Category: Class 7

Students can Download Chapter 9 Rational Numbers Ex 9.2, 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 9 Rational Numbers Ex 9.2

Question 1.
Find the sum :

i)

Solution:

ii)

Solution:
L.C.M of 3 and 5 is 15

iii)

Solution:
LCM of 10 and 15 is 30

iv)

Solution:
LCM of 11 and 9 is 99

v)

Solution:
L.C.M of 19 and 57 is 57

vi)

Solution:

vii)

Solution:

L.C.M of 3 and 5 is 15

Question 2.
Find:
i)

Solution:

ii)

Solution:

iii)

Solution:

L.C.M of 13 and 15 is 195

iv)

Solution:

v)

Solution:

L.C.M. of 9 and 1 is 9
Thus,

Question 3.
Find the product:
i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

Question 4.
Find the value of:
i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

vii)

Solution:

Students can Download Chapter 14 Symmetry Ex 14.3, 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 14 Symmetry Ex 14.3

Question 1.
Name any two figures that have both line symmetry and rotational symmetry.
Solution:
The circle and an equilateral triangle have both line symmetry and rotational symmetry.

Question 2.
Draw, wherever possible, a rough sketch of

i) a triangle with both line and rotational symmetries of order more than 1.
Solution:
An equilateral triangle has 3 lines of symmetry and rotational symmetry of order 3.

ii) a triangle with only symmetry and to rotational symmetry of order more than 1.
Solution:

Isosceles triangle has only one line symmetry but no rotational symmetry of order more than 1.

iii) a quadrilateral with a rotational symmetry of order more than 1 but not a line symmetry.
Solution:
A parallelogram has no line symmetry but has a rotational symmetry of order 2.

iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Solution:
An isosceles trapezium has one line of symmetry of order more than 1.

Question 3.
If a figure has two or more lines of symmetry, should it have rotational symmetry of order more than 1 ?
Solution:
Yes

Question 4.
Fill in the blanks :

Question 5.
Name the quadrilaterals which have both line and rotational symmetry of order more than 1.
Solution:
i) Square
ii) Rectangle

Question 6.
After rotating by 600 about a centre, a figure looks exactly the same as its original position. At what other angles will this happen for the figure ?
Solution:
The figure is regular hexagon. The other angles are 120°, 180°, 240°, 300°and 360°.

Question 7.
Can we have a rotational symmetry of order more than 1 whose angle of rotation is
Solution:
i) 45° ?
yes

ii) 17° ?
No, It is possible

Students can Download Chapter 11 Perimeter and Area Ex 11.4, 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 11 Perimeter and Area Ex 11.4

Question 1.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Found the area of the path. Also find the area of the garden in hectare.
Solution:

Area of the garden = l × b
Length = l = 90 m
Breath = b = 75 m
∴ Area of the garden EFG ||)} = 90 × 75
= 6750 sq mtrs
1 Hectare = 10,000 sqmts.

Area of the outer skirt of the garden = l × b
length = 90 + 5 + = 100 m
breadth = 75 + 5 + 5 = 85 m
∴ Area = 100 × 85 = 8,500sqm

Area of the path = 0.850 – 0.675 = 0.175
0.175 hectare or 1750 sq mts

Question 2.
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Solution:
length = l = 125 breadth = b = 65 m
width of a path = 3 m
Area of a rectangular park = l × b = 125 × 65
= 8125 sq mts
Area of ABCD = l × b
l = 125 + 3 + 3 = 131 b = 65 + 3 + 3 = 71
∴ Area = 131 × 71 = 9301 sq mts
Area of the path = Area of ABCD – Area
of park 9301 – 8125 = 1176 sq mts

Question 3.
A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:

Area of cardboard = l × b
l = 8 cm b = 5 cm Margin = 1.5 cm
Area of picture = l × b
l = 8 = (1.5 + 1.5) = 5 cms b = 5 = (1.5 + 1.5) = 2 cms
Area = l × b = 5 × 2
Area of margin = Area of Cardboard – Area of picture
= 40 – 10 = 30 sq cms

Question 4.
A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find the area of the verandah: the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2. The area of the verandah with the room. (Fig KLMN)
Solution:
l = length = 5.5 + 2.25 + 2.25 = 10 m
b = breadth = 4 + 2.25 + 2.25 = 8.5 m
∴ Area 10 × 8.5
= 85 sq cms.

