Tag: NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5.

• The Triangle and its Properties Class 7 Ex 6.1
• The Triangle and its Properties Class 7 Ex 6.2
• The Triangle and its Properties Class 7 Ex 6.3
• The Triangle and its Properties Class 7 Ex 6.4
• The Triangle and its Properties Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 6
Chapter Name	The Triangle and its Properties
Exercise	Ex 6.5
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

Question 1.
PQR is a triangle right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Solution:
QR² = 10² + 24² By Pythagoras Property
⇒ = 100 + 576 = 676
⇒ QR = 26 cm.

Question 2.
ABC is a triangle right-angled at C. If AB – 25 cm and AC = 7 cm, find BC.

Solution:
AC² + BC² = AB² By Pythagoras Property
⇒ 7² + BC² = 25²
⇒ 49 + BC² = 625
⇒ BC² = 625 – 49
⇒ BC² = 576
⇒ BC = 24 cm.

Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution:
Let the distance of the foot of the ladder from the wall be a m. Then,

Hence, the distance of the foot of the ladder from the wall is 9 m.

Question 4.
Which of the following can be the sides of a right triangle ?

1. 2.5 cm, 6.5 cm, 6 cm.
2. 2 cm, 2 cm, 5 cm.
3. 1.5 cm, 2 cm, 2.5 cm.

In the case of right-angled triangles, identify the right angles.
Solution:
1. 2.5 cm, 6.5 cm, 6 cm We see that
(2.5)² + 6² = 6.25 + 36 = 42.25 = (6.5)²
Therefore, the given lengths can be the sides of a right triangle. Also, the angle between the lengths, 2.5 cm and 6 cm is a right angle.

2. 2 cm, 2 cm, 5 cm
∵ 2 + 2 = 4 \(\ngtr\) 5
∴ The given lengths cannot be the sides of a triangle
The sum of the lengths of any two sides of a triangle is greater than the third side

3. 1.5 cm, 2 cm, 2.5 cm We find that
1.5² + 2² = 2.25 + 4 = 6.25 = 2.5²
Therefore, the given lengths can be the sides of a right triangle.
Also, the angle between the lengths 1.5 cm and 2 cm is a right angle.

Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Solution:
AC = CD Given
In right angled triangle DBC, DC² = BC² + BD²
by Pythagoras Property = 5² + 12² = 25 + 144 = 169

⇒ DC = 13 ⇒ AC = 13
⇒ AB = AC + BC = 13 + 5 = 18
Therefore, the original height of the tree = 18 m.

Question 6.
Angles Q and R of a ∆ PQR are 25° and 65°. Write which of the following is true:

(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²
Solution:
(ii) PQ² + RP² = QR² is true.

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Solution:
In right-angled triangle DAB, AB² + AD² = BD²
⇒ 40² + AD² = 41² ⇒ AD² = 41² – 40²
⇒ AD² = 1681 – 1600
⇒ AD² = 81 ⇒ AD = 9
∴ Perimeter of the rectangle = 2(AB + AD) = 2(40 + 9) = 2(49) = 98 cm
Hence, the perimeter of the rectangle is 98 cm.

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Solution:
Let ABCD be a rhombus whose diagonals BD and AC are of lengths 16 cm and 30 cm respectively.
Let the diagonals BD and AC intersect each other at O.
Since the diagonals of a rhombus bisect each other at right angles. Therefore
BO = OD = 8 cm,
AO = OC = 15 cm,
∠AOB = ∠BOC
= ∠COD = ∠DOA = 90°
In right-angled triangle AOB.
AB² = OA² + OB²
By Pythagoras Property
⇒ AB² = 15² + 8²
⇒ AB² = 225 + 64
⇒ AB² = 289
⇒ AB = 17cm
Therefore, perimeter of the rhombus ABCD = 4 side = 4 AB = 4 × 17 cm = 68 cm
Hence, the perimeter of the rhombus is 68 cm.

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2.

