Prasanna

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 ਗਰਾਫ਼ਾਂ ਬਾਰੇ ਜਾਣਕਾਰੀ Ex 15.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 ਗਰਾਫ਼ਾਂ ਬਾਰੇ ਜਾਣਕਾਰੀ Exercise 15.3

1. ਢੁੱਕਵੇਂ ਪੈਮਾਨੇ ਦਾ ਪ੍ਰਯੋਗ ਕਰਦੇ ਹੋਏ, ਹੇਠਾਂ ਲਿਖੀਆਂ ਸਾਰਣੀਆਂ ਵਿਚ ਦਿੱਤੀਆਂ ਗਈਆਂ ਰਾਸ਼ੀਆਂ ਦੇ ਲਈ ਗਰਾਫ਼ ਬਣਾਉ :

ਪ੍ਰਸ਼ਨ (a).
ਸੇਬਾਂ ਦਾ ਮੁੱਲ :

ਹੱਲ:
ਦਿੱਤੀ ਗਈ ਸਾਰਣੀ ਹੈ :

ਦਿੱਤੀ ਗਈ ਸਾਰਣੀ ਦਾ ਗਰਾਫ਼ ਚਿੱਤਰ ਵਿਚ ਦਰਸਾਏ ਅਨੁਸਾਰ ਹੈ ।

ਪ੍ਰਸ਼ਨ (b).
ਕਾਰ ਦੁਆਰਾ ਤੈਅ ਕੀਤੀ ਗਈ ਦੂਰੀ :

(i) 7.30 ਵਜੇ ਸਵੇਰ ਅਤੇ 8 ਵਜੇ ਸਵੇਰ ਦੇ ਅੰਤਰਾਲ ਵਿਚ ਕਾਰ ਦੁਆਰਾ ਕਿੰਨੀ ਦੂਰੀ ਤੈਅ ਕੀਤੀ ਗਈ ?
(ii) ਕਾਰ ਦੀ 100 km ਦੂਰੀ ਤੈਅ ਕਰ ਲੈਣ ’ਤੇ ਸਮਾਂ ਕੀ ਸੀ ?
ਹੱਲ:
ਕਾਰ ਦੁਆਰਾ ਤੈਅ ਕੀਤੀ ਗਈ ਦੂਰੀ

ਦਿੱਤੀ ਗਈ ਸਾਰਣੀ ਦਾ ਗਰਾਫ ਚਿੱਤਰ ਵਿਚ ਦਰਸਾਏ ਅਨੁਸਾਰ ਹੈ ।
(i) 20 Km
(ii) ਸਵੇਰੇ 7.30 ਵਜੇ

ਪ੍ਰਸ਼ਨ (c).
ਜਮਾਂ ਧਨ ‘ਤੇ ਸਾਲਾਨਾ ਵਿਆਜ

(i) ਕੀ ਗਰਾਫ ਮੂਲ ਬਿੰਦੂ ਤੋਂ ਲੰਘਦਾ ਹੈ ?
(ii) ਗਰਾਫ ਤੋਂ ਤੋਂ 2500 ਦਾ ਸਾਲਾਨਾ ਵਿਆਜ ਪਤਾ ਕਰੋ ।
(iii) ਤੋਂ 280 ਵਿਆਜ ਪ੍ਰਾਪਤ ਕਰਨ ਦੇ ਲਈ ਕਿੰਨਾ ਧਨ ਜਮਾਂ ਕਰਨਾ ਹੋਵੇਗਾ ?
ਹੱਲ:

(i) ਹਾਂ।
(ii) ₹ 200
(iii) ₹ 3500

2. ਸਾਰਣੀਆਂ ਦੇ ਲਈ ਗਰਾਫ਼ ਖਿੱਚੋ :

ਪ੍ਰਸ਼ਨ (i).

ਕੀ ਇਹ ਰੇਖੀ ਗਰਾਫ਼ ਹੈ ?
ਹੱਲ:
ਦਿੱਤੀ ਗਈ ਸਾਰਣੀ ਹੈ :

ਦਿੱਤੀ ਗਈ ਸਾਰਣੀ ਦਾ ਗਰਾਫ਼ ਚਿੱਤਰ ਵਿਚ ਦਰਸਾਏ ਅਨੁਸਾਰ ਹੈ :

ਹਾਂ, ਇਹ ਰੇਖੀ ਗਰਾਫ ਹੈ

ਪ੍ਰਸ਼ਨ (ii).

ਕੀ ਇਹ ਰੇਖੀ ਗਰਾਫ਼ ਹੈ ?
ਹੱਲ:
ਦਿੱਤੀ ਗਈ ਸਾਰਣੀ ਹੈ :

ਦਿੱਤੀ ਗਈ ਸਾਰਣੀ ਦਾ ਗਰਾਫ਼ ਚਿੱਤਰ ਵਿਚ ਦਰਸਾਏ ਅਨੁਸਾਰ ਹੈ।
ਨਹੀਂ, ਇਹ ਰੇਖੀ ਗਰਾਫ਼ ਨਹੀਂ ਹੈ ।

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 ਗਰਾਫ਼ਾਂ ਬਾਰੇ ਜਾਣਕਾਰੀ Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 ਗਰਾਫ਼ਾਂ ਬਾਰੇ ਜਾਣਕਾਰੀ Exercise 15.2

ਪ੍ਰਸ਼ਨ 1.
ਹੇਠ ਦਿੱਤੇ ਬਿੰਦੂਆਂ ਨੂੰ ਇਕ ਵਰਗੀਕ੍ਰਿਤ ਕਾਗਜ਼ (Graph Sheet) ‘ਤੇ ਅੰਕਿਤ ਕਰੋ ਅਤੇ ਜਾਂਚੋ ਕਿ ਕੀ ਇਹ ਸਾਰੇ ਇਕ ਸਰਲ ਰੇਖਾ ‘ਤੇ ਸਥਿਤ ਹਨ ?
(a) A (4, 0), B (4, 2), C (4, 6), D (4, 25)
(b) P (1, 1), Q (2, 2), R (3, 3), S (4, 4)
(c) K (2, 3), L (5, 3), M (5, 5), N (2,5).
ਹੱਲ:

(a) ਬਿੰਦੂ A, B, C ਅਤੇ D ਇਕ ਰੇਖਾ ਉੱਤੇ ਸਥਿਤ ਹਨ ।
(b) ਬਿੰਦੂ P, Q, R ਅਤੇ S ਇਕ ਰੇਖਾ ਉੱਤੇ ਸਥਿਤ ਹਨ ।
(c) ਬਿੰਦੂ K, L, M ਅਤੇ N ਇਕ ਰੇਖਾ ਉੱਤੇ ਸਥਿਤ ਨਹੀਂ ਹਨ ।

ਪ੍ਰਸ਼ਨ 2.
ਬਿੰਦੂਆਂ (2, 3) ਅਤੇ (3, 2) ਵਿਚੋਂ ਲੰਘਦੀ ਹੋਈ ਇਕ ਸਰਲ ਰੇਖਾ ਖਿੱਚੋ । ਉਹਨਾਂ ਬਿੰਦੂਆਂ ਦੇ ਨਿਰਦੇਸ਼ ਅੰਕ ਲਿਖੋ ਜਿਹਨਾਂ ਤੇ ਇਹ ਰੇਖਾ -ਧੁਰੇ ਅਤੇ y-ਧੁਰੇ ਨੂੰ ਕੱਟਦੀ ਹੈ ।
ਹੱਲ:

ਇਹ ਰੇਖਾ x-ਧੁਰੇ ਨੂੰ (5, 0) ਅਤੇ y-ਧੁਰੇ ਨੂੰ (0, 5) ਉੱਤੇ ਕੱਟੇਗੀ ।

ਪ੍ਰਸ਼ਨ 3.
ਗਰਾਫ਼ ਵਿਚ ਬਣਾਏ ਗਏ ਚਿੱਤਰਾਂ ਵਿਚ ਹਰੇਕ ਦੇ | ਸਿਖਰਾਂ ਦੇ ਨਿਰਦੇਸ਼ ਅੰਕ ਲਿਖੋ ।
ਹੱਲ:

O (0, 0)
A (2, 0)
B (2, 3)
C (0, 3)
P (4, 3)
Q (6, 1)
R (6, 5)
S (4, 7)
K (10, 5)
L (7, 7)
M (10, 8)

ਪ੍ਰਸ਼ਨ 4.
ਹੇਠ ਲਿਖੇ ਕਥਨਾਂ ਵਿਚ ਕਿਹੜੇ ਸੱਚ ਹਨ ਅਤੇ ਕਿਹੜੇ ਝੂਠ ? ਝੂਠ ਨੂੰ ਠੀਕ ਕਰੋ ।
(i) ਕੋਈ ਬਿੰਦੂ ਜਿਸਦਾ -ਨਿਰਦੇਸ਼ ਅੰਕ ਸਿਫਰ ਹੈ, ਅਤੇ y-ਨਿਰੇਦਸ਼ ਅੰਕ ਸਿਫਰ ਨਹੀਂ ਹੈ, y-ਧੁਰੇ ਤੇ ਸਥਿਤ ਹੁੰਦਾ ਹੈ ।
(ii) ਕੋਈ ਬਿੰਦੂ ਜਿਸਦਾ y-ਨਿਰਦੇਸ਼ ਅੰਕ ਸਿਫਰ ਹੈ, ਅਤੇ r-ਨਿਰਦੇਸ਼ ਅੰਕ 5 ਹੈ, y-ਧੁਰੇ ਤੇ ਸਥਿਤ ਹੁੰਦਾ ਹੈ ।
(iii) ਮੁਲ ਬਿੰਦੁ ਦੇ ਨਿਰਦੇਸ਼ ਅੰਕ (0, 0) ਹੈ ।
ਹੱਲ:
(i) ਸੱਚ
(ii) ਖੂਨ, (5, 0) x-ਧੁਰੇ ਉੱਤੇ ਸਥਿਤ ਹੋਵੇਗਾ ।
(iii) ਸੱਚ !

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 ਗਰਾਫ਼ਾਂ ਬਾਰੇ ਜਾਣਕਾਰੀ Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 ਗਰਾਫ਼ਾਂ ਬਾਰੇ ਜਾਣਕਾਰੀ Exercise 15.1

ਪ੍ਰਸ਼ਨ 1.
ਹੇਠਾਂ ਲਿਖੇ ਗਰਾਫ਼, ਕਿਸੇ ਹਸਪਤਾਲ ਵਿਚ ਇਕ ਰੋਗੀ ਦਾ ਪ੍ਰਤੀ ਘੰਟੇ ਲਿਆ ਗਿਆ ਤਾਪਮਾਨ ਦਰਸਾਉਂਦਾ ਹੈ :
(a) ਦੁਪਹਿਰ 1 ਵਜੇ ਰੋਗੀ ਦਾ ਤਾਪਮਾਨ ਕੀ ਸੀ ?