Area of room = l × b
l = 5.5 m = 5.5 × 4 N
∴ Area of verandah = Area of KLMN – Area of room
= 85 – 22 = 63 sq mts
Cost of cementing the floor = Rs 200 / sq m
Total cost of cementing = 63 × 200 = Rs 12,600

Question 5.
A path 1 m wide is built along the border and inside a square garden of side 30m. Find :
i) the area of the path
ii) the cost of planting grass in the remaining portion of the garden at the rate of ₹ 40 per m².
Solution:

Area of KLMN = side × side = 30 × 30
Area of ABCD = 30 – (1 + 1) = 28
= 28 × 28
= 784 sq mts
Area of the path = Area of KLMN – Area of ABCD
Cost of planting grass = Rs 40 / sq m
∴ Total cost of planting = 784 × 40
= Rs 31.360

Question 6.
Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Solution:

Area of PQRS = l x b = 700 × 300
= 2,10,000 sq m
Area of cross Roads = Area of KLMN abcd – 1, 2, 3, 4
= 300 × 10 + 700 × 10 – 10 × 10
= 3000 + 7000 – 100
= 10,000 – 100 = 9,900 sq mts
Area of the park excluding the cross roads = 2,10,000 – 9,900
= 2,00,100 sq mts.

Question 7.
Through a rectangular filed of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 in. Find :
i) The area covered by the roads.

Area covered by the roads = Area of klmn + Area of abcd – Area of 1, 2, 3, 4
Area of abcd = 90 × 3 = 270 sq mtrs
Area of klmn = 60 × 3 = 180 sq mts.

= 180 + 270 – 9 = 441 sq mts.

ii) the cost of constructing the roads at the rate of ₹ 110 per m².
Cost of constructing the roads = Rs 110/sq m
∴ Total cost of constructing the roads
= 441 × 110 = Rs. 48.510

Question 8.
Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure)
and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown).
Did she have any cord left? (r = 3.14)
Solution:

Radius of the circular pipe} r = 4 cm
Cirumference of the circular pipe } 2πr
=2 × 3.14 × 4
= 25.12 cms
Perimeter of the square box = 4 × side
= 4 × 4
= 16 cms
yes, the length of the cord left = 25.12 – 16
= 9.12 cms

Question 9.
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find
Solution:

i) the area of the whole land
Area of the whole land = l × b
l = 10 mts = 10 × 5
b = 5 mts = 50 sq mts

ii) the area of the flower bed
Area of the flower bed = πr²
r = 2m = 3.14 × 2 × 2
= 12.56 sq mts

iii) the area of the lawn excluding the area of the flower bed
Solution:
The area of the lawn excluding the circular flower bed } = 50 – 12.56
= 37.44 sq m

iv) the circumference of the flower bed.
The circumference of the circular flower bed} = 2πr
= 2 × 3.14 × 2
= 12.56 mts.

Question 10.
In the following figures, find the area of the shaded portions :

Solution:
1) Area of the rectangle ABCD = l × b
l = 18 cms = 18 × 10
b = 10 cms = 180 sq cms

= 30 sq cms

= 40 sq cms
Area of the shaded position = Area of rectangle ABCD – (AreaofAEF+AreaofBCE)
= 180 – (30 + 40)
= 180 – 70 = 1110 sq cms

ii) Area of the square PQRS = side × side
= 20 × 20
= 400 sq cms.

= 50 sq cm

= 100 sq cms

= 100 sq cm
∴ Area of shaded region = 400 – (50 + 100 – 100)
= 400 – 250
= 150 sq cms.

