Maths Chapter 4 Factors and Multiples

   

Students can Download Maths Chapter 4 Factors and Multiples Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 5 Maths Chapter 4 Factors and Multiples

KSEEB Class 5 Maths Factors and Multiples Ex 4.1

What is the factor of 50? The factors of 50 are 1, 2, 5,10, 25, …

Question 1.
Circle the multiples of 4 with blue colour, cross the multiples of 6 with red colour and underline the multiples of 9 with a pencil.
Answer:
KSEEB Solutions for Class 5 Maths Chapter 4 Factors and Multiples 1
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 & 100 → Multiples of 4
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90 & 96 → Multiples of 6
9, 18, 27, 36, 45, 54, 63, 72, 81, 90 & 99 → Multiples of 9

KSEEB Solutions

Question 2.
Circle the multiples of 7 in the following numbers.
7, 13, 14, 21, 22, 35, 36, 42 and 45
Answer:
7, 14, 21, 35, 42

Question 3.
Circle the multiples of 12 in the following numbers.
6,12,18, 24, 30, 36, 42, 48, 54, 60, 66, 72
Answer:
12, 24, 36, 48, 60, 72

KSEEB Solutions

Question 4.
Write the multiples of 2 between the numbers 50 and 60.
Answer:
52, 54, 56, 58
G : 52/2 = 26

Question 5.
Write the multiples of 15 between the numbers 50 and 100.
Answer:
60, 75, 90
G : 15 × 4 = 60

Question 6.
Write five multiples of the following numbers.
Answer:
5 multiples of 15 are 15, 30, 45, 60 & 75.
5 multiples of 17 are 17, 34, 51, 68 & 85.
5 multiples of 19 are 19, 38, 57, 76 & 95
5 multiples of 23 are 23, 46, 69, 92 & 115.

KSEEB Solutions for Class 5 Maths Chapter 4 Factors and Multiples 2

KSEEB Solutions

The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24! 24 has 8 factors in total.

Question 7.
Find which of the following numbers are factors of 24?
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22 ಮತ್ತು 24
Answer:
24 – 1 × 24
24 – 4 × 6
24 – 2 × 12
24 – 3 × 8
1, 2, 3, 4, 6, 8, 12 & 24

Question 8.
Write any two factors of the following numbers.
6, 18, 28, 36, 42, 48
Answer:
Any two factors of 6 are 2 & 3
Any two factors of 28 are 2 & 14
Any two factors of 42 are 2 & 21
Any two factors of 18 are 2 & 9
Any two factors of 36 are 3 & 12
Any two factors of 48 are 2 & 24

KSEEB Solutions

Factoring Calculator will help you find the factors 18 i.e. 1, 2, 3, 6, 9, 18 the numbers that when divided results in a whole number and a zero remainder.

Question 9.
Write all the factors of the following numbers.
9, 13, 20, 26, 40
Answer:
All the factors of 9 are 1,3 & 9 All the factors of 13 are 1 & 13 All the factors of 20 are 1, 2, 4, 5, 10 & 20
20 = 1 × 20
2 × 10 = 20
4 × 5 = 20
5 × 4 = 20
10 × 2 = 20
20 × 1 = 20
All the factors of 26 are 1 ,2, 13 & 26
All the factors of 40 are 1, 2, 4, 5, 8, 10, 20 & 40

KSEEB Solutions

Question 10.
Write the factor tree for the following numbers:
Answer:

KSEEB Solutions for Class 5 Maths Chapter 4 Factors and Multiples 3

Prime factors of 84 are 2×2, 3, 7.

Question 11.
Complete the following factor tree by writing missing numbers.

KSEEB Solutions for Class 5 Maths Chapter 4 Factors and Multiples 4KSEEB Solutions for Class 5 Maths Chapter 4 Factors and Multiples 5
Answer:

(i)

KSEEB Solutions for Class 5 Maths Chapter 4 Factors and Multiples 6

(ii)

KSEEB Solutions for Class 5 Maths Chapter 4 Factors and Multiples 7

The prime factors of 38 are 2 and 19.

KSEEB Solutions

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.

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