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PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.7

March 7, 2024March 6, 2024 by Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7

1. Solve, dividing decimal number by 10, 100 or 1000 in the following :

Question (i).
2.7 ÷ 10
Answer:
2.7 ÷ 10 = \(\frac{27}{10} \times \frac{1}{10}\)
= \(\frac{27}{100}\)
= 0.27

Question (ii).
3.35 ÷ 10
Answer:
3.35 ÷ 10 = \(\frac{335}{100} \times \frac{1}{10}\)
= \(\frac {335}{1000}\)
= 0.335

Question (iii).
0.15 ÷ 10
Answer:
0.15 ÷ 10 = \(\frac{15}{100} \times \frac{1}{10}\)
= \(\frac {15}{1000}\)
= 0.015

Question (iv).
32.7 ÷ 10
Answer:
32.7 ÷ 10 = \(\frac{327}{10} \times \frac{1}{10}\)
= \(\frac {327}{100}\)

Question (v).
5.72 ÷ 100
Answer:
5.72 ÷ 100 = \(\frac{572}{100} \times \frac{1}{100}\)
= \(\frac {572}{10000}\)
= 0.0572

Question (vi).
23.75 ÷ 100
Answer:
23.75 ÷ 100 = \(\frac{2375}{100} \times \frac{1}{100}\)
= \(\frac {2375}{10000}\)
= 0.2375

Question (vii).
532.73 ÷ 100
Answer:
532.73 ÷ 100 = \(\frac{53273}{100} \times \frac{1}{100}\)
= \(\frac {53273}{10000}\)
= 5.3273

Question (viii).
1.321 ÷ 100
Answer:
1.321 ÷ 100 = \(\frac{1321}{1000} \times \frac{1}{100}\)
= \(\frac {1321}{10000}\)
= 0.01321

Question (ix).
2.5 ÷ 1000
Answer:
2.5 ÷ 1000 = \(\frac{25}{10} \times \frac{1}{1000}\)
= \(\frac {25}{10000}\)
= 0.0025

Question (x).
53.83 ÷ 1000
Answer:
53.83 ÷ 1000 = \(\frac{5383}{100} \times \frac{1}{1000}\)
= \(\frac {5383}{100000}\)
= 0.05383

Question (xi).
217.35 ÷ 1000
Answer:
217.35 ÷ 1000 = \(\frac{21735}{100} \times \frac{1}{1000}\)
= \(\frac {21735}{100000}\)
= 0.21735

Question (xii).
0.2 ÷ 1000
Answer:
0.2 ÷ 1000 = \(\frac{2}{10} \times \frac{1}{1000}\)
= \(\frac {2}{10000}\)
= 0.0002

2. Solve, dividing decimal number by whole number.

Question (i).
7.5 ÷ 5
Answer:
7.5 ÷ 5 = \(\frac{75}{10} \times \frac{1}{5}\)
= \(\frac {15}{10}\)
= 1.5

Question (ii).
16.9 ÷ 13
Answer:
16.9 ÷ 13 = \(\frac{169}{10} \times \frac{1}{13}\)
= \(\frac {13}{10}\)
= 1.3

Question (iii).
65.4 ÷ 6
Answer:
65.4 ÷ 6 = \(\frac{654}{10} \times \frac{1}{6}\)
= \(\frac {109}{10}\)
= 10.9

Question (iv).
0.121 ÷ 11
Answer:
0.121 ÷ 11 = \(\frac{121}{1000} \times \frac{1}{11}\)
= \(\frac {11}{1000}\)
= 0.011

Question (v).
11.84 ÷ 4
Answer:
11.84 ÷ 4 = \(\frac{1184}{100} \times \frac{1}{4}\)
= \(\frac {296}{100}\)
= 2.96

Question (vi).
47.6 ÷ 7
Answer:
47.6 ÷ 7 = \(\frac{476}{10} \times \frac{1}{7}\)
= \(\frac {68}{10}\)
= 6.8

3. Solve, dividing the decimal number by decimal number

Question (i).
3.25 ÷ 0.5
Answer:
3.25 ÷ 0.5 = \(\frac{325}{100} \div \frac{5}{10}\)
= \(\frac{325}{100} \times \frac{10}{5}\)
= \(\frac {65}{10}\)
= 6.5

Question (ii).
5.4 ÷ 1.2
Answer:
5.4 ÷ 1.2 = \(\frac{54}{10} \div \frac{12}{10}\)
= \(\frac{54}{10} \times \frac{10}{12}\)
= \(\frac {9}{2}\)
= 4.5

Question (iii).
26.32 ÷ 3.5
Answer:
26.32 ÷ 3.5 = \(\frac{2632}{100} \div \frac{35}{10}\)
= \(\frac{2632}{100} \times \frac{10}{35}\)
= \(\frac {752}{100}\)
= 7.52

Question (iv).
2.73 ÷ 13
Answer:
2.73 ÷ 13 = \(\frac{273}{100} \times \frac{10}{13}\)
= \(\frac {21}{10}\)
= 2.1

Question (v).
12.321 ÷ 11.1
Answer:
12.321 ÷ 11.1 = \(\frac{12321}{1000} \div \frac{111}{10}\)
= \(\frac{12321}{1000} \times \frac{10}{111}\)
= \(\frac {111}{100}\)
= 1.11

Question (vi).
0.0018 ÷ 0.15
Answer:
0.0018 ÷ 0.15 = \(\frac{18}{10000} \div \frac{15}{100}\)
= \(\frac{18}{10000} \times \frac{100}{15}\)
= \(\frac {12}{1000}\)
= 0.012

4. 25 steel chairs were purchased by a school for ₹ 11,883.75. Find the cost of one steel chair.
Answer:
Cost Price of 25 steel chairs = ₹ 11,883.75
Cost Price of 1 steel chair = ₹ 11,883.75 ÷ 15
= ₹ \(\frac{11,88375}{100} \times \frac{1}{15}\)
= ₹ \(\frac {47535}{100}\)
= ₹ 475.35

5. A car covers a distance of 276.75 km in 4.5 hours. What is the average speed of the car ?
Answer:
Total Distance covered = 276.75 km
Time taken = 4.5 hours
Average speed of car = \(\frac{Distance}{Time}\)
= \(\frac{276.75}{4.5}\)
= \(\frac{27675}{100} \times \frac{10}{45}\)
= \(\frac {615}{10}\)
= 61.5 km/hr.