(b) ਰੋਗੀ ਦਾ ਤਾਪਮਾਨ 38.5° c ਕਦੋਂ ਸੀ ?
(c) ਇਸ ਪੂਰੇ ਅੰਤਰਾਲ ਵਿਚ ਰੋਗੀ ਦਾ ਤਾਪਮਾਨ ਦੋ ਵਾਰ ਇਕ ਸਮਾਨ ਹੀ ਸੀ । ਇਹ ਦੋ ਸਮੇਂ, ਕਿਹੜੇ-ਕਿਹੜੇ ਹਨ ?
(d) 1.30 ਵਜੇ ਦੁਪਹਿਰ ਰੋਗੀ ਦਾ ਤਾਪਮਾਨ ਕੀ ਸੀ ? ਇਸ ਸਿੱਟੇ ‘ਤੇ ਤੁਸੀਂ ਕਿਸ ਤਰ੍ਹਾਂ ਪਹੁੰਚੋਗੇ ?
(e) ਕਿਹੜੇ ਅੰਤਰਾਲਾਂ ਵਿਚ ਰੋਗੀ ਦਾ ਤਾਪਮਾਨ ‘ਵੱਧਣ ਦੇ ਰੁਝਾਨ ਨੂੰ ਦਰਸਾਉਂਦਾ ਹੈ ?
ਹੱਲ:
(a) 36.5° C
(b) 12 ਵਜੇ ਦੋਪਹਿਰ
(c) 1 ਵਜੇ ਦੁਪਹਿਰ, 2 ਵਜੇ ਦੁਪਹਿਰ
(d) ਦੁਪਹਿਰ 1.30 ਵਜੇ ਰੋਗੀ ਦਾ ਤਾਪਮਾਨ 36.5°C ਸੀ । ਦੁਪਹਿਰ 1 ਵਜੇ ਤੋਂ ਦੁਪਹਿਰ 2 ਵਜੇ ਦੇ ਵਿਚਲੇ x-ਅਕਸ਼ ਉੱਤੇ ਸਥਿਤ ਬਿੰਦੁ ਦੁਪਹਿਰ 1 ਵਜੇ ਅਤੇ ਦੁਪਹਿਰ 2 ਵਜੇ ਨੂੰ ਦਰਸਾਉਂਣ ਵਾਲੇ ਬਿੰਦੂਆਂ ਨਾਲ ਸਮਦੂਰੀ ਹੈ, ਇਸ ਲਈ ਇਹ ਦੁਪਹਿਰ 1 ਵਜੇ 30 ਮਿੰਟ ਦਾ ਸਮਾਂ ਦਰਸਾਵੇਗਾ ।
(e) 9 ਵਜੇ ਸਵੇਰ ਤੋਂ 10 ਵਜੇ ਸਵੇਰ ਤੱਕ, 10 ਵਜੇ ਸਵੇਰ ਤੋਂ 11 ਵਜੇ ਸਵੇਰ ਤੱਕ, 2 ਵਜੇ ਦੁਪਹਿਰ ਤੋਂ 3 ਵਜੇ ਸ਼ਾਮ ਤੱਕ ਰੋਗੀ ਦੇ ਤਾਪਮਾਨ ਵੱਧਣ ਦਾ ਰੁਝਾਨ ਦਰਸਾਉਂਦੇ ਹਨ ।

ਪ੍ਰਸ਼ਨ 2.
ਇਕ ਨਿਰਮਾਣ ਕੰਪਨੀ ਦੇ ਵੱਖ-ਵੱਖ ਸਾਲਾਂ ਵਿਚ ਕੀਤੀ ਗਈ ਵਿਕਰੀ ਹੇਠਾਂ ਦਿੱਤੇ ਗਰਾਫ਼ ਦੁਆਰਾ ਦਰਸਾਈ ਗਈ ਹੈ :

(a) (i) ਸਾਲ 2002 ਵਿਚ
(ii) ਸਾਲ 2006 ਵਿਚ ਕਿੰਨੀ ਵਿਕਰੀ ਸੀ ?
(b) (i) ਸਾਲ 2003 ਵਿਚ
(ii) ਸਾਲ 2005 ਵਿਚ ਕਿੰਨੀ ਵਿਕਰੀ ਸੀ ?
(c) ਸਾਲ 2002 ਅਤੇ ਸਾਲ 2006 ਵਿੱਚ ਵਿਕਰੀ ਵਿਚ ਕਿੰਨਾ ਅੰਤਰ ਸੀ ?
(d) ਕਿਸ ਅੰਤਰਾਲ ਵਿਚ ਵਿਕਰੀ ਦਾ ਇਹ ਅੰਤਰ ਸਭ ਤੋਂ ਜ਼ਿਆਦਾ ਸੀ ?
ਹੱਲ:
(a) (i) ਸਾਲ 2002 ਵਿਚ ਵਿਕਰੀ 4 ਕਰੋੜ ਰੁਪਏ ਸੀ ।
(ii) ਸਾਲ 2006 ਵਿਚ ਵਿਕਰੀ 8 ਕਰੋੜ ਰੁਪਏ ਸੀ ।
(b) (i) ਸਾਲ 2003 ਵਿਚ ਵਿਕਰੀ 7 ਕਰੋੜ ਰੁਪਏ ਸੀ ।
(ii) ਸਾਲ 2005 ਵਿਚ ਵਿਕਰੀ 8 ਕਰੋੜ ਰੁਪਏ (ਲਗਪਗ ਸੀ ।
(c) ਸਾਲ 2002 ਅਤੇ ਸਾਲ 2006 ਵਿਚ ਵਿਕਰੀ ਵਿਚ ਅੰਤਰ
= 8 ਕਰੋੜ ਰੁ: – 4 ਕਰੋੜ ਰੁ: : 4 ਕਰੋੜ ਰੁ:
(d) ਸਾਲ 2005 ਵਿਚ ਵਿਕਰੀ ਦਾ ਅੰਤਰ ਸਭ ਤੋਂ ਜ਼ਿਆਦਾ ਸੀ ।

ਪ੍ਰਸ਼ਨ 3.
ਬਨਸਪਤੀ-ਵਿਗਿਆਨ ਦੇ ਇਕ ਪ੍ਰਯੋਗ ਵਿਚ, ਸਮਾਨ ਪ੍ਰਯੋਗਸ਼ਾਲਾ ਪਰਿਸਥਿਤੀਆਂ ਵਿਚ ਦੋ ਪੌਦੇ A ਅਤੇ B ਉਗਾਏ ਗਏ ।ਤਿੰਨ ਹਫ਼ਤਿਆਂ ਤੱਕ ਉਹਨਾਂ ਦੀਆਂ ਉੱਚਾਈਆਂ ਨੂੰ ਹਰ ਹਫਤੇ ਦੇ ਅੰਤ ਵਿਚ ਮਾਪਿਆ ਗਿਆ ਨਤੀਜਿਆਂ ਨੂੰ ਹੇਠਾਂ ਲਿਖੇ ਗਰਾਫ਼ ਵਿਚ ਦਰਸਾਇਆ ਗਿਆ ਹੈ :

(a) (i) 2 ਹਫ਼ਤੇ ਬਾਅਦ
(ii) 3 ਹਫ਼ਤੇ ਬਾਅਦ ਪੌਦੇ A ਦੀ ਉੱਚਾਈ ਕਿੰਨੀ ਸੀ ?
(b) (i) 2 ਹਫ਼ਤੇ ਬਾਅਦ
(ii) 3 ਹਫ਼ਤੇ ਬਾਅਦ
ਪੌਦੇ B ਦੀ ਉੱਚਾਈ ਕਿੰਨੀ ਸੀ ?
(c) ਤੀਸਰੇ ਹਫ਼ਤੇ ਵਿਚ ਪੌਦੇ A ਦੀ ਉੱਚਾਈ ਕਿੰਨੀ ਵਧੀ ?
(d) ਦੁਸਰੇ ਹਫ਼ਤੇ ਦੇ ਅੰਤ ਤੋਂ ਤੀਸਰੇ ਹਫ਼ਤੇ ਦੇ ਅੰਤ ਤਕ ਪੌਦੇ B ਦੀ ਉੱਚਾਈ ਕਿੰਨੀ ਵਧੀ ?
(e) ਕਿਸ ਹਫ਼ਤੇ ਵਿਚ ਪੌਦੇ A ਦੀ ਉੱਚਾਈ ਸਭ ਤੋਂ ਜ਼ਿਆਦਾ ਵਧੀ ?
(f) ਕਿਸ ਹਫ਼ਤੇ ਵਿਚ ਪੌਦੇ B ਦੀ ਉੱਚਾਈ ਸਭ ਤੋਂ ਘੱਟ ਵਧੀ ?
(g) ਕੀ ਕਿਸੇ ਹਫ਼ਤੇ ਵਿਚ ਦੋਨਾਂ ਪੌਦਿਆਂ ਦੀ ਉੱਚਾਈ ਬਰਾਬਰ ਸੀ ? ਪਛਾਣੋ ।
ਹੱਲ:
(a) (i) 7 cm (ii) 9 cm
(b)(i) 7 cm (ii) 10 cm.
(c) 2 cm
(d) 3 cm
(e) ਦੂਸਰੇ ਹਫ਼ਤੇ ਵਿਚ ।
(f) ਪਹਿਲੇ ਹਫ਼ਤੇ ਵਿਚ ।
(g) ਦੂਸਰੇ ਹਫ਼ਤੇ ਦੇ ਅੰਤ ਵਿਚ ।

ਪ੍ਰਸ਼ਨ 4.
ਹੇਠਾਂ ਦਿੱਤਾ ਗਰਾਫ਼, ਕਿਸੇ ਹਫ਼ਤੇ ਦੇ ਹਰੇਕ ਦਿਨ ਦੇ ਲਈ ਪੂਰਵ ਅਨੁਮਾਨਿਤ ਅਤੇ ਅਸਲ ਤਾਪਮਾਨ ਦਰਸਾਉਂਦਾ ਹੈ :
(a) ਕਿਸ ਦਿਨ ਪੂਰਵ ਅਨੁਮਾਨਿਤ ਤਾਪਮਾਨ ਅਤੇ ਵਾਸਤਵਿਕ ਤਾਪਮਾਨ ਸਮਾਨ ਹਨ ?
(b) ਹਫ਼ਤੇ ਵਿਚ ਵੱਧ ਤੋਂ ਵੱਧ ਪੂਰਵ ਅਨੁਮਾਨਿਤ ਤਾਪਮਾਨ ਕੀ ਸੀ ?
(c) ਹਫ਼ਤੇ ਵਿਚ ਘੱਟ ਤੋਂ ਘੱਟ ਅਸਲ ਤਾਪਮਾਨ ਕੀ ਸੀ ?
(d) ਕਿਸ ਦਿਨ ਅਸਲ ਤਾਪਮਾਨ ਅਤੇ ਪੂਰਵ ਅਨੁਮਾਨਿਤ ਤਾਪਮਾਨ ਵਿਚ ਅੰਤਰ ਵੱਧ ਤੋਂ ਵੱਧ ਸੀ ।

ਹੱਲ:
(a) ਮੰਗਲਵਾਰ, ਸ਼ੁਕਰਵਾਰ, ਐਤਵਾਰ ।
(b) 35°C
(c) 15°C
(d) ਵੀਰਵਾਰ ।

5. ਹੇਠਾਂ ਦਿੱਤੀ ਸਾਰਣੀ ਨੂੰ ਦੇਖ ਕੇ ਇਕ ਰੇਖੀ ਗਰਾਫ਼ ਬਣਾਉ :

ਪ੍ਰਸ਼ਨ (a).
ਵੱਖ-ਵੱਖ ਸਾਲਾਂ ਵਿਚ ਕਿਸੇ ਪਹਾੜੀ ਸ਼ਹਿਰ ਵਿਚ ਬਰਫ਼ ਪੈਣ ਦੇ ਦਿਨਾਂ ਦੀ ਸੰਖਿਆ :

ਹੱਲ:
(a) ਵੱਖ-ਵੱਖ ਸਾਲਾਂ ਵਿਚ ਕਿਸੇ ਪਹਾੜੀ ਨਗਰ ਵਿਚ ਬਰਫ਼ ਪੈਣ ਦੇ ਦਿਨਾਂ ਦੀ ਸੰਖਿਆ


x-ਧੁਰੇ ਤੇ ਸਾਲ ਦਿਖਾਏ ਗਏ ਹਨ ਅਤੇ y-ਧੁਰੇ ਤੇ ਦਿਨ ਦਿਖਾਏ ਗਏ ਹਨ !