Question 11.
Find the area of the quadrilateral ABCD.
Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC

Solution:
Area of the quadrilateral ABCD =
Area of the ∆ ABC + Area of the ∆ ACD

∴ Area of quadrilateral ABCD = 33 + 33
= 66 sq cms

Students can Download Chapter 9 Rational Numbers Ex 9.1, 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 9 Rational Numbers Ex 9.1

Question 1.
List five rational numbers between:
a) -1 and 0
Solution:

we see that

∴ Five rational numbers between -1 and 0 are

ii) -2 and -1
Solution:

we see that

∴ Five Rational Numbers between -2 and -1 are

iii)

Solution:

we see that

Thus five rational numbers between

iv)

Solution:

we see that

Thus 5 Rational Numbers in between \(\frac{1}{2}\) and \(\frac{2}{3}\) are

Question 2.
Write four more rational numbers in each of the following patterns :
i)

Solution:

Thus we came to know the pattern in the above numbers ie., both numerator and denominator should be multiplied by same whole number.
∴ The four more rational numbers would be

ii)

Solution:

Thus we observe the pattern in these numbers.
The other numbers would be,

 

iii)

Solution:

Thus we observe the pattern in these numbers. The other numbers would be,

iv)

Solution:

Thus we observe the pattern in these numbers
The other numbers would be

Question 3.
Give four rational numbers equivalent to :
i) \(\frac{-2}{7}\)
Four rational numbers equivalent to \(\frac{-2}{7}\) are

ii) \(\frac{5}{-3}\)
Four rational numbers equivalent to \(\frac{5}{-3}\) are

iii) \(\frac{4}{9}\)
Solution:
Four rational numbers equivalent to \(\frac{4}{9}\) are

Question 4.
Draw the number line and represent the following rational numbers on it :
i) \(\frac{3}{4}\)
Solution:

ii) \(\frac{-5}{8}\)
Solution:

iii) \(\frac{-7}{4}\)
Solution:

iv) \(\frac{7}{8}\)
Solution:

Question 5.
The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.
Solution:

The rational number represented by

The rational number represented by

The rational number represented by

The rational number represented by

The rational number represented by P Q R S is equal to respectively.

Question 6.
Which of the following pairs represent the same rational number ?
i)

Solution:
\(\frac{-7}{21}\) is negative rational number and \(\frac{3}{9}\) is a positive Rational Number. Therefore this pair does not represent the same rational number.

ii)

Solution:

∴ The above pair of numbers represent the same rational number.

iii)

Solution:

∴ The above pair of numbers represent the same rational number.

iv)

Solution:

∴ The above pair of numbers represent the same rational number.

v)

Solution:

∴ The above pair of numbers represent the same rational number.

vi)

Solution:
\(\frac{1}{3}\) is a positive rational number and \(\frac{-1}{9}\) is a negative Rational Number.
Therefore this pair does not represent the same rational number.

vii)

Solution:
is a positive rational number and \(\frac{5}{-9}\) is a negative rational number.

Question 7.
Rewrite the following rational numbers in the simplest form :

i) \(\frac{-8}{6}\)
The H. C. F of 6 and 8 is 2.
Then its simplest form would be,

ii) \(\frac{25}{45}\)
The H. C. F of 25 and 45 is 5.
Then its simplest form would be,

iii) \(\frac{-44}{72}\)
The H. C. F of 44 and 72 is 4.
Then its simplest form would be

iv)

The H. C. F of 8 and 10 is 2.
Then Its simplest form would be

Question 8.
Fill in the boxes with the correct symbol out of >, <, and =.

i)

Solution:
\(\frac{-5}{7}\) is a negative rational number
\(\frac{2}{3}\) is a positive rational number.

ii)

Solution:
To make the equal denominator

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

vii)

Solution:
0 is greater than the negative numbers,

Question 9.
Which is greater in each of the following

i)

Solution:
L.C.M of 3 & 2 is 6

 

ii)

L. C. M of 6 and 3 is 6

iii)

Solution:
L. C M of 4 and 3 is 12

 

iv)