6. Multiple Choice Questions :

Question (i).
27.5 ÷ 10 = ?
(a) 275
(b) 0.275
(c) 2.75
(d) None of these.
Answer:
(c) 2.75

Question (ii).
The value of 1.5 ÷ 3 is :
(a) 5
(b) 0.05
(c) 0.5
(d) 4.5.
Answer:
(c) 0.5

Question (iii).
The average of decimal number 1.1, 2.1 and 3.1 is :
(a) 2.5
(b) 1.1
(c) 2.1
(d) 6.3.
Answer:
(c) 2.1

7. On dividing a decimal number by 100, the decimal point is shifted to the left by one place. (True/False)
Answer:
False.

PSEB 12th Class History Notes Chapter 9 Guru Tegh Bahadur Ji and His Martyrdom

March 7, 2024March 6, 2024 by Bhagya

This PSEB 12th Class History Notes Chapter 9 Guru Tegh Bahadur Ji and His Martyrdom will help you in revision during exams.

PSEB 12th Class History Notes Chapter 9 Guru Tegh Bahadur Ji and His Martyrdom

Early Career:

Guru Tegh Bahadur Ji was born on April 1, 1621 A.D. at Amritsar.
The name of his father was Guru Hargobind Ji and the name of his mother was Nanaki.
Guru Tegh Bahadur Ji was imparted education by Baba Buddha Ji and Bhai Gurdas Ji.
Guru Tegh Bahadur Ji was married to Gujari, the daughter of Lai Chand, a resident of Kartarpur.
On the instructions of his father, Guru Tegh Bahadur Ji lived at Bakala for 20 years.
It was Makhan Shah Lubana who found him out and the Sikhs accepted him as their Guru.
He assumed Guruship in 1664 A.D.

Travels of Guru Tegh Bahadur Ji:

After assuming Guruship, Guru Tegh Bahadur Ji undertook the travels of several areas within and outside Punjab.
The main objective of these travels was to propagate Sikhism and spread the message of truth and love.
First, Guru Tegh Bahadur Ji travelled to Aipritsar, Walla, Ghukewali, Khadur Sahib, Goindwal Sahib, Tarn Taran, Khemkaran, Kiratpur Sahib, Bilaspur, etc. in 1664 A.D.
After this, he travelled to Eastern India i.e. Saifabad, Dhamdhan, Delhi, Mathura, Brindaban, Agra, Kanpur, Banaras, Gaya, Patna, Dhaka, and Assam.
In 1673 A.D. Guru Tegh Bahadur Ji travelled to the Malwa and Bangar regions of Punjab for the second time.
These travels led to the glory of Guru Tegh Bahadur Sahib and Sikhism.

Martyrdom of Guru Tegh Bahadur Ji:
The main facts associated with the martyrdom of Guru Tegh Bahadur Ji are as follows:

Causes:

The enmity between the Sikhs and Mughals was increasing.
Aurangzeb was an orthodox Sunni Muslim.
Naqshbandis tried to instigate Aurangzeb against the Sikhs.
Ram Rai was using all measures to get hold of Guruship.
Kashmiri Pandits pleaded help from Guru Sahib for their defence.

Martyrdom:

Guru Tegh Bahadur Ji along with his three companions.
Bhai Mati Das Ji, Bhai Sati Das Ji, and Bhai Dayala Ji were brought before the court of Delhi on November 6, 1675 A.D.
They were asked to embrace Islam which they outrightly refused.
After his three companions were martyred, Guru Tegh Bahadur Ji was also martyred on November 11, 1675 A.D. at Chandni Chowk in Delhi.

Importance:

The martyrdom of Guru Tegh Bahadur Ji sent a wave of hatred and revenge in the whole of Punjab against the Mughal empire.
Hinduism was protected against extinction.
Guru Gobind Singh Ji was inspired to establish the Khalsa Panth.
It marked the beginning of the tradition of sacrifice among the Sikhs for the protection of Sikhism.
It sounded the death knell of the Mughal Empire.

PSEB 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.4

March 7, 2024March 6, 2024 by Bhagya

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

Question 1.
Verify that the number given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :
(i) 2x³ + x² – 5x + 2; \(\frac{1}{2}\), 1, -2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Solution:
(i) Let p (x) = 2x³ + x² – 5x + 2
Compare it with ax³ + bx² + cx + d
a = 2, b = 1, c = -5, d = 2
Now, P(\(\frac{1}{2}\)) = 2\(\left(\frac{1}{2}\right)^{3}\) + \(\left(\frac{1}{2}\right)^{2}\) – 5\(\frac{1}{2}\) + 2

= \(\frac{1}{4}+\frac{1}{4}-\frac{5}{2}\)

= \(\frac{1+1-10+8}{4}=\frac{0}{4}\) = 0
∴ \(\frac{1}{2}\) is zero of p (x)
and p(1) = 2(1)³ + (1)² – 5 (1) + 2
= 2 + 1 – 5 + 2 = 5 – 5 = 0
∴ 1 is zero of p (x).
Also, p(-2) = 2(-2)³ + (- 2)² -5(-2) + 2
= -16 + 4 + 10 + 2 = – 16+ 16 = 0
∴ – 2 is zero of p (x).
From above discussion, it is clear that \(\frac{1}{2}\), 1, – 2 are the zeroes of given polynomial.
Let these zeroes are
α = \(\frac{1}{2}\), β = 1, γ = – 2
Now, α + β + γ = \(\frac{1}{2}\) + 1 + (-2)
= \(\frac{1}{2}\) + 1 – 2
= \(\frac{1+2-4}{2}\)
= –\(\frac{1}{2}\) = –\(\frac{b}{a}\)