ਪ੍ਰਸ਼ਨ (b).
ਵੱਖ-ਵੱਖ ਸਾਲਾਂ ਵਿਚ ਇਕ ਪਿੰਡ ਵਿਚ, ਪੁਰਸ਼ਾਂ ਅਤੇ ਇਸਤਰੀਆਂ ਦੀ ਸੰਖਿਆ (ਹਜ਼ਾਰਾਂ ਵਿਚ)

ਹੱਲ:
ਵੱਖ-ਵੱਖ ਸਾਲਾਂ ਵਿਚ ਇਕ ਪਿੰਡ ਵਿਚ, ਪੁਰਸ਼ਾਂ ਅਤੇ ਇਸਤਰੀਆਂ ਦੀ ਸੰਖਿਆ ਹਜ਼ਾਰਾਂ ਵਿਚ)

ਨੋਟ : ………… ਪੁਰਸ਼
_________ ਇਸਤਰੀਆਂ

ਪ੍ਰਸ਼ਨ 6.
ਇਕ ਡਾਕੀਆ ਕਿਸੇ ਸ਼ਹਿਰ ਦੇ ਕੋਲ ਹੀ ਪੈਂਦੇ ਇਕ ਕਸਬੇ ਵਿਚ ਇਕ ਵਪਾਰੀ ਕੋਲ ਪਾਰਸਲ ਪਹੁੰਚਾਉਣ ਲਈ ਸਾਈਕਲ ਤੇ ਜਾਂਦਾ ਹੈ । ਵੱਖ-ਵੱਖ ਸਮਿਆਂ ਤੇ ਸ਼ਹਿਰ ਤੋਂ ਉਸਦੀ ਦੂਰੀ ਹੇਠਾਂ ਦਿੱਤੇ ਗਰਾਫ਼ਾਂ ਦੁਆਰਾ ਦਰਸਾਈ ਗਈ ਹੈ ।
(a) x-ਧੁਰੇ ਤੇ ਸਮਾਂ ਦਰਸਾਉਣ ਦੇ ਲਈ ਕੀ ਪੈਮਾਨਾ ਵਰਤਿਆ ਗਿਆ ਹੈ ?
(b) ਉਸਨੇ ਪੂਰੀ ਯਾਤਰਾ ਦੇ ਲਈ ਕਿੰਨਾ ਸਮਾਂ ਲਿਆ ?
(c) ਵਪਾਰੀ ਦੇ ਥਾਂ ਦੀ ਸ਼ਹਿਰ ਤੋਂ ਦੂਰੀ ਕਿੰਨੀ ਹੈ ?
(d) ਕੀ, ਡਾਕੀਆ ਰਸਤੇ ਵਿਚ ਕਿਤੇ ਰੁਕਿਆ ? ਵਿਸਥਾਰ ਨਾਲ ਦੱਸੋ ?
(e) ਕਿਸ ਅੰਤਰਾਲ ਵਿਚ ਉਸਦੀ ਚਾਲ ਸਭ ਤੋਂ ਜ਼ਿਆਦਾ ਸੀ ?
ਹੱਲ:
(a) 4 : ਇਕਾਈਆਂ = 1 ਘੰਟਾ
(b) 3\(\frac{1}{2}\) ਘੰਟੇ

(c) 22 km
(d) ਹਾਂ, ਇਹ ਗਰਾਫ ਖਿਤਿਜ ਭਾਗ ਨੂੰ ਦਰਸਾਉਂਦਾ ਹੈ । (10 ਵਜੇ ਸਵੇਰੇ ਤੋਂ 10.30 ਵੱਜੇ ਸਵੇਰ ਤੱਕ)
(e) 8 ਵਜੇ ਸਵੇਰੇ ਅਤੇ 9 ਵਜੇ ਸਵੇਰ ਦੇ ਵਿਚ

ਪਸ਼ਨ 7.
ਹੇਠਾਂ ਦਿੱਤੇ ਗਰਾਫਾਂ ਵਿਚ ਕਿਹੜੇ-ਕਿਹੜੇ ਗਰਾਫ ਸਮੇਂ ਅਤੇ ਤਾਪਮਾਨ ਦੇ ਵਿਚ ਸੰਭਵ ਹਨ ? ਤਰਕ ਦੇ ਨਾਲ ਆਪਣੇ ਉੱਤਰ ਦਿਓ ।

ਹੱਲ:
(iii) ਸੰਭਵ ਨਹੀਂ ਹੈ । ਕਿਉਂਕਿ ਇੱਕ ਹੀ ਸਮੇਂ ਵਿੱਚ ਤਾਪਮਾਨ ਦੇ ਵੱਖ-ਵੱਖ ਮੁੱਲ ਨਹੀਂ ਹੋ ਸਕਦੇ ।

Punjab State Board PSEB 5th Class Welcome Life Book Solutions Chapter 9 Right Solution to the Problem Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Welcome Life Chapter 9 Right Solution to the Problem

Welcome Life Guide for Class 5 PSEB Right Solution to the Problem Textbook Questions and Answers

(a) Let’s fly a kite

Today I am reading to you an incident of Hon’ble Teacher Mr. Sandeep Singh Bahlvi, published in ‘Bal Magazine’ …………..

You may have commonly seen flying kites. Many of you children will also be flying kites on holidays. During Childhood, I used to be like you. I also used to fly kites.

A day before spring, me and my friend Nima were flying kites. Other people were also flying kites on the roofs of the houses. We both were having fun.

Suddenly the string got entangled from Nimme. I wanted to leave more string but the string got entangled. All the fun was gritty. The kite could not be saved and got cut. I got very angry at Nimme. Nimme vented his anger on the wheel.

Now I think, that it was not of Nimme’s fault. There was no fault of wheel either. The fault was of the entanglement in the string, of the confusion. So, if there was no entangle¬ment then the kite would have kept flying in the sky.

Oral Questions:

Question 1.
From where did the teacher read the story to the children?
Answer:
Form the child magazine.

Question 2.
Who all were flying the kite?
Answer:
Author (Shri Sandeep Singh Bahlavi) and Nima.

Question 3.
Why was the children’s fun got spoiled?
Answer:
The kite was cut due to the entanglement of the string.

Question 4.
Whose fault was it?
Answer:
Entanglement of the string.

The children were told that Problems are part of life.

They are in every man’s life. They are for me also but there is a solution to every problem

So let’s sing
If the question is not understood, if the elder
child threatens, if any worry disturbs
Go to your teacher and get rid of the confusion
Neither be scared nor scare anyone.

(b) The solution to every problem

A picture of the Cabinet (Raj Darbar) will be displayed in front of the children. The faces shown in this picture will be anxious.

Five children will recite five verses
The first child – Let’s look at this picture The sitting king, the standing minister.

The second child – The problem is really serious. Sisters and brothers all together.
The third child – The whole assembly is in trouble. The courtier is astonished/ in wonder
The fourth child – All the gifted. Wise great scientists are welcome.

Fifth child – The solution to the problem will be found. Disaster will subside.

So kids – from this poetry lesson we have come to the conclusion that:
1. Even the greatest person can have a trouble/confusion.
2. In times of difficulty/confusion there is a need to discuss together.
3. Logic and Argument can solve complexities

Question 1.
Write about one of your problems in two lines.
Answer:
When it rains, all of my books and I get wet and so I have to take leave on that day.

Question 2.
What did you do to resolve your confusion?
Answer:

1. I started keeping my school bag in the big bag of polythene.
2. I used my father’s old raincoat while going to school oil rainy days. In this way, I solved my problem.

(C) Teacher’s note

Savijot’s is a bright student of class five. He has been frustrated for the last two months. His interest in studies has also decreased. His peers have also noticed his stress. He also remains absent from school.

Savijot’s conversation with the teacher :

Teacher: What’s the matter, you are no longer interested in Reading Corner?
Savijot’s : Sir I am short of time.

Teacher: How is that possible? At first, you were at the forefront reading every new book.
Savijot’s : Earlier it was fine sir, now it has become difficult due to household work.

Teacher : But tell me the reason?
Savijot’s : My father has been ill for the last three months.

Teacher : What happened?
Savijot’s : Sir, TB has been detected from the test report.

Teacher : Don’t be afraid, there is a cure for this disease. Patients get free medicine from government hospitals.
Savijot’s : Yes sir, we have just started treatment from there.

Teacher : Both the treatment and precaution are important.
Savijot : OK sir.

Teacher : Don’t worry about your studies. Your syllabus will be completed. Don’t be afraid of anything. Let me know if there is any need. You are our wise student.

Precautions: TB is a contagious disease. When caring for a patient, you should put a mask on your face. The patient’s clothes should be washed every day and the bed sheets should be changed. The patient should be given a balanced diet.

Oral Questions:

Question 1.
Why was Savijot’s upset?
Answer:
His father was suffering from T.B.

Question 2.
What was his father’s ailment?
Answer:
From T.B.

Question 3.
What did the teacher tell her?
Answer:
Do not be afraid, this disease has its treatment. Patients on get free medicine form government hospitals.

Question 4.
Name two infectious diseases
Answer:
T.B. and Corona.

Write some lines about Savijot’s in your diary

PSEB 5th Class Welcome Life Guide Right Solution to the Problem Important Questions and Answers

Multiple Choice Questions :

Question 1.
From which magazine was the published incident told by reading?
(a) From Child Magazine
(b) From Newspaper
(c) From Book
(d) From Novel.
Answer:
(a) From Child Magazine.

Question 2.
The incident, which was told, was written by whom?
(a) Gurdial Singh
(b) Sandeep Singh Bahlavi
(c) Amrita Singh
(d) Dushyant Kumar.
Answer:
(b) Sandeep Singh Bahlavi.

Question 3.
With whom was the author angry?
(a) With Nima
(b) With Himself
(c) With Kite
(d) With Neighbour.
Answer:
(a) With Nima.

Question 4.
The Kite was cut, who was responsible for this?
(a) Nima
(b) Charkhari
(c) Entanglement
(d) None of the above.
Answer:
(c) Entanglement.

Fill in the blanks :

1. Savijot is an intelligent ………………….. of class fifth.
2. The ………………….. of Savijot was suffering from the disease of T.B.
3. T.B. is a ………………….. disease.
Answer:
1. Student
2. father
3. communicable.