Solution:
(∵ All positive numbers are greater than all negative numbers.)

v)

Solution:

Question 10.
Write the following rational numbers in ascending order :
i)

-3 < -2 < -1
∴ Ascending Order is

ii)

Solution:
L. C. M of 3 & 9 is 9

iii)

Solution:

Denominator same, So the numerators are
-42 < -21 < -12

Students can Download Chapter 14 Symmetry Ex 14.2, 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 14 Symmetry Ex 14.2

Question 1.
Which of the following figures have rotational symmetry of order more than 1:
Solution:

(a), (b), (d), (e), and (f) because these objects look exactly the same more than 1 number of times in a complete turn.

Question 2.
Give the order of rotational symmetry for each figure :
Solution:

fig a has order 2
a) This object while rotating at two positions 180° and 360° it looks exactly the same. Therefore it has rotational symmetry of order 2.

b) This object in full turn, these are two positions 180° and 360° it looks exactly the same. Therefore it has rotational symmetry of order 2.

fig b has order 2

c) This object while rotating there are three positions 120° and 240° and 360° the object looks exactly the same. Therefore it has rotational symmetry of order 3. fig c has order 3

d) In full turn, this object will rotate and three are four positions such as 90°, 180° 270°, and 360° the figure looks exactly the same. Therefore it has rotational symmetry of order 4.

e) In a full turn, this object will rotate and there are four positions such as 90°, 180° 270°, and 360°, the figure looks exactly the same. Therefore it has rotational symmetry of order 4.

fig e has order 4

f) The figure is a regular pentagon. In a full turn are five positions at 72°, 144°, 216°, 288°, and 360° the figure looks exactly the same. Therefore it has rotational symmetry of order 5.

fig f has order 5

g) In a full turn, there are 6 positions. i.e., 60°, 120°, 180°, 240°, 300°, and 360° the figure looks exactly the same. Therefore it has rotational symmetry of order 6.

fig g has order 6

h) This figure while rotating there are 3 positions the figure looks exactly the same at 120°, 240°, and 360°. Therefore it has rotational symmetry of order 3.

fig h has order 3

Students can Download Chapter 6 The Triangles and Its Properties Ex 6.4, 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 6 The Triangles and Its Properties Ex 6.4

Question 1.
Is it possible to have a triangle with the following sides ?
i) 2 cm, 3 cm, 5 cm.
Solution:
greatest side is 5 cms.
2 + 3 = 5
Here sum of the side is equal to third side.
So it is impossible to construct the triangle by given measures.
Because the property of a triangle says that the sum of the two sides of a triangle is greater than the length of the third side.

ii) 3 cm, 5 cm, 7 cm
Solution:
The greatest side is 7 cm.
3 + 6 = 9 cms > 7
6 + 7 = 13 > 3
7 + 3 = 10 > 6
∴ It is possible to construct the triangle with the sides of 3, 6 and 7 cms.

iii) 6 cm, 3 cm, 2 cm
6 + 3 = 9 > 2
3 + 2 = 5 > 6
2 + 6 = 8 > 3
∴ It is not possible to construct the triangle with the sides of 6, 3 and 2 cms.

Question 2.
Take any point ‘O’ in the interior of a triangle PQR. Is

i) OP + OQ > PQ
Solution:
Yes. Sum of the length of any two sides of a triangle is greater than the length of the 3 side.

ii) OQ + OR > QR
Solution:
Yes, Sum of the length of any two sides of a triangle is greater than the length of the 3^rd side.

iii) OR + OP > RP
Solution:
Yes, Sum of the length of any two sides of a triangle is greater than the length of the 3^rd side.

Question 3.
AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM ?
(Consider the sides of triangles ∆ ABM and ∆ AMC.)
Solution:

Consider the triangle ABM
In ∆ ABM
AB + BM > AM — 1 (Sum of the length of any 2 sides of a triangle is greater that the 3^rd side.)
In ∆ ACM
CM + CA > AM — 2 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
Add 1 and 2
(AB + BM) + (CM + CA) > AM + AM
AB + (BM + CM) + CA > 2AM
AB + BC + CA > 2AM (yes, it is proved.)

