PSEB 7th Class Social Science Notes Chapter 19 Democracy – Representative Institutions

This PSEB 7th Class Social Science Notes Chapter 19 Democracy – Representative Institutions will help you in revision during exams.

PSEB 7th Class Social Science Notes Chapter 19 Democracy – Representative Institutions

→ Modern Democracy: It is a representative democracy. The reason is that modem states are large in size and their population is more.

→ In such a situation, the whole of the population cannot participate in the administration directly. So it chooses representatives which run the government.

→ Right to Vote: It is the right of the people to cast their votes or to choose their representatives.

→ In India ‘One Person One Vote’ principle gives way to ‘Universal Adult Franchise’.

→ Secret Ballot: The modem elections are fought through secret ballot.

PSEB 7th Class Social Science Notes Chapter 19 Democracy - Representative Institutions

→ It means that every citizen casts his vote by his own sweet will. He cannot be compelled to disclose his vote cast.

→ Candidate: The person who fights for election is called a candidate.

→ They are of two types: one belonging to a political party and the other having no relation with any political party.

→ Election Process: The elections are conducted under the supervision of the Election Commission.

→ A special process is adopted for this which includes a declaration of the election date, filing nominations, examination of nominations, campaigning, voting, counting, and declaration of results.

→ Election Symbol: Every political party has a special symbol.

→ Even independent candidates are given symbols.

→ These symbols are given by the election commission and this helps to identify the candidate in a better way.

→ Election Campaign: It is the most decisive part of the election process.

→ The public meetings are conducted, manifestoes are declared, promises on posters are pasted everywhere and the public is given information about the policies of the political parties if voted to power.

PSEB 7th Class Social Science Notes Chapter 19 Democracy - Representative Institutions

→ Election Manifesto: Every political party tries to tell the public what will it do if voted to power. This is called an election manifesto.

→ Importance of free and fair election: The Election Commission ensures that the elections should be free and fair, only then the right candidates can be elected.

→ The public gets a capable and popular govt, and the democracy becomes strong.

→ Political Parties: People coming together for the attainment of identical political objectives make political parties.

→ Functions of Political Parties: Making public opinion, educating the people politically, contesting the elections, framing the government, criticizing the government, creating coordination among the public and government are the main functions of the political parties.

→ Single Party, Two-Party and Multiparty System: In India, there is a multiparty system because there are more than two parties contesting the elections.

PSEB 7th Class Social Science Notes Chapter 19 Democracy - Representative Institutions

→ Role of opposition: The opposition controls the activities of the government by criticizing it and stops the government from becoming a dictator.

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 1.
Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
Answer:
(x + 4) (x + 10)
= (x)2 + (4 + 10)x + (4)(10)
= x2 + 14x + 40

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(ii) (x + 8) (x – 10)
Answer:
(x + 8) (x – 10)
= (x)2 + (8 – 10)x + (8)(- 10)
= x2 – 2x – 80

(iii) (3x + 4) (3x – 5)
Answer:
(3x + 4) (3x – 5)
= (3x)2 + (4 – 5) (3x) +(4) (- 5)
= 9x2 – 3x – 20

(iv) \(\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)\)
Answer:
\(\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)\)
= (y2)2 – \(\left(\frac{3}{2}\right)^{2}\)
= y4 – \(\frac{9}{4}\)

(v)(3 – 2x) (3 + 2x)
Answer:
(3 – 2x) (3 + 2x)
= (3)2 – (2x)2
= 9 – 4x2

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

Question 2.
Evaluate the following products without multiplying directly:
(i) 103 × 107
Answer:
103 × 107
= (100 + 3) (100 + 7)
= (100)2 + (3 + 7) (100) + (3) (7)
= 10000 + 1000 + 21
= 11,021

(ii) 95 × 96
Answer:
95 × 96
= (90 + 5) (90 + 6)
= (90)2 + (5 + 6) (90) + (5) (6)
= 8100 + 990 + 30 = 9120

OR

95 × 96
= (100 – 5) (100 – 4)
= (100)2 + (- 5 – 4) (100) + (- 5) (- 4)
= 10000 – 900 + 20
= 10020 – 900 = 9120