Tick Right (✓) or Wrong (✗) :

1. T.B. is not a communicable disease.
2. The patients get free medicine of T.B. from government hospitals.
3. The kite is cut. It was the fault of Nima.
Answer:
1. ✗
2. ✓
3. ✗

Mind Mapping :

Answer:

Match the following :

1. Basant (a) Reading comer
2. Kite was cut (b) Communicable disease
3. T.B. (c) Kite
4. Book Reading (d) Entanglement of the string.
Answer:
1. (c)
2. (d)
3. (b)
4. (a).

Short Answer Type Questions

Question 1.
If kite is cut, who is responsible for it?
Answer:
Entanglement of the string.

Question 2.
What was the problem in Savijot’s house?
Answer:
His father was suffering from T.B.

Question 3.
Where is T.B. treated?
Answer:
At government hospitals, without any cost.

Question 4.
Which type of food should be given to the patient of T.B.?
Answer:
He should be given balanced diet.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ InText Questions

ਕੋਸ਼ਿਸ਼ ਕਰੋ :

ਗੁਣਨਖੰਡ ਬਣਾਉ :

ਪ੍ਰਸ਼ਨ (i).
12x + 36
ਹੱਲ:
12x + 36 ਅਸੀਂ ਜਾਣਦੇ ਹਾਂ ਕਿ :
12x = 2 × 2 × 3 × x
ਪਦਾਂ 36 = 2 × 2 × 3 × 3
ਇਹਨਾਂ ਦੋਨਾਂ ਪਦਾਰਥ 2, 2 ਅਤੇ 3 ਸਾਂਝੇ ਗੁਣਨਖੰਡ ਹਨ।
ਇਸ ਲਈ 12x + 36 = (2 × 2 × 3 × x) + (2 × 2 × 3 × 3)
= 2 × 2 × 3 × [(x) + (3)]
= 12 × (x + 3) (ਪਦਾਂ ਨੂੰ ਮਿਲਾਉਣ ਤੇ)
= 12 (x + 3) (ਲੋੜੀਂਦਾ ਗੁਣਨਖੰਡ ਰੂਪ)

ਪ੍ਰਸ਼ਨ (ii).
22y – 33z
ਹੱਲ:
22y – 337
ਅਸੀਂ ਜਾਣਦੇ ਹਾਂ ਕਿ :
22y = 2 × 11 × y
33z = 3 × 11 × z
ਇਹਨਾਂ ਦੋਨਾਂ ਪਦਾਂ ਵਿਚ 11 ਸਾਂਝੇ ਗੁਣਨਖੰਡ ਹਨ ।
ਹੁਣ, 22y – 33z = 2 × 11 × y – 3 × 11 × z
= 11 × [(2 × y) – (3 × z)]
= 11 (2y – 3z)

ਪ੍ਰਸ਼ਨ (iii).
14pq + 35pqr.
ਹੱਲ:
14pq + 35pqr
ਅਸੀਂ ਜਾਣਦੇ ਹਾਂ ਕਿ :
14pq = 2 × 7 × p × q
35pqr = 5 × 7 × p × q × y
ਇਹਨਾਂ ਦੋਨਾਂ ਪਦਾਂ ਵਿਚ 7, p, q ਸਾਂਝੇ ਗੁਣਨਖੰਡ ਹਨ ।
14pq + 35pqr = 2 × 7 × p × q + 5 × 7 × p × q × r
= 7 × p × q [(2) + (5 + r)]
= 7pq (2 + 5r)

ਕੋਸ਼ਿਸ਼ ਕਰੋ :

ਭਾਗ ਦਿਉ :

ਪ੍ਰਸ਼ਨ (i).
24xy²z³ ਨੂੰ 6yz² ਨਾਲ
ਹੱਲ:
24xy²z³ = 2 × 2 × 2 × 3 × x × y × y × z × z × z
6yz² = 2 × 3 × y × z × z.
ਇਸ ਲਈ, (24xy²z³) ÷ (6yz²)
= \(\frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z}\)
= 2 × 2 × x × y × z = 4xyz

ਪ੍ਰਸ਼ਨ (ii).
63a²b⁴c⁶ ਨੂੰ 7a²b⁴c³ ਨਾਲ
ਹੱਲ:
63a²b⁴c⁶
= 3 × 3 × 7 × a × a × b × b × b × b × c × c × с × с × с × с
7a²b²c³
=7 × a × a × b × b × c × c × c
63a²b⁴c⁶ ÷ 7a²b²c³
= \(\frac{3 \times 3 \times 7 \times a \times a \times b \times b \times b \times b \times c \times c \times c \times c \times c \times c}{7 \times a \times b \times b \times c \times c \times c}\)
= 3 × 3 × b × b × c × c × c
= 9b²c³

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Exercise 14.4

ਹੇਠ ਲਿਖੇ ਗਣਿਤਿਕ ਕਥਨਾਂ ਵਿਚ ਗ਼ਲਤੀ ਪਤਾ ਕਰਕੇ ਉਸਨੂੰ ਸਹੀ ਕਰੋ :

ਪ੍ਰਸ਼ਨ 1.
4 (x – 5) = 4x – 5
ਹੱਲ:
L.H.S. = 4 (x – 5)
= 4 × x – 4 × 5
= 4x – 20.
ਜਦੋਂ ਅਸੀਂ ਕਿਸੇ ਵਿਅੰਜਕ ਨੂੰ ਉਸਦੇ ਕੋਸ਼ਠਕ ਦੇ ਬਾਹਰ ਲਿਖੇ ਅਚਲ ਜਾਂ ਚਲ) ਨਾਲ ਗੁਣਾ ਕਰਦੇ ਹਾਂ, ਤਾਂ ਵਿਅੰਜਕ ਦੇ ਹਰੇਕ ਪਦ ਨਾਲ ਉਸ ਅਚਰ (ਜਾਂ ਚਰ) ਨੂੰ ਗੁਣਾ ਕੀਤਾ ਜਾਂਦਾ ਹੈ ।
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
4 (x – 5) = 4x – 20

ਪ੍ਰਸ਼ਨ 2.
x (3x + 2) = 3x² + 2
ਹੱਲ:
L.H.S. = x (3x + 2) = x × 3x + x × 2
= 3x² + 2x
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ
x (3x + 2) = 3x² + 2x

ਪ੍ਰਸ਼ਨ 3.
2x + 3y = 5xy
ਹੱਲ:
L.H.S. = 2x + 3y = 2x + 3y
ਦੋ ਅਲੱਗ-ਅਲੱਗ ਚਰਾਂ ਨੂੰ ਜੋੜਿਆ ਨਹੀਂ ਜਾ ਸਕਦਾ ।
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ : 2x + 3y = 2x + 3y.

ਪ੍ਰਸ਼ਨ 4.
x + 2x + 3x = 5x
ਹੱਲ:
L.H.S. = x + 2x + 3x = 6
ਕਿਸੇ ਪਦ ਦੇ ਗੁਣਾਂਕ 1 ਨੂੰ ਦਰਸਾਇਆ ਨਹੀਂ ਜਾਂਦਾ ਹੈ। ਪਰ ਬਰਾਬਰ ਪਦਾਂ ਨੂੰ ਜੋੜਦੇ ਸਮੇਂ ਅਸੀਂ ਇਸਨੂੰ ਜੋੜ ਵਿਚ ਸ਼ਾਮਿਲ ਕਰਦੇ ਹਾਂ ।
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ : x + 2x + 3x = 6x

ਪ੍ਰਸ਼ਨ 5.
5y + 2y + y – 7y = 0
ਹੱਲ:
L.H.S. = 5y + 2y + y – 7y = 8y – 7y = y
ਕਿਸੇ ਪਦ ਦੇ ਗੁਣਾਂਕ 1 ਨੂੰ ਹਮੇਸ਼ਾ ਦਰਸਾਇਆ ਨਹੀਂ ਜਾਂਦਾ, ਪਰੰਤੁ ਬਰਾਬਰ ਪਦਾਂ ਨੂੰ ਜੋੜਦੇ ਸਮੇਂ ਅਸੀਂ ਇਸਨੂੰ ਜੋੜ ਵਿਚ ਸ਼ਾਮਿਲ ਕਰਦੇ ਹਾਂ ।
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ
5y + 2y + y – 7y = y

ਪ੍ਰਸ਼ਨ 6.
3x + 2x = 5x²
ਹੱਲ:
L.H.S. = 3x + 2x = 5
ਜਦੋਂ ਦੋ ਬਰਾਬਰ ਪਦਾਂ ਦੇ ਵਿਚ ਜਮਾਂ (+) ਦਾ ਨਿਸ਼ਾਨ ਹੋਵੇ ਤਾਂ ਉਹਨਾਂ ਨੂੰ ਜੋੜਿਆ ਜਾਂਦਾ ਹੈ ।
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
3x + 2 = 5x

ਪ੍ਰਸ਼ਨ 7.
(2x)² + 4 (2x + 7) = 2x² + 8x + 78.
ਹੱਲ:
LH.S. = (2x)² + 4 (2x) + 7
= 4x² + 8x + 7
ਜਦੋਂ ਅਸੀਂ ਕਿਸੇ ਇਕ-ਪਦੀ ਦਾ ਵਰਗ ਕਰਦੇ ਹਾਂ ਤਾਂ ਸੰਖਿਆਤਮਕ ਗੁਣਾਂ ਅਤੇ ਹਰੇਕ ਗੁਣਾਂ ਅਤੇ ਹਰੇਕ ਗੁਣਨਖੰਡ ਦਾ ਵਰਗ ਕੀਤਾ ਜਾਂਦਾ ਹੈ । ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
(2x)² + 4 (2x) + 7 = 4x² + 8x + 7

ਪ੍ਰਸ਼ਨ 8.
(2x)² + 5x = 4x + 5x = 9x
ਹੱਲ:
L.H.S. = (2x)² + 5x = 4x² + 5x
ਜਦੋਂ ਅਸੀਂ ਕਿਸੇ ਇਕ-ਪਦ ਦਾ ਵਰਗ’ ਕਰਦੇ ਹਾਂ ਤਾਂ ਸੰਖਿਆਤਮਕ ਗੁਣਾਂ ਅਤੇ ਹਰੇਕ ਗੁਣਨਖੰਡ ਦਾ ਵਰਗ ਕੀਤਾ ਜਾਂਦਾ ਹੈ ।
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
(2x)² + 5x = 4x² + 5x

ਪ੍ਰਸ਼ਨ 9.
(3x + 2)² = 3x² + 6x + 4
ਹੱਲ:
L.H.S. (3x + 2)² = (3x)² + 2 × 3x × 2 + (2)²
= 9x² + 12x + 4