Question 4.
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?
Solution:

Consider the ∆ ABC
AB + BC > CA — 1 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
In ∆ ACD
AD + CD > AC — 2 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
Adding 1 and 2
AB + BC + CD + DA > 2AC — 3
In ∆ ABD
AB + DA > BD — 4 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
In ∆ BCD
BC + CD > BD — 5 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
Adding 4 and 5
AB + DA + BC + CD > 2BD — 6
Adding 3 and 6

∴ AB + BC + CD + DA > AC + BD (yes, it is proved)

Question 5.
ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD) ?
Solution:

In ∆ AOB OA + OB > AB — 1 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
In ∆ BOC
OB + OC > BC — 2 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
In ∆ COD
OC + CD > CD — 3 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
In ∆ AOD
OA + OD > AD — 4 (Sum of the length of any 2 sides of a triangle is greater than the 3^rd side.)
Add all the 4

∴ AB + BC + CD + DA< 2 (AC + BD) (proved)

Question 6.
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Solution:
Let ‘x’ cm to the length of the 3^rd side.
According to the sum of the length of any 2 sides of a triangle is greater than 3^rd side.
∴ 12 + 15 >x
27 > x
or x < 27 15 + x > 12
x > 12 – 15 = -3
x + 12 > 15
x > 15 – 12 = 3
∴ x > -3 and x > 3 we should take x > 3
∴ The length of 3^rd side of a triangle should be in between 3 cm and 27 cms.

Students can Download Chapter 14 Symmetry Ex 14.1, 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 14 Symmetry Ex 14.1

Question 1.
Copy the figures with punched holes and find the axes of symmetry for the following :

Solution:

 

The dotted lines are the axes of symmetry for the above figs.

Question 2.
Given the line(s) of symmetry, find the other hole(s) :

Solution:

Question 3.
In the following figures, the mirror line (i.e., the line of symmetry) is given as a dotted line. Complete each figure performing reflection in the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image). Are you able to recall the name of the figure you complete ?
Solution:

Solution:

The dotted lines are the axes of symmetry for the above figs.

Question 2.
Given the line(s) of symmetry, find the other hole(s) :

Solution:

Question 3.
In the following figures, the mirror line (i.e., the line of symmetry) is given as a dotted line. Complete each figure performing reflection in the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image). Are you able to recall the name of the figure you complete ?

Solution:

Question 4.
The following figures have more than one line of symmetry. Such figures are said to have multiple lines of symmetry.

Identify multiple lines of symmetry, if any, in each of the following figures:
a)

Solution:

b)

Solution:

 

c)

Solution:

 

d)

Solution:

 

e)

Solution:

f)

Solution:

 

g)

Solution:

h)

Solution:

Question 5.
Copy the figure given here :
Solution:
Take any one diagonal as a line of symmetry and shade a few more squares to make the figure symmetric about a diagonal. Is there more than one way to do that? will the figure be symmetric about both the diagonals?
Yes, there is more than one way to make the figure symmetric.

i) Let us take the diagonal BD and shade the squares as shown in the figure to make the figure symmetric about BD.
ii) Similarly the figure is symmetric about the diagonals.
iii) The figure is symmetric about EF and GH also.

Question 6.
Copy the diagram and complete each shape to be symmetric about the mirror line(s)
Take any one diagonal as a line of symmetry and shade a few more squares to make the figure symmetric about both the diagonals?
Solution:

Question 7.
State the number of lines of symmetry for the following figures.
Solution:
a)

3 lines of symmetry.

b)

1 line of symmetry

c)

Zero or No lines of symmetry.

d)

4 lines of symmetry

e)

2 lines of symmetry

f)

2 lines of symmetry

g)

Zero or No lines of symmetry

h)

Zero or No lines of symmetry

i)

6 lines of symmetry

j)

Many or Infinite number of symmetry.