αβ + βγ + γα = (\(\frac{1}{2}\)) (1) + (-1) (-2) + (-2) (\(\frac{1}{2}\))
= \(\frac{1}{2}\) – 2 – 1
= \(\frac{1-4-2}{2}\)
= \(-\frac{5}{2}\) = \(\frac{c}{a}\)

αβγ = (\(\frac{1}{2}\)) (1) (-2)
= (\(-\frac{2}{2}\))
= (\(-\frac{d}{a}\))
From above discussion, it is clear that there is a relationship between zeroes and coefficients.

(ii) Let p (x) = x³ – 4x² + 5x – 2.
Compare it with ax³ + bx² + cx + d
∴ a = 1, b = – 4, c = 5, d = – 2
Now p(2) = (2)³ – 4 (2)² + 5 (2) – 2
= 8 – 16 + 10 – 2
= 18 – 18 = 0 2 is zero of p (2).
and p( 1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2 = 6 – 6 = 0
From above discussion it is clear that 2, 1, 1 are the zeroes of given polynomial. Ans.
Let these zeroes are
α = 2, β = 1, γ = 1
Now, α + β + γ = 2 + 1 + 1 = 4
= \(\frac{-(-4)}{1 \cdot}=\frac{-b}{a}\)
αβ + βγ + γα = (2) (1) + (1) (1) + (1) (2) = 2 + 1 + 2 = 5
= \(\frac{5}{1}=\frac{c}{a}\)

αβγ = (2) (1) (1) = 2
= \(\frac{-(-2)}{1}=\frac{-d}{a}\)
From above discussion, it is clear that there is a relationship between zeroes and coefficient.

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.
Solution:
The general expression of cubic polynomial are
ax³ + bx² + cx + d.
Let α, β, γ be its zeroes
α + β + γ = Sum of zeroes = 2
αβ + βγ + γα = Sum of product of zeroes = – 7
αβγ = Product of zeroes = – 14
∴ ax³ + bx² + cx + d
= k [(x – α) (x – β) (x – γ)] where k is any constant.
= k [x³ – (α + β + γ) x² + (αβ + βγ + γα)x – αβγ]
= k [x³ – 2x² – 7x + 14] [Using (1)]
For different values of k, we get different cubic polynomial.

Question 3.
If the zeroes of the polynomial x³ – 3x² + x +1 are a – b, a, a + b, find a and b.
Solution:
Let p (x) = x³ – 3x² + x + 1 and its zeroes are a – b, a, a + b.
a – b is zero of p (x) ………(Given)
∴ p (a – b) = 0
or (a – b)³ – 3 (a – b)² + (a – b) + 1 = 0 or
[a³ – b³ – 3a²b + 3ab²] – 3 [a² + b² – 2ab] + a – b + 1 = 0
or a³ – b³ – 3a²b + 3ab² – 3a² – 3b²
+ 6ab + a – b + 1 = 0 ….(1)
and a is zero of p (x) …………(Given)
∴ p (a) = 0
or a³ – 3a³ + a + 1 = 0 ………….(2)
Also, a + b is zero of p (x) …(Given)
∴ p (a + b) = 0
or (a + b)³ – 3 (a + b)² + (a + b) + 1 = 0
or (a³ + b³ + 3a²b + 3ab²) – 3 (a² + b³ + 2ab) + a + b – 1 = 0 or
a³ + b³ + 3a²b + 3ab² – 3a² – 3b³ – 6ab + a + b + 1 = 0 …………….(3)
Adding (1) and (3), we get :
2a³ + 6ab² – 6a² – 6b² + 2a + 2 = 0 or
a³+ 3ab² – 3a² – 3b² + a + 1 = 0 or
(a³ – 3a³ + a + 1) + (3ab² – 3b²) = 0 or
0 + 3b² (a – 1) = 0 [Using (2)]
or a – 1 = 0
or a = 1 ………(4)
From (3) and (4), we get:
(1)³ + b³ + 3(1)²b + 3(1 )b² – 3(1)² – 3b² – 6 (1) b + 1 + b + 1 = 0 or
1 + b³+ 3b + 3b² – 3 – 3b² – 6b + b + 2 = 0
or b³ – 2b = 0 or
b (b² – 2) = 0 or
b² – 2 = 0 or
b² = 2
or b = ±√2
Hence a = 1, b = ±√2

Another Solution:
Given that three zeroes of polynomial x³ – 3x² + x + 1 are a – b, a, a + b respectively
Now, sum of zeroes = (a – b) + a + (a + b) = a – b + a + a + b = 3a
But, sum of zeroes using coefficient = \(\frac{-\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}\)
= \(\frac{-(-3)}{1}\) = 3
∴ 3a = 3 or a = 1
Also, product of zeroes = (a – b) . a . (a + b) = (a² – b²) a
Putting the value of a, we get :
= (1² – b²) . 1
= 1 – b²
But, product of zeroes using coefficient = \(=\frac{-\text { Constant term }}{\text { Coefficient of } x^{3}}\)
= \(\frac{-1}{1}\) = -1
1 – b² = – 1
– b² = – 1 – 1
– b² = – 2 or
b² = 2
b = ±√2
Hence a = 1 and b = ±√2