ਪ੍ਰਸ਼ਨ 10.
x = – 3 ਮੁੱਲ ਭਰਨ ‘ਤੇ ਪ੍ਰਾਪਤ ਹੁੰਦਾ ਹੈ
(a) x² + 6x + 4 ਤੋਂ (-3)² + 5 (-3) + 4 = 9 + 2 + 4 = 15 ਪ੍ਰਾਪਤ ਹੁੰਦਾ ਹੈ ।
(b) x² – 5xr + 4 ਤੋਂ (-3)² – 5 (-3) + 4 = 9 – 15 + 4 = – 2 ਪ੍ਰਾਪਤ ਹੁੰਦਾ ਹੈ ।
(c) x² + 5x ਤੋਂ (-3)² + 5 (-3) = -9 – 15 = – 24 ਪ੍ਰਾਪਤ ਹੁੰਦਾ ਹੈ ।
ਹੱਲ:
(a) x² + 5x + 4 ਨਾਲ (-3)² + 5(-3) + 4
= 9 – 15 + 4 = 13 – 15 = – 2
(b) x² – 5x + 4 ਨਾਲ (-3)² – 5(-3) + 4 = 9 + 15 + 4 = 28
(c) x² + 5x ਨਾਲ (-3)² + 5 (-3) = 9 – 15 = -6

ਪ੍ਰਸ਼ਨ 11.
(y – 3)² = y² – 9
ਹੱਲ:
L.H.S. (y – 3)² = (y)² + 2 × y × (3) + (-3)²
= y² – 6 + 9
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
(y – 3)² = y² – 6y + 9

ਪ੍ਰਸ਼ਨ 12.
(z + 5)² = z² + 25
ਹੱਲ:
L.H.S. = (z + 5)² = (z)² + 2 × 2 × 5 + (5)²
= z² + 10z + 25
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
(z + 5)² = z² + 10z + 25

ਪ੍ਰਸ਼ਨ 13.
(2a + 3b) (a – b) = 2a²2 – 3b²
ਹੱਲ:
L.H.S. = (2a + 3b) (a – b) = 2a(a – b) + 3b (a – b)
2a² – 2ab + 3ab – 3b² = 2a² + ab – 3b²
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
(2a + 3b) (a – b) = 2a² + ab – 3b²

ਪ੍ਰਸ਼ਨ 14.
(a + 4) (a + 2) = a² + 8
ਹੱਲ:
L.H.S. = a +4) (a + 2) = a (a + 2) + 4
(a + 2) = a² + 2a + 4a + 8 = a² + 6a + 8.
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
(a + 4) (a + 2) = a² + 6a + 8

ਪ੍ਰਸ਼ਨ 15.
(a – 4) (a – 2) = a² – 8
ਹੱਲ:
L.H.S. = (a – 4) (a – 2) = a (a – 2) – 4(a – 2)
= a² – 2a – 4a + 8 = a² – 6a + 8
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
(a – 4) (a – 2) = a² – 6a + 8

ਪ੍ਰਸ਼ਨ 16.
\(\frac{3 x^{2}}{3 x^{2}}\) = 0
ਹੱਲ:
LH.S. = \(\frac{3 x^{2}}{3 x^{2}}\) = 1
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
\(\frac{3 x^{2}}{3 x^{2}}\) = 1

ਪ੍ਰਸ਼ਨ 17.
\(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + 1 = 2
ਹੱਲ:
L.H.S : \(\frac{3 x^{2}+1}{3 x^{2}}\) = \(\frac{3 x^{2}}{3 x^{2}}\) + \(\frac{1}{3 x^{2}}\)
= 1 + \(\frac{1}{3 x^{2}}\)
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
\(\frac{3 x^{2}+1}{3 x^{2}}\) = \(\frac{3 x^{2}}{3 x^{2}}\) + \(\frac{1}{3 x^{2}}\) = 1 + \(\frac{1}{3 x^{2}}\)

ਪ੍ਰਸ਼ਨ 18.
\(\frac{3x}{3x+2}\) = \(\frac{1}{2}\)
ਹੱਲ:
L.H.S. = \(\frac{3x}{3x+2}\)
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
\(\frac{3x}{3x+2}\) = \(\frac{3x}{3x+2}\)

ਪ੍ਰਸ਼ਨ 19.
\(\frac{3}{4x+3}\) = \(\frac{1}{4x}\)
ਹੱਲ:
L.H.S. = \(\frac{3}{4x+3}\)
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
\(\frac{3}{4x+3}\) = \(\frac{3}{4x+3}\)

ਪ੍ਰਸ਼ਨ 20.
\(\frac{4x+5}{4x}\) = 5
ਹੱਲ:
L.H.S. = \(\frac{4x+5}{4x}\) = \(\frac{4x}{4x}\) + \(\frac{5}{4x}\)
= 1 + \(\frac{5}{4x}\)
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
\(\frac{4x+5}{4x}\) = \(\frac{4x}{4x}\) + \(\frac{5}{4x}\) = 1 + \(\frac{5}{4x}\)

ਪ੍ਰਸ਼ਨ 21.
\(\frac{7x+5}{5}\) = 7x
ਹੱਲ:
L.H.S. = \(\frac{7x+5}{5}\) = \(\frac{7x}{5}\) + \(\frac{5}{5}\)
= \(\frac{7x}{5}\) + 1
ਇਸ ਲਈ, ਸਹੀ ਕਥਨ ਹੈ :
\(\frac{7x+5}{5}\) = \(\frac{7x}{5}\) + \(\frac{5}{5}\) = \(\frac{7x}{5}\) + 1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Exercise 14.3

1. ਹੇਠਾਂ ਲਿਖਿਆਂ ਦੀ ਵੰਡ ਕਰੋ :

ਪ੍ਰਸ਼ਨ (i).
28x⁴ ÷ 56x
ਹੱਲ:
28x⁴ ÷ 56x = \(\frac{28 x^{4}}{56 x}\)
= \(\frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x}\)
= \(\frac{1}{2}\)x³

ਪ੍ਰਸ਼ਨ (ii).
– 36y³ ÷ 9y²
ਹੱਲ:
– 36y³ ÷ 9y² = \(\frac{-36 y^{3}}{9 y^{2}}\)
= \(\frac{2 \times 2 \times 3 \times 3 \times y \times y \times y}{3 \times 3 \times y \times y}\)
= -2 × 2 × y = -4y

ਪ੍ਰਸ਼ਨ (iii).
66pq²r³ ÷ 11qr²
ਹੱਲ:
66pq²r³ ÷ 11qr² = \(\frac{66 p q^{2} r^{3}}{11 q r^{2}}\)
= \(\frac{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r}\)
= 2 × 3 × p × q × r
= 6pqr

ਪ੍ਰਸ਼ਨ (iv).
34x³y³z³ ÷ 51xy²z³
ਹੱਲ:
34x³y³z³ ÷ 51xy²z³ = \(\frac{34 x^{3} y^{3} z^{3}}{51 x y^{2} z^{3}}\)
= \(\frac{2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z}{3 \times 17 \times x \times y \times y \times z \times z \times z}\)
= \(\frac{2 \times x \times x \times y}{3}\) = \(\frac{2}{3}\)x²y

ਪ੍ਰਸ਼ਨ (v).
12a⁸b⁸ ÷ (-6a⁶b⁴).
ਹੱਲ:
12a⁸b⁸ ÷ (-6a⁶b⁴) = \(\frac{12 a^{8} b^{8}}{-6 a^{6} b^{4}}\)
= \(\begin{gathered}
2 \times 2 \times 3 \times a \times a \times a \times a \times a \times a \times \\
\frac{a \times a \times b \times b \times b \times b \times b \times b \times b \times b}{-2 \times 3 \times a \times a \times a \times a} \\
\times a \times a \times b \times b \times b \times b
\end{gathered}\)
= -2 × a × a × b × b × b × b
= -2a²b⁴

2. ਦਿੱਤੇ ਹੋਏ ਬਹੁਪਦ ਨੂੰ ਦਿੱਤੀ ਹੋਈ ਇਕ ਪਦੀ ਨਾਲ ਭਾਗ ਦਿਉ :

ਪ੍ਰਸ਼ਨ (i).
(5x² – 6x) ÷ 3
ਹੱਲ:
5x² – 6x = 5 × x × x – 2 × 3 × x
= x × (5 × x – 2 × 3)
= x × (5x – 6)
ਇਸ ਲਈ, (5x² – 6x) ÷ 3x = \(\frac{x \times(5 x-6)}{3 \times x}\)
= \(\frac{1}{3}\)(5x – 6)

ਪ੍ਰਸ਼ਨ (ii).
(3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
ਹੱਲ:
3y⁸ – 4y⁶ + 5y⁴ = y⁴(3y⁴ – 4y² + 5)
ਇਸ ਲਈ, (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
= \(\frac{\left(3 y^{8}-4 y^{6}+5 y^{4}\right)}{y^{4}}\) = \(\frac{y^{4}\left(3 y^{4}-4 y^{2}+5\right)}{y^{4}}\)
= 3y⁴ – 4y² + 5

ਪ੍ਰਸ਼ਨ (iii).
8 (x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z²
ਹੱਲ:
8 (x³y²z² + x²y³z² + x²y²z³) .
= 2 × 2 × 2 × x² × y² × z² (x+y + z)
ਇਸ ਲਈ, 8 (x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z²
= \(\frac{2 \times 2 \times 2 \times x^{2} \times y^{2} \times z^{2} \times(x+y+z)}{2 \times 2 \times x^{2} \times y^{2} \times z^{2}}\)
= 2 (x + y + z)

ਪ੍ਰਸ਼ਨ (iv).
(x³ + 2x² + 3x) ÷ 2x
ਹੱਲ:
x³ + 2x² + 3x = x × x × x + 2 × x × x + 3 × x
= x × (x × x + 2 × x + 3)
= x × (x² + 2x + 3)
ਇਸ ਲਈ, (x³ + 2x² + 3x) ÷ 2x
= \(\frac{x \times\left(x^{2}+2 x+3\right)}{2 x}\) = \(\frac{1}{2}\)(x² + 2x + 3)

ਪ੍ਰਸ਼ਨ (v).
(p³q⁶ – p⁶q³) ÷ p³q³.
ਹੱਲ:
p³q⁶ – p⁶q³ = p³ × q³ × q³ – p³ × p³ × q³ = p³ × q³ × (q³ – p³)
ਇਸ ਲਈ, (p³q⁶ – p⁶q³) ÷ p³q³
= \(\frac{p^{3} \times q^{3} \times\left(q^{3}-p^{3}\right)}{p^{3} \times q^{3}}\) = q³ – p³

3. ਹੇਠਾਂ ਲਿਖਿਆ ਦੀ ਵੰਡ ਕਰੇ :

ਪ੍ਰਸ਼ਨ (i).
(10x – 25) ÷ 5
ਹੱਲ:
10x – 25 = 2 × 5 × x -5 × 5
= 5 (2 × x – 5)
= 5 × (2x – 5)
ਇਸ ਲਈ, (10x – 25) ÷ 5 = \(\frac{10 x-25}{5}\)
= \(\frac{5×(2x-5)}{5}\) = 2x – 5

ਪ੍ਰਸ਼ਨ (ii).
(10x – 25) ÷ (2x – 5)
ਹੱਲ:
10x – 25 = 2 × 5 × x – 5 × 5
= 5 × (2 × x – 5) = 5 × (2x – 5)
ਇਸ ਲਈ, (10x – 25) ÷ 2x – 5
\(\frac{10 x-25}{2x-5}\) = \(\frac{5×(2x-5)}{(2x-5)}\)