Question 8.
What letters of the English alphabet have reflectional symmetry (i.e. symmetry related to mirror reflection) about.
a. a vertical mirror
A H I M O T U V W X Y
These letter have vertical symmetry

b. a horizontal mirror
B C D E H I O X
These letter have horizontal symmetry.

c. both horizontal and vertical mirrors
H I O X
These have both horizontal and vertical symmetry

Question 9.
Give three examples of shapes with no line of symmetry.
Solution:
1) Quadrilateral
2) Scalene triangle
3) Parallelogram
Have no line of symmetry.

Question 10.
What other names can you give to the line of symmetry of
a) an isosceles triangle?
b) a circle?
Solution:
a) The line of symmetry in an isosceles triangle is also known as Median.
b) The line of symmetry in a circle is also called Diameter.

Students can Download Chapter 6 The Triangles and Its Properties Ex 6.3, 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 6 The Triangles and Its Properties Ex 6.3

Question 1.
Find the value of the unknown x in the following diagrams :

Solution:
i) ∠A + ∠B + ∠C = 180°
x° + 50° + 60° = 180° (By Angle sum property of a triangle)
∠x + 110° = 180°
∠x = 180° – 110°
∠x = 70°

ii) ∠P + ∠Q + ∠R = 180°
90° + 30° + ∠x = 180° (By Angle sum property of a triangle)
120° + ∠x = 180°
∠x = 180° – 120° = 60°

iii) ∠X + ∠Y + ∠Z = 180°
Solution:
30° + 110° + ∠x = 180° (By Angle sum property of a triangle)
140° + ∠x = 180°
∴ ∠x = 180° – 140° = 40°

iv) 50° + x + x = 180°
50° + 2x = 180°
(By Angle sum property of a triangle)
2x = 180° – 50° = 130°
x = \(\frac{180^{\circ}}{2}\)
∴ ∠x = 60°

v) x + x + x = 180°
(By Angle sum property of a triangle)
3x = 180°
x = \(\frac{180^{\circ}}{2}\)
∴ x = 60°

vi) x + 2x + 90° = 180° (ByAngle sum property of a triangle)
3x = 180°

Question 2.
Find the values of the unknowns x and y in the following diagrams:

Solution:
i) x + 50° = 120° [By the exterior angle property of a triangle]
x= 120° – 50° = 7o°
By Angle sum property of a triangle
x + y + 50° = 180°
x + y = 180° – 50° = 130°
70° + y = 130°
y = 130° – 70° = 60°
∴ x = 70°, y = 60°

ii) 50° + y° + x° = 180°
∠y = 80° [vertically opposite angles]
50° + 80° + x° = 180° (Angle sum property of a triangle)
130° + x = 180°
x° = 180° – 130° = 50°
[∴ x = 50°, y = 80°]

iii) ∠x = 50° + 60° = 110° [By exterior – angle property of a triangle]
y + 50° + 60° = 180° ( By Angle sum property of a triangle)
y + 110° = 180°
∴ y = 180° – 110° = 70°
∴ y = 70°, x = 110°

iv) ∠x = 60° [vertically opposite angles]
30° + x° + y° = 180°
(By Angle sum property of a triangle)
30° + 60° + y° = 180°
90° + y° = 180°
∴ y°= 180° – 90° = 90°
∴ x = 60° y = 90°

v) ∠y = 90° [vertically opposite angles]
∠x + ∠x + ∠y = 180°
(Angle sum property of a triangle)
2x + y° = 180°
2x + 90° = 180°
2x = 180° – 90°= 90°

∴ ∠x = 45° & ∠y = 90°

vi) ∠x ∠y [vertically opposite angles]
∠x + ∠x + ∠y = 180°
(Angle sum property of a triangle)
2x + ∠y = 180°
2x + x° = 180° (∵ y = x)
3x = 180°

∴ x = 60° & y = 60°

Students can Download Chapter 13 Exponents and Powers Ex 13.3, 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 13 Exponents and Powers Ex 13.3