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(iii) 104 × 96
Answer:
104 × 96
= (100 + 4)(100-4)
= (100)2 – (4)2
= 10000 – 16 = 9984

Question 3.
Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
Answer:
9x2 + 6xy + y2
= (3x)2 + 2(3x) (y) + (y)2
= (3x + y)2 = (3x + y) (3x + y)

(ii) 4y2 – 4y + 1
Answer:
4y2 – 4y + 1
= (2y)2 – 2(2y)(1) + (1)2
= (2y – 1)2 = (2y – 1) (2y – 1)

(iii) x2 – \(\frac{y^{2}}{100}\)
Answer:
x2 – = (x)2 – \(\left(\frac{y}{10}\right)^{2}\)
= \(\left(x+\frac{y}{10}\right)\) \(\left(x-\frac{y}{10}\right)\)

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

Question 4.
Expand each of the following using suitable identities:
(i) (x + 2y + 4z)2
Answer:
(x + 2y + 4z)2
= (x)2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2(2y)(4z) + 2(4z)(x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

(ii) (2x – y + z)2
Answer:
(2x – y + z)2
= [2x + (- y) + z]2
= (2x)2 + (- y)2 + (z)2 + 2 (2x)(- y) + 2(- y)(z) + 2(z) (2x)
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx

(iii) (- 2x + 3y + 2z)2
Answer:
(- 2x + 3y + 2z)2
= [(- 2x) + 3y + 2z]2
= (- 2x)2 + (3y)2 + (2z)2 + 2(- 2x)(3y) + 2(3y)(2z) + 2(2z)(- 2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx

(iv) (3a – 7b – c)2
Answer:
(3a – 7b – c)2
= [3a + (- 7b) + (- c)]2
= (3a)2 + (- 7b)2 + (- c)2 + 2 (3a) (- 7b) + 2(- 7b) (- c) + 2(- c) (3a)
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(v)(- 2x + 5y – 3z)2
Answer:
(- 2x + 5y – 3z)2
= [(-2x) + 5y + (-3z)]2
= (- 2x)2 + (5y)2 + (- 3z)2 + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx

(vi) [\(\frac{1}{4}\)a – \(\frac{1}{2}\)b + 1]2
Answer:
PSEB 9th Class Maths Solutions Chapter 2 Polynomial Ex 2.5 1

Question 5.
Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Answer:
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz ;
= (2x)2 + (3y)2 + (- 4z)2 + 2 (2x) (3y) + 2(3y) (- 4z) + 2(- 4z) (2x)
= [2x + 3y + (- 4z)]2
= (2x + 3y – 4z)2
= (2x + 3y – 4z) (2x + 3y – 4z)

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(ii) 2x2 + y2 + 8z2 – 2√2 xy + 4√2 yz – 8xz
Answer:
2x2 + y2 + 8z2 – 2√2 xy + 4√2 yz – 8 xz
= (- √2x)2 + (y)2 + (2√2 z)2 + 2 (- √2x) (y) + 2(y) (2√2 z) + 2(2√2 z) (- √2 x)
= [(- √2 x) + y + 2√2 z]2
= (- √2x + y + 2√2 z)2
= (- √2x + y + 2√2 z) (- √2 x + y + 2√2 z)

Question 6.
Write the following cubes in expanded form:
(i) (2x + 1)3
Answer:
(2x + 1)3
= (2x)3 + (1)3 + 3(2x) (1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3+ 1 + 12x2 + 6x
OR
(2x + 1)3
= (2x)3 + 3(2x)2(1) + 3(2x) (1)2 + (1)3
= 8x3 + 12x2 + 6x + 1

(ii) (2a – 3b)3
Answer:
(2a – 3b)3
= (2a)3 – (3b)3 – 3 (2a) (3b) (2a – 3b)
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(iii) \(\left[\frac{3}{2} x+1\right]^{3}\)
Answer:
PSEB 9th Class Maths Solutions Chapter 2 Polynomial Ex 2.5 2