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± √3, find other zeroes.
Solution:
Given that two zeroes are (2 + √3) and (2 – √3)
∴ [x – (2 + √3)] [x – (2 – √3)] are factors of given polynomial.
Now, [x – (2 + √3)] [x – (2 – √3)]
= x² – [2 – √3 + 2 + √3]x + [(2 + √3) + (2 – √3)]
= x² – 4x + [(2)² – (√3)²]
= x² – 4x + 1
∴ (x² – 4x + 1) is factor of given polynomial. Now, apply division algorithm to given polynomial and (x² – 4x + 1)

PSEB 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.4 1

∴ x⁴ – 6x³ – 26x² + 138x – 35
= (x² – 4x + 1) (x² – 2x – 35)
= (x² – 4x + 1) [x² + 5x – 7x – 35]
S = – 2, P = – 35
= (x² – 4x + 1) [x (x + 5) – 7 (x + 5)]
= (x² – 4x + 1) (x + 5).(x – 7)
Now, other zeroes of polynomials are given by :
x + 5 = 0 Or x – 7 = 0
x = – 5 Or x = 7
∴ the zeroes of the given fourth degree polynomial are :
2 + √3, 2 – √3, -5, 7.

Question 5.
If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
Given that, polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k then remainder comes out to be x + a
So, first of all we divide,
x⁴ – 6x³ + 16x² – 25x + 10 with x² – 2x + k and find quotient and remainder.

PSEB 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.4 2

∴ by division algorithm for Polynomials x⁴ – 6x³ + 16x² – 25x +10
= (x² – 2x + k) [x² – 4x + (8 – k) + [(- 9 + 2k) x + (10 – 8k + k²]
Quotient = x² – 4x + (8 – k) and
Remainder = (- 9 + 2k) x + (10 – 8k + k²)
But, remainder = x + a ….(Given)
∴ (- 9 + 2k) x + (10 -8k + k²) = x + a
Compare the like coefficients, we get :
-9 + 2k = 1 Or
2k = 1 + 9
2k = 10
k = \(\frac{10}{2}\) = 5 or

10 – 8k + k² = a
Putting the value of k, we get
10 – 8 × 5 + (5)² = a
10 – 40 + 25 = a
-5 = a
a = -5
Hence k = 5 and a = -5

PSEB 12th Class History Notes Chapter 8 Guru Har Rai Ji and Guru Har Krishan Ji

March 7, 2024March 6, 2024 by Bhagya

This PSEB 12th Class History Notes Chapter 8 Guru Har Rai Ji and Guru Har Krishan Ji will help you in revision during exams.

PSEB 12th Class History Notes Chapter 8 Guru Har Rai Ji and Guru Har Krishan Ji

Guru Har Rai Ji:

Guru Har Rai Ji was born on 30th January 1630 A.D. at Kiratpur Sahib.
He was the grandson of Guru Hargobind Ji.
He was married to Sulakhni, the daughter of Daya Ram of Anup city.
He assumed Guruship on March 8, 1645 A.D.
He established three centres for the propagation of Sikhism which he called ‘Bakhshishes’.
Dara, the son of the Mughal emperor Shah Jahan often came for the darshan of Guru Har Rai Ji.
Guru Har Rai Ji helped Dara against Aurangzeb.
On being summoned to Delhi by Aurangzeb, Guru Har Rai Ji sent his son Ram Rai to Delhi.
Because, he deliberately misinterpreted the Gurbani in Delhi Darbar, he was declared unworthy of Guruship by Guru Har Rai Ji.
He appointed his younger son Har Krishan as his successor.
He immersed in Eternal Light on October 6, 1661 A.D.

Guru Har Krishan Ji:

He was born on July 7, 1656 A.D. at Kiratpur Sahib.
The name of his father was Guru Har Rai Sahib and the name of his mother was Sulakhni.
He assumed Guruship at a small age of 5 years in 1661 A.D.
He is often remembered as the Bal Guru.
He was summoned by Aurangzeb to Delhi.
He went to Delhi in 1664 A.D. where he tirelessly served the people suffering from smallpox and cholera.
He himself fell a victim to smallpox.
He immersed in Eternal Light on March 30, 1664 A.D. at Delhi.
He uttered the words “Baba Bakala” before breathing his last which meant that the next Guru of the Sikhs is at Baba Bakala.

PSEB 12th Class History Notes Chapter 7 Guru Hargobind Ji and Transformation of Sikhism

March 7, 2024March 6, 2024 by Bhagya

This PSEB 12th Class History Notes Chapter 7 Guru Hargobind Ji and Transformation of Sikhism will help you in revision during exams.

PSEB 12th Class History Notes Chapter 7 Guru Hargobind Ji and Transformation of Sikhism

Early Career:

Guru Hargobind Ji was born on 19 June 1595 A.D. at village Wadali in Distt. Amritsar.
The name of his father was Guru Arjan Dev Ji and the name of his mother was Ganga Devi.
Five sons and one daughter were born to him He assumed the Gurgaddi in 1606 A.D.

New Policy of Guru Hargobind Ji:
The main features of the New Policy of Guru Hargobind Ji are as follow:

Causes:

The Mughal emperor Jahangir could not tolerate the flourishing of any other religion except Islam.
Jahangir got Guru Arjan Dev Ji martyred in 1606 A.D.
Guru Arjan Dev Ji himself had instructed Guru Hargobind Ji to adopt the New Policy.

Features:

Guru Hargobind Ji wore the two swords known as Miri and Piri.
Guru Hargobind Ji organized an army.
He instructed the Sikhs to bring him offerings of arms and horses.
He undertook the construction of the Akal Takht Sahib.
He adopted several paraphernalia of sovereignty and assumed royal symbols.
The city of Amritsar was fortified The fort of Lohgarh was constructed.
Guru Ji made several changes in his daily life.

Importance:

The Sikh saints became soldiers.
The Sikhs were united.
The propagation and preaching of Sikhism were enhanced.
The relations between the Sikhs and the Mughals were strained.
The New Policy established the basis of the Khalsa Panth.

Guru Hargobind Ji and Jahangir:

Jahangir took Guru Hargobind Ji as a prisoner in 1606 A.D.
He was kept in the fort of Gwalior.
There is a diverse opinion among the historians regarding the period of imprisonment of Guru Ji.
When Guru Ji was freed, then he insisted on the freedom of the other 52 imprisoned kings.
For this reason, Guru Hargobind Ji began to be called Bandi Chhod Baba.
After the release, the relations between Guru Hargobind Ji and Jahangir took a friendly turn.

Guru Hargobind Ji and Shah Jahan:

In 1628 A.D., when Shah Jahan became the Mughal emperor, the relations between the Mughals and the Sikhs got strained again.
Shah Jahan’s fanaticism sent a wave of resentment among the Sikhs.
In 1634 A.D., the first battle between the Mughals and the Sikhs was fought at Amritsar.
The Sikhs were victorious in the battle.
The battles fought between the Sikhs and the Mughals at Lahira, Kartarpur, and Phagwara were won by the Sikhs.
These victories spread the fame of Guru Hargobind Ji far and wide.

Immersed in Eternal Light:

In 1635 A.D., Guru Hargobind Sahib built a new town called Kiratpur Sahib.
He spent the last ten years of his life here.
Before his death, Har Rai Ji was nominated by him as his successor.
Guru Hargobind Ji was immersed in Eternal Light on March 3, 1645 A.D.

PSEB 12th Class History Notes Chapter 6 Guru Arjan Dev Ji and His Martyrdom

March 7, 2024March 6, 2024 by Bhagya

This PSEB 12th Class History Notes Chapter 6 Guru Arjan Dev Ji and His Martyrdom will help you in revision during exams.

PSEB 12th Class History Notes Chapter 6 Guru Arjan Dev Ji and His Martyrdom

Early Career and Difficulties:

Guru Arjan Dev Ji was born on April 15, 1563 A.D. at Goindwal Sahib.
The name of his father was Guru Ram Das Ji and the name of his mother was Bibi Bhani Ji.
He was married to Ganga Devi, the daughter of Krishan Chand of village Mou of Phillaur.
He assumed Guruship in 1581 A.D.
After his assumption of Guruship, his elder brother Prithi Chand adopted an attitude of open defiance.
He had to face opposition from the Naqshbandis and the Brahmans.
Diwan Chandu Shah of Lahore was also against him.

Development of Sikhism under Guru Arjan Dev Ji:

During his pontificate, Guru Arjan Dev Ji undertook multifarious tasks for the development of Sikhism.
During his Guruship, he got Sri Harmandir Sahib constructed. Its construction work was completed in 1601 A.D.
In 1590 A.D. Tarn in 1593 A.D. Kartarpur and in 1595 A.D. Sri Hargobindpur were founded.
The compilation of the Adi Granth Sahib Ji was the greatest accomplishment of Guru Arjan Dev Ji. This great task was accomplished in 1604 A.D.
Guru Arjan Dev Ji contributed significantly towards the Masand system.
Guru Arjan Dev Ji encouraged the Sikhs the trade horses with the Arab countries.
He kept the doors of Sikhism open by nominating his successor.

Martyrdom of Guru Arjan Dev Ji:
A few facts related to the martyrdom of Guru Arjan Dev Ji are as follows:

Causes:

Jahangir was an orthodox Sunni Muslim.
He could not tolerate the growing prosperity of Sikh Panth.
Prithi Chand, the elder brother of Guru Arjan Dev Ji had started hatching conspiracies to acquire the Gurgaddi.
Chandu Shah, the Diwan of Lahore wanted to take revenge for his insult from Guru Sahib Ji.
The Naqshbandis also instigated Jahangir against Guru Arjan Dev Ji.
The help extended to Shahzada Khusrau by Guru Arjan Dev Ji became the immediate cause of his martyrdom.

Martyrdom:

On Jahangir’s instructions, Guru Arjan Dev Ji was taken prisoner on 24th May 1606 A.D.
He was asked to pay a fine of rupees 2 lakhs which was refused.
Guru Arjan Dev Ji was martyred at Lahore on 30th May 1606 A.D.

Importance:

The martyrdom of Guru Arjan Dev Ji is considered to be an important incident in Sikh history.
Because he was the first Sikh Guru who gave his martyrdom he was called ‘Shaheedan de Sirtaj’.
As a consequence of this martyrdom, there came about a transformation in the form of Sikhism.
Guru Hargobind Ji decided to adopt a new policy of Miri and Piri.
The Sikhs began to unite in a single band.
The relations between the Sikhs and Mughals grew tense.
The era of Mughal atrocities bega.
Sikhism gained greater popularity than before.

PSEB 12th Class History Notes Chapter 5 Development of Sikhism Under Guru Angad DevJi, Guru Amar Das Ji and Guru Ram Das Ji

March 7, 2024March 6, 2024 by Bhagya

This PSEB 12th Class History Notes Chapter 5 Development of Sikhism Under Guru Angad DevJi, Guru Amar Das Ji and Guru Ram Das Ji will help you in revision during exams.

PSEB 12th Class History Notes Chapter 5 Development of Sikhism Under Guru Angad DevJi, Guru Amar Das Ji and Guru Ram Das Ji

Early Career of Guru Angad Dev Ji:

Guru Angad Dev Ji was born in a village named Matte Di Sarai on 31st March 1504 A.D.
The original name of Guru Angad Dev Ji was Bhai Lehna Ji.
The name of his father was Pheru Mai and the name of his mother was Sabhrai Devi.
At the age of 15 years, he was married to Bibi Khivi Ji, the daughter of Devi Chand of the same village.
In due course of time, he was blessed with two daughters Bibi Amro and Bibi Anokhi, and two sons Datu and Dasu.
Once when he set out for a pilgrimage of Jwalamukhi, he met Guru Nanak Dev Ji at Kartarpur.
So overwhelmed was he by his personality and teachings, that he decided to become Guru Nanak Dev Ji’s disciple.
Guru Nanak Dev Ji was greatly impressed by his true devotion and tireless service and appointed him as his successor on 7th September 1539 A.D.

Development of Sikhism under Guru Angad Dev Ji:

Guru Angad Dev Ji popularised the Gurmukhi script.
He also collected the hymns of Guru Nanak Dev Ji.
Guru Angad Dev Ji summoned Bhai Bala, a devotee of Guru Nanak Dev Ji, and got a Janam Sakhi written on Guru Nanak Dev Ji’s life.
Many historians did not agree with this view.
Guru Angad Dav Ji expanded the Langar system.
Guru Angad Dev Ji more effectively organised the institution of Sangat.
He denounced the Udasi sect thereby leading to the preservation of the purity and originality of the Sikh religion.
He founded new towns named Goindwal Sahib near Khadur Sahib.
He gave his blessings to Humayun thereby restoring the cordial relations between the Mughals and the Sikhs.

Immersed in Eternal Light:

Foreseeing his end, Guru Angad Dev Ji appointed Guru Amar Das Ji as his successor.
He immersed, in Eternal Light on 29th March 1552 A.D.

Early career and Difficulties of Guru Amar Das Ji:

Guru Amar Das Ji was born in Basarke village of Amritsar on 5th May 1479 A.D.
The name of his father was Tej Bhan Bhalla.
At the age of 24 years, he was married to Mansa Devi, the daughter of Devi Chand.
He was blessed with two sons Baba Mohan and Baba Mohri and two daughters Bibi Dani and Bibi Bhani.
At the age of 62, he became the disciple of Guru Angad Dev Ji.
He assumed Guruship in March 1552 A.D.
He was 73 years old at that time.
When Guru Amar Das Ji succeeded to the Guruship, he had to face opposition from both the sons of Guru Angad Dev Ji, Dasu, and Datu.
Baba Sri Chand who was the elder son of Guru Nanak Dev Ji had founded the Udasi sect also opposed him.
He also faced stiff opposition from the Muslims of Goindwal Sahib.

Development of Sikhism under Guru Amar Das Ji:

Guru Amar Das Ji retained Guruship from 1552 to 1574 A.D. Goindwal Sahib was the centre of activities of Guru Amar Das Ji.
The first significant step undertaken by Guru Amar Das Ji for the development of Sikhism was the construction of a Baoli with 84 steps at Goindwal Sahib.
He developed the langar institution.
He established the Manji system (22 Manjis) to convey the message of Sikhism in far-off areas.
Guru Amar Das Ji denunciated the Udasi sect.
Guru Amar Das Ji vehemently opposed the prevalent’social malpractices.
Guru Amar Das Ji introduced new (special) rituals for the Sikhs to be observed on the occasions of birth, marriage, and death.
Akbar’s visit to Goindwal Sahib in 1568 A.D. helped to establish friendly relations between the Sikhs and the Mughals.

Immersed in Eternal Light:

Guru Amar Das Ji appointed Bhai Jetha Ji, his son-in-law as his successor in 1574 A.D.
Guru Amar Das Ji was immersed in Eternal Light on September 1, 1574 A.D.

The early career of Guru Ram Das Ji:

Guru Ram Das Ji was born at Chuna Mandi in Lahore on 24th September 1534 A.D.
His childhood name was Bhai Jetha Ji.
His father’s name was Hari Das and his mother’s name was Daya Kaur.
Impressed by the personality of Guru Amar Das Ji, he became Guru Amar Das Ji’s disciple.
In 1553 A.D. he was married to Bibi Bhani, the youngest daughter of Guru Amar Das Ji.
He assumed Guruship in 1574 A.D. He was the 4th Guru of the Sikhs.

Development of Sikhism under Guru Ram Das Ji:

Guru Ram Das Ji established Ramdaspura or Amritsar in 1577 A.D.
He initiated the digging work of two serovars (tanks) Amritsar and Santokhsar.
The Masand system was introduced by him for the propagation of Sikhism and to collect money from the Sikhs.
The reconciliation between the Sikhs and the Udasis proved to be a milestone in Sikh history during the pontificate of Guru Ram Das Ji.
He continued the Sangat, Pangat, and Manji systems.
He further consolidated the friendly relations with the Mughal emperor Akbar.

Immersed in Eternal Light:

Before Guru Ram Das Ji was immersed in Eternal Light, he nominated his youngest son Arjan Dev as his successor.
He immersed in Eternal Light on 1st September 1581 A.D.

PSEB 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.2

March 6, 2024March 5, 2024 by Bhagya

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x -4
Solution:
(i) Given Quadratic polynomial,
x² – 2x – 8
∵ [S = -2, P = -8]
= x² – 4x + 2x – 8
= x (x – 4) + 2 (x – 4)
= (x – 4) (x + 2)
The value of x² – 2x – 8 is zero,
iff (x – 4) = 0 or (x + 2) = 0
iff x = 4 or x = – 2
Therefore, zeroes of x2 – 2x – 8 are – 2 and 4.
Now, Sum of zeroes = (- 2) + (4) = 2
= \(\frac{-(-2)}{1}=-\frac{(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)
Product of zeroes = (- 2) (4) = – 8
= \(\frac{-8}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
Hence, relationship between the zeroes and the coefficient are verified.

(ii) Given quadratic polynomial,
4s² – 4s + 1
= 4s² – 2s – 2s + 1
∵ [S = -4, P = 4 × 1]
= 2s (2s – 1) – 1 (2s – 1)
= (2s – 1) (2s – 1)
The value of 4s² – 4s + 1 is zero
iff (2s – 1) = 0 or (2s – 1) = 0
iff s = \(\frac{1}{2}\) or s = \(\frac{1}{2}\)
Therefore, zeroes of 4s² – 4s + 1 are \(\frac{1}{2}\) and \(\frac{1}{2}\)

Now, sum of zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1
= \(\frac{-(-4)}{4}=\frac{-(\text { Coefficient of } s)}{\left(\text { Coefficient of } s^{2}\right)}\)

Product of Zeroes = \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.

(iii) Given quadratic polynomial,
6x² -3 – 7x
= 6x² – 7x – 3
∵ [S = – 7, P = 6x-3=-18]
= 6x² -9x + 2x-3
= 3x (2x – 3) + 1 (2x – 3)
= (2x – 3) (3x + 1)
The value of 6x² – 3 – 7x is zero
iff (2x – 3) = 0 or 3x + 1 = 0
iff x = \(\frac{3}{2}\) or x = –\(\frac{1}{3}\)
Therefore, zeroes of 6x² – 3 – 7x are \(\frac{3}{2}\) and –\(\frac{1}{3}\)
Now, Sum of zeroes = \(\frac{1}{3}\)

= \(\frac{3}{2}+\left(\frac{-1}{3}\right)\)

= \(\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}\)

= \(\frac{7}{6}=\frac{-(-7)}{6}\)

= \(\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)

Product of zeroes = \(\left(\frac{3}{2}\right)\left(\frac{-1}{3}\right)\)

= \(\frac{-3}{6}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

Hence, relationship between the zeroes and the coefficients are verified.

(iv) Given quadratic polynomial,
4u² + 8u = 4u(u + 2)
The value of 4u² + 8 u is zero
iff 4u = 0 or u + 2 = 0
iff u = 0 or u = – 2
Therefore, zeroes of 4u² + 8M are 0 and – 2
Now, Sum of zeroes = 0 + (- 2) = -2
= \(\frac{-8}{4}\)
= \(-\frac{(\text { Coefficient of } u)}{\text { Coefficient of } u^{2}}\)

Product of zeroes = (0) (- 2) = 0
= \(\frac{0}{4}=\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.

(v) Given quadratic polynomial,
t² – 15 = t² – (√15)²
= (t – √15) (t + √15)
The value of t² – 15 is zero
iff t – √15 = 0 or t + √15 = 0
iff t = √15 or t = – √15
Therefore, zeroes of t² – 15 are – √15 and √15.
Now, Sum of zeroes = -√15 + √15 = 0 = \(\frac{0}{1}\)
= \(\frac{-(\text { Coefficient of } t)}{\text { Coefficient of } t^{2}}\)

Product of zeroes = (-√15) (√15) = – 15 = \(-\frac{15}{1}\)

= \(\frac{0}{1}\)
= \(=\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.

(vi) Given quadratic polynomial,
3x² – x – 4
= 3x² + 3x – 4x – 4
= 3x (x + 1) – 4 (x + 1)
∵ [S = – 1, P = 3 x – 4 = – 12]
= (x + 1) (3x – 4)
The value of 3x² – x – 4 is zero
iff (x + 1) = 0 or 3x – 4 = 0
iff x = -1 or x = \(\frac{4}{3}\)
Therefore, zeroes of 3x² – x – 4 are – 1 and \(\frac{4}{3}\)

Now, Sum of zeroes = – 1 + \(\frac{4}{3}\)
= \(\frac{-3+4}{3}=\frac{1}{3}\)
= \(\frac{-(-1)}{3}=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\)
Product of zeroes = (- 1) \(\frac{4}{3}\)
= \(\frac{-4}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac{1}{4}\), -1
(ii) √2, \(\frac{1}{3}\)
(iii) 0, √5
(iv) 1, 1
(v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
(vi) 4, 1. [MQP 2015]
Solution:
(i) Given that, sum of zeroes and products of zeroes of given polynomial \(\frac{1}{4}\) are -1 respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = \(\frac{1}{4}\)
and αβ = Product of zeroes = – 1
Now, ax² + bx + c = k (x – α) (x – β) where k is any constant
= k [x² – (α + β)x + αβ]
= k[x² – \(\frac{1}{4}\)x + (-1)]
= k[x² – \(\frac{1}{4}\)x – 1]
for different value of k, we get different quadratic polynomials.

(ii) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are √2 and \(\frac{1}{3}\) respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = √2
and αβ = Product of zeroes = \(\frac{1}{3}\)
Now, ax² + bx + c = k (x – α) (x – β) where k is any constant
= k[x² – (α + β) x + αβ]
= k[x² – √2x + \(\frac{1}{3}\)]of k, we get different quadratic polynomial.
for different values of k, we get different quadratic polynomial.

(iii) Given that, sum of zeroes and products of zeroes of given quadratic polynomial are 0 and √5 respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 0
and αβ = Product of zeroes = √5
Now, ax² + bx + c = k(x – α) (x – β)
where k is any constant
= k [x² – (a + (α+ β)x + αβ)
= k[x² – 0x + √5]
= k[x² + √5]
for different values of k, we get different quadratic polynomial.

(iv) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 1 and 1 respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 1 and
αβ = Product of zeroes = 1
Now, ax² + bx + c = k(x – α) (x – β)
where k is any constant.
= k [x² – (α + β)x + αβ]
= k [x² – 0x + √5]
= it [x² – x + √5]
for different values of k, we get different quadratic polynomial.

(v) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are \(-\frac{1}{4}\) and \(\frac{1}{4}\) respectively.
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β
α + β = Sum of zeroes = –\(\frac{1}{4}\)
and αβ = Product of zeroes = \(\frac{1}{4}\)
Now ax² + bx + c = k (x – α) (x – β) where k is any constant
= k[x² – (α + β) x + αβ]
= k[x² – (-\(\frac{1}{4}\))x + \(\frac{1}{4}\)]
= k[x² + \(\frac{1}{4}\)x + \(\frac{1}{4}\)]
For different values of k, we get different quadratic polynomial.

(vi) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 4 and 1 respectively. Let the quadratic polynomial be
ax² + bx + c and its zeroes be α and β
α + β = sum of zeroes = 4 and
αβ = Product of zeroes = 1
Now, ax2 + bx + c = k(x – α) (x – β) where k is any constant
= k [x² – (α + β) x + αβ]
= k [x² – 4x + 1]
For different values of k, we get different quadratic polynomials.

PSEB 12th Class History Notes Chapter 3 Political, Social and Economic Conditions of the Punjab in the beginning of the 16th Century

March 6, 2024March 5, 2024 by Bhagya

This PSEB 12th Class History Notes Chapter 3 Political, Social and Economic Conditions of the Punjab in the beginning of the 16th Century will help you in revision during exams.

PSEB 12th Class History Notes Chapter 3 Political, Social and Economic Conditions of the Punjab in the beginning of the 16th Century

Political Condition:

The political condition of Punjab was quite deplorable.
Punjab was under the Sultanate of Delhi which was under the Lodhi Sultans at that time.
In 1469 A.D. the Sultan of Delhi, Bahlol Lodhi appointed Tatar Khan Lodhi as the Governor of Punjab.
Tatar Khan Lodhi was killed during an unsuccessful revolt against the Lodhi Sultan.
In 1500 A.D. the new Lodhi Sultan, Sikandar Lodhi appointed Daulat Khan Lodhi as the Governor of Punjab.
As soon as Ibrahim Lodhi became the new Sultan, Daulat Khan Lodhi started hatching conspiracies against him.
Daulat Khan Lodhi invited Babar to invade India.
Babar invaded Punjab five times between 1519 and 1526 A.D.
During his fifth invasion, Babar defeated Daulat Khan Lodhi to establish control over Punjab.
On 21st April 1526 A.D., Babar defeated Ibrahim Lodhi in the First Battle of Panipat.
Consequently, Punjab slipped from the hands of the Lodhi dynasty into the hands of the Mughals.

Social Condition:

At the beginning of the 16th century, the social condition of Punjab was deplorable.
Society was divided into two major sects the Hindus and the Muslims Related to the ruling elite class, the Muslims had several special privileges.
The Muslim society was divided into upper, middle, and lower classes.
The condition of Muslim women was pitiable.
The Hindus were in majority but they were deprived of their rights.
Hindu society was divided into several castes and sub-castes.
The elite class of society ate delicious foods and wore expensive clothes.
The lower classes wore ordinary clothes and ate frugal meals.
At that time hunting, polo, animal fights, chess, dance, music, and cards, etc. were sources of entertainment.
Education was imparted in mosques, madrasas, and temples.

Economic Condition:

The economic condition of Punjab was very good.
The main occupation of the people of Punjab was agriculture.
The main crops grown here were wheat, barley, maize, rice, and sugarcane.
It had rich harvests Industry was the other main occupation of the people.
The most important industry was the textile industry.
Besides this, there were other industries that manufactured leather goods, arms, utensils, toys, and articles of ivory.
The occupation of cattle rearing was also prevalent.
The domestic and foreign trade of Punjab was also quite prosperous.
The foreign trade of Punjab was carried on with countries like Afghanistan, Iran, Arabia, Syria, Tibet, and China, etc.
Lahore and Multan were the two most popular towns in Punjab.
Due to low prices, ordinary people also enjoyed a good standard of living.

PSEB 12th Class Sociology Notes Chapter 12 Social Issues: Old Age and Disability

March 6, 2024March 5, 2024 by Bhagya

This PSEB 12th Class Sociology Notes Chapter 12 Social Issues: Old Age and Disability will help you in revision during exams.

PSEB 12th Class Sociology Notes Chapter 12 Social Issues: Old Age and Disability

Old Age Homes:

Homes are made for elders so that they can live happily over there.

Disability:

It means the consequences of impairment in terms of functional performance and activity.

Old Age:

That stage of life starts after 60 years, which no one likes, and in which one faces many physical and other problems.

Gerontology:

Field of science which studies the process of aging.

→ Old age is a necessary and natural part of human life and all humans have to pass through this phase.

→ No one wants to go through this phase because an individual faces many physical problems in this.

→ One needs to depend upon others in this phase.

→ According to United Nations, people above the age of 60 years come under the category of old age.

→ Almost in all the western countries, age of 60-65 years is kept to take pension benefits and other facilities.

→ Symbols of old age are visible quite early such as tooth decay, grey hair, hunched back, hearing loss, slow pace of the walk, vision impairment, etc.

→ At this age, one also faces many physical problems such as heart problems, high blood pressure, arthritis, sugar, etc.

→ In old age, one faces many problems such as health-related issues, economic insecurity, home-related problems, social problems, mental problems, own role-related problems, etc. Due to stress and tensions, death comes very quickly.

→ Old age problems can be removed in many ways such as by making old age homes, by starting welfare programmes for them, by creating easy jobs for them, by taking care of them, by providing better health facilities, by making strict laws, etc.

→ In almost the entire world, around 100 crore people are there who are facing disability in one way or the other.

→ The meaning of disability is any type of physical problem such as impairment, handicap, the inability of listening, etc.

→ Disability can be of many types such as locomotor disability, visual disability, hearing disability, mental disability, speech disability.

→ There can be many reasons for a disability such as a disease, birth-based problems, malnutrition, congenital factor, stress, accident, etc.

→ Disabled persons have to face many problems such as segregation and isolation, poverty, social oppression, etc.

→ The problems of disabled persons can be cured in many ways such as by providing better health facilities, by removing discrimination, by giving them education in normal schools, by arousing consciousness among people, etc.