Question 1.
Write the following numbers in the expanded forms :
279404, 3006194, 2806196, 120719, 20068
Solution:
i) 279404
= 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4 × 1
= 2× 10⁵ + 7× 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰

ii) 3006194
= 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4 × 1
= 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰

iii) 2806196
= 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10¹ + 6 × 10⁰
= 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰

iv) 120719
= 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9 × 1
= 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰

v) 20068
= 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1
= 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Question 2.
Find the number from each of the following expanded forms :
a. 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
b. 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰
c. 3 × 10⁴ + 7 × 10² + 5 × 10⁰
d. 9 × 10⁵ + 2 × 10² + 3 × 10¹
Solution:
a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 80,000 + 6000 + 0 + 40 + 5
= 86,045

b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰
= 400000 + 5000 + 300 + 2
= 4,05,302

c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰
= 40000 + 500 + 5
= 130,7051

d) 9 × 10⁵ + 2 × 10² + 3 × 10¹
= 900000 + 200 + 30
= 9,00,230

Question 3.
Express the following numbers in standard form:
(i) 5,00,00,000
(ii) 70,00,000
(iii) 3,18,65,00,000
(iv) 3,90,878
(v) 39087.8
(vi) 3908.78
Solution:
i) 5,00,00,000
5 × 10⁷

ii) 70,00,000
7 × 10⁶

iii) 3,18,65,00,000
3.1865 × 10⁹

iv) 3,90,878
3.90878 × 10⁵

v) 39087.8
3.90878 × 10³

vi) 3908.78
3.90878 × 10³

Question 4.
Express the number appearing in the following statements in standard form.
Solution:
a) The distance between Earth and Moon is 384,000,000 m.
= 3.84 × 10⁸ m

b) Speed of light in vacuum is 300,000,000 m/s
= 3 × 10⁸ m/s

c) Diameter of the Earth is 1,27,56,000 m.
1.2756 × 10⁷ m

d) Diameter of the Sun is
1,400,0, 000m.
1.4 × 10⁹m

e) Number of average stars in a galaxy = 100,000,000m.
= 1 × 10¹¹ m

f) The universe is estimated to be about 12,000,000,000 years old.
= 1.2 × 10¹⁰ years old

g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000, 000, 000 m.
= 3 × 10²⁰ m

h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
= 6.023 × 10²² molecules

i ) The earth has 1,353,000,000 cubic km of sea water.
= 1.353 × 10⁹ cubic km

j) The population of India was about 1,027,000000 in March, 2001.
= 1.027 × 10⁹

Students can Download Chapter 13 Exponents and Powers Ex 13.2, 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 13 Exponents and Powers Ex 13.2

Question 1.
Using laws of exponents, simplify and write the answer in exponential form :
(i) 3² × 3⁴ × 3⁸
(ii) 6¹⁵ ÷ 6¹⁰
(iii) a³ × a²
(iv) 7^x × 7²
(v) (5²)³ ÷ 5²
(vi) 2⁵ × 5⁵
(vii) (a⁴ × b⁴)
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) × 2³
(x) 8^t ÷ 8²
Solution:
i) 3² × 3⁴ × 3⁸
= 3^{2 + 4 + 8} = 3¹⁴ [∵ a^m × aⁿ = a^{m + n}]

ii) 6¹⁵ ÷ 6¹⁰
= 6^{15 – 10} = 6⁵ [∵ a^m ÷ aⁿ = a^{m – n}]

iii) a³ × a²
= a^{3 + 2} = a⁵ [∵ a^m × aⁿ = a^{m + n}]

iv) 7^x × 7²
= 7^{x + 2} [∵ a^m × aⁿ = a^{m + n}]

v) (5²)³ ÷ 5³
= 5^{2 × 3} ÷ 5³ = 5^{6 – 3} [(a^m)ⁿ = a^mn
∵ a^m ÷ aⁿ = a^{m – n}]

vi) 2⁵ × 5⁵
= (2 × 5)⁵ = 10⁵ [∵ a^m × b^m = (ab)^m]