(iv) \(\left[x-\frac{2}{3} y\right]^{3}\)
Answer:
PSEB 9th Class Maths Solutions Chapter 2 Polynomial Ex 2.5 3

Question 7.
Evaluate the following using suitable identities:
(i) (99)3
Answer:
(99)3 = (100 – 1)3
= (100)3 – (1)3 – 3 (100) (1) (100 – 1)
= 1000000 – 1 – 300(99)
= 1000000 – 1 – 29700
= 9,70,299

(ii) (102)3
Answer:
(102)3 = (100 + 2)3
= (100)3 + (2)3 + 3(100) (2) (100 + 2)
= 1000000 + 8 + 600(102)
= 1000000 + 8 + 61200
= 10,61,208

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(iii) (998)3
Answer:
(998)3 = (1000 – 2)3
= (1000)3 – (2)3 – 3(1000) (2) (1000 – 2)
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 + 12000
= 994000000 + 12000 – 8
= 994012000 – 8
= 99,40,11,992

Question 8.
Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
Answer:
8a3 + b3 + 12a2b + 6ab2
= (2a)3 + (b)3 + 3 (4a2) (b) + 3 (2a) (b2)
= (2a)3 + (b)3 + 3 (2a)2 (b) + 3 (2a) (b2)
= (2a + b)3
= (2a + b) (2a + b) (2a + b)

(ii) 8a3 – b3 – 12a2b + 6ab2
Answer:
8a3 – b3 – 12a2b + 6ab2
= (2a)3 + (- b)3 + 3 (4a2) (- b) + 3 (2a) (b2)
= (2a)3 + (- b)3 + 3 (2a)3 (- b) + 3 (2a) (- b)2
= (2a – b)3
= (2a – b) (2a – b) (2a – b)

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(iii) 27 – 125a3 – 135a + 225a2
Answer:
27 – 125a3 – 135a + 225a2
= (3)3 + (- 5a)3 + 3(9) (- 5a) + 3 (3) (25a2)
= (3)3 + (- 5a)3 + 3(3)2 (- 5a) + 3 (3) (- 5a)2
=(3 – 5a)3.
= (3 – 5a)(3 – 5a) (3 – 5a)

(iv) 64a3 – 27b3 – 144a2b + 108ab2
Answer:
64a3 – 27b3 – 144a2b + 108ab2
= (4a)3 + (- 3b)3 + 3(16a2)(-3b) + 3(4a) (9b2)
= (4a)3 + (- 3b)3 + 3(4a)2(- 3b) + 3(4a)(- 3b)2
= (4a – 3b)3
= (4a – 3b) (4a – 3b) (4a – 3b)

(v) 27p3 – \(\frac{1}{216}\) – \(\frac{9}{2}\)p2 + \(\frac{1}{4}\)p
Answer:
PSEB 9th Class Maths Solutions Chapter 2 Polynomial Ex 2.5 4

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

Question 9.
Verify:
(i) x3 + y3 = (x + y) (x2 – xy + y2)
Answer:
R.H.S. = (x + y) (x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3
= L.H.S.

(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Answer:
R.H.S. = (x – y) (x2 + xy + y2)
= x(x2 + xy + y2) – y(x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3
= L.H.S.

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

Question 10.
Factorise each of the following:
[Hint: See Question 9]
(i) 27y3 + 125z3
Answer:
27y3 + 125z3
We know, a3 + b3 = (a + b) (a2 – ab + b2)
Replacing a by 3y and b by 5z, we get
(3y)3 + (5z)3 = (3y + 5z) [(3y)2 – (3y)(5z) + (5z)2]
∴ 27y3 + 125z3 = (3y + 5z) (9y2 – 15yz + 25z2)

OR

We know, a3 + b3 = (a + b)3 – 3ab(a + b)
Replacing a by 3y and b by 5z, we get
(3y)3 + (5z)3 = (3y + 5z)3 – 3 (3y) (5z) (3y + 5z)
∴ 27y3 + 125z3 = (3y + 5z) [(3y + 5z)3 – 45yz]
= (3y + 5z)(9y3 + 30yz + 25z2 – 45yz)
= (3y + 5z) (9y2 – 15yz + 25z2)

(ii) 64m3 – 343n3
Answer:
We know, a3 – b3 = (a – b) (a2 + ab + b2)
Replacing a by 4m and b by 7n, we get
(4m)3 – (7n)3 = (4m – 7n) [(4m)3 + (4m) (7n) + (7n)2]
∴ 64m3 – 343n3 = (4m – 7n) (16m2 + 28mn + 49n2)

OR

We know. a3 – b3 = (a – b)3 + 3ab(a – b)
Replacing a by 4m and b by 7n, we get
(4m)3 – (7n)3 = (4m – 7n)3 + 3 (4m) (7n)(4m – 7n)
= (4m – 7n) [(4m 7n)2 + 84mn]
= (4m – 7n) (16m2 – 56mn + 49n2 + 84mn)
= (4m – 7n) (16m2 + 28mn + 49n2)

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

Question 11.
Factorise: 27x3 + y3 + z3 – 9xyz
Answer:
We know, a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
Replacing a by 3x, b by y and c by z, we get
(3x)3 + (y)3 + (z)3 – 3 (3x) (y) (z) = (3x + y + z) [(3x)2 + (y)2 + (z)2– (3x) (y) – (y) (z) — (z) (3x)]
∴ 27x3 + y3 + z3 – 9xyz = (3x + y + z) (9x2 + y2 + z2– 3xy – yz – 3zx)

Question 12.
Verify that x3 + y3 + z3 – 3xyz = \(\frac{1}{2}\)(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2)
Answer:
R.H.S. = \(\frac{1}{2}\)(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
= \(\frac{1}{2}\) (x + y + z) (x2 – 2xy + y2 + y2 – 2yz + z2 + z2 – 2zx + x2)
= \(\frac{1}{2}\) (x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= x(x2 + y2 + z2 – xy – yz – zx) + y(x2 + y2 + z2 – xy – yz – zx) + z(x2 + y2 + z2 – xy – yz – zx)
= x3 + xy2 + xz2 – x2y – xyz – zx2 + x2y + y3 + yz2 – xy2 – y2z – xyz + x2z + y2z + z3 – xyz – yz2 – z2x
= x3 + y3 + z3 – 3xyz
= L.H.S.

Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Answer:
We know the Idendity
x3 + y3 + z3 – 3xyz = (x + y + z) (x2+ y3 + z2 – xy – yz – zx)
If x + y + z = 0. we get
x3 + y3 + z3 – 3xyz = (0) (x2 + y2 + z2 – xy – yz – zx)
∴ x3 + y3 + z3 – 3xyz = 0
∴ x3 + y3 + z3 = 3xyz

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

Question 14.
Without actually calculating the cubes, find the value of each of the following:
(i) (- 12)3 + (7)3 + (5)3
Answer:
Taking a = -12, b = 7 and c = 5, we get
a + b + c = (- 12) + 7 + 50.
Now, If a + b + c = 0, then a3 + b3 + c3 = 3abc.
∴ (- 12)3 + (7)3 + (5)3 = 3(- 12) (7) (5)
= (- 36) (35)
= – 1260

(ii) (28)3 + (- 15)3 + (- 13)3
Answer:
Talking a = 28, b = – 15 and c = – 13, we get
a + b + c = 28 + (- 15) + (- 13) = 0.
Now, If a + b + c = 0, then a3 + b3 + c3 = 3abc.
∴ (28)3 + (- 15)3 + (- 13)3 = 3 (28) (- 15) (- 13)
= (84)(195)
= 16,380

Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area: 25a2 – 35a + 12
Answer:
We know, area of a rectangle = length × breadth
Hence, two factors of area can give possible expressions for length and breadth. So here, we will try to obtain two factors of the expression of area.
25a2 – 35a + 12 = 25a2 – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
=(5a – 4) (5a – 3)
Thus, the length and breadth of the rectangle are (5a – 3) and (5a – 4) respectively.
Note : Traditionally, length > breadth in a rectangle.

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(ii) Area: 35y2 + 13y – 12
Answer:
We know, area of a rectangle = length × breadth
Hence, two factors of area can give possible expressions for length and breadth. So here, we will try to obtain two factors of the expression of area.
35y2 + 13y – 12 = 35y2 + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
= (5y + 4) (7y – 3)
Thus, the length and breadth of the rectangle are (7y – 3) and (5y + 4) respectively.

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
Answer:
(i) Volume: 3x2 – 12x
Answer:
We know, volume of a cuboid = length × breadth × height
Hence, three factors of volume can give possible expressions for length, breadth and height.
So here, we will try to obtain three factors of the expression of volume.
3x2 – 12x = 3x(x – 4)
= 3 × x × (x – 4)
Thus, one possible answer for the dimensions of the cuboid is 3. x and (x – 4).
Note: Other possible answers can be given as 1. 3x and (x – 4) or 1, x and (3x – 12).

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.5

(ii) Volume: 12ky2 + 8ky = 20k
Answer:
We know, volume of a cuboid = length × breadth × height
Hence, three factors of volume can give possible expressions for length, breadth and height.
So here, we will try to obtain three factors of the expression of volume.
12ky2 + 8ky – 20k = 4k (3y2 + 2y – 5)
= 4k (3y2 – 3y + 5y – 5)
= 4k [3y (y – 1) + 5(y – 1)1]
= 4k (y – 1) (3y + 5)
Thus, one possible answer for the dimensions of the cuboid is 4k, (y – 1) and (3y + 5).

PSEB 7th Class Social Science Notes Chapter 18 Democracy and Equality

This PSEB 7th Class Social Science Notes Chapter 18 Democracy and Equality will help you in revision during exams.

PSEB 7th Class Social Science Notes Chapter 18 Democracy and Equality

→ Democracy: It means the rule of the general public. The power to administer lies with the public.

→ Freedom and Equality: These are two pillars of democracy.

→ Lawful Administration: It means where the administration runs according to the predetermined Constitution.

PSEB 7th Class Social Science Notes Chapter 18 Democracy and Equality

→ The chief of the state cannot apply his will over the public and the public reserves the right to question his activities.

→ Direct Democracy: Here the public itself manages the administration which is not possible in this age.

→ Indirect Democracy: Here the administration is run by people’s representatives.

→ So the election process is more important in an indirect democracy.

→ Democracy and Political Parties: Political parties control the Govt.

→ The opposition criticizes the government and controls its activities.

→ Referendum: Its role is major for the success of democracy.

→ Elections: Special place of elections in a democracy.

PSEB 7th Class Social Science Notes Chapter 18 Democracy and Equality

→ People change the incapable government through elections.

→ Adult Franchise: All adults have the right to vote and it is based on the law of equality.

→ Dictatorship: The power of administration is concentrated in a single hand or in a few hands.

→ Elections are not held in such countries and discretionary powers are exercised by the dictator.

PSEB 7th Class Social Science Notes Chapter 17 India in the Eighteenth Century

This PSEB 7th Class Social Science Notes Chapter 17 India in the Eighteenth Century will help you in revision during exams.

PSEB 7th Class Social Science Notes Chapter 17 India in the Eighteenth Century

→ Rise of new states: In the 18th century, whatever remained of the Mughal empire, many new states emerged.

→ The main were Bengal, Hyderabad, Avadh, Punjab, Mysore, and Maratha state.

→ Marathas: The most powerful group of India after the decline of Mughals.

→ Ashtha Pradhan: The council of eight ministers in the period of Shivaji.

→ Peshwa: Chief of Ashtha Pradhan.

PSEB 7th Class Social Science Notes Chapter 17 India in the Eighteenth Century

→ Punjab: In Punjab, the Gurujis established Sikh Panth. After the 10th Guru Sri Guru Gobind Singh Ji, Banda Bahadur established a Sikh state in Punjab.

→ Bengal: In Bengal, the Mughal Subedar Murshid Quli Khan established an independent state. At last, the English captured it.

→ Avadh: The founder of the free state of Avadh was Saadat Khan. The nawabs of Avadh gave birth to “Lucknow Culture”.

→ Mysore: In Mysore, Hyder Ali founded an independent state. He snatched the authority from Mysore’s Hindu King Nanjaraj.

PSEB 7th Class Social Science Notes Chapter 16 The Development of Regional Cultures

This PSEB 7th Class Social Science Notes Chapter 16 The Development of Regional Cultures will help you in revision during exams.

PSEB 7th Class Social Science Notes Chapter 16 The Development of Regional Cultures

Language and Literature (Sultanate period):

  • During this period, language and literature were greatly developed.
  • A new language Urdu developed by a mixture of Hindi and Persian.
  • Many Muslim scholars studied ancient Hindu scriptures.
  • In this period, many important books were written in the Hindi language.
  • Chand Bardai wrote ‘Prithvi Raj Raso’, Malik Mohammed Jaisi wrote ‘Padmawat’, Jaidev wrote ‘Geet Govind’ and Kalhan wrote ‘Raj Tarangini’.

Mughal Period Literature: Tuzuk-i-Babri’, ‘Humayun Nama’, ‘Akbar Nama’, ‘Ain-i-Akbari’, ‘Badshah Nama’, etc.

PSEB 7th Class Social Science Notes Chapter 16 The Development of Regional Cultures

Punjabi Literature:

  • During the medieval period, the holy writings of Guru Sahebs and other Punjabi poets enriched the Punjabi culture.
  • Sri Guru Granth Sahib, Dasham Granth, the writings of Bhai Gurdas Ji came in this period.

Art of Painting:

  • There was the development of the art of painting as well.
  • Abdus Samad, Mir Sayyad Ali, Sanwaldas, Jagannath, Tarachand, etc. showed their skill with paintbrushes.
  • All these painters were during the times of Akbar. Akbar respected them.
  • Similarly, Jahangir also respected painters in his court.
  • Mohammad Murad, Ustad Mansur, Agha Raza, Mohammad Nadir were his famous painters.

PSEB 7th Class Social Science Notes Chapter 16 The Development of Regional Cultures

Art of Music:

  • The Mughal period did not lack in the art of music.
  • Babar was a good poet.
  • He created poetry and songs.
  • During the times of Akbar,’ ‘Sangeet Samrat Tansen’ and ‘Baiju Bawara’ gave a new standard to the art of music.
  • Aurangzeb was very much against music.
  • During his period this art declined.

PSEB 7th Class Social Science Notes Chapter 15 Religious Developments

This PSEB 7th Class Social Science Notes Chapter 15 Religious Developments will help you in revision during exams.

PSEB 7th Class Social Science Notes Chapter 15 Religious Developments

→ Main Religions of the World: Hinduism, Buddhism Jainism, Zoroastrianism, Christianity, Islam, and Sikhism.

→ Islam: This religion was founded by Prophet Hazrat Mohammad.

PSEB 7th Class Social Science Notes Chapter 15 Religious Developments

→ The main scripture of Islam is The Quran’, which includes the directions about different aspects of life.

→ Sikhism: This religion was founded by Sri Guru Nanak Devji. After him, nine Gurujis developed the Sikh Panth.

→ The main scripture of this religion is ‘Sri Guru Granth Sahibji’.

→ Sufism: Sufism is the liberal form of the Islamic religion. The followers of this religion were known as ‘Pirs’.

→ They used black coloured blankets, which were called Sufi, so these Pirs were called Sufis.

→ Bhakti Movement: To do away with the religious and social evils, there was a Bhakti movement, in India during the medieval period.

→ The main feature of this movement was that it was not propagated by anyone great man, but by different holy men in different areas.

→ But still, the principles of the Bhakti movement were the same everywhere.

PSEB 7th Class Social Science Notes Chapter 15 Religious Developments

→ Hinduism: The main classification was Shaiv and Vaishnavs.

→ Shaivism was founded by Shankaracharyaji in the 9th century.

→ The main founders of Vaishnavism were Sri Ramanandji and Chaitanya Mahaprabhuji.

PSEB 7th Class Social Science Notes Chapter 14 Tribes, Nomad and Settled Societies

This PSEB 7th Class Social Science Notes Chapter 14 Tribes, Nomad and Settled Societies will help you in revision during exams.

PSEB 7th Class Social Science Notes Chapter 14 Tribes, Nomad and Settled Societies

→ Tribal Society: Tribal Society is a society that lives very much away from the modern system and exists in forests, valleys, and mountains.

→ They never like to interfere in anyone’s internal matters and never like to be interfered with by anyone.

→ Against caste system: During the medieval period, a number of tribal societies emerged and they did not obey the rules and customs of the caste system.

PSEB 7th Class Social Science Notes Chapter 14 Tribes, Nomad and Settled Societies

→ Many occupations: Tribal societies had the main occupation of agriculture but there were few other occupations available for them like hunting, food gathering and pastoralism, etc.

→ Many tribes: During the medieval period, many tribes existed in all parts of the Indian subcontinent.

→ Some of them were the Bhils, Gonds, Ahoms, Kuki’s Koli’s, Kui, Oraon, etc.

→ Life of the Nomadic groups: Nomads, during the medieval period, lived on the rearing of animals.

→ They went far away to graze their animals.

→ They made both ends with animals breeding.

→ They moved from one place to another for selling their goods by loading on the animals.

→ The Aborn Community: They ruled over Assam. They had come from China.

→ The name of their first ruler was ‘Sufaka’. They defeated many local tribes.

PSEB 7th Class Social Science Notes Chapter 14 Tribes, Nomad and Settled Societies

→ The Gond Tribe: It is a tribe in the middle part of India.

→ They are based in states like western Orissa, eastern Maharashtra, Chhattisgarh, and Madhya Pradesh.

→ They established many states from the 15th to 18th century. Rani Durgawati was a famous Gond ruler.

PSEB 7th Class Maths MCQ Chapter 8 Comparing Quantities

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 8 Comparing Quantities MCQ Questions

Multiple Choice Questions

Question 1.
Find the ratio of ₹ 10 to 10 paise.
(a) 1 : 1
(b) 100 : 1
(c) 1000 : 1
(d) 1000 : 10
Answer:
(b) 100 : 1

Question 2.
The ratio of ₹ 5 to 50 Paise is :
(a) 5 : 50
(b) 1 : 10
(c) 10 : 1
(d) 50 : 5.
Answer:
(c) 10 : 1

Question 3.
The ratio of 15 kg to 210 g is :
(a) 15 : 210
(b) 15 : 21
(c) 500 : 7
(d) 7 : 500.
Answer:
(c) 500 : 7

Question 4.
The Percentage of \(\frac {12}{16}\) is :
(a) 25%
(b) 12%
(c) 75%
(d) 16%
Answer:
(c) 75%

PSEB 7th Class Maths MCQ Chapter 8 Comparing Quantities

Question 5.
Convert \(\frac {5}{4}\) into percent.
(a) 100%
(b) 125%
(c) 75%
(d) 16%
Answer:
(b) 125%

Question 6.
Convert 12.35 into percent.
(a) 12.35%
(b) 123.5%
(c) 1235%
(d) 1.235%
Answer:
(c) 1235%

Question 7.
What percent part of figure is shaded ?
PSEB 7th Class Maths MCQ Chapter 8 Comparing Quantities 1
(a) 30%
(b) 50%
(c) 60%
(d) 20%
Answer:
(c) 60%

Question 8.
15% of 250 is :
(a) 250
(b) 375
(c) 37.5
(d) 3750
Answer:
(c) 37.5

PSEB 7th Class Maths MCQ Chapter 8 Comparing Quantities

Fill in Blanks :

Question 1.
25% of 120 litres is ………….. litres.
Answer:
30

Question 2.
The ratio of 4 km to 300 m is …………..
Answer:
40

Question 3.
The price at which an article is purchased is called …………..
Answer:
Cost price

Question 4.
If the selling price of an article is less than to cost prices then there is …………..
Answer:
loss

Question 5.
The symbol ………….. stands for percent
Answer:
%

PSEB 7th Class Maths MCQ Chapter 8 Comparing Quantities

Write True or False

Question 1.
The Ratio 1 : 5 and 2 : 15 are equivalent. (True/False)
Answer:
False

Question 2.
A ratio remains unchanged, if both of its terms are multiplied or divided by the same number. (True/False)
Answer:
True

Question 3.
If the selling price of an article is more than its cost price then there is a profit. (True/False)
Answer:
True

Question 4.
If cost price and selling price both are equal then three is profit. (True/False)
Answer:
False

Question 5.
Profit loss percentage is calculated on the cost price. (True/False)
Answer:
True

PSEB 7th Class Social Science Notes Chapter 13 Towns, Traders and Craftsmen

This PSEB 7th Class Social Science Notes Chapter 13 Towns, Traders and Craftsmen will help you in revision during exams.

PSEB 7th Class Social Science Notes Chapter 13 Towns, Traders and Craftsmen

→ Development of Cities: After the development of agriculture, the villages came into existence. The villages expanded and the cities developed.

→ Types of Cities: Cities were of many types such as capitals, temple cities, port cities, trade cities, etc.

→ Traders and Artisans: During the medieval period, the Indian artisans made high-quality and high-standard goods.

→ The traders exported these goods and India became ‘The Golden Sparrow’.

→ Surat: Surat was an important industrial and trading town of the medieval period.

PSEB 7th Class Social Science Notes Chapter 13 Towns, Traders and Craftsmen

→ Lahore: Lahore is located in modern Pakistan.

→ It remained the capital of Punjab for a long time.

→ Amritsar: It is the biggest and the holiest religious place of the Sikhs.

→ In the medieval period, it was an important trading town.

→ Sri Harmandar Sahib located in Amritsar is world-famous.

PSEB 7th Class Social Science Notes Chapter 12 Monumental Architecture

This PSEB 7th Class Social Science Notes Chapter 12 Monumental Architecture will help you in revision during exams.

PSEB 7th Class Social Science Notes Chapter 12 Monumental Architecture

Monuments from 800 A.D. to 1200 A.D:

  • Many beautiful temples were built from 800 A.D. to 1200 A.D.
  • The Tejpal Temple of Mount Abu, Mahadev Temple in Khajuraho, and the Sun Temple of Konark is worth watching.
  • The walls, roofs, pillars, and doors of temples were decorated with beautiful statues.

Temples of South India:

  • From 800 A.D. to 1200 A.D., the most important temple is Brihdeshwar Temple.
  • It was constructed by Chola king, Raja Raja I.
  • Another temple Gangaikondcholpuram built by Chola king Rajendra I is also very famous.
  • The temples built by the Pallava dynasty are famous for their unique beauty.
  • These temples have been cut out of rocks.
  • A very famous temple is the Mahabalipuram temple.
  • The biggest temple built by Pallava rulers is Kailash Temple.

PSEB 7th Class Social Science Notes Chapter 12 Monumental Architecture

The architecture of the Sultanate Period:

  • The Turks and Afghans brought with themselves a new style of architecture.
  • A new art style was born with the remix of their art and the Indian building art.
  • In this style, many buildings were built with pillars, domes, and minarets.
  • These things were used in Mosques, Palaces, Tombs, and other buildings.

Buildings of Sultans:

  • Qutab-ud-din Aibak built many mosques in Delhi and Ajmer.
  • Iltutmish completed Qutab Minar.
  • It is the highest minaret in India.
  • Alauddin Khalji built a mosque in Delhi on the tomb of Nizamuddin Aulia.
  • He built ‘Ilahi Darwaja’ near Qutab Minar.
  • During the Tughlaq Dynasty, another famous building is the Tomb of Tughlaq Shah.

PSEB 7th Class Social Science Notes Chapter 12 Monumental Architecture

The Mughal Buildings:

  • The Mughals had a real taste for beautiful buildings.
  • During the time of Akbar, the Agra Fort and Fatehpur Sikri are especially famous.
  • The Taj Mahal built by Shah Jahan is world-famous for its beauty.
  • Besides, he also built Moti Masjid and Jama Masjid in Agra and Delhi as well as the Red Fort in Delhi.