ਪ੍ਰਸ਼ਨ (iii).
10y (6y + 21) ÷ 5 (2y + 7)
ਹੱਲ:
10y (6y + 21) = 2 × 5 × y × (2 × 3 × y + 3 × 7)
= 2 × 5 × y × 3 × (2 × y + 7)
= 2 × 5 × 3 × y (2y + 7)
ਇਸ ਲਈ, 10y (6y + 21) ÷ 5 (2y + 7)
= \(\frac{10 y(6 y+21)}{5(2 y+7)}\)
= \(\frac{2 \times 5 \times 3 \times y \times(2 y+7)}{5 \times(2 y+7)}\)
= 2 × 3 × y = 6y

ਪ੍ਰਸ਼ਨ (iv).
9x²y² (3z – 24) ÷ 27xy (z – 8)
ਹੱਲ:
9x²y² (3z – 24)
= 3 × 3 × x × x × y × y × (3 × 2 = 3 × 8)
= 3 × 3 × 3 × x × x × y × y × (z – 8)
ਇਸ ਲਈ, 9x²y²(3z – 24) = 27xy (z – 8)
= \(\frac{9 x^{2} y^{2}(3 z-24)}{27 x y(z-8)}\)
= \(\frac{3 \times 3 \times 3 \times x \times x \times y \times y \times(z-8)}{3 \times 3 \times 3 \times x \times y \times(z-8)}\)
= x × y = xy

ਪ੍ਰਸ਼ਨ (v).
96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6).
ਹੱਲ:
96abc (3a – 12) (5b – 30)
= 2 × 2 × 2 × 2 × 2 × 3 × a × b × c × (3 × a – 3 × 4) (5 × b – 5 × 6)
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × a × b × c × (a – 4) × 5 × (b – 6)
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × a × b × c (a – 4) (b – 6)
ਇਸ ਲਈ, 96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)
= \(\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)}\)
= \(\begin{gathered}
2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \\
\times a \times b \times c \times(a-4) \times(b-6) \\
\hline 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times(a-4)(b-6)
\end{gathered}\)
= 2 × 5 × a × b × c.
= 10abc

4. ਨਿਜਦੋਸ ਆਨੁਮਾਰ ਭਾਗ ਕਰੇ :

ਪ੍ਰਸ਼ਨ (i).
5 (2x + 1) (3x + 5) ÷ (2x + 1)
ਹੱਲ:
5 (2x + 1) (3x + 5) ÷ (2x + 1)
= \(\frac{5(2x+1)(3x+5)}{2x+1}\)
= 5 (3x + 5)

ਪ੍ਰਸ਼ਨ (ii).
26xy (x + 5) (y – 4) ÷ 13x (y – 4)
ਹੱਲ:
26xy (x + 5) (y – 4) ÷ 13x (y – 4)
= \(\frac{26 x y(x+5)(y-4)}{13 x(y-4)}\)
= \(\frac{2 \times 13 \times x \times y \times(x+5) \times(y-4)}{13 \times x \times(y-4)}\)
= 2 × y × (x + 5) = 2y (x + 5)

ਪ੍ਰਸ਼ਨ (iii).
52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
ਹੱਲ:
52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)
= \(\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r) \times(r+p)}\)
= \(\frac{2 \times 2 \times 13 \times p \times q \times r \times(p+q) \times(q+r) \times(r \times p)}{2 \times 2 \times 2 \times 13 \times p \times q \times(q+r) \times(r+p)}\)
= \(\frac{1}{2}\) × r × (p + q) = \(\frac{1}{2}\)r(p + q)

ਪ੍ਰਸ਼ਨ (iv).
20 (y + 4) (y² + 5y + 3) ÷ 5 (y + 4)
ਹੱਲ:
20 (y + 4) (y² + 5y + 3) ÷ 5 (y + 4)
= \(\frac{20(y+4)\left(y^{2}+5 y+3\right)}{5 \times(y+4)}\)
= \(\frac{5 \times 4 \times(y+4) \times\left(y^{2}+5 y+3\right)}{5 \times(y+4)}\)
= 4 × (y² + 5y + 3) = 4 (y² + 5y + 3)

ਪ੍ਰਸ਼ਨ (v).
x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
ਹੱਲ:
x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
= \(\frac{x \times(x+1) \times(x+2) \times(x+3)}{x \times(x+1)}\)
= (x + 2) (x + 3)

5. ਵਿਅੰਜਕ ਦੇ ਗੁਣਨਖੰਡ ਕਰੋ ਅਤੇ ਨਿਰਦੇਸ਼ ਅਨੁਸਾਰ ਭਾਗ ਕਰੇ :

ਪ੍ਰਸ਼ਨ (i).
(y² + 7y + 10) ÷ (y + 5)
ਹੱਲ:
(i) y² + 7y + 10
= y² + 2y + 5y + 10
= y(y + 2) + 5 (y + 2)
= (y + 2) (y + 5)
ਇਸ ਲਈ, (y² + 7y + 10) ÷ (y + 5)
= \(\frac{(y+2)(y+5)}{(y+5)}\) = y + 2

ਪ੍ਰਸ਼ਨ (ii).
(m² – 14m – 32) ÷ (m + 2)
ਹੱਲ:
m² – 14m – 32
= m² – 16m + 2m = 32
= m (m – 16) + 2 (m – 16)
= (m – 16) (m + 2)
ਇਸ ਲਈ, (m² – 14m – 32) ÷ (m + 2)
= \(\frac{m^{2}-14 m-32}{m+2}\)
= \(\frac{(m-16)(m+2)}{(m+2)}\)
= m – 16

ਪ੍ਰਸ਼ਨ (iii).
(5p² – 25p + 20) ÷ (p – 1)
ਹੱਲ:
5p² – 25p + 20 = 5 × (p² – 5p + 4)
= 5 × [p² – 4p – p + 4]
= 5 × [p (p – 4) – 1 (p – 4)]
= 5 × (p – 4) (p – 1)
ਇਸ ਲਈ, 5p² – 25p + 20 ÷ (p – 1)
= \(\frac{5 p^{2}-20 p+20}{p-1}\)
= \(\frac{5 \times(p-4)(p-1)}{(p-1)}\)
= 5 × (p – 4) = 5(p – 4)

ਪ੍ਰਸ਼ਨ (iv).
4yz (z² + 6z – 16) ÷ 2y (z + 8)
ਹੱਲ:
4yz (z² + 6z – 16) = 2 × 2 × y × z × (z² + 8z – 27 – 16) = 2 × 2 × y × z × [z (z + 8) – 2 (z + 8)]
= 2 × 2 × y × z × (z + 8) (z – 2)
ਇਸ ਲਈ, 4yz (z² + 6z – 16) ÷ 2y (2 + 8)
= \(\frac{4 y z\left(z^{2}+6 z-16\right)}{2 y(z+8)}\)
= \(\frac{2 \times 2 \times y \times z \times(z+8)(z-2)}{2 \times y \times(z+8)}\)
= 2 × z × (z – 2)
= 2z (z – 2)

ਪ੍ਰਸ਼ਨ (v).
5pq (p² – q²) ÷ 2p (p + q)
ਹੱਲ:
5pq (p² – q²) = 5 × p × q × (p + q) (p – q)
ਇਸ ਲਈ, 5pq (p² – q²) ÷ 2p (p + q)
= \(\frac{5 p q\left(p^{2}-q^{2}\right)}{2 p(p+q)}\)
= \(\frac{5 \times p \times q \times(p+q)(p-q)}{2 \times p \times(q+q)}\)
= \(\frac{5}{2}\)q (p – q)

ਪ੍ਰਸ਼ਨ (vi).
12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
ਹੱਲ:
12xy (9x² – 16y²) = 2 × 2 × 3 × x × y × [(3x)² – (4y)²]
= 2 × 2 × 3 × x × y × (3x + 4y) (3x – 4y)
ਇਸ ਲਈ, 12xy (9x² – 16y²) ÷ 4y (3x + 4y)
= \(\frac{12 x y\left(9 x^{2}-16 y^{2}\right)}{4 x y(3 x+4 y)}\)
= \(\frac{2 \times 2 \times 3 \times x \times y \times(3 x+4 y)(3 x-4 y)}{2 \times 2 \times x \times y \times(3 x+4 y)}\)
= 3 (3x – 4y)

ਪ੍ਰਸ਼ਨ (vii).
39y³(50y² – 98) ÷ 26y²(5y + 7)
ਹੱਲ:
39y³(50y² – 98)
= 3 × 13 × y × y × y × (2 × 5 × 5 × y × y – 2 × 7 × 7)
= 3 × 13 × y × y × y × 2 × (5y × 5y – 7 × 7)
= 2 × 3 × 13 × y × y × y × (5y + 7) (5y – 7)
ਇਸ ਲਈ, 39y (50y³ – 98) ÷ 26y²(5y + 7)
= \(\frac{39 y^{3}\left(50 y^{2}-98\right)}{26 y^{2}(5 y+7)}\)
= \(\frac{2 \times 3 \times 13 \times y \times y \times y \times(5 y+7)(5 y-7)}{2 \times 13 \times y \times y \times(5 y+7)}\)
= 3 × y × (5y – 7) = 3y (5y – 7)

Punjab State Board PSEB 5th Class Welcome Life Book Solutions Chapter 8 Self-Defence Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Welcome Life Chapter 8 Self-Defence

Welcome Life Guide for Class 5 PSEB Self-Defence Textbook Questions and Answers

(a) Through the game of Karate

Self-defense means protection of oneself. If anybody attacks you how will you defend yourself from that attack?

Children, do you know, in last few days girls were given training of Karate in the school? It was given so that girls can defend themselves after learning Karate. Never depend on anybody.

Oral Questions:

Question 1.
What is self-defense?
Answer:
The meaning of self-protection is to protect ourselves.

Question 2.
Have you seen a game of Karate?
Answer:
Yes, I have seen.

Question 3.
Why Karate is needed?
Answer:
Karate is the art of fighting with hands without using any weapon. We do not keep weapons all the time with us and so there is a need to learn Karate for using hands as weapons.

Teacher will encourage other children for self-defense by action of some girls.

(b) Through the game of Fencing

Fencing is a very traditional game of the world. In old days many battles were won on the basis of swordsmanship. This game needs very hard work to be perfect. The ground of fencing is called “piste”. This game is also played on indoor ground or outdoor ground.

Both the swordsmen attack each other and use shield for their defence. You must have watched the skill/art of fencing in the serials like “The Ramayana’and The Mahabharataon T.V.

Oral Questions:

Question 1.
What is called the ground of fencing?
Answer:
The ground of swordplay is called Piste.

Question 2.
Where did you see fencing?
Answer:
In the programmes of T.V. like Ramayana and Mahabharata.

Question 3.
Which thing is used to defend attack of fencing?
Answer:
For this, shield is used.

(c) Through the game of “Gatka’/stick fighting

With the help of a poster, Teacher will tell the students how you can defend yourself during the game of Gatka and swordsmanship, It will be told to the students in simple way.

Teacher will tell that for sports where you require sportsmanship, you also need self -confidence and these achievements can not be attained without self¬confidence.

(Gatka is the art of war of Punjabis in which it is art of fighting against enemy during cease fire. Any male/female can take its training. Nihang Sikhs are perfect in this art. Art of weapon is one of the unlimited arts. Gatka is the one of these styles which is played in Punjab and many parts of Northern India. Highlights of this can be seen on HolaMohalla at Anandpur Sahib. Game of Gatka is a game of self-defence as similar to Japan’s game Karate. A Small shield and 3.5 palm long stick is used in Gatka (stick-fighting).

Oral Question:

Question 1.
Where have you seen the game of Gatka or stick-fighting?
Answer:
On the occasion of Hola-Mohalla at Shri Anandpur Sahib.

Question 2.
Who can play the game of Gatka or stick-fighting?
Answer:
Any man or woman can play this sport.

(d) Good Touch, Bad Touch

Teacher will give information to the students about good-bad touch with the help of the poster of body parts.

Oral Questions:

Question 1.
How do you feel when anybody touches you?
Answer:
If some of our family members touches us then we feel good but if some outsider touches us, we feel bad and get angry also.

Question 2.
What will you do when anybody touches you?
Answer:
We will try to prevent ourselves from him and will also inform our parents.

Question 3.
How do you deal in the market/ fairs?
Answer:
It is required to visit market/fairs very carefully so that nobody can come near us. They can touch us with bad intention and can steal our things also. So there is a need to be careful by ourselves and should remain with our parents.

(e) Conversation

Teacher : “Doyougoto the market /fairs or festivals and religious places?
Girls : “Yes, madam.

Teacher : “You must collide with each other at such places?
Girls : “There is a lot of scrimmages while watching glimpses.

Teacher : “There are molesters at such places who tease or molest girls, young girls or women.
Girls : (They are listening quietly while looking at teacher’s face).

Teacher : “If someone touches you except your parents, how do you feel?
Girls : “(Answer collectively)” Very bad.

Teacher : “Well done ! If you feel bad then that touch is not good then what should do?
Girls : All Quiet.

Teacher : “Never be frightened. Do not get nervous also. Tell your parents and start shouting. Do you understand my point?

Oral Question :

Question 1.
Do you go to see glimpses?
Answer:
Yes, a number of times.

Question 2.
Should you trust on strangers?
Answer:
No, not at all.

Question 3.
What will you do if someone touches you?
Answer:
We will tell our parents and create noise.

Teacher will interact with girls in a friendly environment and guide how to protect oneself from bad touch of anybody.

(f) Well Done! Minni

Mini reads in class 5. She is the monitor of the class. She is a good Kho- Kho player. In school, children play games like Kho-Kho and Kabaddi. Boys and girls play together in it.

It was Sunday.The Ramayana serial was going on. Mini was sitting quietly in front of the TV. Mother asked, “Mini is anything hurting?”

“No, Mom “while switching off the TV, Mini answered in a quiet voice.

“Then why is my butterfly sitting withered today?” Mother put her hand on her shoulder and said.

“Mom tell me! “After thin king for a while, Mini said, “Yes, say my love!” Mom said holding her hand lovingly!

“Now you put your hand on my shoulder then nothing happened.” Mini was talking intermittently as if remembering something.

(Mummy was looking surprisingly at her face)

“Yesterday while playing, when a boy put his hand on my shoulder, I didn’t like it. And after that I couldn’t play.” Minnie looked at her mother for answer.

This was the reason for your silence. The mother hugged Minnie and said, “My queen daughter, when I touch you, whether it is your shoulder, mouth or hand, nothing happens to you because I am your mother. You know this touch since your childhood So you like it.

Oh mom! Minnie said hugging. “Dear Minnie! Do you feel bad when a stranger touches you at these places!” The mother asked by looking at Minnie.

“Too much!” Minnie made a bad face and said, “Yes, dear! That means that touch is not good, it’s bad”. Mom taught,” Dear Minnie, if any touch that doesn’t feel good, makes you uncomfortable, is bad. If someone touches you in a bad way, tell your teacher, Father or me, but don’t be silent.

“Don’t worry mother! Your Mini is not so weak to let go the wrong toucher, as lama player” Mini said. She said laughingly while switching on T.V. “Well done Minnie!” The mother said taking Mini in her lap.

Oral questions :

Question 1.
Do you listen to stories from your elders?
Answer:
Yes, a number of times.

Question 2.
Which game do you play in the school?
Answer:
Kho-Kho and Kabaddi.

Question 3.
Do you support at home?
Answer:
Yes.

Question 4.
Which serial do you watch on TV?
Answer:
Ramayana.

(The teacher will inspire the girls with a story.)

(G) Protection from Pandemic

Any disease that affects people from different countries all over the world is called pandemic. For example Plague, Chicken pox, Cholera and Corona etc. Nowadays, Corona has turned into a pandemic.

Teacher will tell the children about Corona virus its symptoms and remedies in detail.

★ Teacher will explain this disease by an activity.For example, make a circle of the children. The children will stand behind each other. The circle will be filled by children.
★ Teacher will ask a child to push. When the child will push another child then the child standing in the front will fall. Thus, all the students standing in the circle will fall down.

(Teacher will tell that Corona is a viral disease and spreads from person to person. Teacher will also tell about prevention of it.)

Teacher will tell the children again to stand in a circle and repeat the same activity. The child will push again. All the children standing in circle will fall down.

Teacher will pick out a child from the circle holding his/her arm. Now children will stop falling.

(Teacher will tell that in this disease social and physical distancing is a must otherwise, this chain will continue. Teacher will tell that in order to break this chain we should have distance from others, wear mask, cover mouth with cloth or handkerchief while sneezing, keep your surroundings clean and wash hands for at least 20 seconds.)

Wear mask
Wash hands properly multiple times
Keep physical/ social distancing

Oral Question :

Question 1.
Do you know about pandemics?
Answer:
Yes, with this a number of people get sick in a very short period. I know the name of that disease – Corona.

Question 2.
Tell me the name of any pandemic?
Answer:
Yes, with this a number of people get sick in a very short period. I know the name of that disease – Corona.

Question 3.
What can we do to be safe from COVID-19?
Answer:
Cover your nose and mouth with masks, wash your hands again and again, keep social distancing at least 1 kilometre far away from each other. Keep your surroundings clean.

PSEB 5th Class Welcome Life Guide Self-Defence Important Questions and Answers

Multiple Choice Questions :

Question 1.
How can we do self-defence?
(a) Karate
(b) Swordplay
(c) Gataka
(d) All are correct.
Answer:
(d) All are correct.

Question 2.
What do we call the ground of swordplay?
(a) Piste
(b) Shield .
(c) Astroturf
(d) None of the above.
Answer:
(a) Piste

Question 3.
The correct statement is :
(a) Self-defence means to protect ourselves.
(b) Swordplay is the oldest sport in the world.
(c) Gataka is a fatal art.
(d) All are correct.
Answer:
(d) All are correct.

Question 4.
The wrong statement is :
(a) Nihang Singhs are experts in the art of Gataka.
(b) Only men can learn Gataka.
(c) Gataka is played in Punjab.
(d) Karate is a weapon-art of Japan.
Answer:
(b) Only men can learn Gataka.

Question 5.
The weapon-art of Karate is related to which country?
(a) Japan
(b) India
(c) Kerala
(d) Norway
Answer:
(a) Japan.

Question 6.
Which of the following is pande-mic?
(a) Cholera
(b) Plague
(c) Corona
(d) All are correct.
Answer:
(d) All are correct.

Question 7.
What should we do to protect ourselves from Corona?
(a) To wear masks
(b) To wash hands again and again
(c) Keep Social distancing
(d) Adopt all the above measures.
Answer:
(d) Adopt all the above measures.

Question 8.
Which game is related to self-defence?
(a) Carom Board
(b) Karate
(c) Hockey
(d) None of the above.
Answer:
(b) Karate.

Fill in the blanks :

1. In Gataka, a of the length …………………………………… of three and a half hands and small shield is used.
2. Karate is a weapon-art of ………………………………….. .
3. If someone …………………………………… us with bad intention then we should tell our parents.
4. Mother asked why her …………………………………… is withered today.
5. The disease of …………………………………… is a pandemic.
Answer:
1. long stick
2. Japan
3. touches
4. butterfly
5. Corona.

Tick Right (✓) or Wrong (✗) :

1. If someone touches us with bad intention then we should create noise.
2. We should not shake hands with one-another to protect ourselves from Corona.
3. Mini was the monitor of fifth class.
Answer:
1. ✓
2. ✓
3. ✓

Mind Mapping :

Answer:

Match the following :

1. Karate – (a) Punjab
2. Gataka – (b) Pandemic
3. Corona – (c) Piste
4. Swordplay – (d) Japan
Answer:
1. (d)
2. (a)
3. (b)
4. (c).

Short Answer Type Questions

Question 1.
Tell the names of some weapon- arts.
Answer:
Karate, Swordplay, Gataka.

Question 2.
Gataka is the art of which place?
Answer:
Gataka is played in Punjab and some areas of North India.

Question 3.
What is used to play Gataka?
Answer:
A long stick of three and a half hands and small shield is used.

Question 4.
Where is the sport of swordplay played?
Answer:
Swordplay can be played in the open ground or closed ground.

Question 5.
What should we need to do to avoid corona pandemic?
Answer:
Wear mask, wash hands again and again, mention distance from each other, don’t shake hands with others, don’t touch the someone things.

Question 6.
If someone touches you with bad intention, then what would you do?
Answer:
We will tell our parents and create noise.

Punjab State Board PSEB 5th Class Welcome Life Book Solutions Chapter 7 Cooperation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Welcome Life Chapter 7 Cooperation

Welcome Life Guide for Class 5 PSEB Cooperation Textbook Questions and Answers

(a) Help and be good :

Story : once a father told his two sons that he wants to see, which one of them is good. I give you one year for this work. During this year you have to do good deeds. After one year , I will see which work you did. Those works will reveal your goodness’. He said.

Following their father’s orders, both of them left to do their works.

The elder son thought that,” I will earn a lot of money . Father will get very happy after looking at my earned money and he will award me of being a better son.”

He opened a small shop in the village and kept on selling cheap products at high rates. He made good money in one year.

The younger son left for the woods, empty-handed. He kept on walking till evening. He met an old man and woman on the way. Both of them were crying.

The younger boy asked old men and women with love, ” Babaji why are you crying ?”

“Our young son died of a poisonous snake bite. He was our only support. Now we will die hungry.” The old man started crying very loudly.

“Babaji , please don’t cry. Consider me as your Son only. I will help you.” The boy took the old man and woman to their house. He looked after their farm and crop for the entire year. When the crop ripened , he left the grains at their house. He remembered that he had to meet his father after one year. He left for his village, after telling the old man and woman that he will come back before the time of planting the next crop.

The father called both the boys together, in the evening and asked about the works done by them. The elder son said, “I had opened a shop and now I have earned a lot of money.”

The younger son said, “Father I have not earned any money. I found an old man and woman crying on the way. They looked like my own parents tome. But by savingtheir lives I do have earned their blessings.”

The father was very happy with the younger son. He said, “son, everybody is earning money in the world but our real earning is our goodness. By doing a good deed , the fame earned by you is far better than any income in the world and is priceless. In this way the younger brother has more goodness than the elderone. He is good.”

Then, putting his hand over elder brother’s head, father said,” Son, you also have to do good deeds like your younger brother and be good like him.”

Questions/Answers :

Question 1.
What test did the father want to take of his sons ?
Answer:
On Goodness.

Question 2.
What work did the elder son do?
Answer:
He opened a shop and earned a lot of money.

Question 3.
What work did the younger son do?
Answer:
He served an old woman and man and also looked after their agriculture.

Question 4.
Which son’s work did the father like and why ?
Answer:
Father liked the profession of younger son because he had served an old man and woman earned blessings.

Question 5.
What did the father explain to both the sons ?

Note : The teacher will make the students do the following activities. More of such activities can be done depending on the time :

Activity :

(a) Elder children can be made to cover the books and notebooks of younger children
(b) Elder children can teach the younger children.

(b) How to help?

(The teacher can share some incident, related to ‘helping out’ from his or from someone’s life or can narrate the following incident orally)

It had been 2-3 days of my teaching in the primary school. Winter season was coming to a close. One day ,it rained a lot and it got cold again. We raised the children not wearing the sweaters and send them home to come back, wearing one. Two girls were still standing.

“Girls, you also go home and come back after wearing sweater”

” we don’t have them sir”

“Never mind if you don’t have one of your uniform, wear any other sweater.” I said again.

” we don’t have any sweater.” One of the two girls said.

“Girls, you kept coming like this all year!”

They used to wrap shawls around them in such a way that it was not visible whether they were wearing a sweater below it or not. Both the girls did not speak a word.

I asked them again,” children you spent the entire year like this. Why did you not tell earlier.? I felt very sad.

Next day I brought two most beautiful sweaters for them. I called the girls to office and said, “Take this girls, I have brought the most beautiful sweaters for you and now in return you both try to fetch good marks for me , yourself and your family.”

When the result of examination was declared Both the girls came at first and second position.

Questions/Answers

Question 1.
Have you ever helped anybody?
Answer:
Yes, I have done.

Question 2.
Tell about any help that you did of anybody?
Answer:
I had helped an old lady to cross the road and had informed through phone to her housemates that she is alright. I had made her reach towards her house near the street.

(C) How to co-operate

Poem

Mother cooks food
I serve it.
Papa come home from work
I take water for him
Papa feels happy
And carries me on his shoulder.
This is how I co-operate

If my grandpa has to go somewhere
Orneedstooil his hair
Orheneedsto wear a new suit
I get his things for him.
This is how I co-operate

If the teacher has to teach something
Ortoputusto study
Needs to check our notebooks
Or requires someone to read the lesson
I read it loudly
This is how I co-operate

Questions/Answers :

Question 1.
How does the child co-operate at home?
Answer:
By giving utensils to mother in the kitchen and water to the father.

Question 2.
How does the child co-operate at school?
Answer:
The child reads the chapter loudly in the class to make others listen.

Question 3.
How does he serve his grandfather?
Answer:
The child serves his grandfather by massaging his hair with oil and giving clothes.

Question 4.
Does he co-operate with his family or with everybody?
Answer:
The child cooperates with all either relatives or other people.

Note : The teacher will tell the students that you also have to co-operate with everybody and tell me who all you cooperated with and how.

PSEB 5th Class Welcome Life Guide Cooperation Important Questions and Answers

Multiple Choice Questions :

Question 1.
In chapter ‘Cooperation’, what did the elder brother think?
(a) To earn money
(b) To go abroad
(c) To remain idle
(d) None of the above.
Answer:
(a) To earn money.

Question 2.
How did the son of old man and woman die?
(a) By becoming sick
(b) By sting of snake
(c) By accident
(d) All are wrong.
Answer:
(b) By sting of snake.

Question 3.
Why did the two girls remain standing?
(a) They had no sweaters.
(b) Their houses were very far.
(c) They Were afraid of going to the home.
(d) All are wrong.
Answer:
(a) They had no sweaters.

Question 4.
Those children, who had hot worn sweaters, were
(a) sent to homes
(b) given punishment
(c) were sent out of the class
(d) All are correct
Answer:
(a) sent to homes.

Fill in the blanks :

1. ‘The first son thought a lot about …………………………………… .
2. That boy took the old man and woman to …………………………………… .
3. Next day, I brought two beautiful …………………………………… and gave to the girls.
Answer:
1. earning money
2. their home
3. sweaters.

Tick Right (✓) or Wrong (✗) :

1. Elder son did not do any work.
2. Younger son helped the old man and woman in agriculture.
3. Father was pleased with the work of other son.
4. Elder chidren should provide help in studies to little children.
Answer:
1. ✗
2. ✓
3. ✓
4. ✓

Mind Mapping :

Answer:

Match the following :

1. Elder Son (a) Serving old man and woman
2. Little Son (b) Children’s going home
3. Raining (c) Having no sweaters
4. Two girls (d) Earning money
Answer:
1. (d)
2. (a)
3. (b)
4. (c).

Short Answer Type Questions

Question 1.
How much time did the father give to both the sons?
Answer:
Father gave the time of one year.

Question 2.
What did the elder son think to do?
Answer:
He thought about earning a lot of money.

Question 3.
What did the elder brother do to earn money?
Answer:
He opened a shop.

Question 4.
Why were old man and woman were weeping on their way?
Answer:
Because their young boy had died by the sting of the snake.

Question 5.
What did the father say about goodness?
Answer:
He said that money is collected by all the people in the world but real earning is your goodness.

Question 6.
Write a few lines about the help being provided in the following pictures.
Answer:

• In the first picture, a child is helping an old man to cross the road.
• In the second picture, two old men are being provided with things of need.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 ਗੁਣਨਖੰਡੀਕਰਨ Exercise 14.1

1. ਦਿੱਤੇ ਹੋਏ ਪਦਾਂ ਵਿਚ ਸਾਂਝਾ ਗੁਣਨਖੰਡ ਪਤਾ ਕਰੋ :

ਪ੍ਰਸ਼ਨ (i).
12x, 36
ਹੱਲ:
12x, 36
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
∴ ਸਾਂਝਾ ਗੁਣਨਖੰਡ = 2 × 2 × 3 = 2

ਪ੍ਰਸ਼ਨ (ii).
2y, 22xy
ਹੱਲ:
2y, 22xy
2y = 2 × y
22xy = 2 × 11 × x × y
∴ ਸਾਂਝਾ ਗੁਣਨਖੰਡ = 2 × y = 2y

ਪ੍ਰਸ਼ਨ (iii).
14pq, 28p²q²
ਹੱਲ:
14pq, 28p²q²
14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 7 × p × p × q × q
∴ ਸਾਂਝਾ ਗੁਣਨਖੰਡ = 2 × 7 × p × q = 14pq

ਪ੍ਰਸ਼ਨ (iv).
2x, 3y², 4
ਹੱਲ:
2x, 3y², 4
2x = 2 × x
3y² = 3 × y × y
4 = 2 × 2
∴ ਸਾਂਝਾ ਗੁਣਨਖੰਡ = 1

ਪ੍ਰਸ਼ਨ (v).
6abc, 24ab², 12a²b
ਹੱਲ:
6abc, 24ab², 12a²b
6abc = 2 × 3 × a × b × c.
24ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
∴ ਸਾਂਝਾ ਗੁਣਨਖੰਡ = 2 × 3 × a × b = 6ab

ਪ੍ਰਸ਼ਨ (vi).
16x³, -4x², 32x
ਹੱਲ:
16x³, -4x², 32x³
16x³ = 2 × 2 × 2 × 2 × x × x × x
-4x² = – 1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ ਸਾਂਝਾ ਗੁਣਨਖੰਡ = 2 × 2 × x × x = 4x

ਪ੍ਰਸ਼ਨ (vii).
10pq, 20qr, 30rp
ਹੱਲ:
10pq, 20qr, 30rp
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
∴ ਸਾਂਝਾ ਗੁਣਨਖੰਡ = 2 × 5 = 10

ਪ੍ਰਸ਼ਨ (viii).
3x²y³, 10x³y², 6x²y²z.
ਹੱਲ:
3x²y³, 10x³y², 6x²y²z
3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
∴ ਸਾਂਝਾ ਗੁਣਨਖੰਡ = x × x × y × y = x²y²

2. ਹੇਠਾਂ ਲਿਖੇ ਵਿਅੰਜਕਾਂ ਦੇ ਗੁਣਨਖੰਡ ਬਣਾਉ

ਪ੍ਰਸ਼ਨ (i).
7x – 42
ਹੱਲ:
7x – 42
= 7 (x – 6)

ਪ੍ਰਸ਼ਨ (ii).
6p – 12q
ਹੱਲ:
6p – 129
= 6 (p – 2q)

ਪ੍ਰਸ਼ਨ (iii).
7a² + 14a
ਹੱਲ:
7a² + 14a
= 7a (a + 2)

ਪ੍ਰਸ਼ਨ (iv).
-16z + 20z³
ਹੱਲ:
-16z + 20z³
= -4z (4 – 5z²)

ਪ੍ਰਸ਼ਨ (v).
20l²m + 30alm
ਹੱਲ:
20l²m + 30alm
= 10lm (2l + 3a)

ਪ੍ਰਸ਼ਨ (vi).
5x²y – 15xy²
ਹੱਲ:
5x²y – 15xy²
= 5xy (x – 3y)

ਪ੍ਰਸ਼ਨ (vii).
10a² – 15b² + 20c²
ਹੱਲ:
10a² – 15b² + 20c²
= 5 (2a² – 3b² + 4c²)

ਪ੍ਰਸ਼ਨ (viii).
-4a² + 4ab – 4ca
ਹੱਲ:
-4a² + 4ab – 4ca
= -4a (a – b + c)

ਪ੍ਰਸ਼ਨ (ix).
x²yz + xy²z + xyz²
ਹੱਲ:
x²yz + xy²z + xyz²
= xyz (x + y + z)

ਪ੍ਰਸ਼ਨ (x).
ax²y + bxy² + cryz.
(ਤਿੰਨਾਂ ਪਦਾਂ ਨੂੰ ਮਿਲਾਉਣ ਤੇ) :
ਹੱਲ:
ax²y + bxy² + cxyz
= xy (ax + by + cz)

3. ਗੁਣਨਖੰਡ ਬਣਾਉ :

ਪ੍ਰਸ਼ਨ (i).
x² + xy + 8x + 8y
ਹੱਲ:
x² + xy + 8x + 8y
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)

ਪ੍ਰਸ਼ਨ (ii).
15xy – 6x + 5y – 2
ਹੱਲ:
15xy – 6x + 5y – 2
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

ਪ੍ਰਸ਼ਨ (iii).
ax + bx – ay – by
ਹੱਲ:
ax + bx – ay – by
= x (a + b) – y (a + b)
= (a + b) (x – y)

ਪ੍ਰਸ਼ਨ (iv).
15pq + 15 + 9q + 25p
ਹੱਲ:
15pq + 15 + 99 + 25p
= 15pq + 25p + 99 + 15
= 5p (3q + 5) + 3 (3q + 5)
= (3q + 5) (5p + 3)

ਪ੍ਰਸ਼ਨ (v).
z – 7 + 7xy – xyz.
ਹੱਲ:
z – 7 + 7xy – xyz
= z – xyz – 7 + 7xy
= z (1 – xy) – 7 (1 – xy)
= (1 – xy) (z – 7)