• Visualising Solid Shapes Class 7 Ex 15.1
• Visualising Solid Shapes Class 7 Ex 15.2
• Visualising Solid Shapes Class 7 Ex 15.3
• Visualising Solid Shapes Class 7 Ex 15.4
• Visualising Solid Shapes Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 15
Chapter Name	Visualising Solid Shapes
Exercise	Ex 15.2
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2

Question 1.
Use isometric dot paper and make an isometric sketch for each one of the given shapes :


Solution:

Question 2.
The dimensions of a cuboid are 5 cm, 3cm and 2cm. Draw three different isometric sketches of this cuboid
Solution:

Question 3.
Three cubes each with 2 cm edge are placed side by side to for,n a cuboid. Sketch an oblique or isometric sketch of this cuboid.
Solution:

Question 4.
Make an oblique sketch for each one of the given isometric shapes :

Solution:

Question 5.
Give (i) an oblique sketch and (ii) an isometric sketch for each of the following:
(a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique?)
(b) A cube with an edge 4 cm long.
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3.

• Visualising Solid Shapes Class 7 Ex 15.1
• Visualising Solid Shapes Class 7 Ex 15.2
• Visualising Solid Shapes Class 7 Ex 15.3
• Visualising Solid Shapes Class 7 Ex 15.4
• Visualising Solid Shapes Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 15
Chapter Name	Visualising Solid Shapes
Exercise	Ex 15.3
Number of Questions Solved	1
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3

Question 1.
What cross-sections do you get when you give a
(i) vertical cut
(ii) horizontal cut to the following solids?
(a) A brick
(b) A round apple
(c) A dice
(d) A circular pipe
(e) An ice cream cone.
Solution:

S.No.	Vertical Cut	Horizontal Cut
(a)	(i) rectangle	(ii) rectangle
(b)	(i) circle	(ii) circle
(c)	(i) square	(ii) square
(d)	(i) circle	(ii) rectangle
(e)	(i) triangle	(ii) circle

We hope the NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4.

• Visualising Solid Shapes Class 7 Ex 15.1
• Visualising Solid Shapes Class 7 Ex 15.2
• Visualising Solid Shapes Class 7 Ex 15.3
• Visualising Solid Shapes Class 7 Ex 15.4
• Visualising Solid Shapes Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 15
Chapter Name	Visualising Solid Shapes
Exercise	Ex 15.4
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4

Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions):

Solution:

1. a circle.
2. a rectangle.
3. a rectangle.

Question 2.
Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these!)

Solution:

1. → Sphere, cylinder
2. → Cube, cuboid (of the square end)
3. → Cone, the prism of the triangular base
4. → Cylinder, the prism of the square base

Question 3.
Examine if the following are true statements:

1. The cube can cast a shadow in the shape of a rectangle.
2. The cube can cast a shadow in the shape of a hexagon.

Solution:

1. True, the cube can cast a shadow in the shape of a rectangle.
2. False, the cube can not cast a shadow in the shape of a hexagon.

We hope the NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2.

• Symmetry Class 7 Ex 14.1
• Symmetry Class 7 Ex 14.3
• Symmetry Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 14
Chapter Name	Symmetry
Exercise	Ex 14.2
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for 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:
Figures (a), (b), (d), (e) and (f) have rotational symmetry of order more than 1.

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


Solution:

(a) → 2
(b) → 2
(c) → 3
(d) → 4
(e) → 4
(f) → 5
(g) → 6
(h) → 3

Mark a point P as shown in figure (i). We see that in a full turn, there are two positions (on rotation through the angles 180° and 360°) when the figure looks exactly the same. Because of this, it has rotational symmetry of order 2.

Mark a point P as shown in figure (i). We see that in a full turn, there are two positions (on rotation through the angles 180° and 360°) when the figure looks exactly the same.

Because of this, it has rotational symmetry of order 2.

Mark a point P as shown in figure (i). We see that in a full turn, there are three positions (on rotation through the angles 120°, 240°, and 360°) when the figure looks exactly the same. Because of this, it has rotational symmetry of order 3.

Mark a point P as shown in figure (i). We see that in a full turn, there are four positions (on rotation through the angles 90°, 180°, 270°, and 360°) when the figure looks exactly the same. Because of this, we say that it has rotational symmetry of order 4.

Similarly,

(e) In a full turn, there are four positions (on rotation through the angles 90°, 180°, 270°, and 360°) when the figure looks exactly the same.

∴ It has rotational symmetry of order 4.

(f) The figure is a regular pentagon. In a full turn, there are five positions (on rotation through the angles 72°, 144°, 216°, 288°, and 360°) when the figure looks exactly the same.

∴ It has rotational symmetry of order 5.

(g) In a full turn, there are six positions (on rotation through the angles 60°, 120°, 180°, 240°, 300°, and 360°) when the figure looks exactly the same.

It has rotational symmetry of order 6.

(h) In a full turn, there are three positions (on rotation through the angles 120°, 240°, and 360°) when the figure looks exactly the same.

∴ It has rotational symmetry of order 3.

We hope the NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3.

• Symmetry Class 7 Ex 14.1
• Symmetry Class 7 Ex 14.2
• Symmetry Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 14
Chapter Name	Symmetry
Exercise	Ex 14.3
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3

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

Question 2.
Draw, wherever possible, a rough sketch of
(i) a triangle with both line and rotational symmetry of order more than 1.
(ii) a triangle with only line symmetry and no rotational symmetry of order more than 1.
(iii) a quadrilateral with rotational symmetry of order more than 1 but not a line symmetry.
(iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Solution:
(i) An equilateral triangle has 3 lines of symmetry and rotational symmetry of order 3.


(ii) An isosceles triangle has only one line symmetry but no rotational symmetry of order more than 1.

(iii) A parallelogram has no line of symmetry but has a rotational symmetry of order 2.

(iv) An isosceles trapezium has one line of symmetry but no rotational 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:
When a figure has two or more lines of symmetry, then the figure should have rotational symmetry of order more than 1.

Question 4.
Fill in the blanks:

Solution:

Question 5.
Name the quadrilaterals which have both line and rotational symmetry of order more than 1.
Solution:
The name of quadrilaterals having both line and rotational symmetry is square.

Question 6.
After rotating by 60° 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 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
(i) 45°?
(ii) 17°?
Solution:
(i) Yes

(ii) No

We hope the NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3.

• Exponents and Powers Class 7 Ex 13.1
• Exponents and Powers Class 7 Ex 13.2
• Exponents and Powers Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 13
Chapter Name	Exponents and Powers
Exercise	Ex 13.3
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3

Question 1.
Write the following numbers in the expand forms :
(i) 279404
(ii) 3006194
(iii) 2806196
(iv) 120719
(v) 20068.
Solution:

Question 2.
Find the number from each of the following expanded forms :

Solution:

Question 3.
Express the following numbers in standard form:

1. 5,00,00,000
2. 70,00,000
3. 3,18,65,00,000
4. 3,90,878
5. 39087.8
6. 3908.78

Solution:

1. 5,00,00,000 = 5 × 10⁷
2. 70,00,000 = 7 × 10⁶
3. 3,18,65,00,000 = 3.1865 × 10⁹
4. 3,90,878 = 3.90878 × 10⁵
5. 39087.8 = 3.90878 × 10⁴
6. 3908.78 = 3.90878 × 10³

Question 4.
Express the number appearing in the following statements in standard form :

1. The distance between Earth and Moon is 384,000,000 m.
2. The speed of light in a vacuum is 300,000,000 miles.
3. The diameter of the Earth is 1,27,56,000 m.
4. Diameter of the Sun is 1,400,000,000 m.
5. In a galaxy, there are on average 100,000,000,000 stars.
6. The universe is estimated to be about 12,000,000,000 years old.
7. The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.
8. 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
9. The earth has 1,353,000,000 cubic km of seawater.
10. The population of India was about 1,027,000,000 in March 2001.

Solution:

1. The mean distance between Earth and Moon is 3.84 × 10⁸ m.
2. The speed of light in a vacuum is 3 × 10⁸ m/s.
3. The diameter of the Earth is 1.2756 × 10⁷ m.
4. The diameter of the Sun is 1.4 × 10⁹ m.
5. In a galaxy, there are on average 1 × 10¹¹ stars.
6. The universe is estimated to be about 1.2 × 10¹⁰ years old.
7. The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 3 × 10²⁰ m.
8. 6.023 × 10²² molecules are contained in a drop of water weighing 1.8 gm.
9. The earth has 1.353 × 10⁹ cubic km of seawater.
10. The population of India was about 1.027 × 10⁹ in March 2001.

We hope the NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2.

• Exponents and Powers Class 7 Ex 13.1
• Exponents and Powers Class 7 Ex 13.3
• Exponents and Powers Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 13
Chapter Name	Exponents and Powers
Exercise	Ex 13.2
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for 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 :

Solution:

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

Solution:

Question 3.
Say true or false and justify your answer:

Solution:

Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768.
Solution:

Question 5.
Simplify

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.4
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind

Solution:

Let the number of digits formed be n, Then, the number of segments required to form n digits is given by the algebraic expression 5n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 1 = 25 + 1 = 26
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 1 = 50 + 1 = 51
3. the number of segments required to form 100 digits of this kind = 5 × 100 + 1 = 500 + 1 = 501.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 3n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 3 × 5 + 1 = 15 + 1 = 16
2. the number of segments required to form 10 digits of this kind = 3 × 10 + 1 = 30 + 1 = 31
3. The number of segments required to form 100 digits of this kind = 3 × 100 + 1 = 300 + 301.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 5n + 2.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 2 = 25 + 2 = 27
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 2 = 50 + 2 = 52
3. the number of segments required to form loo digits of this kind = 5 × 100 + 2 = 500 + 2 = 502.

Question 2.
Use the given algebraic expression to complete the table of number patterns.

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.3
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of :

1. m – 2
2. 3m – 5
3. 9 – 5m
4. 3m² – 2m – 7
5. \(\frac { 5m }{ 2 } \) – 4

Solution:

Question 2.
If p = – 2, find the value of :

1. 4p + 7
2. – 3p² + 4p + 7
3. – 2p³ – 3p² + 4p + 7.

Solution:

Question 3.
Find the value of the following expressions, when x = – 1 :

1. 2x – 7
2. – x + 2
3. x² + 2x + 1
4. 2x² – x – 2.

Solution:

Question 4.
If a = 2, b = – 2, find the value of :

1. a² + b²
2. a² + ab + b²
3. a² – b².

Solution:

Question 5.
When a = 0, b = – 1, find the value of the given expressions :

1. 2a + 2b
2. 2a² + b² + 1
3. 2a²b + 2ab² + ab
4. a² + ab + 2.

Solution:

Question 6.
Simplify the expressions and find the value ifx is equal to 2.

1. x + 7 + 4 (x – 5)
2. 3 (x + 2) + 5x – 7
3. 6x + 5 (x – 2)
4. 4 (2x – 1) + 3x + 11.

Solution:

Question 7.
Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.

1. 3x – 5 – x + 9
2. 2 – 8x + 4x + 4
3. 3a + 5 – 8a + 1
4. 10 – 3b – 4 – 5b
5. 2a – 2b – 4 – 5 + a.

Solution:

Question 8.
(i) If z = 10, find the value of z³ – 3(z – 10).
(ii) If p = -10, find the value of p² – 2p – 100.
Solution:

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Solution:

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3 2(a² + ab) + 3 – ab.
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.2
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms:

1. 21b – 32 + 7b – 20b
2. – z² + 13z² – 5z + 7z³ – 15z
3. p – (p – q) – q – (q – p)
4. 3a – 2b – ab – (a – b + ab) + 3ab + b – a
5. 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
6. (3y² + 5y – 4) – (8y – y² – 4).

Solution:

Question 2.
Add:

Solution:

Question 3.
Subtract:

1. -5y² from y²
2. 6xy from – 12xy
3. (a – b) from (a + b)
4. a (b – 5) from b (5 – a)
5. -m² + 5mn from 4m² – 3mn + 8
6. -x² + 10x – 5 from 5x – 10
7. 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
8. 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq.

Solution:

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 76 + 16?
Solution:

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:

Question 6.
(a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:

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