vii) a⁴ × b⁴
= a⁴ × b⁴ = (ab)⁴ [∵ a^m × b^m = (ab)^m]

viii) (3⁴)³
= 3^{4 × 3} = 3¹² [(a^m)ⁿ = a^mn]

ix) (2²⁰ ÷ 2¹⁵)
(2²⁰ ÷ 2¹⁵) × 2³ = 2^{20 – 15} × 2³ [a^m ÷ aⁿ = a^{m – n}]
= 2⁵ × 2³
= 2^{5 + 3} = 2⁸ [a^m × aⁿ = a^{m + n}]

x) 8^t ÷ 8²
= 8^{t – 2} [∵ a^m ÷ aⁿ = a^{m – n}]

Question 2.
Simplify and express each of the following in exponential form :
i)

Solution:

ii) ((5²)³ × 5⁴) ÷ 5⁷
= (5^{2 × 3} × 5⁴) ÷ 5⁷ ((a^m)ⁿ = a^mn)
= 5⁶ × 5⁴ ÷ 5⁷
= 5^{6 + 4} ÷ 5⁷ = 5^{10 – 7} = 5³

iii) 25⁴ ÷ 5³
= (5²)⁴ ÷ 5³ = 5^{2 × 4} ÷ 5³
= 5⁸ ÷ 5³ = 5^{8 – 3} = 5⁵ [a^m ÷ aⁿ = a^{m – n}

iv)


= 7^{2 – 1} × 11^{8 – 3} [a^m ÷ aⁿ = a^{m – n}]
= 7¹ × 11⁵ = 7 × 11⁵

vi) 2⁰ + 3⁰ + 4⁰
= 1 + 1 + 1 = [3] [a⁰ = 1]

vii) 2⁰ × 3⁰ × 4⁰
= 1 × 1 × 1 = 1

viii) (3⁰ + 2⁰) × 5⁰
= (1 + 1) × 1 = 2 × 1 = 2

ix)

= (2a)²

x)

= a ^{5 – 3} × a⁸
= a² × a⁸ a^{2 + 8}

xi)

xii) (2³ × 2)²
= (2³ × 2¹)²
= (2⁴)²
= 2^{4 × 2} = 2⁸

Question 3.
Say true or false and justify your answer:
i) 10 × 10¹¹ = 100¹¹
L.H.S = 10 × 10¹¹ = 10^{1 + 11} = 10¹² RHS
100¹¹ = (10 × 10)¹¹
= (10²)¹¹
= 1o² × 11 = 10²²
∴ 10¹² ≠ 10²²
So 10 × 10¹¹ ≠ 100¹¹ ∴ It is false.

ii) 2³ > 5²
= 2 × 2 × 2 > 5 × 5
= 8 > 25
∴ 2³ > 5² ∴ It is false.

iii) 2³ × 3² = 6⁵
= 2 × 2 × 2 × 3 × 3
= 6 × 6 × 6 × 6 × 6
= 72 = 7776
∴ 2³ × 3² ≠ 6⁵ ∴ It is false.

iv) 3⁰ = (1000)⁰
1 = 1
∴ 3⁰ = (1000)⁰ ∴ It is true

Question 4.
Express each of the following as a product of prime factors only in exponential form :

i) 108 x 192

= 2 × 2 × 3 × 3 × 3 2 × 2 × 2 × 2 × 2 × 2 × 3
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 2⁸ × 3⁴

It is the product of prime factor

ii) 270
270 = 2 × 3 × 3 × 3 × 5
= 2 × 3³ × 5

It is required prime factor.

iii) 729 × 64
729 = 3 × 3 × 3 × 3 × 3 × 3
64 = 2 × 2 × 2 × 2 × 2 × 2
∴ 729 × 64 = 3⁶ × 2⁶

It is the required prime factor.

iv) 768
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 2⁸ × 3¹ or = 2⁸ × 3

It is a required prime factor.

Question 5.
Simplify :
i)

Solution:

ii)

Solution:

iii)

Solution: