Shreya

Get Updated Bihar Board Class 9th English Book Solutions in PDF Format and download them free of cost. Bihar Board Class 9 English Book Solutions Prose Chapter 3 A Silent Revolution Questins and Answers provided are as per the latest exam pattern and syllabus. Access the topics of Panorama English Book Class 9 Solutions Chapter 3 A Silent Revolution through the direct links available depending on the need. Clear all your queries on the Class 9 English Subject by using the Bihar Board Solutions for Chapter 3 A Silent Revolution existing.

Panorama English Book Class 9 Solutions Chapter 3 A Silent Revolution

If you are eager to know about the Bihar Board Solutions of Class 9 English Chapter 3 A Silent Revolution Questins and Answers you will find all of them here. You can identify the knowledge gap using these Bihar Board Class 9 English Solutions PDF and plan accordingly. Don’t worry about the accuracy as they are given after extensive research by people having subject knowledge alongside from the latest English Textbooks.

Bihar Board Class 9 English A Silent Revolution Text Book Questions and Answers

A. Work in small groups and discuss these questions:

A Silent Revolution Class 9 Bihar Board Question 1.
How do you send your message to your friends and relatives?
Answer:
Nowadays I send my message to my friends and relatives on a mobile phone. Sometimes the message is sent through a Short Messaging Service Centre (SMSC) run by the service provider.

A Silent Revolution Class 9 Questions And Answers Bihar Board  Question 2.
Have you ever sent an SMS? How did you write your message? How did you send it?
Answer:
Yes, I have sent an SMS. For this, I have to register the network service centre into my handset. The message service centre sent the message where I wanted to send it

B. Answer the following questions briefly:

A Silent Revolution Class 9 In Hindi Bihar Board Question 1.
What does SMS stand for?
Answer:
SMS stands for Short Messaging Service.

A Silent Revolution In Hindi Bihar Board Question 2.
How was it conceived?
Answer:
It was conceived as a part of the Global System for mobile communication (GSM) digital standard.

Silent Revolution Meaning In Hindi Bihar Board Question 3.
What ability does it have?
Answer:
It has the ability to send and receive a text message on a mobile phone.

Class 9 English Chapter 3 Bihar Board Question 4.
How are messages sent and routed?
Answer:
Messages are sent and routed through a short message service centre (SMSC) run by the service provider.

Class 9 English Chapter 3 Question Answer Bihar Board Question 5.
What ensures that the message is delivered at the destination mobile even if it is switched off or out of the coverage area?
Answer:
The messages are routed through a short messaging service centre (SMSC) run by the service provider. This ensures that the message is delivered at the destination mobile even if it is switched off or out of the coverage area.

Class 9th English Chapter 3 Bihar Board Question 6.
What does SMSC stand for?
Answer:
SMSC stands for Short Messaging Service Centre.

Class 9 English Chapter 3 Answers Bihar Board Question 7.
What is its function?
Answer:
The SMSC stores the message and forwards it when the mobile is switched on or enters the network.

English Chapter 3 Class 9 Bihar Board Question 8.
On what account may the delivery of a message be delayed?
Answer:
The delivery of a message may be delayed due to con¬gestion.

English Ch 3 Class 9 Bihar Board Question 9.
What is the beauty of SMS?
Answer:
The beauty of SMS is that messages can be sent and received even while making voice calls.

Question 10.
What is a voice call?
Answer:
A voice call is a sound that takes over a dedicated radio channel.

Question 11.
How does it differ from SMS?
Answer:
A Voice call differs from SMS in that it was voice whereas SMS uses text. Again it takes over a dedicated radio channel for the duration of the call while SMS travels over and above the radio channel using the signalling path.

C. Long Answer Type Questions

Question 1.
How has SMS brought about a silent revolution?
Answer:
The launch of SMS opened a new vista in the field of text communication, providing a new easy way, to the people in communicating. This way SMS has brought about a silent revolution.

Question 2.
How do you think that SMS has now become the most preferred option for communication?
Answer:
SMS has now become the most preferred option for communication no doubt. It is very cheap and easy. It is available with mobile phone which has the keypads attached to them. It has its beauty. The beauty of SMS is that messages can be sent and received even while making voice calls. For this cause, it has opened a new era in the field of text communication. So I think that all about SMS.

Question 3.
What is the attractive feature of SMS? How does a voice call differ from SMS?
Answer:
The attractive feature of SMS is that it ensures that the message is delivered at the destination mobile even if it is switched off or out of the coverage area. The SMSC stores the message and forwards it when the mobile is switched on or enters the network.
A voice call differs from SMS in that a voice call takes over a dedicated radio channel for the duration of the call While the short messages travel over and above the radio channel using the signalling path. The voice call can not be completed if the mobile phone is switched off or out of network.

Question 4.
“The launch of SMS has opened a new vista in the field of text communication.” Explain.
Answer:
The launch of SMS has opened a new vista in the field of text communication. Really it has changed the pattern of communication. It provides a new easy way for the people to communicate their messages. People can type their message using their small handset keypads and send it at the destination even if it is switched off or out of network area. No doubt SMS is a miracle.

Comprehension Based Questions with Answers

1. Short Messaging Service or SMS was conceived as a part of the Global System for Mobile Communication (GSM) digital standard. It is the ability to send and receive text messages (alphanumeric) on a mobile phone. SMS, like e-mail, is a store and forward service that utilizes gateways to send messages from senders to the recipients.
However, messages are not sent directly from the sender to the receiver but are routed through a Short Messaging Service Centre, (SMSC) run by the service provider. This ensures that the message is delivered at the destination mobile even if it is switched off or out of the coverage area. The SMSC stores the message and forwards it when the mobile is switched on or enters the network. Normally, messages are delivered instantly but at times there can be a delay of some hours due to congestion.

Questions:

1. Name the lesson and its author.
2. What is the full form of SMS, SMSC and GSM?
3. How are the messages sent?
4. What happened when the receiver’s mobile is switched off?
5. Find a word from the passage which means ‘imagined’.

Answers:

1. The name of the lesson is A silent Revolution and its author is Kunal Varma.
2. Full form of SMS is Short Message Service and for SMSC is Short Message Service Centre and for GSM is Global System for Mobile Communication.
3. The message is not sent directly but are delivered from SMSC which run by the service provider.
4. The SMSC stores the message and forwards it when the mobile is switched on or enters the network.
5. Alphanumeric.

2. The beauty of SMS is that messages can be sent and received even while making voice calls. This is possible because a voice call takes over a dedicated radio channel for the duration of the call, while the short messages travel over and above the radio channel using the signalling path. The process of sending messages and reading them generally varies from handset to handset. However, confirmation of message delivery is immediate and there is always an alert signal to convey the arrival of a message. SMS messages are immediate but not simultaneous like the Instant Messaging Service, which allows virtual real-time text conversations with people who are simultaneously logged on to the Internet.

Questions:

1. What is the beauty of SMS?
2. Why do the messages vary?
3. How are SMS messages?
4. Find the word from the text which means ‘devoted’.

Answers:

1. The beauty of SMS is that messages can be sent and received even while making a voice call.
2. The process of sending messages and reading them generally varies from handset to handset.
3. SMS messages are immediate but not simultaneous like the Instant Messaging Services which allows virtual real-time text conversation with people who are logged on to the Internet.
4. Dedicated.

3. Access to SMS is generally free and a beginner has only to register the network service centre into his/her handset. The message service centre number for BSNL is +919434099997. The launch of SMS opened a new vista in the field of text communication, providing a new easy way for the people to communicate. The limitation of characters (160 for the GSM networks at present) or the tedious process of typing from the small handset keypads failed to deter the spirit of the enthusiasts. The SMS revolution that took roots in Europe slowly spread to other parts of the globe, especially Asia. From the first short message, believed to have been sent in December 1992 from a PC to a mobile phone on the Vodafone GSM network in the UK, SMS has come a long way today.

Questions:

1. What has to do a beginner for SMS?
2. What has opened a new era?
3. What is the limitation of characters for the GSM network at present?
4. When did the SMS revolution take place?

Answers:

1. A beginner has only to register the network service centre into his handset for SMS.
2. The launch of SMS has opened a new era in the field of text communication.
3. The limitation of characters for GSM networks at present is 160.
4. The SMS revolution that took roots in Europes, when first short message have been sent in December 1992 from a PC to a mobile phone on the Vodafone GSM network in the UK.

4. Judging by its success, at present not many would believe that SMS had a very silent beginning. Not even the cellular operators could comprehend the potential of this sleepy tech¬nology initially and cared little to advertise it as an attraction for mobile users. However, all that is history now. To-day every market player, from cellular operators to mobile handset manufactures, is keen to capture its share of the pie. Nokia recently launched the first Hindi compatible handsets 3350, to give its users the option of sending messages in Hindi. Buoyed by the success of SMS, the industry is now preparing for the more advanced MMS or multimedia messaging service, which would enable pictures, sounds and longer formatted texts to be sent to other MMS-enabled terminals or e-mail addresses via the mobile.

Questions:

1. What would not be believed about SMS?
2. What is the condition of mobile these days?
3. What do you know about Hindi handset?
4. What are possible in the near future?

Answers:

1. SMS’S success at present not many would believe that SMS had a very silent beginning.
2. These days every market player, from cellular operators to mobile handset manufactures is keen to capture its share of the pie.
3. Nokia recently launched the first Hindi compatible handsets 3350 to its users the option of sending a message in Hindi.
4. The industry is now preparing for the more advanced MMS or multi-media messaging service which would enable pictures, sounds and longer formatted text to be sent to other MMS.

Hope you liked the article on Bihar Board Solutions of Class 9 English Chapter 3 A Silent Revolution Questi0ns and Answers and share it among your friends to make them aware. Keep in touch to avail latest information regarding different state board solutions instantly.

Bihar Board Class 8 Maths Solutions Chapter 9 बीजीय व्यंजक Text Book Questions and Answers.

BSEB Bihar Board Class 8 Maths Solutions Chapter 9 बीजीय व्यंजक

Bihar Board Class 8 Maths बीजीय व्यंजक Ex 9.1

बीजीय व्यंजक Class 8 Bihar Board प्रश्न 1.
जोडिए-
(a) xy, 3xy
(b) x² + 3x, 2x + 9
(c) x², y²
(d) 7x – 8x
(e) 8a, -2a, 7a, 2b
(f) 8x, -2x, -6x
(g) 2.3x, 1.7x
(h) \(\frac{2}{3}\)x, \(\frac{1}{3}\)x, -x
उत्तर
(a) xy + 3xy
= (3 + 1) xy
= 4xy
(b) x² + 3x + (2x + 9)
x² + 3x + (2x + 9)
= x² + 3x + 2x + 9
= x² + (3 + 2)x + 9
= x² + 5x + 9
(c) x² + y² = x² + y²

(d) 7x + (-8x)
= 7x – 8x
= (7 – 8)x
= -x
(e) 8a + -2a + 7a + 2b
= 8a + 7a + 2b – 2b
= (8 + 7) a
= 15a
(f) 8x + (-2x) + (-6x)
= 8x – 2x – 6x
= (8 – 2 – 6)x
= (8 – 8) x
= 0x
= 0
(g) 2.3x + 1.7x
= (2.3 + 1.7) x
= (4.0) x
= 4x

Class 8 Bihar Board Math Solution प्रश्न 2.
पहले व्यंजक में से दूसरे को घटाइए-
(a) 22x, 10x
(b) 17xy, 19xy
(c) a² + 1, -2a
(d) 8x, -8x
(e) 7xy, 7xy
(f) 7.3x, 1.3x
(g) -6x + y + 4z – 8, -2y + x – 5z + 8
(h) \(\frac{x}{2}-\frac{x}{4}, \frac{x}{3}\)
उत्तर
(a) 22x – 10x = (22 – 10)x = 12x
(b) 17xy – 19xy
= (17 – 19)xy
= -2xy
(c) a² + 1 + – 2a = a² – 2a + 1
(d) 8x – (-8x) = 8x + 8x = 16x
(e) 7xy – 7xy
= (7 – 7)xy
= 0xy
= 0

(f) 7.3x – 1.3x = (7.3 – 1.3)x
= (6.0)x
= 6x
(g) -6x + y + 4z – 8 – (-2y + x – 5z + 8)
= -6x + y + 4z – 8 + 2y – x + 5z – 8
= -6x – x + y + 2y + 4z + 5z – 8 – 8
= -7x + 3y + 9z – 16

Bihar Board 8th Class Math Solution प्रश्न 3.
सरल कीजिए
(a) 2x – 3y – 7x + 2x – y + 2
(b) 5y³ – 3y² + 2y – 1 + 2y² + 6y – 5
(c) 6a – 3b + c – 6a + 3b + 7c
(d) 8x² + 5xy + 3y² + 3x² + 2xy – 6y²
उत्तर
(a) 2x – 3y – 7x + 2x – y + 2
= 2x – 7x + 2x – 3y – y + 2
= 4x – 7x – 4y + 2
= -3x – 4y + 2
(b) 5y³ – 3y² + 2y – 1 + 2y² + 6y – 5
= 5y³ – 3y² + 2y² + 2y + 6y – 5 – 1
= 5y³ – y² + 8y – 6
(c) 6a – 3b + c – 6a + 3b + 7c
= 6a – 6a – 3b + 3b + c + 7c
= 8c
(d) 8x² + 5xy + 3y² + 3x² + 2xy – 6y²
= 8x² + 3x² + 3y² – 6y² + 5xy + 2xy
= 11x² – 3y² + 7xy

Bihar Board Class 8 Math Solution प्रश्न 4.
यदि किसी त्रिभुज की भुजाएँ x + 1, x + 2 एवं x + 3 हैं तो इसकी परिमिति क्या होगी?
उत्तर
त्रिभुज की परिमिति = भुजाओं का योग = a + b + c
a = x + 1, b = x + 2, c = x + 3
त्रिभुज की परिमिति = x + 1 + x + 2 + x + 3
= 3x + 6
= 3(x + 2)

Class 8 Maths Bihar Board प्रश्न 5.
यदि किसी वर्ग की एक भुजा x – 7 है तो उसकी परिमिति ज्ञात कीजिए।
उत्तर
वर्ग की परिमिति = 4 × भुजा
= 4 × (x – 7)
= 4x – 28

बीजीय व्यंजक कक्षा 8 गणित Bihar Board प्रश्न 6.
रहीम की उम्र x + 6 वर्ष और महेश की उम्र y वर्ष है, दोनों की उम्र का योग और अंतर क्या होगा?
उत्तर
रहीम की उम्र = x – 6
महेश की उम्र = y
उम्र का यांग = x – 6 + y
उम्र का अंतर = x – y – 6

Bihar Board Math Solution Class 8 प्रश्न 7.
किसी आयत की दो आसन्न भुजाएँ क्रमशः x² + 2x + 1 एवं x² – 2x + 1 हैं तो आयत की परिमिति क्या होगी?
उत्तर
आयत की परिमिती = 2 (ल. + चौ.)
= 2(x² + 2x + 1 + x² – 2x + 1)
= 2(2x² + 2)
= 4x² + 4
= 4(x² + 1)

Bihar Board Class 8 Math Solution In Hindi प्रश्न 8.
किसी त्रिभुज की दो भुजाएँ क्रमश: x², y² हैं। यदि परिमिति x² + y² + z²हो तो त्रिभुज की तीसरी भुजा ज्ञात कीजिए।
उत्तर
∆ की परिमिती = भुजाओं का योग = x² + y² + z²
प्रश्न से,
पहली भुजा = x²
दूसरी भुजा = y²
∆ की तीसरी भुजा = x² + y² + z² – (x² + y²)
= x² + y² + z² – x² – y²
= z²

Bihar Board Class 8 Maths बीजीय व्यंजक Ex 9.2

Bihar Board Class 8 Math प्रश्न 1.
गुणनफल ज्ञात कीजिए
(a) 8x × (-2)
(b) -3x × -3x²y
(c) 6mn × 7np
(d) 4p³ × 3p³
(e) x²y × xyz
(f) 2.5x × 4x
(g) 2.5x × 2.5y
(h) \(\frac {1}{2}\)x × \(\frac {1}{2}\)y
(i) \(\frac {1}{2}\)xy × 2xy
(j) 2x × 2x² × 2x³
(k) -3x²y × (-6) × 7xy
उत्तर
(a) 8x × (-2) = -16x
(b) -3x × -3xy = -9x³y
(c) 6mn × 7np = 6 × 7 × m × n × n × p = 42mn²p
(d) 4p³ × 3p³ = 4 × 3 × p³ × p³ = 12p⁶
(e) x²y × xy² = x² × x × y × y × z = x³y²z
(f) 2.5x × 4x = 2.5 × 4 × x × x = 10x²
(g) 2.5x × 2.5y = 2.5 × 2.5 × x × y = 6.25xy
(h) \(\frac {1}{2}\)x × \(\frac {1}{2}\)y
= \(\frac {1}{2}\) × \(\frac {1}{2}\) × x × y
= \(\frac {1}{4}\)xy
(i) \(\frac {1}{2}\)xy × 2xy
= \(\frac {1}{2}\) × 2 × xy × xy
= x²y²
(j) 2x × 2x² × 2x²
= 2 × 2 × 2 × x × x² × x²
= 8x⁵
(k) -3x²y × -6 × 7xy
= -3 × -6 × 7 × x² × x × y × y
= +126x²y²

Class 8 Math Solution Bihar Board प्रश्न 2.
किसी आयत की आसन्न भुजाएँ क्रमश: 6p²q² एवं 2pq हैं तो आयत का क्षेत्रफल क्या होगा?
उत्तर
आयत का क्षेत्रफल = ल० × चौ०
= 6p²q² × 2pq
= 6 × 2 × p² × p × q² × q
= 12p³q³

Bihar Board 8 Class Math Solution प्रश्न 3.
यदि किसी वर्ग की भुजा √2 x²y² है तो वर्ग का क्षेत्रफल क्या होगा?
उत्तर
वर्ग की भुजा = √2 x²y²
वर्ग का क्षे० = (भु०)²
= (√2 x²y²)²
= √2 × √2 × x² × x² × y² × y²
= 2x⁴y⁴

Bihar Board Class 8 Math Solution In Hindi Pdf Download प्रश्न 4.
किसी त्रिभुज का आधार 7xyz एवं संगत शीर्षलंब 2x है तो त्रिभुज का क्षेत्रफल क्या होगा?
उत्तर
∆ का आधार = 7xyz
शीर्षलंब = 2x
∆ का क्षेत्रफल = \(\frac {1}{2}\) × आधार × शीर्ष
= \(\frac {1}{2}\) × 7xyz × 2x
= 7x²yz

Bihar Board Solution Class 8 Math प्रश्न 5.
समबाहु त्रिभुज का क्षेत्रफल ज्ञात कीजिए यदि उसकी भुजा 3x है।
उत्तर
समबाहु ∆ का क्षे = \(\frac{\sqrt{3}}{4}\) × भुजा²
= \(\frac{\sqrt{3}}{4}\) × (3x)²
= \(\frac{\sqrt{3}}{4}\) × 9x²
= \(\frac{9 \sqrt{3}}{4} x^{2}\)

Bihar Board Class 8 Maths Solutions प्रश्न 6.
उस घन का आयतन क्या होगा जिसकी कोर 6a हो?
उत्तर
घन का कोर = 6a
घन का आयतन = (कोर)³
= (6a)³
= 216a³

Bihar Board Solution.Com Class 8 प्रश्न 7.
यदि एक कलम का मूल्य x²y हो तो y²x कलम का मूल्य क्या होगा?
उत्तर
एक कलम का मू० = x²y
y²x का मू० = x²y × y²x = x³y³

Bihar Board Class 8th Math Solution प्रश्न 8.
यदि कोई व्यक्ति \(\frac{x^{2}}{2}\) km/h की चाल से चल रहा हो तो 2 घंटे में वह कितनी दूरी तय कर लेगा?
उत्तर
व्यक्ति की चाल = \(\frac{x^{2}}{2}\) km/h
समय = 2 घंटे
दूरी = चाल × समय
= \(\frac{x^{2}}{4}\) × 4 km
= x² km

Bihar Board Class 8 Maths बीजीय व्यंजक Ex 9.3

Class 8 Maths Bihar Board Solution प्रश्न 1.
दिए गए बीजीय व्यंजकों का गुणा कीजिए-
(a) (4a – 5b) × (2a – 6b)
(b) (1.5x – 0.5y) × (1.5x + 0.5y)
(c) (\(\frac{1}{2}\)pq – \(\frac{3}{2}\)q) × (pq – q)
(d) (a + b) × (3x – y)
(e) (a²b² – c²d²) × (a²b² + c²d²)
(f) (2a + 2b + c) (a + b – c²)
उत्तर
(a) (4a – 5b) × (2a – 6b)
= 4a × 2a – 4a × 6b – 5b × 2a + 5b × 6b
= 8a² – 24ab – 10ba + 30b²
= 8a² – 34ab + 30b²
(b) (1.5x – 0.5y) × (1.5x + 0.5y)
= 1.5x × 1.5x + 1.5x × 0.5y – 0.5y × 1.5x – 0.5y × 0.5y
= 2.25x² + 0.75xy – 0.75xy – 0.25y²
= 2.25x² – 0.25y²
= (1.5x)² – (0.5y)²

(d) (a + b) × (3x – y)
= a × 3x – a × y + b × 3x – b × y
= 3ax – ay + 3bx – by
= 3ax – ay + 3bx – by
(e) (a²b² – c²d²) × (a²b² + c²d²)
= a²b² × a²b² + a²b² × c²d² – c²d² × a²b² – c²d² × c²d²
= a⁴b⁴ + a²b²c²d² – c²d²a²b² – c⁴d⁴
= a⁴b⁴ – c⁴d⁴
(f) (2a + 2b + c) (a + b – c²)
= 2a × a + 2a × b – 2a × c² + 2b × a + 2b × b – 2b × c² + c × a + c × b – c × c²
= 2a² + 2ab – 2ac² + 2ab + 2b² – 2bc² + ac + bc – c³
= 2a² + 4ab – 2ac² + 2b² – 2bc² + ac + bc – c³

प्रश्न 2.
सरल कीजिए-
(a) (a – b)(a + b) – (a + b)(a + b)
(b) (a² – b) (a – b²) + (a – b)²
(c) (2.3x – 1.7y) (2.3x + 1.7y + 5) – 5.29x² + 2.89y²
(d) (a + b)² – (a – b)²
(e) (x + y + z) × (x + y + z)
(f) (a – b) (b – c) + (b – c) (c – a) + (c – a) (a – b)
उत्तर
(a) (a – b)(a + b) – (a + b) (a + b)
= a × a + a × b – b × a – b × b – (a × a + a × b + b × a + b × b)
= a² + ba – ab – b² – (a² + ab + ba + b²)
= a² – b² – a² + ab + ba + b²
= 2ab
(b) (a² – b) (a – b²) + (a – b)²
= a² × a – a² × b² – b × a + b × b² + a² – 2ab + b²
= a³ – a²b² – ab + b³ + a² – 2ab + b²
= a³ – a²b² – 3ab + a² + b² + b³
(c) (2.3x – 1.7y) (2.3x + 1.7y + 5) – 5.29x² + 2.89y²
= (2.3x × 2.3x) + (2.3x × 1.7y) + (2.3x × 5) – (7.7y × 2.3x) – (1.7y × 1.7y) – (1.7y × 5) – 5.29x² + 2.89y²
= 5.29x² + 3.91xy + 11.5x – 5.29xy – 2.89y² – 8.5y – 5.29x² + 2.89y²
= 3.91xy + 11.5x – 5.29xy – 8.5y
= 11.5x – 8.5y – 1.38xy
(d) (a + b)² – (a – b)²
= a² + 2ab + b² – (a² – 2ab + b²)
= a² + 2ab + b² – a² + 2ab – b²
= 4ab
(e) (x + y + z) (x + y + z)
= x × x + x × y + x × z + y × x + y × y + y × z + z × x + z × y + z × z
= x² + xy + xz + yx + y² + yz + xz + zy + z²
= x² + 2xy + 2xz + 2zy + y² + z²
(f) (a – b) (b – c) + (b – c) (c – a) + (c – a) (a – b)
= a × b – a × c – b × b + b × c + b × c – b × a – c × c + c × a + c × a – c × b – a × a + a × b
= ab – ac – b² + bc + bc – ba – c² + ac + ac – cb – a² + ab
= ab + bc + ac – b² – c² – a²

प्रश्न 3.
किसी त्रिभुज का आधार एवं संगत शीर्षलम्ब क्रमशः (x + y)² एवं (x + y)² हैं तो उसका क्षेत्रफल क्या होगा?
उत्तर
क्षेत्रफल = \(\frac {1}{2}\) × आधार × शीर्ष
= \(\frac {1}{2}\) × (x + y)² × (x – y)²
= \(\frac {1}{2}\) × x² + 2xy + y² (x² – 2xy + y²)
= \(\frac {1}{2}\) × (x² × x² – x² × 2xy + x² × y² + 2xy × x² – 2xy × 2xy + 2xy × y² × x² – 2xy × y² + y² × y²)
= \(\frac {1}{2}\) × (x⁴ – 2x³y + x²y² + 2x³y – 4x²y² + 2xy³ + x²y² – 2xy³ + y⁴)
= \(\frac {1}{2}\) × (x⁴ + 2x²y² – 4x²y² + y⁴)
= \(\frac {1}{2}\) × (x⁴ – 2x²y² + y⁴)

प्रश्न 4.
आयत की लम्बाई उसकी चौड़ाई से (x + y) इकाई अधिक है। यदि चौड़ाई z इकाई हो तो आयत की लम्बाई व क्षेत्रफल के लिए व्यंजक लिखिए।
उत्तर
आयत की चौ० = z
आयत की ल० = x + y + z
आयत का क्षे० = ल० × चौ०
= (x + y + z) × z
= xz + yz + z²

प्रश्न 5.
यदि किसी लड़की ने (x + y) रु. प्रति किलो की दर से (m + n) किलोग्राम आलू एवं y रुपये प्रति किलोग्राम की दर से (m – n) किलो टमाटर खरीदे तो उसके कुल कितनी राशि देनी होगी?
उत्तर
आलू की कीमत = (x + y) × (m + n) = xm + xn + ym + yn
टमाटर की कीमत = y × (m – 1) = ym – yn
कुल कीमत = xm + xn + ym + yn + ym – yn
= xm + xn + 2ym

प्रश्न 6.
पिता की उम्र उसके पुत्र की उम्र के (m + n) गुणा है। यदि पुत्र की उम्र (x² – y²) वर्ष हो तो पिता की उम्र के लिए व्यंजक लिखिए।
उत्तर
पुत्र की उम्र = (x² – y²)
पिता की उम्र = (m + n) (x² – y²)
= x²m – my² + nx² – ny²

Bihar Board Class 8 Maths बीजीय व्यंजक Ex 9.4

प्रश्न 1.
उचित सर्वसमिकाओं का उपयोग कर दिए गए व्यंजकों का गुणनफल प्राप्त कीजिए-
(a) (5x + 7y)²
(b) (a + \(\frac{a}{2}\))²
(c) (1.5x + 2.5y)²
(d) (x + \(\frac{1}{x}\))²
(e) (0.4a – 0.5b) (0.4a – 0.5b)
(f) \(\left(\frac{1}{3} a+\frac{2}{3} b\right)\left(\frac{1}{3} a+\frac{2}{3} b\right)\)
(g) (y² – y) (y² – y)
(h) (pqr – 3) (pqr + 3)
(i) (2x + 3) (2x – 5)
(j) (3.5x – y) (3.5x – y)
(k) \(\left(\frac{x}{2}-\frac{y}{2}\right)^{2}\)
(l) \(\left(\frac{1}{x}-\frac{1}{y}\right)^{2}\)
(m) \(\left(x-\frac{1}{x}\right)^{2}\)
उत्तर
(a) (5x + 7y)²
(a + b)² = a² + 2ab + b²
(5x + 7y)² = (5x)² + 2 × 5x × 7y + (7y)²
= 25x² + 70xy + 49y²
(b) (a + \(\frac{a}{2}\))²
= a² + 2 × a × \(\frac{a}{2}\) + (\(\frac{a}{2}\))²
= a² + a² + \(\frac{a^{2}}{4}\)
(c) (1.5x + 2.5y)² = (1.5x)² + 2 × 1.5x × 2.5y + (2.5y)²
= 2.25x² + 7.5xy + 6.25y²
(d) \(\left(x+\frac{1}{x}\right)^{2}=x^{2}+2 \times 2 \times \frac{1}{2}+\left(\frac{1}{x}\right)^{2}\)
= x² + 2 + \(\frac{1}{x^{2}}\)
(e) (0.4a – 0.5b) (0.4a – 0.5b)
= 0.4a × 0.4a – 0.4a × 0.5b – 0.5b × 0.4a – 0.5b × 0.5b
= 0.16a² – 0.2ab – 0.2ab – 0.25b²
= 0.16a² – 0.4ab – 0.25b²

(g) (y² – y) (y² – y)
= y² × y² – y² × y – y × y² + y × y
= y⁴ – y³ – y³ + y²
= y⁴ – 2y³ – y²
(h) (pqr – 3) (pqr + 3)
(a + b)(a – b) = a² – b²
(pqr – 3) (pqr + 3) = (pqr)² – 3²
= p²q²r² – 9
(i) (2x + 3) (2x – 5) = 2x × 2x – 2x × 5 + 3 × 2x – 3 × 5
= 4x² – 10x + 6x – 15
= 4x² – 4x – 15
(j) (3.5x – y) (3.5x – y)
= 3.5x × 3.5x – 3.5x × y – 3.5x × y + y × y
= 12.25x² – 3.5xy – 3.5xy + y²
= 12.25x² – 7xy + y²

प्रश्न 2.
सरल कीजिए-
(a) (x² + y²)²
(b) (3a – 5b)² – (3a + 5b)²
(e) (xyz + xy)² – 2x²y²z

उत्तर
(a) (x² + y²)²
= (x²)² + 2(x²) (y²) + (y²)²
= x⁴ + 2x²y² + y⁴
(b) (3a – 5b)² – (3a + 5b)²
= (3a)² – 2(3a) (5b) + (5b)² – {(3a)² + 2(3a) (5b) +(5b)²}
= 9a² – 30ab + 25b² – {9a² + 30ab + 25b²}
= 9a² – 30ab + 25b² – 9a² – 30ab – 25b²
= -60ab
(c) (xyz – xy)² – 2x²y²z
= (xyz)² – 2(xyz) (xy) + (xy)² – 2x²y²z
= x²y²z² + 2x²y²z² + x²y² – 2x²y²z
= x²y²z² + x²y²

प्रश्न 3.
सर्वसमिकाओं के उपयोग से निम्नलिखित मान ज्ञात कीजिए
(a) 812
(b) (999)
(c) (52)
(d) (498)
(e) (5.5)²
(f) 191 × 209
(g) 10.5 × 9.5
(h) (101)² – (99)²
(i) (1.5)² – (0.5)²
उत्तर
(a) 812
= (80 + 1)²
= (80)² + 2 × 80 × 1 + 1²
= 6400 + 160 + 1
= 6561
(b) (999)² = (1000 – 1)²
= (1000)² – 2 × 1000 × 1 + 1²
= 1000000 – 2000 + 1
= 998001
(c) (52)²
= (50 + 2)²
= (50)² + 2 × 50 × 2 + 2²
= 2500 + 200 + 4
= 2704
(d) (498)²
= (500 – 2)²
= (500)² – 2 × 50 × 2 + 2²
= 250000 – 2000 + 4
= 250000 – 1096
= 248904
(e) (5.5)² = (6 – 0.5)²
= 6² – 2 × 6 × 0.5 + (0.5)²
=36 – 6 + 0.25
= 36.25 – 6
= 30.25
(f) 191 × 209
= (200 – 9) × (200 + 9)
= (200)² – 9²
= 40000 – 81
= 39919
(g) 10.5 × 9.5
= (10 + 0.5) × (10 – 0.5)
= (10)² – (0.5)²
= 100 – 0.25
= 99.75
(h) (101)² – (99)²
= (101 + 99) (101 – 99)
= (200) (2)
= 400
(i) (1.5)² – (0.5)²
= (1.5 + 0.5) (1.5 – 0.5)
= 2 × 1
= 2

प्रश्न 4.
सर्वसमिका (x + a) (x + b) = x² + (a + b)x + ab का उपयोग कर निम्नलिखित का गुणनफल एवं मान ज्ञात कीजिए-
(a) (x + 3y) (x + 5y)
(b) (3x + 7) (3x + 5)
(c) (x – 5)(x + 4)
(d) (2x – 7) (2x – 9)
(e) 52 × 53
(f) 3.1 × 3.2
उत्तर
(a) (x + 3y) (x + 5y)
= x² + (3y + 5y)x + 3y × 5y
= x² + (8y)x + 15y²
= x² + 8yx + 15y²
(b) (3x + 7) (3x + 5)
= (3x)² + (7 + 5)3x + 7 × 5
= 9x² + 36x + 35
(c) (x – 5) (x + 4)
= x² + (-5 + 4)x + (-5) (4)
= x² – x – 20
(d) (2x – 7) (3x – 9)
= (2x)² + (-7 – 9) 2x + (-7) (-9)
= 4x² + 32x + 63
(e) 52 × 53
= (50 + 2) (50 + 3)
= (50)² + (2 + 3) 50 + 2 × 3
= 2500 + 250 + 6
= 2756
(f) 3.1 × 3.2
= (3 + 0.1) (3 + 0.2)
= 3² + (0.1 + 0.2)3 +0.1 × 0.2
= 9 + (0.3)3 + 0.02
= 9 + 0.9 + 0.02
= 9.92

Get Updated Bihar Board Class 9th English Book Solutions in PDF Format and download them free of cost. Bihar Board Class 9 English Book Solutions Poem 3 Blow, Blow, Thou Winter Wind Questions and Answers provided are as per the latest exam pattern and syllabus. Access the topics of Panorama English Book Class 9 Solutions Poem 3 Blow, Blow, Thou Winter Wind through the direct links available depending on the need. Clear all your queries on the Class 9 English Subject by using the Bihar Board Solutions for Poem 3 Blow, Blow, Thou Winter Wind existing.

Panorama English Book Class 9 Solutions Poem 3 Blow, Blow, Thou Winter Wind

If you are eager to know about the Bihar Board Solutions of Class 9 English Poem 3 Blow, Blow, Thou Winter Wind Questions and Answers you will find all of them here. You can identify the knowledge gap using these Bihar Board Class 9 English Solutions PDF and plan accordingly. Don’t worry about the accuracy as they are given after extensive research by people having subject knowledge alongside from the latest English Textbooks.

Bihar Board Class 9 English Blow, Blow, Thou Winter Wind Text Book Questions and Answers

A. Work in small groups and answer the following questions orally:

Blow, Blow, Thou Winter Wind Poem Question Answer Bihar Board Question 1.
Why do you wear woollen clothes in Winter?
Answer:
We wear woollen clothes in winter to protect ourselves from cold.

Blow Blow Thou Winter Wind Poem Question Answer Bihar Board Question 2.
How much do you like this season?
Answer:
I like it very much. Because it is good for health and suitable for work. It is a season of fruits and flowers so it is charming.

Blow Blow Thou Winter Wind Questions And Answers Bihar Board Question 3.
Which is your favourite season?
Answer:
The winter season is my favourite season.

B.1. Answer the following questions very briefly:

Blow, Blow, Thou Winter Wind Question Answer Bihar Board Question 1.
Why does the poet ask the wind to blow?
Answer:
As the wind is not so painful than man’s ingratitude.

Blow Blow Thou Winter Wind Question Answer Bihar Board Question 2.
Why does the poet call the winter wind not so unkind as man’s ingratitude?
Answer:
Because the poet has suffered a lot from ungrateful men.

Blow Blow Thou Winter Wind Bihar Board Question 3.
What makes the poet say Thy tooth is not so keen?
Answer:
The biting winter wind does not hurt the poet as does the brother’s ingratitude which he looks upon the as fierce animal with keen teeth.

Blow, Blow, Thou Winter Wind Meaning In Hindi Bihar Board Question 4.
Explain the mood of the poet when he says “Heigh-ho! sing, heigh-ho! unto the green holly”
Answer:
The mood of the poet when he says “Heigh-ho! sing, heigh-ho unto the green holly” is one of bitterness. This is quite manifest in the line that follows. Most friendship is feigning most loving mere folly.

Blow, Blow, Thou Winter Wind Meaning Bihar Board Question 5.
Explain the use of the word ‘warp’ in the second stanza.
Answer:
The use of the word ‘warp’ in the second stanza suggests freezing of water.

Blow Blow Thou Winter Wind Poem Summary Bihar Board Question 6.
How is nature not so cruel as a man?
Answer:
Nature, in the form of winter bites, but is not so cruel as a man. Its stings are less hurting than the stings of man.

C.1. Long Answer Questions:

Blow Blow Thou Winter Wind Shakespeare Bihar Board Question 1.
The speaker’s tragic mood is very pronounced in the poem. Elaborate.
Answer:
The speaker is a banished king. He is a victim of in-gratitude. He is banished by his younger brother whom he loved and trusted the most. The ingratitude of his younger brother has made him sceptic and he suspects every human relation. Even friendship seems folly to him. This idea has been pronounced in the poem. He is in the forest, it is winter. A chilly winter wind is blowing. It is pinching and painful but not more pinching than the ingratitude man.

Blow Blow Thou Winter Wind Meaning In Hindi Bihar Board Question 2.
What does the poet mean to say “Most friendship is feigning, most loving mere folly”? Explain.
Answer:
The Duke is banished by his most loving brother. He is in the forest. Here he finds that winter wind is. not the enemy of man. There the life is artificial men are restless they are cheated by their own. They quarrel and try to injure one another. There is no peace, joy and life are discontented. Here everyone wants to cheat. Even friendship and love are created here.

Blow Blow Thou Winter Wind Summary In Hindi Bihar Board Question 3.
Why and how is the severe winter kinder than an ungrateful person?
Answer:
The winter sky and cold winter blowing in it are though chilly but not biting as an ungrateful person. Severe winter wind hurt physically but the deeds of an ungrateful man hurt mentally.

Blow, Blow, Thou Winter Wind Poem Pdf Bihar Board Question 4.
Describe how the poet has conveyed the feelings of an afflicted man.
Answer:
The poet has conveyed the feelings of an afflicted man in an expressive mood. He has given a real picture of human nature. When someone loves blindly with anyone, he must be deceived. Faith has a limit. More faith means more pain. And no faith means no pain. Ingratitudeness of a man makes us think that all are the same.

Question 5.
Summarise this poem in about 100 words.
Answer:
See the Summary in English.

C. 2. Group Discussion

Question 1.
Gratitude is a mark of civility.
Answer:
Gratitude is a mark of civility is said truly. Man is rational. It is the best creation of the creator. God has provided us with wit and intelligence. Man has made him separate from others. Man has a society, which has a certain way of life. So man has formed some sort of civilities such as manners, and gratitudes. If someone does good to someone, he must be obliged to him. It gives as manners and way of life to live in society successfully. If someone is cheated by his own, the cheater is called ingratitude: Ingratitude is a sin.

Question 2.
Everything is fair in love and war.
Answer:
It is a popular saying “Everything is fair in love and war.” The Mahabharata war is the best example of regarding love this is also true. Love is love if it is gained the purpose is served. The love between Laila and Majnu, Sri and Farhad are always rememberable. They sacrificed their life for the sake of love. A love does so only in order to put emphasis on love. A lover or beloved is to do everything for his or her love.

Comprehension Based Questions with Answers

1. Blow, Blow, Thou Winter Wind
Blow, blow, thou winter wind,
Thou art not so unkind
As man’s ingratitude;
Thy tooth is not so keen.
Because thou art not seen,
Although thy breath is rude.
Heigh-ho! sing, heigh-ho! unto this green holly;
Most friendship is feigning, most loving mere folly:

Questions:

1. Name the poem and its poet.
2. Who is not so unkind? “As man’s ingratitude”?
3. Why is the tooth of the winter wind not so keen?
4. What does Shakespeare say about friendship and love?
5. What does the expression? “Thy breath is rude”, mean?

Answers:

1. The name of the poem is Blow Blow’Thou Winter Wind and its poet is William Shakespeare.
2. The winter wind is not so unkind as man’s ingratitude.
3. The tooth of the winter wind is not so keen because the winter wind is not seen.
4. Shakespeare says that most friendship is feigning and most loving is mere folly.
5. It means that the winter wind blows with great force.

2. Then, heigh-ho! the holly!
This life is most jolly.
Freeze, freeze, thou bitter sky,
Thou dust not bite so nigh As benefits forgot:
Though thou the waters warp,
Thy sting is not so sharp As friend remembered not.
Heigh-ho! sing, heigh-ho! unto the green holly:
Most friendship is feigning, most loving mere folly:
Then, heigh-ho! the holly!
This life is most jolly.

Questions:

1. Who is being addressed in these lines?
2. How does the winter sky not bite so bitterly as the friend’s forgetfulness?
3. Which words in the passage suggests cold?
4. What is meant by the expression “the waters warp”?

Answers:

1. The poet is addressing the cold sky in which the winter wind blows.
2. The winter sky and the cold wind blowing in it are so bitterly biting as an ungrateful man. A man forgets the good deeds, of his friend. So, the bitter cold of the sky for windy is not so painful as a friend forgetting his friend.
3. The word “Freeze” in the passage suggests cold.
4. The expression “the waters warp” means that the winter creates waves in the seas.

Hope you liked the article on Bihar Board Solutions of Class 9 English Poem 3 Blow, Blow, Thou Winter Wind Questions and Answers and share it among your friends to make them aware. Keep in touch to avail latest information regarding different state board solutions instantly.

Get Updated Bihar Board Class 9th English Book Solutions in PDF Format and download them free of cost. Bihar Board Class 9 English Book Solutions Prose Chapter 4 Too Many People Too Few Trees Questins and Answers provided are as per the latest exam pattern and syllabus. Access the topics of Panorama English Book Class 9 Solutions Chapter 4 Too Many People Too Few Trees through the direct links available depending on the need. Clear all your queries on the Class 9 English Subject by using the Bihar Board Solutions for Chapter 4 Too Many People Too Few Trees existing.

Panorama English Book Class 9 Solutions Chapter 4 Too Many People Too Few Trees

If you are eager to know about the Bihar Board Solutions of Class 9 English Chapter 4 Too Many People Too Few Trees Questins and Answers you will find all of them here. You can identify the knowledge gap using these Bihar Board Class 9 English Solutions PDF and plan accordingly. Don’t worry about the accuracy as they are given after extensive research by people having subject knowledge alongside from the latest English Textbooks.

Bihar Board Class 9 English Too Many People Too Few Trees Text Book Questions and Answers

A. Work in small groups and discuss the relationship between population and pollution. You may include these points in your discussion:

1. Population explosion.
2. Its effect on the development of the country.
3. More people, more land.
4. Deforestation
Answer:
These days the environmental pollution has increased so much, this is a population explosion. This is a problem that worries us and the Government of our country. The government are discussing overpopulation explosion. It is a great hindrance on the development of the country. Man is doing harm to the environment through the growth of population, pollution and deforestation. Human population has grown rapidly. Man has cut down forests to build houses factories and cornfield. More people need more land to live and to cultivate.. Deforestation leads to air pollution, as the trees protect soil, so the cutting down of trees leads to washing away of topsoil as well.

B.1.1. Write ‘T’ for true and ‘F’ for false statements:

1. Throughout most of human existence, the number of births was slightly higher than the number of deaths.
2. More people will need even less food than they need now.
3. With more people, both town and country become more crowded.
4. Higher population density is also not likely to exacerbate crime, ethnic conflict and warfare.
5. Population size and rates of growth are key elements in environmental change.

Answer:

1. – T
2. – F
3. – T
4. – F
5. – T

B. 1.2. Answer the following questions very briefly:

Too Many People Too Few Trees Is Bihar Board Class 9 Question 1.
For how long has the global population been rapidly going up?
Answer:
The global population has been rapidly going up during the last 3 centuries.

Too Many People Too Few Trees Bihar Board Class 9 Question 2.
What happens when the population goes up?
Answer:
When the population goes up pollution of rivers, lakes, air, drinking water and soil also goes up. The quality of life and its value continues to erode.

Too Many People Too Few Tree Is Bihar Board Class 9 Question 3.
How many Americans die each day of asthma?
Answer:
Fourteen Americans die each day of Asthma aggravated by air pollution.

B. 2.1. Complete the following sentences on the basis of the lesson:

1. The more we have, the better __________
2. History and common sense tell us that we _________
3. As the population grows, more and more people are forced _________
4. Forest covered around 40% of the earth’s __________
5. Humanity can continue to fell trees, cross its finger, and ___________

Answer:

1. off we are.
2. can control population growth.
3. to convert forest into farmlands
4. total land area.
5. hope for the best.

B.2.2. Answer the following questions very briefly:

Too Many People, Too Few Trees Is Bihar Board Class 9 Question 1.
Name the countries in which the population growth has been slowed down remarkably?
Answer:
Countries like China, Thailand and Egypt the rate of population growth has slowed down remarkably.

Too Many People Too Few Trees Is A Bihar Board Class 9 Question 2.
The productivity and general health of the world’s forest are threatened. How?
Answer:
The productivity and general health of the world’s forest are threatened by such things as the greenhouse effect, ozone layer depletion, airborne pollution, and acid rain.

Too Many People Too Few Tree Bihar Board Class 9 Question 3.
What hampers the ability of the biosphere to sustain life?
Answer:
The eventual consequences of deforestation may be damage due to quality of life on earth reduction in a number of life forms that share the planet with us and thus hampering the ability of the biosphere to sustain life.

Too Many Pcople Too Few Trees Is Bihar Board Class 9 Question 4.
How does deforestation in Nepal affect India?
Answer:
Deforestation in Nepal affects India. As the trees of Nepal are cut down, its topsoil is gradually being lost and its rains are likelier to cause devastating floods in India.

C. Long Answer Type Questions

Bihar Board Class 9 English Book Solution  Question 1.
Why have human populations always been in flux?
Answer:
The human population have always been in flux, for example, every day some people die while others are born. A few hundred years ago, however, the situation began to change, especially in the industrialized world. With advances in nutrition, sanitation and health, people live longer and” more of them reach reproductive age. So the human population has been rapidly going up.

Bihar Board Solution Class 9 English Question 2.
What does the writer mean by reproductive age? How do people reach this age?
Answer:
The writer means to say that by reproductive age means capable of giving birth to a child through the sexual relationship between man and woman. This age begins at the age of 15 and above at about 50 years. But with advances in nutrition, sanitation, and health, people live longer and more of them reach reproductive age up to 60 and above.

Bihar Board 9th Class English Book Solution Question 3.
What is human-made pollution? How has it affected America? How will it affect your locality?
Answer:
Human-made pollution is due to deforestation depletion of no renewable resources and ozone layer depletion and the greenhouse effect. These all are human-made pollution. It has badly affected America. According to an estimate, about 60,000 Americans die each year from respiratory diseases. Which are caused by human-made pollution? Our locality is also being affected by human-made pollu¬tion. People throw refuse and debris everywhere at in the street, the Drainage system is also defective It is really frightening problem. It may cause diseases.

Panorama English Book Answers Class 9 Question 4.
Population size and rates of growth are a key element in environmental change. Explain with any two examples from your own society.
Answer:
Population size and rates of growth are key elements of environmental change. More people require more food and thus deforestation occurs and the green fields are lost. In our village, a family has to cut down his orchard to build new houses there. Now that place looked’barren, no tree is seen there. Another example is that in Patna, in my locality a few years ago there were farms people do cultivate in them but the whole of the locality filled with roads and houses. Now it looks a housing colony. Farms vanished.

Panorama English Book Class 9 Pdf Download Question 5.
How do countries like Germany, Switzerland, China, Thailand and Egypt manage to ‘reverse’ or slow down population growth? What does ‘reverse’ mean here? How has it been possible?
Answer:
These countries managed to ‘reverse’ or slow down population growth. This has been possible because of several factors that include modernization, literacy, media campaigns, readily available family planning and contraceptive, equal economic, educational and legal opportunities for women.

Panorama English Book Class 9 Solutions Question 6.
What do you mean by environment?
Answer:
Environment means the surroundings in which man lives. It includes all conditions surrounding the man. Physical environment means the land on which we live and the air we breathe. It also means the food we eat and the water we drink. Similarly, social environment relates to man in relation to society. If a man causes harm to the environment he causes harm to himself.

Panorama English Reader Part 1 Class 9 Solutions Question 7.
How does pollution harm people?
Answer:
These days the environment pollution has increased so much that human existence has started facing danger. Trees have en cut. Various chemical factories emit poisonous smoke. This contains carbon dioxide and carbon monoxide gases. These are very, dangerous for human health. Besides this burning of fossil fuel in various vehicles emits poisonous substances. All this has made life unbearable that’s why metropolitan cities have become a death trap. Respiratory diseases have increased manifold. Soon people living in them will have to wear masks. So, it is the call of the hour that this increasing pollution of the biosphere is checked at the earliest as our cities are the most polluted. The authorities should take the necessary step in this direction.

Bihar Board 9th Class English Book Question 8.
What is the Chipko movement? How is it useful?
Answer:
After independence,.the Chipko (hug-a-tree) movement came into being. The village people hug a tree in the Himalayan forest of Uttar Pradesh when it is to be cut. This movement forced the state Government to think about the forests. The result is that cutting of trees for a commercial purpose is now banned above a hight of 1000 m. However, the Forest Department did not plant trees where it had cut them. But thanks to Nature and the strength of these women. The forest has again come up. These have reduced the fury of the flood and soil erosion.

Comprehension Based Questions with Answers

1. Human populations have always been in flux for the simple reason that every day some people die while others are born. Throughout most of human existence, the number of births was slightly higher than the number of deaths; consequently, world populations grew at a very slow rate. A few hundred years ago, however, the situation began to change, especially in the industrialized world. With advances in nutrition, sanitation, and health, people live longer and more of them reach reproductive age. Thus, for the first time in our species existence, the balance between the number of death and births has been significantly disturbed. Consequently, during the last three centuries or so, the global human population has been rapidly going up. Every year, in fact, the world’s population grows by more than 80 million people. It is, for instance, sobering to recall that for every eleven human beings alive now, only one was alive in the year 1950.

Questions:

1. Name the lesson and its author.
2. Why have human population always been in flux?
3. When did the situation begin to change?
4. How do people live longer today?
5. How does the world’s population grow every year?
6. Find out the word in the passage which means ‘continuous flow’.

Answers:

1. The name of the lesson is Too many People too few Trees and its author is Moti Nisan.
2. Human populations have always been in flux for the simple reason that every day some people die while others are born.
3. A few hundred years ago, however, the situations began to change especially in the industrialized.world with advances in nutrition, sanitation and health.
4. With advances in nutrition sanitation and health, people live longer and more of them reach reproductive age.
5. More than 80 million people grow every year and thus it adds to the world’s population.
6. Flux.

2. On first sight, it may appear that, when it comes to something as valuable as a human being, the more we have, the better off we are. In some ways, this is true. All things being equal, more people are likely to generate more inventions, more technological breakthroughs, and more corporate profits. But. taken as a whole, most ecologists are convinced that the world is already overpopulated. Human populations cannot continue to grow indefinitely for the simple reason that the world itself is finite. More people will need own more food than they need now, and therefore, the process of deforestation will continue so that, eventually. wild trees will vanish. As the population goes up, so docs pollution of rivers, lakes, air, drinking water and soil. With more people, both town and country become more crowded. The quality of life, and the value we place on human life, will continue to erode. When the population is stable, increases in such things as food production, number of physicians, or hospitals are often tantamount to improved quality of life, but such increases often fail to keep pace with population growth. Higher population density is also likely to exacerbate crime, ethnic conflicts, and warfare.

Questions:

1. Name the essay and its writer?
2. What does the author think about the population?
3. What is the reason for discontinuing the population?
4. How will the process of deforestation continue?
5. What is the possibility when population goes high?
6. What will be when the population is stable?
7. Find the word in the passage which means ‘make bitter’.

Answers:

1. The name of the essay is Too many people, too Few Trees and its writer is Moti Nisan.
2. The author thinks about the population that on first sight, it may appear that more people will do more work such as inventions etc. but as a whole, the world is already overpopulated.
3. Human populations cannot continue to grow indefinitely for the simple reason that the world itself is finite.
4. More people will need even more food than they need now, and therefore, the process of deforestation will continue.
5. As the population goes up, so does pollution of rivers, lakes, air, drinking water and soil. With more people, both town and country become more crowded. The quality of life and the value we place on human life will continue to erode.
6. When the population is stable, increases in such things as food production, quality of life.
7. Exacerbate.

3. The American government, to take another example, estimates that some 60,000 Americans die each year from respiratory diseases which are in turn caused by human-made pollution. Fourteen Americans die each day of asthtna aggravated by air pollution three times the incidence of just twenty years ago. Needless to say, the situation in cities like Los Angeles, Kathmandu, Mexico, and Shanghai is even worse. In all these cases, the situation could be considerably improved by controlling pollution and population. Moreover, the world, as we have seen, faces such frightening problems as desertification, depletion of nonrenewable resources (e.g. petrol, natural gas, helium), acid rain, loss of wild species, ozone layer depletion, and the greenhouse effect. A United Nations 1993 document puts it this way: “Population size and rates of growth are key elements in environmental change. At any level of development, increased populations increase energy use. resource consumption and environmental stress”. So, the more people the world has, more severe these problems are likely to become.

Questions:

1. How many Americans die each year from respiratory diseases?
2. What is the position of Kathmandu regarding population?
3. What is the frightening problem before the world?
4. What are the key elements in. environment change?

Answers:

1. 60,000 Americans die each year from respiratory diseases.
2. The situation of Kathmandu regarding population is worse, that is high.
3. The world has frightening problems as desertification, depletion of nonrenewable resources, acid rain, loss of wild species, ozone layer depletion and the greenhouse effect.
4. Population size and rates of growth are key elements in environmental change.

4. Thus a large and rapidly growing population make decisive contributions to all environmental problems. In the long run, efforts to save the biosphere depend impart on our species ability to roll back its numbers. Yet, there is a bright side to this otherwise grim tale. History and common sense tell us that we can control population growth. The German and Swedish population, for example, defy world trends and are actually declining. In such overpopulated countries like China, Thailand, and Egypt the rate of population growth has solved down remarkable, thanks to concerted government action. How do these countries manage to reverse, or slow down, population growth? Many factors account for these remarkable declines: modernization, literacy, media campaigns, readily available family planning and contraceptives. equal economic, educational, and legal opportunities for Women. Human beings thus know how to control their numbers. What they have been lacking so far as the resolve to make use of this knowledge.

Questions:

1. What are the growing population doing and how can bio-sphere be saved?
2. Which countries defy the trends of population growth?
3. What are the factors to slow down the population?
4. Find the word in the passage which mean planned activities’. .

Answers:

1. Large and rapidly growing population make decisive contributions to all environmental problems. The bio-sphere can be saved if the human species are able to roll I back its numbers.
2. Germany and Sweden defy the world trends of population growth.
3. The factors that account for the remarkable slow down of population in China, Thailand and Egypt are modernization, literacy, media campaigns, readily available
   contraceptives and equal economic, educational and legal opportunities for women.
4. Campaign.

5. Let us move to another long term problem: the state of the world’s trees. Owing to rapid population growth, poverty, and other factors, many third world people are forced to move into harvest, clear, burn, or, cultivate tropical forest. Thus, population pressures – along with new technologies and the affluent lifestyle of some people exacerbate the problem of deforestation. A country like Nepal has just so much arable land. So, as the population grows, more and more people are forced to convert forests into farmlands. They must also cut down more and more trees for fuel. The people of rich countries are also guilty. To satisfy Westerners’ insatiable demands for hamburgers, more and more tropical rain forests in countries like Brazil are cleared and converted to pastures. Some rich people also buy mahogany furniture, newspaper, and other paper products in vast quantities.lt are frightening to recall, for instance, how many trees must be felled to just produce the Sunday edition of the New York Times! Many forests are also damaged by pollution, tourism, construction of houses and factories, and similar practices. Moreover, the productivity and general health of the world’s forests are threatened by such things as the greenhouse effect, ozone layer depletion, airborne pollution, and acid rain.

Questions:

1. Why are people forced to move into the harvest?
2. Why does deforestation take place?
3. What is frightening to recall?
4. What is threatening for human beings?
5. Find the word in the passage which mean “land for grazing of cattle.”

Answers:

1. Owing to rapid population growth, poverty and other factors, many people are forced to move into the harvest.
2. Population pressures long with new technologies and the affluent lifestyle of some people exacerbate the problem of deforestation.
3. It is frightening to recall, for instance how many trees
   must be felled to just produce the Sunday edition of the ‘New York Times’.
4. The productivity and general health of the world’s forest is threatened by such things as the Green House Effect, Ozone Layer depletion, airborne pollution and acid rain.
5. Pastures.

6. The deforestation crisis is not new. Many earlier civilizations including those on Middle East, New Mexico, and Easter Island, precipitated their own decline through over-population and deforestation. The difference is that we are destroying our forests faster, and on a larger scale, than ever before. Earlier in this century, forests covered around 40% of the earth’s total land area. By this century’s end, that figure will stand at about 25%. The destruction of the forest, in turn, contributes to such things as the greenhouse effect, irreversible loss of many thousands of species of plants and animals, landslides, soil erosion, siltation of rivers and dams, droughts, and weather extremes. For instance, as the trees of Nepal are cut down, its topsoil is gradually being lost and its rains are likelier to cause devastating floods in India and Bangladesh. The eventual consequences of massive and ongoing deforestation are uncertain, but they are likely to damage the quality of life on earth, reduce the number of life forms that share the planet with us, and hamper the ability of the biosphere to sustain life. Humanity can continue to fell trees, cross its fingers, and hope for the best. Or it can take hold of its future and reverse the process of deforestation.

Questions:

1. What happened to earlier civilization?
2. What does the destruction of forest contribute?
3. What is the cause of flood in India and in Bangladesh?
4. What may be the eventual consequences of deforestation?
5. Find the word from the passage which means “hinder”.

Answers:

1. Many earlier civilizations, including those of middle East Mexico, and Easter Island precipitated their own decline. This was only due to overpopulation and deforestation.
2. The destruction of the forest, in turn, contributes to such things as the greenhouse effect, irreversible loss of many thousands of species of plants and animals, landslides soil erosion, siltation of rivers and dams, droughts and weather extremes.
3. As the trees of Nepal are cut down, its topsoil is gradually being lost and its rains are likelier to cause devastating floods in India and Bangladesh.
4. The eventual consequences of deforestation may be the damage to the quality of life on earth, reduction in the number of life forms that share the planet with us and hampering the ability of the biosphere to sustain life.
5. Hamper.

Hope you liked the article on Bihar Board Solutions of Class 9 English Chapter 4 Too Many People Too Few Trees Questins and Answers and share it among your friends to make them aware. Keep in touch to avail latest information regarding different state board solutions instantly.

Bihar Board Class 7 Maths Solutions Chapter 1 पूर्णांक की समझ Text Book Questions and Answers.

BSEB Bihar Board Class 7 Maths Solutions Chapter 1 पूर्णांक की समझ

Bihar Board Class 7 Maths पूर्णांक की समझ Ex 1.1

Bihar Board Class 7 Math Solution In Hindi प्रश्न 1.
निम्नलिखित के बीच की सभी पूर्णांक संख्याएँ लिखिए।
(a) -5 और 5
हल :
-4, -3, -2, -1, 0, 1, 2, 3, 4

(b) -2 और 8
हल :
-1, 0, 1, 2, 3, 4, 5, 6, 7

(c) -6 और -2
हल :
-5, -4, –

(d) -4 और -10
हल :
-5, -6, -7, -8, -9

Bihar Board Class 7 Math Solution प्रश्न 2.
निम्नलिखित में से प्रत्येक में बड़े पूर्णांक पर घेरा (0) लगाएं।
(a) -20, 4
(b) -15, -8
(c) 0, -5
(d) -20, -7
(e) 25, -2
(f) -20, -18
हल :
(a) 4
(b) -8
(c) 0
(d) -7
(e) 25
(f) -18

Bihar Board 7th Class Math Solution प्रश्न 3.
रिक्त स्थानों में उचित चिह्न (>, < और =) को भरिए।
(a) -6 > -8
(b) 4 > 0
(c) -5 < 2
(d) -50 = -54 + 4
(e) 25 = 25
(f) 4 – 15 > 2 – 20
हल :
(d) -50 = -54 + 4
-50 = -50
(∵ – + + = – होता है)
(f) 4 – 15 > 2 – 20
= -11 > -18

Bihar Board Solution Class 7 Math प्रश्न 4.
नीचे दिए गए पूर्णांकों को बढ़ते क्रम में लिखिए।
(a) -8, 12, -5, 15, 20, -2
हल :
-8, -5, -2, 12, 15, 20

(b) 5, 0, -2, -4, -15, 8
हल :
-15, -4, -2, 0, 5, 8

Bihar Board Class 7 Math Book Solution प्रश्न 5.
नीचे दिए गए पूर्णांकों की अगली पूर्णांक संख्या बताइए।
(a) -18
(b) 15
(c) -20
(d) 18
(e) -5
हल :
(a) -17
(b) 16
(c) -19
(d) 19
(e) -4

Class 7 Bihar Board Math Solution प्रश्न 6.
नीचे दिए गए पूर्णांकों के पहले का पूर्णांक बताइए।
(a) 25
(b) -59
(c) -55
(d) -26
(e) +100
हल :
(a) 24
(b) -60
(c) -56
(d) -27
(e) 99

Bihar Board Math Solution Class 7 प्रश्न 7.
रिक्त स्थान भरिए।
(i) (-5) + (2) = -3
(2) + (-5) = -3
क्या (-5) + (2) = 2 + (-5) है? हाँ
कुछ अन्य पूर्णांकों से संख्याएँ लेकर सारणी को पूरा कीजिए एवं जाँचिए-
हल :

Bihar Board Class 7 Maths Solutions प्रश्न 8.
रिक्त स्थानों की पूर्ति करें।
(i) (-a) + (6) = (6)+ (-a………)
(ii) -8 + 8 = 0
हल :
-8 + 8 = 0
(iii) (2) + [a + (-6)] = [2 + a) + (5)]
हल :
2 + [9 – 6] = 11
2 + 3 = 11 – 6
5 = 5
(iv) 15 + 0 = 15

Class 7 Math Solution Hindi Medium Bihar Board प्रश्न 9.
निम्नलिखित को जोड़िए।
(a) -15 में -18 को।
हल :
-33 (∵ – + – = + होता है और यहाँ – का चिह्न बड़े अंक के सामने है।)
(b) -20 में 17 को
हल :
-3 (∵ – + + = – होता है और – का चिह्न बड़े अंक के सामने है।)
(c) +24 में -16 को
हल :
8 (∵ + + – = – होता है और यहाँ – का चिह्न छोटे अंक के सामने है।)
(d) -8 में 5 को
हल :
-3 (∵ – + + = – होता है और यहाँ – का चिह्न बड़े अंक के सामने हैं।)

Class 7 Maths Bihar Board प्रश्न 10.
निम्नलिखित को घटाइए।
(a) -15 में से -5 को
हल :
-15 – (-5) = 15 + 5 = -10
(b) 25 में से -75 को
हल :
25 – (-75) = 25 + 75 = 100
(c) -8 में से -16 को
हल :
-8 – (-16) = -8 + 16 = 8
(d) -20 में से -18 को
हल :
-20 – (-18) = -20 + 18 = -2

Bihar Board Class 7 Math Book Pdf प्रश्न 11.
निम्नलिखित गुणों का एक-एक उदाहरण दीजिए।
(a) क्रम विनिमेय गुण (नियम)
हल :
दो पूर्णांक संख्याओं का योगफल एवं उनके उल्टे क्रम का योगफल समान होते हैं, इस गुण को क्रम विनिमेय गुण कहते हैं। .
उदाहरण-
(-4) + (-6) = -10
(-6) + (-4) = -10
अर्थात् (-4) + (-6) = (-6) + (-4)

(b) साहचर्य गुण
हल :
पूर्णांकों के लिए योग सहचारी होता है।
पूर्णांक a, b और c के लिए हम कह सकते हैं कि
a + (b + c) = (a + b) + c.

(c) संवरक गुण
हल :
दो पूर्णांकों का अंतर एक पूर्णांक संख्या होती है, इसे घटाव का संवरक गुण कहते हैं।
उदाहरण-
a और b दो पूर्णांक संख्याएँ हैं, तो a – b भी एक संख्या होगी।

(d) योज्य तत्समक
हल :
किसी पूर्णांक a के लिए a + 0 = a एवं 0 + a = a को योज्य तत्समक गुण कहते हैं।
उदाहरण-
3 + 0 = 3
पुन: 0 + 3 = 3
0 + 3 = 3 + 0

Class 7 Maths Bihar Board Solution प्रश्न 12.
ऐसा पूर्णांक युग्म लिखिए जिसका-
(a) योग -8
हल :
-4 + (-4)
= -4 – 4
= -8

(b) अन्तर -18 है
हल :
18 – 36
= -18

(c) योग 0 है
हल :
15 + (-15)
= 15 – 15
= 0

Bihar Board Class 7 Maths पूर्णांक की समझ Ex 1.2

Bihar Board Class 7 Maths प्रश्न 1.
गुणा कीजिए
(a) 225 × (-4)
हल :
225 × (-4) = -900

(b) (-405) × (-5)
हल :
(+405) × (+5) = 2025

(c) (-80) × (-50)
हल :
(+80) × (+50)
= 4000

(d) (-11) × 15
हल:
(-11) × 15
= -165

(e) (-3) × 35 × (-10)
हल :
(-3) × 35 × (-10) = +105 × (+ 10) = 1050

(f) (-25) × 0.
हल : 0

(g) (-4) × (-4) × (-4) × (-4)
हल :
(+4) × (+4) × (+4) × (+4)
= 4 × 4 × 4 × 4
= 256

(h) (-2) × (-2) × (-2)
हल :
(+2) × (+2) × (-2)
= 2 × 2 × (-2)
= 4 × (-2)
= -8

(i) (-20) × (+5) × (-25) × (-5)
हल :
(+20) × (+15) × (+25) × (+5)
= 20 × 15 × 25 × 5
= 37500

(j) -50 × 5 × (-20)
हल :
-50 × 5 × (-20) = +250 × (+20) = 5000

प्रश्न 2.
निम्नलिखित में सत्य और असत्य को चुनें-
(i) 18 × (-2) = (-2) × 18
हल :
18 × (-2) = (-2) × 18
-36 = -36 (सत्य है)

(ii) -38 × 1 = 38
हल :
-38 × 1 = 38 (असत्य है)

(iii) (-20) × (-5) = (-5) × (-20)
हल :
(+20) × (+5) = (+5) × (+20)
100 = 100 (सत्य है)

(iv) 43 × 0 = 43
हल :
43 × 0 = 43 (सत्य है)

(v) 1 × -425 = -425
हल :
1 × -425 = -425 (सत्य है)

(vi) -1 × 25 = -25
हल :
-1 × 25 = -25 (सत्य है)

(vii) [(-2) × (-12)] × -24 = (-2) × [(-12) × (-24)]
हल :
[(-2) × (-12)] × -24 = (-2) × [(-12) × (-24)]
(+2) × (+12) × -24 = (-2) × (+12) × (+24)
24 × -24 = (-2) × 288
-576 = -576 (सत्य है)

(viii) (-5) × (2 + 3) = (-5) × 2 + (-5) × 3
हल :
(-5) × (2 + 3) = (-5) × 2 + (-5) × 3
(-5) × 5 = (-5) × 2 + (-15)
-25 = -10 + -15
-25 = -25 (सत्य है)

प्रश्न 3.
निम्नलिखित के सामने उसके उचित गुण को लिखिए-
(i) -25 × (8 + 2) = (-25) × 8 + (-25) × 2
हल :
वितरण गुण।

(ii) (-8) × (4) = (-4) × (-8)
हल :
गुणन की क्रमविनिमेयता।

(iii) (20 × 30) × 40 = 20 × (30 × 40)
हल :
गुणन का साहचर्य गुण।

(iv) -2 × 10 = 20
हल :
संवरक गुण।

(v) -5 × 1 = -5
हल :
गुणा का तत्समक गुण।

प्रश्न 4.
सत्यापित करें-
(i) 42 × (-5) = -5 × 42
हल :
42 × (-5) = -5 × 42
-210 = -210
LHS = RHS (साबित हुआ)

(ii) 25 × (28 + 2) = 25 × 28 + 25 × 2
हल :
25 × (28 + 2) = 25 × 28 + 25 × 2
25 × (30) = 700 + 50
750 = 750
LHS = RHS (साबित हुआ)

(iii) (50 × 60) × 70 = 50 × (60 × 70)
हल :
(50 × 60) × 70 = 50 × (60 × 70)
3000 × 70 = 50 × (4200)
210000 = 210000
LHS = RHS (साबित हुआ)

(iv) (-24) × (5 × 2) = (-24 × 5) × 2
हल :
(-24) × (10) = (-120) × 2
-240 = -240
LHS = RHS (साबित हुआ)

प्रश्न 5.
किस पूर्णांक में (-1) का गुणा करने पर गुणनफल निम्न प्राप्त होते हैं-
(i) 20
हल :
-20
(+1) × (+20) = 20

(ii) -45
हल :
45
(-1) × (45) = -45

(iii) 0
हल :
0
(-1) × (0) = 0

(iv) 1
हल :
-1
(-1) × (1) = -1

(v) -50
हल :
50
(-1) × (-50) = +50

प्रश्न 6.
-4 × 0 = 0 के आधार पर सत्यापित कीजिए कि दो ऋण पूर्णांकों का गुणनफल धनात्मक पूर्णांक होता है।
हल :
(+4) × (+4) = 16
हम कह सकते हैं कि दो ऋण पूर्णांकों का गुणनफल धनात्मक पूर्णांक होता है।

प्रश्न 7.
सरल कीजिए-
(i) (-7) × 5 + (-7) × 11
हल :
(-7) × 5 + (-7) × 11.
= (-35) + (-77)
= -112

(ii) 675 × (-5) – 5 × (-675)
हल :
675 × (-5) – 5 × (-675)
= + 3375 – (+3375)
= 0

(iii) 8 × (50 – 4)
हल :
8 × (50-4)
= 8 × (46)
= 368

(iv) 5 × 27 × (-4)
हल :
5 × 27 × (-4)
= 135 × (-4)
= -540

(v) 987 × 98
हल :
987 × 98
= 96726

(vi) -57 × (-19) + 57
हल :
-57 × (-19) + 57
= 1083 + 57
= 1140

प्रश्न 8.
निम्नलिखित सारणी को पूरा करें-

हल :

प्रश्न 9.
निम्नलिखित में से कौन सत्य और कौन असत्य है?
(i) -20 का विपरीत 20 है।
हल :
सत्य

(ii) किसी पूर्णांक का विपरीत प्राप्त करने के लिए उसमें शून्य से गुणा करते हैं।
हल :
असत्य

(iii) 5 ऋण पूर्णांकों का गुणनफल धन पूर्णांक होता है।
हल :
असत्य

(iv) चार ऋण पूर्णांकों का गुणनफल धन पूर्णांक होता है।
हल :
सत्य

(v) -4 × 1 = -4
हल :
सत्य

(vi) -5 × 0 = 0
हल :
सत्य

प्रश्न 10.
निम्नलिखित में पूर्णांकों के गुणा में सही कथनों के आगे सही का निशान लगाएँ तथा गलत कथनों को सही करके लिखें।
(i) (+2) × (-3) = -6
हल :
सही है (✓)

(ii) (4) × (+8) = +32
हल :
सही है (✓)

(iii) (-2) × (-2) = +4
हल :
सही है (✓)

(iv) (+3) × (+4) = -12.
हल :
असत्य है (✗)

प्रश्न 11.
कमरे का तापमान = 40°C
प्रति घंटे कम किया गया तापमान = 5°C
10 घंटे बाद कमरे का तापमान = 10 × 5°C = 50°C.
40°C – 50°C = -10°C
कमरे का तापमान 10 घंटे बाद -10°C अर्थात् 0° से भी 10° नीचे होगा।

प्रश्न 12.
दस ग्रश्नांवाले एक कक्षा टस्ट में ग्रत्येक् सही उत्तर के तिए (-2) औक दिए जाते क्रेंय्य गयो प्रश्नों के तिए शून्य प्तिया जाता तै।
हल :
सही उत्तर के लिए दिए गए अंक = 5
गलत उत्तर के लिए दिए गए अंक = (-2)
(a) मोहन चार प्रश्नों के सही उत्तर देता है = 4 × 5 = 20 अंक
छह प्रश्नों के गलत उत्तर देता है = (-2) × 6 = -12
प्राप्त किए गए अंक = 20 – (-12) = 18
मोहन को 18 अंक प्राप्त होते हैं।

(b) रेशमा ने पाँच उत्तर सही दिए = 5 × 5 = 25 अंक
रेशमा ने पाँच उत्तर गलत दिए = 5 × (-2) = -10
रेशमा द्वारा प्राप्त किए गए अंक = 25 – (-10) = 15.
रेशमा को 15 अंक प्राप्त हुए।

(c) होना ने दो प्रश्नों के सही उत्तर दिए = 2 × 5 = 10
होना ने पाँच प्रश्नों के गलत उत्तर दिएं = 5 × (-2) = -10
हीना द्वारा प्राप्त किए गए अंक = 10 – (-10) = 20

प्रश्न 13.
सफेद सीमेंट बेचने पर प्रति बोरी लाभ = 8 रु. प्रति बोरी
स्लेटी सीमेंट बेचने पर प्रति बोरी हानि = 5 रु. प्रति बोरी
(a) 5000 बोरियाँ स्लेटी सीमेंट की बेची गयी = 5000 × 5 = 25000
3000 बोरियाँ सफेद सीमेंट की बेची गयी = 3000 × 8 = 24000
सीमेंट कंपनी को लाभ हुआ या हानि = 25000 – 24000
सीमेंट कंपनी को लाभ हुआ हानि = 1000 रु.
सीमेंट कंपनी 1000 रु. की हानि हुई।

(b) सफेद सीमेंट की बोरियों की संख्या = 6400
अर्थात् 6400 × 8 = 51200 रु० का लाभ।
स्लेटी सीमेंट में प्रत्येक बोरी पर हानि = 5 रु०
बोरियों की संख्या = \(\frac{51200}{5}\) = 10240

प्रश्न 14.
प्रदे भिद्युत कम्मगी प्रत्येक रंगीग टी॰वी॰ पर 80 रुपये का लाभ कमाता है और प्रत्येक रेश्रेत्ररटर पर 60 रूपये का चानि चाता ढ़े।
हल :
(a) लाभ = 5000 × 80 = 400000
हानि = 4000 × 60 = 240000
कुल लाभ = 160000

(b) 4000 टेलीविजन बेचने पर हानि = 4000 × 60 = 240000
रंगीन टी०वी० की संख्या = \(\frac{240000}{80}\) = 3000

Bihar Board Class 7 Maths पूर्णांक की समझ Ex 1.3

प्रश्न 1.
हल कीजिए-

प्रश्न 2.
निम्नलिखित गुणा की क्रिया को भाग की दो प्रक्रिया में बदलिए-
(a) 5 × 8 = 40
हल :
(i) 40 ÷ 5 = 8
(ii) 40 ÷ 8 = 5

(b) -4 × -6 = 24
हल :
(i) 24 ÷ (-4)= -6
(ii) 24 ÷ (-6) = -4

(c) -12 × 9 = -108
हल :
(i) -108 ÷ (-12) = 9
(ii) -108 ÷ (9) = -12

(d) -4 × -12 = 48
हल :
(i) 48 ÷ (-4) = -12
(ii) 48 ÷ (-12) = -4

(e) -10 × 8 = -80
हल :
(i) -80 ÷ (-10) = 8
(ii) -80 ÷ 8 = -10

प्रश्न 3.
खाली जगह में उपर्युक्त पूर्णांक भरिए.

हल :
(i) (+8) × (+12) = 96
(ii) (8) × (-9) = -72
(iii) 24 ÷ (-4) = -6
(iv) -80 ÷ 10 = -8
(v) -48 ÷ 6 = -8

प्रश्न 4.
निम्नलिखित में कौन सत्य तथा कौन असत्य है? हल कीजिए और बताइए।

प्रश्न 5.
दोपहर 12 बजे से रात 12 बजे के बीच का समय = 12
तापमान में घटाव = 12 × 2 = 24°
दोपहर 12 बजे तापमान = +10°C
रात 12 बजे तापमान = +10 – 24°
शून्य डिग्री से 8°C नीचे होने के लिए तापमान में कमी = +10 + 8 = 18°
समय = \(\frac{18}{2}\) = 9 घंटे

प्रश्न 6.
कुल तय की गई दूरी = 10m + 350m = 360m
6m तय करने में
360m = \(\frac{360}{6}\) = 60 मि० = 1 घंटा

Bihar Board Class 7 Maths पूर्णांक की समझ Ex 1.4

प्रश्न 1.
सरल कीजिए

प्रश्न 2.
कोष्ठकों का प्रयोग कर निम्नलिखित प्रश्नों को गणितीय रूप दीजिए
(i) (5 + 15) ÷ 8
(ii) 69 ÷ (4 × 6 – 1)
(iii) (24 – 4) ÷ 5 = 4
(iv) 124 ÷ {(25 + 5) + 1}
(v) [{(2 × 4) – 2} × 9] ÷ 6

प्रश्न 3.
सरल कीजिए-
(i) 50 + {15 – 5 + (8 – 2)}
हल :
50 + {15 – 5 + (8 – 2)}
= 50 + {15 – 5 + 6}
= 50 + {15 + 1}
= 50 + 16
= 66

(ii) 8 [6 + 2 {5 – 4(5 – 8}]
हल :
8 [6 + 2 {5 – 4(5 – 8)}]
= 8[6 + 2{5 – 4(-3)}]
= 8[6 + 2{5 + 12}]
= 8[6 + 2{17}]
= 8{6 + 34}
= 8 × 40
= 320

(iii) 12 ÷ \(\overline{6-2}\) + 10
हल :
12 ÷ \(\overline{6-2}\) + 10
= 12 ÷ 4 + 10
= 3 + 10
= 13

(iv) 15 + [2 – 3 – {2 (5 – 4 + 1)}]
हल :
15 + [2 – 3 – {2 (5 – 4 + 1)}]
= 15 + [2 – 3 – {2 (6 – 4)}]
= 15 + [2 – 3 – {2 (2)}]
= 15 + {2 – 3 – 4]
= 15 + [2 – 7]
= 15 + [-5]
= 15 – 5
= 10

(v) 103 – [144 ÷ (12 × 12) + 5 + 12 ÷ \(\overline{6-2}\) + 10]
हल :
103 – [144 ÷ (12 × 12) + 5 + 12 ÷ \(\overline{6-2}\) + 10]
= 103 – [144 ÷ (12 × 12) + 5 + 12 ÷ 4 + 10]
= 103 – [144 ÷ (144) + 5 + \(\frac{12}{4}\) + 10]
= 103 – [\(\frac{144}{144}\) + 5 + 3 + 10]
= 103 – [1 + 18]
= 103 – 19
= 84

(vi) 5 [5 – {5 – (5 – 5 – 5)}]
हल :
5 [5 – {5 – (5 – 5 – 5)}]
= 5 [5 – {5 – (5 – 10)}]
= 5 [5 – {5 – (-5)}]
= 5 [5 – {5 + 5}]
= 5 [5 – {10}]
= 5[5 – 10]
= 5 × -5
= -25

(vii) 15 – (-3) (4 – 4) ÷ {5 + (6) × (-3)}
हल :
15 – (-3)(4 – 4) ÷ {5 + (6) × (-3)}
= 15 – (-3) (4 – 4) ÷ {5 – 18}
= 15 – (-3) (0) ÷ (-13)
= 15 – \(\frac{0}{-13}\)
= 15 – 0
= 15

(viii) (-6) + (-6) ÷ 2 – [(-5) × (-1) – 2 (4 – 2)]
हल :
(-6) + (-6) ÷ 2 – [(-5) × (-1) – 2(4 – 2)]
= -6 – 6 ÷ 2 – [5 – {2(2)}]
= -12 ÷ 2 – [5 – 4]
= -12 ÷ 2 – [1]
= -12 ÷ 2 – 1
= -12 ÷ 1
= -12

Bihar Board Class 7 Maths Solutions Chapter 10 राशियों की तुलना Text Book Questions and Answers.

BSEB Bihar Board Class 7 Maths Solutions Chapter 10 राशियों की तुलना

Bihar Board Class 7 Maths राशियों की तुलना Ex 10.1

Bihar Board Class 7 Math Solution In Hindi प्रश्न 1.
अनुपात ज्ञात कीजिए-
(a) 3 किग्रा. का 600 ग्रा. से।
(b) 2 घंटे का 30 मिनट से।
(c) 340 सेमी. का 4 मीटर से।
(d) 75 रुपये का 200 पैसे से।
हल :

Bihar Board 7th Class Math Solution प्रश्न 2.
निम्नलिखित अनुपातों का सरलतम रूप लिखिए-
(a) 45 : 60
(b) 144 : 84
(c) 184 : 12
हल :

Bihar Board Class 7 Math Solution प्रश्न 3.
उपेन्द्र का वेतन 42000 रु. प्रतिमाह है और वह प्रतिमाह 6000 रु. आयकर में जमा करते हैं। ज्ञात कीजिए-
(a) आय का आयकर के साथ अनुपात
(b) आयकर का आय के साथ अनुपात
(c) क्या ये दोनों अनुपात तुल्य है?
हल :
कुल वेतन = 42000
(a) आय का आयकर के साथ अनुपात = \(\frac{42000}{6000}\) = 7 : 1
(b) आयकर का आय के साथ अनुपात = \(\frac{6000}{42000}\) = 1 : 7
(c) नहीं, यह दोनों अनुपात द्रव्य नहीं है।

Bihar Board Class 7 Math प्रश्न 4.
एक रिबन (Ribbon) की लम्बाई 10 मीटर एवं इसकी चौड़ाई 25 सेमी. है। निम्न का अनुपात ज्ञात कीजिए-
(a) लम्बाई का चौड़ाई के साथ
(b) चौड़ाई का लम्बाई के साथ
(c) क्या दोनों अनुपात तुल्य है?
हल :
कुल लम्बाई = 10 मी०
चौड़ाई = 25 cm
(a) कोल का चौ० के साथ अनपात = \(\frac{10 \times 100}{25 \mathrm{cm}}\) = \(\frac{40}{1}\) = 40 : 1
(b) चौ० का ल० के साथ अनुपात = \(\frac{25}{10 \times 100}\) = \(\frac{1}{40}\) = 1 : 40
(c) नहीं, ये द्रव्य नहीं है।

Bihar Board Math Solution Class 7 प्रश्न 5.
निम्नलिखित अनुपातों का दो तुल्य अनुपात ज्ञात कीजिए-
(a) 3 : 7
(b) 4 : 9
हल :

Bihar Board Class 7 Math Book Solution प्रश्न 6.
यदि किसी समानुपात के प्रथम तीन पद 3, 5 तथा 12 है तो चौथा पद ज्ञात कीजिए।
हल :
प्रथम तीन अनुपात = 3, 5, 12
3 : 5 :: 12 : x
\(\frac{3}{5}=\frac{12}{x}\)
x = \(\frac{12 \times 5}{3}\) = 20

Class 7 Bihar Board Math Solution प्रश्न 7.
यदि 3 : x :: 9 : 15 हो तो x का मान ज्ञात कीजिए।
हल :
3 : x :: 9 : 15
\(\frac{3}{x}=\frac{9}{15}\)
x = \(\frac{3 \times 15}{9}\) = 5

Math Class 7 Bihar Board प्रश्न 8.
बाजार में केले 18 रु. प्रति दर्जन बिक रहे है, तो 10 केलों का मूल्य क्या होगा?
हल :
1 दर्जन अर्थात् 12 केले = 18
1 केला का मू = \(\frac{18}{12}\)
10 केलों का मू० = \(\frac{18}{12}\) × 10 = 15 रु०

Bihar Board Class 7 Math Book प्रश्न 9.
मिठाई बनाने में चीनी और खोये का अनुपात 3:7 रखा जाये तो 12 किग्रा. चीनी की मिठाई बनाने हेतु कितने खोये की आवश्यकता होगी?
हल :
3 : 7 :: 12 : x
\(\frac{3}{7}=\frac{12}{x}\)
x = \(\frac{12 \times 7}{3}\) = 28

Class 7 Math Bihar Board प्रश्न 10.
एक मोटरसाइकिल 2 ली. में 120 किमी. दूरी तय करता है। बताइए 300 किमी. दूर तक जाने में कितने लीटर पेट्रोल की आवश्यकता होगी?
हल :
2 ली = 120 किमी
x = 300
x = \(\frac{300 \times 2}{120}\) = 5 ली.

Bihar Board Class 7th Math Solution प्रश्न 11.
एक मकान का 4 माह का किराया 10,000 रुपये हैं तो पूरे वर्ष का किराया बताइए।
हल :
4 माह का किराया = 10000
12 = x
x = \(\frac{10000 \times 12}{4}\) = 30000

Bihar Board Class 7 Maths राशियों की तुलना Ex 10.2

राशियों की तुलना कक्षा 7 प्रश्न 1.
दी गई भिन्न संख्याओं को प्रतिशत में बदलिए।
(a) \(\frac{3}{10}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{5}\)
(d) \(\frac{5}{8}\)
(e) \(\frac{7}{12}\)
हल :

Class 7 Math Solution Bihar Board प्रश्न 2.
दी गई दशमलव भिन्नों को प्रतिशत में बदलिए।
(a) 0.45
(b) 1.25
(c) 3.2
(d) 0.375
हल :

Bihar Board Math Class 7 प्रश्न 3.
दिए गए प्रतिशत को साधारण व दशमलव भिन्नों में बदलिए और अपने उत्तर को सरलतम रूप में लिखिए।
(a) 25%
(b) 18%
(c) 12\(\frac{3}{4}\)%
(d) 60%
हल :

Class 7 Math Solution Hindi Medium Bihar Board प्रश्न 4.
नीचे दिए गए चित्रों का कितना प्रतिशत भाग छायांकित है?

हल :
(a) \(\frac {3}{8}\)
(b) \(\frac {3}{5}\)
(c) \(\frac {7}{14}\) = \(\frac {1}{2}\)

7th Class Math Bihar Board प्रश्न 5.
एक व्यक्ति की मासिक आय 7000 रु. है तथा वह 1400 रु. प्रतिमाह बचाता है तो वह अपनी आय का कितना प्रतिशत वह खर्च करता है?
हल :
कुल आय = 7000
प्रतिमाह बचाता है = 1400
प्रतिमाह खर्च करता है.-

आप का 80% खर्च करता है तो बचाता है-

Bihar Board Solution Class 7 Math प्रश्न 6.
एक आदमी अपनी आय का चौथाई भाग भोजन पर, 25% शिक्षा पर तथा 22% किराया पर खर्च करता है। यदि वह 266 रु. बचाता है तो उसकी आय क्या है?
हल :
चौथाई भाग भोजन पर = 25%
शिक्षा पर = 25%
किराया पर = 22%
आय का खर्च = 72%
वह अपनी आय का (100 – 72)% बचाता है।
अर्थात् वह 28% बचाता है।
x का 28% = 266
x = \(\frac{266 \times 100}{28}\) = 950
कुल आय = 950

प्रश्न 7.
एक शहर की जनसंख्या प्रतिवर्ष 5% बढ़ जाती है। यदि उसकी वर्तमान जनसंख्या 5,14,700 है तो अगले वर्ष इसकी जनसंख्या क्या होगी।
हल :
प्रतिवर्ष 5% बढ़ती है-
अगले वर्ष की जनसंख्या = \(\frac{514700 \times 110}{100}\) = 56617000

प्रश्न 8.
किसी विद्यालय के छात्र संघ के निर्वाचन में अध्यक्ष पद के लिए दो छात्रों में सीधी टक्कर थी। यदि विजयी छात्रा को कुल 55% वोट मिले और वह 70 मतों से विजयी हुआ तो कुल कितने वैध मत पड़े और पराजित प्रत्याशी को कितने वोट मिले?
हल :
माना कुल मत = x

पराजित प्रत्याशी के वोट = \(700 \times \frac{45}{100}\) = 315

प्रश्न 9.
एक कुर्सी और एक टेबल दोनों की कुल कीमत 2800 रु. है। यदि कुर्सी की कीमत टेबल की कीमत से 40% कम है तो कुर्सी की कीमत बताइए।
हल :
माना कि टेबल की कीमत = x

कुर्सी की कीमत = 1750 – 1750 × \(\frac{40}{100}\) = 1750 – 700 = 1050

Bihar Board Class 7 Maths राशियों की तुलना Ex 10.3

प्रश्न 1.
निर्देशानुसार रिक्त स्थान की पूर्ति कीजिए-

हल :
(ii) क्रय मू० = 700
कि मु. = 679
हानि = 21 रु.
हानि% = \(\frac{21 \times 100}{700}\) = 3%
(iii) क्र० मू० = 300
वि० मू = 324
लाभ = वि. मू – क्र. मू० = (324 – 300) = 24
लाभ% = \(\frac{24 \times 100}{300}\) = 8%
(iv) क्र० मू० = 110
वि. मू = 88
हानि = (110 – 88) = 22
हानि% = \(\frac{22 \times 100}{110}\) = 20%

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिए-

हल :
(ii) क्र. मू० = 500
लाभ = 25
वि. मू० = 525
लाभ% = \(\frac{25 \times 100}{525}=4 \frac{16}{21}\)
(iii) क्र. मृ = 630
हानि = 70
वि० मू० = 560
हानि% = \(\frac{70 \times 100}{630}=\frac{100}{9}=11 \frac{1}{9}\)
(iv) क. मू० = 400
हानि = 40
वि० मू० = 360
हानि = \(\frac{40 \times 100}{400}\) = 10%

प्रश्न 3.
रिक्त स्थानों की पूर्ति कीजिए-

हल :
(ii) वि. मू० = 1400
लाभ = 280
क्र० मू० = 1120
लाभ% = \(\frac{280 \times 100}{1120}\) = 25%
(iii) वि० मू० = 950
हानि = 50
क्र० मू० = 1000
हानि% = \(\frac{50 \times 100}{950}\) = 5\(\frac{5}{19}\)%
(iv) वि. मू० = 375
लाभ = 25
क्र मू = 350
लाभ% = \(\frac{25 \times 100}{350}\) = \(\frac{50}{7}=7 \frac{1}{7}\)

प्रश्न 4.
एक वस्तु का क्रय मूल्य 80 रु. है और वह वस्तु 25% के लाभ पर बेची गई तो लाभ और विक्रय मूल्य बताइए।
हल :
क्रय मू० = 80
लाभ = 25%


लाभ = वि. मू० – क्र० मू. = 100 – 80 = 20

प्रश्न 5.
कोई मशीन 7% की हानि पर 837 रु. में बेची गई तो उसका क्रय मूल्य निकालिए।
हल :
वि. मू० = 837
हानि = 7%

प्रश्न 6.
किसी वस्तु की 72 रु. में बेचने से 10% की हानि होती है। बताइए कि उस वस्तु को कितने में बेचने 20% का लाभ होगा?
हल :
72 रु. में बेचने पर 10% की हानि होती है।

प्रश्न 7.
एक रेडियो को 880 रु. में बेचने से 10% लाभ होता है तो बताइए कि यदि उसे 760 रु. में बेचा जाय तो बेचने वाला कितने प्रतिशत के लाभ या हानि में रहेगा?
हल :
वि० मू० = 880
लाभ = 10%

प्रश्न 8.
एक कुर्सी 20% हानि पर 240 रु. में बिकती है। यदि विक्रय मूल्य 10% बढ़ जाए तो बताइए कि कितने प्रतिशत की हानि होगी?
हल :
वि. मू० = 240
हानि = 20%
क्र. मू = \(\frac{240 \times 100}{80}\) = 300
वि० मू. 10% बढ़ जाए तो = 240 + 24 = 264
हानि = 300 – 264 = 36
हानि = \(\frac{36 \times 100}{300}\) = 12%

प्रश्न 9.
एक व्यापारी ने 1 रुपये के 5 की दर से 1000 नींबू खरीदकर एक रुपये के 4 की दर से बेच दिया तो उसका लाभ प्रतिशत ज्ञात कीजिए।
हल :
1 रु. में 5 आम
1000 आम = \(\frac{1000}{5}\) = 200 रु०
1000 आम का क्र० मू० = 200 रु०
बेचा 1 रु. में 4 आम
अर्थात् 1000 आम का वि० मूळ = \(\frac{1000}{4}\) = 250
लाभ = 250 – 200 = 50
लाभ% = \(\frac{50 \times 100}{200}\) = 25%

प्रश्न 10.
एक दुकानदार ने दो साइकिलें 1100 रु. प्रति साइकिल के हिसाब से बेची। एक पर उसे 10% का लाभ एवं दूसरे पर उसे 20% की हानि हुई। बताइए उसे लाभ हुआ कि नहीं? लाभ या हानि प्रतिशत में ज्ञात कीजिए।
हल :
प्रत्येक साइकिल का क्र. मू० = 1100
दो साइकिलों का क्र० मू० = 1100 × 2 = 2200
एक पर उसे 10% का लाभ हआ।
वि० म० = \(\frac{1100 \times 110}{100}\)
दूसरे पर 20% की हानि हुई-
वि० म० = \(\frac{1100 \times 80}{100}\) = 880
कुल वि. मू० = 1210 + 880 = 2090 उसे लाभ नहीं हुआ।
उसे हानि हुई है।
हानि मू० = 2200 – 2090 = 110
हानि% = \(\frac{110 \times 100}{2200}\) = 5%

Bihar Board Class 7 Maths राशियों की तुलना Ex 10.4

प्रश्न 1.
750 रु. का 9% वार्षिक व्याज की दर से 6 वर्षो का ब्याज ज्ञात कीजिए तथा मिश्रधन भी निकालिए।
हल :
मू० = 750 (P = 750)
दर = 9% (R = 9%)
समय = 6 (T = 6)

मिश्रधन = 750 + 405 = 1155

प्रश्न 2.
500 रु. का 1 रु. 50 पैसे प्रति सैकड़े प्रतिमाह की दर से 15 महीने का ब्याज ज्ञात कर्रे।
हल :
मूलधन = 500 = 5 सेकेण्ड
1.50 प्रति सैंकड़े/प्रतिमाह
500 रु. = 5 × 1.50/प्रतिमाह = 7.50/प्रतिमाह
15 माह = 7.50 × 15 = 112.50

प्रश्न 3.
कितने प्रतिशत वार्षिक ब्याज की दर से कोई मूलधन 4 वर्षों में अपना सवा गुना हो जाएगा?
हल :
माना कि० मूलधन = 100
सवा गुना = 125
समय = 4
ब्याज = 125 – 100 = 25

प्रश्न 4.
कितने प्रतिशत ब्याज की दर से 450 रु. तीन वर्षों में 504 रु. हो जाएगा?
हल :
मू० = 450
समय = 3 वर्ष
ब्याज = 504 – 450 = 54

प्रश्न 5.
यदि कोई मिश्रधन 5 वर्षों में मूलधन का \(\frac {5}{4}\) हो जाता है तो ब्याज की दर ज्ञात कीजिए।
हल :
समय = 5 वर्ष
माना कि मू = 100

प्रश्न 6.
कितने वर्षों में 5% वार्षिक ब्याज की दर से 600 रु. का मिश्रधन 700 रु. हो जाएगा?
हल :
मू० = 600
ब्याज = 700 – 600 = 100
दर = 5%

प्रश्न 7.
कितने समय में 6\(\frac {1}{2}\)% वार्षिक ब्याज की दर से कोई धन दुगुना हो जाएगा?
हल :
माना कि मू० धन = 100
मिश्रधन (दुगुना) = 200
ब्याज = 200 – 100 = 100
दर = 6.5% (6\(\frac {1}{2}\))

प्रश्न 8.
12% वार्षिक ब्याज की दर से कौन सा धन 5 वर्षों में 400 रु. हो जाएगा?
हल :
दर = 12%
समय = 5
माना कि ब्याज = 400
मू० = \(\frac{400 \times 100}{12 \times 5}=\frac{8000}{12}\) = 666.66

प्रश्न 9.
कितना धन 5% वार्षिक ब्याज की दर से 8 वर्षों में 560 रु. हो जाएगा?
हल :
माना कि मू० = 100
Rs = 5.
T = 8
तो ब्याज = \(\frac{100 \times 8 \times 5}{100}\) = 40
जब मू० 100 तो मिश्रधन = 140 {100 + 40}
जब मू० x तो मिश्रधन = 560
140 × x = 100 × 560
x = \(\frac{100 \times 560}{140}\) = 400

प्रश्न 10.
कितने धन का 6% वार्षिक ब्याज की दर से 2.5 वर्ष में वही ब्याज होगा जो 400 रु. का 5 वार्षिक ब्याज की दर से 3 वर्षों में होगा?
हल :
प्रश्नानुसार,

Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड Text Book Questions and Answers.

BSEB Bihar Board Class 8 Maths Solutions Chapter 14 गुणनखंड

Bihar Board Class 8 Maths गुणनखंड Ex 14.1

गुणनखंड ज्ञात कीजिए Class 8 Bihar Board प्रश्न 1.
दिए गए पदों में सार्व (उभयनिष्ठ) गुणनखंड ज्ञात कीजिए-
(a) 9y, 27
(b) 5x, 25x
(c) 7ab, -14ab
(d) -16x²y², -x²y²z²
(e) 17x, 102y
(f) 11xyz, 100z
(g) a²bc, ab²c, abc²
(h) 2x, 3y, 5z
(i) 20x²y², 30y²z², 40z²x²
(j) 2x (a + b)(b + c), x (a + b)
उत्तर
(a) 9y = 3 × 3 × y
27 = 3 × 3 × 3
सार्व उभयनिष्ठ गुणनखण्ड = 3 × 3 = 9

(b) 5x = 5 × x
25x = 5 × 5 × x
सार्व गुणनखण्ड = 5x

(c) 7ab = 7 × a × b
-14ab = -2 × 7 × a × b
सार्व गुणनखण्ड = 7ab

(d) -16x²y² = -2 × 2 × 2 × 2 × x × x × y × y
x²y²z² = -x × x × y × y × z × z
सार्व गुणनखण्ड = x²y²

(e) 17x = 17 × x
102y = 6 × 17 × y
सार्व गुणनखण्ड = 17

(f) 11xyz = 11 × x × y × z
100z = 2 × 2 × 5 × 5 × z
सार्व गुणनखण्ड = z

(g) a²bc = a × a × b × c
ab²c = a × b × b × c
abc² = a × b × c × c
सार्व गुणनखंड = a × b × c = abc

(h) 2x = 2 × x × 1
3y = 3 × y × 1
5z = 5 × z × 1
सार्व गुणनखंड = 1

(i) 20x²y² = 2 × 2 × 5 × x × x × y × y
30y²z² = 2 × 3 × 5 × y × y × z × z
40z²x² = 2 × 2 × 2 × 5 × z × z × x × x
सार्व गुणनखण्ड = 2 × 5 = 10

(j) 2x (a + b) (b + c) = 2 × x × (a + b) × (b + c)
x (a + b) = x × (a + b)
सार्व गुणनखण्ड = x(a + b)

Gunankhand Class 8 Bihar Board प्रश्न 2.
दिए गए उदाहरण के आधार पर खाली जगह को भरिए-

उत्तर

गुणनखंड कक्षा 8 Bihar Board प्रश्न 3.
निम्नलिखित का गुणनखण्ड ज्ञात कीजिए-
(a) 12x² – 15y² – 24x²z²
(b) -6a² + 36a – 24ab
(c) 3a² + ab + 9a + 3b
(d) 6ab – 4b + 6 – 9a
(e) ab² + a²b + ac + bc
(f) a²bc + b²ca + c²ab + a + b + c
(g) a(b – c) + d(c – b)
(h) 3y(y + 3) + 6y(3y + 9)
(i) a³ – 3a² + a – 3
(j) ab² – bc² – ab + c²
(k) xy(a² + b²) + ab (x² + y²)
उत्तर
(a) 12x² – 15y² – 24x²z²
= 3 × 2 × 2 × x × x – 3 × 5 × y × y – 2 × 2 × 2 × 3 × x × x × z × z
= 3(4x² – 5y² – 8x²z²)

(b) -6a2 + 36a – 24ab
= -2 × 3 × a × a + 2 × 2 × 3 × 3 × a – 2 × 2 × 2 × 3 × a × b
= -6a (a – 6 + 4b)

(c) 3a² + ab + 9a + 3b
= a (3a + b) + 3(3a + b)
= (3a + b) (a + b)

(d) 6ab – 4b + 6 – 9a
= 2b (3a – 2) – 3 (3a – 2)
= (3a – 2) (2b – 3)

(e) ab² + a²b + ac + bc
= ab (b + a) + c (b + a)
= (b + a) (ab + c)

(f) a²bc + b²ca + c²ab + a + b + c
= abc (a + b + c) + 1 (a + b + c)
= (a + b + c) (abc + 1)

(g) a(b – c) + d (c – b)
= ab – ac + dc – db
= a(b – c) – b(b – c)
= (a – b) (b – c)

(h) 3y (y + 3) + 6y (3y + 9)
= 3y² + 9y + 18y² + 54y
= 21y² + 63y
= 21y (y + 3)

(i) a³ – 3a² + a – 3
= a²(a – 3) + 1 (a – 3)
= (a – 3) (a² + 1)

(j) ab² – bc² – ab + c²
= b(ab – c²) – 1(ab – c²)
= (ab – c²) (b – 1)

(h) xy(a² + b²) + ab (x² + y²)
= xya² + xyb² + abx² + aby²
= ay(ax + by) + bx(ax + by)
= (ay + bx) (ax + by)

Bihar Board Class 8 Maths गुणनखंड Ex 14.2

Bihar Board Class 8 Math Solution प्रश्न 1.
निम्नलिखित व्यंजकों का गुणनखंड ज्ञात कीजिए-
(a) 1 + 2x + x²
(b) a²b² – 6abc + 9c²
(c) 1 – (a – b)²
(d) 16 (a – b)² – 9 (a + b)
(e) (x + y)² – 10 (x + y) + 25
(f) (a + b)² – 4ab
(g) 4x² – y²+ 4y – 4
(h) 9x² – \(\frac{n^{2}}{4}\)
(i) a² + a + 4 + 3a
(j) x² + 6x + 8
(k) y² – 13y + 30
(l) x² + 9x – 22
उत्तर
(a) 1 + 2x + x²
= 1 + x + x + x²
= 1(1 + x) + x(1 + x)
= (1 + x)(1 + x)

(b) a²b² – 6abc + 9c²
= a²b² – 3abc – 3abc + 9c²
= ab(ab – 3c) – 3c(ab – 3c)
= (ab – 3c) (ab – 3c)

(c) 1 – (a – b)²
= 1 – (a² – 2ab – b²)
= 1 – a² + 2ab + b²
= (1 – a – b)(1 + a – b)

(d) 16(a – b)² – 9 (a + b)²
= 16(a² – 2ab + b²) – 9 (a² + 2ab + b²)
= 16a² – 32ab + 16b² – 9a² – 18ab – 9b²
= 7a² + 7b² – 50ab
= 7a² – 50ab + 7b²
= 7a² – ab – 49ab + 7b²
= a(7a – b) – 7b (7a – b)
= (7a – b)(a – 7b)

(e) (x + y)² – 10(x + y) + 25
= (x + y)² – 5(x + y) – 5(x + y) + 25
= (x + y) [(x + y) – 5] – 5 [(x + y) – 5]
= (x + y – 5) (x + y – 5)

(f) (a + b)² – 4ab
= a² + 2ab + b² – 4ab
= a² – 2ab + b²
= a² – ab – ab + b²
= a(a – b) – b(a – b)
= (a – b)(a – b)

(g) 4x² – y² + 4y – 4
= 4x² – (y + 2)²
= (2x – y + 2) (2x + y – 2)

(h) 9x² – \(\frac{n^{2}}{4}\)
= (3x)² – \(\left(\frac{n}{2}\right)^{2}\)
= \(\left(3 x-\frac{n}{2}\right)\left(3 x+\frac{n}{2}\right)\)

(i) a² + a + 4 + 3a
= a² + 4a + 4
= a² + 2a + 2a + 4
= a(a + 2) + 2(a + 2)
= (a + 2) (a + 2)

(j) x² + 6x + 8
= x² + 4x + 2x + 8
= x(x + 4) + 2(x + 4)
= (x + 4) (x + 2)

(k) y² – 13y + 30
= y² – 10y – 3y + 30
= y(y – 10) – 3 (y – 10)
= (y – 10) (y – 3)

(l) x² + 9x – 22
= x² + 11x – 2x – 22
= x(x + 11) – 2(x + 11)
= (x + 11)(x – 2)

Bihar Board 8th Class Math Solution प्रश्न 2.
निम्नलिखित व्यंजकों का गुणनखण्ड कीजिए-
(a) x² – 6x – 135
(b) 8(x + y)³ – 50(x + y)
(c) 4x² + 9y² + 12xy – 1
(d) 75 – x² + 10x
(e) 12a² – 27
(f) ax² – bx² + by² – ay²
उत्तर
(a) x² + 6x – 135
= x² – 15x + 9x – 135
= x (x – 15) + 9 (x – 15)
= (x – 15) (x + 9)

(b) 8(x + y)³ – 50(x + y)
= 2 (x + y) (2x + 2y – 5) (2x + 2y + 5)
= 2 (x + y) (2x + 2y – 5) (2x + 2y + 5)

(c) 4x² + 9y² + 12xy – 1
= (2x)² + (3y)² + 2(2x) (3y) – 1
= (2x + 3y)² – 12
= (2x + 3y + 1) (2x + 3y – 1)

(d) 75 – x² + 10x
= -x² + 10x + 75
= -x² + 15x – 5x + 75
= x (x – 15) – 5 (x – 15)
= (x – 15) (x – 5)

(e) 12a² – 27
= 3 (4a² – 9)
= 3 [(2a)² – 32]
= 3 (2a – 3) (2a + 3)

(f) ax² – bx² + by² – ay²
= x² (a – b) + y² (b – a)
= (x² – y²) (a – b)
= (x + y) (x – y) (a – b)

Bseb Class 8 Math Solution प्रश्न 3.
निम्नलिखित व्यंजकों का गुणनखंडन कीजिए-
(a) 16x⁴ – 81y⁴
(b) x⁴ – 1
(c) x⁴ – (x – y)⁴
(d) 9x² – 4y² – 3x + 2y
(e) (x + y) + 4 (x + y)² + 4x + 4y
उत्तर
(a) 16x⁴ – 81y⁴
= ((2x)²)² – ((3y)²)²
= (2x + 3y)² – (2x – 3y)²
= (4x² + 12xy + 9y²) (4x² – 12xy + 9y²)
= (2x – 3y) (2x + 3y) (4×2 + 9y2)

(b) x⁴ – 1
= (x²)² – (1²)²
= (x² + 1) (x² – 1)
= (x – 1) (x + 1) (x² + 1)

(c) x⁴(x – y)⁴
= (x²)² – ((x – y)²)²
= (x² – x – y) (x² – 2xy + y²)²
= y(2x – y) (2x² – 2xy + y²)

(d) 3x² – 4y² – 3x + 2y
= (3x)² – (2y)² – 3x + 2y
= (3x + 2y) (3x – 2y) – (3x + 2y)
= (3x + 2y) (3x + 2y – 1)

(e) (x + y)³ + 4(x + y) + 4x + 4y
= x³ + 3x²y + 3xy² + y² + 4 (x² + 2xy + y²) + 4x + 4y
= x³ + 3x²y + 3xy² + y² + 4x² + 8xy + 4y² + 4x + 4y
= (x + y) (x + y + z) (x + y + z)

Bihar Board Class 8 Maths गुणनखंड Ex 14.3

Class 8 Math Bihar Board प्रश्न 1.
निम्नलिखित का भाग कीजिए
(a) -2x²yz का 4xyz से
(b) \(-\frac{1}{2}\) का \(\frac{x}{2}\) से
(c) (3x²)⁵ का (9x²)³ से
(d) (7x⁵)² × (3y⁵)⁵ का 27y³ से
(e) 8x⁶y⁶ का -4x⁴y⁶ से
उत्तर

Gunankhand Math Class 8 प्रश्न 2.
दिए गए बहुपद को एकपदी से भाग कीजिए-
(a) (5m³ – 30m²) ÷ 5m
(b) (12x⁴ – 6x²) ÷ (-3x²)
(c) (5x² – 15x) ÷ (x – 3)
(d) (6x⁴ + 9x³ – 12x²) ÷ 3x²
उत्तर

Class 8 Maths Bihar Board प्रश्न 3.
(a) (a² + 8a + 16) ÷ (a + 4)
(b) {(a + b)² – 4ab} ÷ (a – b)²
(c) (a⁴ – b⁴) ÷ (a² – ab)
(d) (x⁴ – 81) ÷ (x² + 9)
(e) 121x² + 16y² – 88xy ÷ 4y – 11x
(f) (x² – x – 30) ÷ (x – 6)
(g) (p² – p + \(\frac{1}{4}\)) ÷ (p – \(\frac{1}{2}\))
(h) (x² – 5xy + 6y²) ÷ (x – 2y)
(i) (27x³ + 3x² – 2x + 8) ÷ (3x – 2)
उत्तर

Bihar Board Class 8 Maths Solutions Chapter 5 वर्ग और वर्गमूल Text Book Questions and Answers.

BSEB Bihar Board Class 8 Maths Solutions Chapter 5 वर्ग और वर्गमूल

Bihar Board Class 8 Maths वर्ग और वर्गमूल Ex 5.1

Bihar Board Class 8 Math Solution प्रश्न 1.
निम्नलिखित सं का वर्ग ज्ञात करें।
(i) 42
(ii) 46
(iii) 58
(iv) 98
(v) 94
(vi) 45
उत्तर

Class 8 Bihar Board Math Solution प्रश्न 2.
निम्नलिखित का वर्ग निकालें।
(i) 25
(ii) 55
(iii) 95
(iv) 105
(v) 115
उत्तर

Bihar Board Solution Class 8 Math प्रश्न 3.
निम्नलिखित संख्याओं में से कौन-सी संख्याएँ पूर्ण वर्ग हैं? जाँच कीजिए।
(i) 256
(ii) 360
(iii) 324
(iv) 400
उत्तर

Bihar Board Math Solution Class 8 प्रश्न 4.
निम्नलिखित संख्याओं में से कौन-कौन पूर्ण वर्ग हैं?
13, 16, 17, 48, 49, 64, 72, 343, 373758
उत्तर
13 = 13 × 1 (पूर्ण वर्ग नहीं है)
16 = 2 × 2 × 2 × 2 = 2 × 2 = 4 (पूर्ण वर्ग है)
17 = 1 × 17 (पूर्ण वर्ग नहीं है)
48 = 2 × 2 × 2 × 2 × 3 (पूर्ण वर्ग नहीं है)
49 = 7 × 7 (पूर्ण वर्ग है)
64 = 2 × 2 × 2 × 2 × 2 = 2 × 2 × 2 (पूर्ण वर्ग है)
72 = 2 × 2 × 2 × 3 × 3 (पूर्ण वर्ग नहीं है)
343 = 7 × 7 × 7 (पूर्ण वर्ग नहीं है)
337358 = 2 × 3 × 7 × 11 × 809 (पूर्ण वर्ग नहीं है)

Bihar Board Class 8 Math Book Solution प्रश्न 5.
निम्नलिखित में कौन सम संख्या के वर्ग हैं?
169, 196, 256, 1296, 6561
उत्तर
दिए गए संख्याओं में 196, 256 तथा 1296 के इकाई स्थान पर सम सं० है।
ये सम संख्या के वर्ग हैं।

Bihar Board 8th Class Math Solution प्रश्न 6.
निम्नलिखित संख्याओं में से कौन-सी पूर्ण वर्ग है?
400, 4000, 330550, 12345600000
उत्तर

Bihar Board Class 8 Maths Solutions प्रश्न 7.
कोष्ठक में सही संख्या लिखें :
(a) 24² – 23² = ______
(b) 102² – 101² = ______
(c) 501² – 500² = ______
(d) 400² – 399² = ______
उत्तर
(a) 24² – 23²
= (24 + 23) (24 – 23)
= 47 × 1
= 47

(b) 102² – 101²
= (102 + 101) (102 – 101)
= 203 × 1
= 203

(c) 501² – 500²
= (501 + 500) (501 – 500)
= 1001 × 1
= 1001

(d) 400² – 399²
= (400 + 399) (400 – 399)
= 799 × 1
= 799

Class 8 Math Solution Bihar Board प्रश्न 8.
निम्नलिखित में कौन-सा त्रिक पाइथागोरस त्रिक है?
(1, 2, 3), (3, 4, 5), (6, 8, 10), (1, 1, 1), (2, 2, 3), (15, 36, 39)
उत्तर
पाइथागोरस त्रिक = a² + b² = c²
(1, 2, 3) = 1² + 2² = 3² = 1 + 4 ≠ 9 (पाइथोगरस त्रिक नहीं है)
(3, 4, 5) = 3² + 4² = 5² = 9 + 16 = 25 = 25 = 25 (पाइथोगोरस त्रिक है)
(6, 8, 10) = 6² + 8² = 10² = 36 + 64 = 100 = 100 = 100 (पाइथागोरस त्रिक है)
(1, 1, 1) = 1² + 1² = 12 = 1 + 1 = 1 = 2 ≠ 1 (पाइथागोरस त्रिक नहीं है)
(2, 2, 3) = 2² + 2² = 3² = 4 + 4 = 9 = 8 ≠ 9 (पाइथागोरस एक त्रिक नहीं है)
(15, 36, 39) = 15² + 36² = 39² = 225 + 1296 = 1521 = 1521 = 1521 (पाइथागोरस त्रिक है)

Bihar Board Class 8th Math Solution प्रश्न 9.
निम्नलिखित प्रतिरूप का प्रेक्षण करके छुटी हुई संख्याओं को ज्ञात करें :

उत्तर

Bihar Board 8 Class Math Solution प्रश्न 10.
विषम संख्याओं के क्रमिक घटाव की क्रिया द्वारा निम्नलिखित संख्याओं की जाँच करें कि कौन-सी संख्या पूर्ण वर्ग संख्या है?
(i) 81
(ii) 121
(iii) 144
(iv) 36
उत्तर
(i) 81
81 – 1 = 80
80 – 3 = 77
77 – 5 = 72
72 – 7 = 65
यह पूर्ण वर्ग संख्या है।
65 – 9 = 56
56 – 11 = 45
45 – 13 = 32
32 – 15 = 17

(ii) 121
121 – 1 = 120
यह पूर्ण वर्ग संख्या है।
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
57 – 17 = 40
40 – 19 = 21
72 – 15 = 57
21 – 21 = 0

(ii) 144
36 – 1 = 35
35 – 3 = 32
32 – 5 = 27
27 – 7 = 20
20 – 9 = 11
11 – 11 = 0
यह भी एक पूर्ण वर्ग है।

(iv) 36
144 – 1 = 143
143 – 3 = 140
140 – 5 = 135
135 – 7 = 128
128 – 9 = 119
119 – 11 = 108
108 – 13 = 95
95 – 15 = 80
80 – 17 = 63
63 – 19 = 44
44 – 21 = 23
23 – 23 = 0
यह भी एक पूर्ण वर्ग है।

Bihar Board Class 8 Math Solution In Hindi Pdf Download प्रश्न 11.
निम्नलिखित संख्याओं में से किन-किन संख्या का वर्ग विषम संख्या होगा?

1. 531
2. 5436
3. 3249
4. 82004

उत्तर

1. 531 – वर्ग विषम होगा
2. 5436 – वर्ग सम होगा
3. 3249 – वर्ग विषम होगा
4. 2004 – वर्ग सम होगा

Bihar Board Class 8 Math Solution In Hindi प्रश्न 12.
योग संक्रिया किये बिना योगफल ज्ञात कीजिए :

1. 1 + 3 + 5 + 7 + 9 + 11
2. 1 + 3 + 5 +…………+ 51
3. 1 + 3 + 5 + 7 +………..+ 101
4. 7 + 9 + 11 + 13 +……….+ 21

उत्तर

1. 1 + 3 + 5 + 7 + 9 + 11 = 36
2. 1 + 3 + 5 +…….+ 51 = (26)² = 676
3. 1 + 3 + 5 + 7 +…….. 101 = (51)² = 2601
4. 7 + 9 + 11 + 13 + …….. + 21 = 112

Bihar Board Class 8 Maths Solution प्रश्न 13.
निम्नलिखित संख्याओं के वर्गों के इकाई के अंक क्या होंगे?

1. 25
2. 64
3. 272
4. 799
5. 5423
6. 2467
7. 5438
8. 99880
9. 43546

उत्तर

1. 25 के वर्ग का इकाई का अंक = 5
2. 64 के वर्ग का इकाई का अंक = 6
3. 272 के वर्ग का इकाई का अंक = 4
4. 799 के वर्ग का इकाई का अंक = 1
5. 5423 के वर्ग का इकाई का अंक = 9
6. 2467 के वर्ग का इकाई का अंक = 9
7. 5438 के वर्ग का इकाई का अंक = 4
8. 99880 के वर्ग का इकाई का अंक = 0
9. 43546 के वर्ग का इकाई का अंक = 6

प्रश्न 14.
निम्नलिखित संख्याएँ स्पष्ट रूप से पूर्ण वर्ग संख्याएँ नहीं हैं, इसका कारण दीजिए।

1. 1052
2. 23457
3. 54328
4. 325473
5. 25000
6. 743522
7. 543000
8. 56430

उत्तर

निम्नलिखित सं० पूर्ण रूप से वर्ग संख्याएँ नहीं है क्योंकि-

1. 1052 – जिन संख्याओं के इकाई स्थान पर 2 हो वो कभी पूर्ण संख्या नहीं कहलाती।
2. 23457 – जिन संख्याओं के अंत में 7 होता है वो कभी पूर्ण वर्ग संख्या नहीं कहलाती।
3. 54328 – जिन संख्याओं के अंत में 54328 होता है वो कभी पूर्ण वर्ग सं॰ नहीं कहलाती।
4. 325473 – जिन संख्याओं की इकाई की सं० 3 हो वे पूर्ण वर्ग संख्याएँ नहीं होती।
5. 25000 – जिन संख्याओं के अंत में शून्य की संख्या विषम हो वे कभी पूर्ण वर्ग सं. नहीं होती।
6. 743522 – जिन संख्याओं के अंत में 2 होता है वे कभी पूर्ण वर्ग सं॰ नहीं होता।
7. 543000 – जिन संख्याओं के अंत में विषम शून्य संख्याएँ होती हैं वे कभी पूर्ण वर्ग संख्याएँ नहीं होती हैं।
8. 56430 – जिन संख्याओं के अंत में विषम शून्य संख्या होती है वे कभी पूर्ण वर्ग सं० नहीं होती है।

Bihar Board Class 8 Maths वर्ग और वर्गमूल Ex 5.2

प्रश्न 1.
निम्नलिखित संख्याओं का वर्गमूल ज्ञात करें
(i) 625
(ii) 900
(iii) 1444
(iv) 3249
(v) 5776
(vi) 10404
(vii) \(\overline{19600}\)
उत्तर

प्रश्न 2.
निम्नलिखित संख्याओं में से प्रत्येक के वर्गमूल में अंकों की संख्या बिना गणना के ज्ञात करें।

1. 81
2. 121
3. 256
4. 4489
5. 361
6. 27225
7. 390625

उत्तर

1. 81 के वर्गमूल में अंक की सं० = 1
2. 121 के वर्गमूल में अंकों की सं० = 42
3. 256 के वर्गमूल में अंकों की सं० = 2
4. 4489 के वर्गमूल में अंकों की सं० = 2
5. 361 के वर्गमूल में अंकों की सं० = 2
6. 27225 के वर्गमूल में अंकों की सं. = 3
7. 390625 के वर्गमूल में अंकों की सं० = 3

प्रश्न 3.
निम्नलिखित भिन्नों का वर्गमूल ज्ञात करें।

उत्तर

प्रश्न 4.
निम्नलिखित दशमलव संख्याओं का वर्गमूल ज्ञात करें।
(i) 2.25
(ii) 6.76
(iii) 156.25
(iv) 9.8596
(v) 31.36
(vi) 1.816
(vii) 0.2916
उत्तर

प्रश्न 5.
निम्नलिखित संख्याओं में से प्रत्येक में सबसे छोटी-से-छोटी संख्या क्या घटज्ञई जाए कि पूर्ण वर्ग संख्या प्राप्त हो जाए। इस प्रकार प्राप्त पूर्ण वर्ग संख्याओं का वर्गमूल भी ज्ञात करें।
(i) 90
(ii) 7581
(iii) 1989
(iv) 3250
(v) 402
(vi) 825
(vii) 4000
(viii) 2509
उत्तर

प्रश्न 6.
निम्नलिखित संख्याओं में से प्रत्येक में न्यूनतम संख्या क्या जोड़ी जाए कि वह एक पूर्ण संख्या बन जाए। इस प्रकार प्राप्त पूर्ण वर्ग संख्याओं का वर्गमूल भी ज्ञात करें।
(i) 130
(ii) 8400
(iii) 6203
(iv) 6412
(v) 525
(vi) 1750
(vii) 252
(viii) 1825
उत्तर

प्रश्न 7.
छः अंकों की वह बड़ी-से-बड़ी संख्या ज्ञात करें जो कि एक पूर्ण वर्ग संख्या है। संख्या का वर्गमूल भी ज्ञात करें।
उत्तर

अभीष्ट सं० = 999999 – 1998 = 998001
वर्गमूल = 999

प्रश्न 8.
चार अंकों की वह बड़ी-से-बड़ी संख्या ज्ञात कीजिए जो कि एक पूर्ण वर्ग संख्या है। प्राप्त वर्ग संख्या का वर्गमूल भी ज्ञात कीजिए।
उत्तर
चार अंकों की बड़ी सं = 9999

अभीष्ट सं० = 9999 – 198 = 9801
वर्गमूल = 99

प्रश्न 9.
छः अंकों की वह छोटी-से-छोटी संख्या ज्ञात करें कि एक पूर्ण वर्ग संख्या हो। इस प्रकार से प्राप्त वर्ग संख्या का वर्गमूल भी ज्ञात कीजिए।
उत्तर
छः अंकों की छोटी सं० = 100000

अभीष्ट सं० = 10000 + 489 = 100489
वर्गमूल = 317

प्रश्न 10.
एक वर्गाकार मैदान का क्षेत्रफल 60025 m² है। एक आदमी साइकिल से 5 metre की चाल से मैदान के चारों ओर चलता है तो कितने समय में वह प्रारंभिक बिन्दु पर आ जाएगा।
उत्तर

वह आदमी 245 से० में प्रारंभिक बिन्दु पर पहुँचेगा।

Bihar Board Class 8 Maths Solutions Chapter 8 राशियों की तुलना Text Book Questions and Answers.

BSEB Bihar Board Class 8 Maths Solutions Chapter 8 राशियों की तुलना

Bihar Board Class 8 Maths राशियों की तुलना Ex 8.1

Bihar Board Class 8 Math Solution प्रश्न 1.
सरल अनुपात ज्ञात कीजिए
अ. 14 मीटर का 7 मीटर 35 सेमी. से
ब. 3 रु. का 80 पै, से
स. 150 किग्रा. का 210 किग्रा. से
द. 2 घंटे का 50 मिनट से
उत्तर
(अ) 14 मीटर का 7 मीटर 35 सं० मी० से

Bihar Board Class 8 Math Solution In Hindi प्रश्न 2.
निम्नलखित अनुपातों को प्रतिशत में परिवर्तित कीजिए-
अ. 3 : 25
ब. 16 : 25
स. 3 : 16
उत्तर

Class 8 Bihar Board Math Solution प्रश्न 3.
60 विद्यार्थियों में से 40 प्रतिशत विद्यार्थियों को विज्ञान विषय . रुचिकर लगता है तो उन विद्यार्थियों की संख्या बताइये जिन्हें विषय में कम रुचि है।
उत्तर
40% विद्यार्थियों को विज्ञान विषय में रुचि है। अर्थात् 60% विद्यार्थियों को विज्ञान विषय में रुचि नहीं है।
सं० = 60 × 60% = \(\frac{60 \times 60}{100}\) = 36

Class 8 Maths Bihar Board प्रश्न 4.
किसी परीक्षा में उत्तीर्ण होने के लिए एक परीक्षार्थी को पूर्णांक के 33 प्रतिशत अंक प्राप्त करने हैं, उसे 225 अंक मिले जो कि 33 प्रतिशत से 6 अंक कम थे। बताइए परीक्षा में पूर्णांक क्या था?
उत्तर
माना कि कुल अंक = x
x का 33% = 225 + 6
\(\frac{x \times 33}{100}\) = 231
x = \(\frac{231 \times 100}{33}\)
x = 700

Bihar Board Math Solution Class 8 प्रश्न 5.
रहीम अपना निवास स्थान 8 बजे सुबह छोड़ देता है और उसी दिन शाम 4 बजे अपने घर लौट आता है, तो 24 घंटे का कितना प्रतिशत वह अपने निवास स्थान पर व्यतीत करता है?
उत्तर
बाहर बिताया कुल सं० = 8 बजे सुबह से शाम के 4 बजे तक = 8 घंटे
% समय = 24 × x% = 8
x = \(\frac{28 \times 100}{24}\) = 33.3%

Bihar Board Solution Class 8 Math प्रश्न 6.
पहली संख्या, दूसरी संख्या से 20% अधिक है तो दूसरी संख्या पहली संख्या से कितना प्रतिशत कम है?
उत्तर
माना पहली सं० = 100 दूसरी से 20% अधिक है।
दूसरी सं० = 80
दूसरी सं० कम है = 100 – 80 = 20
\(80 \times \frac{x}{100}\) = 20
x = \(\frac{20 \times 100}{80}\) = 25%

Bihar Board Class 8 Maths राशियों की तुलना Ex 8.2

Bihar Board 8th Class Math Solution प्रश्न 1.
रोहित एक पुराना अलमीरा 6700 रुपये में खरीदकर उस पर 300 रु. उसके मरम्मत में खर्च करता है। उसके बाद उसे वह 7500 रु० में बेच देता है। उसका लाभ या हानि प्रतिशत ज्ञात कीजिए।
उत्तर
पुराने आलमीरे की कीमत = 6700
मरम्मत में खर्च = 300
क्र.मू. = 7000
लाभ = वि० मू० – क्र० मू० = 7500 – 7000 = 500
लाभ% = \(\frac{500}{7000} \times 100=\frac{50}{7}=7 \frac{1}{7} \%\)

Bihar Board Class 8 Math प्रश्न 2.
प्रत्येक के लिए x, y, z का मान ज्ञात करें।

उत्तर
(i) क्र०मू० = 1500 + 320 = 1820
वि०मू० = क्र-मू० + लाभ = 1820 + 280 = 2100
लाभ% = \(\frac{280}{1820}\) × 100 = \(\frac{400}{26}\) = 15.38%

(ii) वि.मू = 1500 + 320 = 1820
हानि% = 10%
क्र०मू = \(\frac{100}{90} \times 1820\) = 2022.22

(iii) खरीद मू = 500
क-मू = 575
ऊपरी व्यय = 575 – 500 = 75/-
लाभ = 125
वि०मू० = क्र०मू० + लाभ = 575 + 125 = 700
लाभ% = \(\frac{125}{575}\) × 100 = \(\frac{500}{23}\) = 21.73%

(iv) क्र मू० = 9000 + 200 = 9200
वि०मू० = 7200
हानि = 9200 – 7200 = 2000
हानि% = \(\frac{2000}{9200} \times 100\) = \(\frac{2000}{92}\) = 21.43%

(v) खरीद मू० = 500 – 100 = 400
लाभ = \(500 \times \frac{20}{100}\) = 100
वि०मू० = क्र०मू० + लाभ% = 500 + 100 = 600

Class 8 Math Bihar Board प्रश्न 3.
एक बिजली के पंखे को 510 रु. में बेचने पर एक दुकानदार को 15 प्रतिशत की हानि उठानी पड़ती है, बताइए दुकानदार ने पंखा कितने में खरीदा? यदि वह पंखे को 630 रु. में बेचे तो उसे कितने प्रतिशत लाभ या हानि होगी?
उत्तर
हानि% = 15%
वि०मू = 510

Bihar Board Class 8th Math Solution प्रश्न 4.
मुकेश स्पोर्ट्स की दुकान से एक फुटबॉल गेंद 20 प्रतिशत के बट्टे पर 192 रु. में खरीदता है तो फुटबाल गेंद का अंकित मूल्य क्या है?
उत्तर
बट्टा = 20%
अंकित मू० = ?
वि० मू० = 192
अंकित मू० = \(\frac{192 \times 100}{80}\) = 240

Bihar Board Class 8 Math Solution In Hindi Pdf Download प्रश्न 5.
एक दुकानदार एक जोड़ी जूते पर 1250 रु. मूल्य अंकित करके ग्राहक को खरीदने पर 20 प्रतिशत की छूट देता है। छूट देने के बाद भी दुकानदार को 25 प्रतिशत का लाभ प्राप्त होता है, तो जूते का क्रय मूल्य क्या है?
उत्तर
बट्टा = 1250 × \(\frac{20}{100}\) = 250
वि०मू० = 1250 – 250 = 1000
1000 में बेचने पर लाभ = 20%
क्र०मू० = 1000 × \(\frac{80}{100}\) = 800

Class 8 Math Solution Bihar Board प्रश्न 6.
सोहन द्वारा एक डिपार्टमेंटल स्टोर से खरीदी गई सामग्री का बिल निम्नानुसार है। बिल की कुल राशि ज्ञात कीजिए।

उत्तर


कुल राशि = 260 + 330 + 273 + 61.20 = 924.20

Bihar Board 8 Class Math Solution प्रश्न 7.
राखी को 250 रु. मूल्य की खेल सामग्री तथा 220 रु. मूल्य के चमड़े का बैग क्रमश 6 प्रतिशत और 12 प्रतिशत बिक्री कर देकर खरीदना पड़ा हो, तो बताइए उसने कुल कितने रुपये चुकाए?
उत्तर

= 220 + 26.40
= 246.40
कुल मू० = 265 + 246.40 = 511.40

Bihar Board Class 8 Maths राशियों की तुलना Ex 8.3

प्रश्न 1.
सुधीर ने एक कोट 4500 रु. मूल्य के खरीदे। उसे बिक्री कर के 6 प्रतिशत अतिरिक्त देने पड़े तो बताइए कि सुधीर ने कोट खरीदने में कुल कितने रुपये लगाए?
उत्तर
कोट का क्रय मू० = 4500
बिक्री कर = 6%
बिक्री कर का मू० = 4500 × \(\frac{6}{100}\) = 270
कुल क्रय मु० = 4500 + 270 = 4770

प्रश्न 2.
एक दुकानदार ने अपनी दुकान से 3 महीने की बिक्री के बाद 4500 रु. वैट के रूप में जमा किया। यदि वैट की दर 4 प्रतिशत हो तो यह बताइए कि उसे कितनी मूल राशि का सामान बेचा?
उत्तर
बैट के रूप में जमा किया = 4500
बैट की दर = 4%
वि.मू. × 4% = 4500
वि०म० = \(\frac{4500 \times 100}{4}\) = 112500/-

प्रश्न 3.
रजिया ने एक दवा विक्रेता के यहाँ से 625 रु. अंकित मूल्य की दवाई खरीदी और उस पर 12 रु. 50 पैसे अतिरिक्त कर दिया। बताइए कि अतिरिक्त कर की दर प्रतिशत क्या थी?
उत्तर
दवा का मू० = 625
बैट मू० = 12.50
वि० मू० × \(\frac{x}{100}\) = 12.50
x = \(\frac{1250 \times 100}{625 \times 100}=\frac{1250}{625}\) = 2%
मिश्रधन = 25000 + 6000 = 31000

प्रश्न 4.
मिश्रधन ज्ञात कीजिए जब ब्याज की गणना प्रतिवर्ष की जाती है-
अ. 7500 रु. पर 2 वर्ष के लिए 6 प्रतिशत वार्षिक ब्याज की दर से।
ब. 25000 रु. पर 3 वर्ष के लिए 8 प्रतिशत वार्षिक ब्याज की दर से।
उत्तर

प्रश्न 5.
चक्रवृद्धि ब्याज ज्ञात कीजिए-
अ. 6000 रु. पर 3 वर्ष के लिए 10 प्रतिशत वार्षिक ब्याज की दर से।
ब. 4000 रु. पर 2 वर्ष के लिए 5 प्रतिशत वार्षिक ब्याज की दर से।
उत्तर

प्रश्न 6.
वह धन ज्ञात कीजिए जो 8 प्रतिशत वार्षिक ब्याज की दर से 2 वर्ष में 7290 रु. हो जाता है।
उत्तर
R = 8%
T = 2
A = 7290

प्रश्न 7.
कितने प्रतिशत वार्षिक ब्याज की दर से 4000 रु. 2 वर्ष में 5290 रु. हो जाता है।
उत्तर
P = 4000
T = 2
A = 5290
C.I = 4000 – 5290 = 1290

प्रश्न 8.
स्वाति एक जमीन का टुकड़ा खरीदने हेतु बैंक से 40960 रु. कर्ज 12½% वार्षिक दर 1½ वर्ष के लिए कर्ज देती है और ब्याज का संयोजन अर्द्धवार्षिक होता है तो स्वाति को कितनी राशि देनी पड़ेगी और उनके द्वारा भुगतान किया गया ब्याज की राशि भी ज्ञात कीजिए।
उत्तर
P = 40960

प्रश्न 9.
रवि ने 32,000 रु. बैंक में जमा किया। उस बैंक द्वारा जमा की गई राशि पर ब्याज के संयोजन तिमाही घोषित हो तो और ब्याज की दर से 5 प्रतिशत वार्षिक हो तो रवि को महीने बाद कितनी राशि प्राप्त होगी।
उत्तर

प्रश्न 10.
5 प्रतिशत वार्षिक दर से बढ़ते हुए वर्ष 2005 के अंत में एक शहर की जनसंख्या 44.10 हो गई। बताइए वर्ष 2003 में इस शहर की जनसंख्या कितनी थी?
उत्तर
आज शहर की जनसंख्या = 4410
दर = 5%
2003 अर्थात् दो वर्ष पहले
T = 2

प्रश्न 11.
एक जनित्र (Generator) का वर्तमान मूल्य 42000 रु. है। यदि उसका अवमूल्यन प्रतिशत वार्षिक हो तो 2 वर्ष बाद उस जनित्र का मूल्य क्या होगा?
उत्तर
वर्तमान मू० = 4200
समय = 2
दर = 5%

Bihar Board Class 8 Maths Solutions Chapter 13 क्षेत्रमिति Text Book Questions and Answers.

BSEB Bihar Board Class 8 Maths Solutions Chapter 13 क्षेत्रमिति

Bihar Board Class 8 Maths क्षेत्रमिति Ex 13.1

Bharti Bhawan Class 8 Math Solution In Hindi प्रश्न 1.
बगल की आकृतियों में एक आयताकार और एक वर्गाकार खेल के मैदान के माप दिए हुए हैं। यदि इनके परिमाप समान हैं तो किस मैदान का क्षेत्रफल अधिक होगा?
उत्तर

वर्ग का परिमाप = 4 × 80 = 320 m²
आयत का परिमाप = 2(60 + 100) = 2 × 160 = 320 m²
वर्ग का क्षे० = (80)² = 6400 m²
आयत का क्षे० = 60 × 100 = 6000 m²
वर्ग का क्षेत्रफल अधिक है।

Bihar Board Class 8 Math Solution प्रश्न 2.
विमला के पास एक आयताकार प्लॉट है (जैसा कि चित्र में दिखाया गया है) वह प्लॉट के बीच में एक वर्गाकार घर बनाना चाहती है। घर के चारों ओर फुलवारी लगवानी है। उसे फुलवारी लगाने में 40 रु. प्रति वर्गमीटर की दर से कितने रुपये खर्च करने होंगे?
उत्तर
प्लॉट का कुल क्षे० = 80 × 60 = 4800 m²
घर का क्षे० = (40)² = 1600 m²
बची हुई जगह = 4800 – 1600 = 3200 m²
फुलवारी की जगह = 3200 m²
कीमत = 3200 × 40 = 128000 m²

Bihar Board Math Solution Class 8 प्रश्न 3.
अमरेश अपने घर के आँगन में ईंट बिछवाना चाहता है। यदि आँगन की लम्बाई 20 मीटर और चौड़ाई 15 मीटर हो तो एक ईंट की लम्बाई 25 सेमी. और 80 सेमी. हो तो उस आँगन में कितने ईंटें लगेंगी? (कच्चा चित्र बिना हल करें)।
उत्तर

आँगर का कुल क्षे० = 20 × 15 = 300 m²
ईंट का क्षे० = 25 × 80 = 2000 cm = 2 m
ईंटों की सं० = \(\frac{300 m^{2}}{2 m}\) = 150 ईंटें

Bihar Board Class 8 Math Solution In Hindi Pdf Download प्रश्न 4.
एक त्रिभुजाकार खेत का क्षेत्रफल 600 वर्गमीटर तथा ऊँचाई 60 मीटर है तो उस खेत का आधार ज्ञात करें।
उत्तर
क्षे० = 600 m²
ॐ = 60 m
Δ का क्षे० = \(\frac {1}{2}\) × आ० × ॐ
600 = \(\frac {1}{2}\) × x × 60
x = \(\frac{600 \times 2}{60}\)
x = 20 m

Class 8 Math Bihar Board प्रश्न 5.
एक धावक को कम से कम दूरी तय करने के लिए निम्न में से किस आकृति पर चक्कर लगाना चाहिए? आप जानते हैं कि सम्पूर्ण वृत्त की परिधि का सूत्र c = 2πr जहाँ r वृत्त की त्रिज्या है।
उत्तर
(a) अर्धवृत्त की परिधि = πr
= \(\frac{22}{7} \times \frac{42}{2}\)
= 66 m
(b) ∆ की परिधि = 3 × भु० = 14 × 3 = 42 m
(c) वृत्त की परिधि = 2πr = 2 × \(\frac{22}{7}\) × 14 = 88m
सबसे कम क्षे० = b

विभिन्न आकृतियों का क्षे०
समलम्ब चतुर्भुज का क्षे० = \(\frac {1}{2}\) × h (b₁ + b₂)
= \(\frac {1}{2}\) × ॐ (समान्तर भुजाओं का योग)
समचतुर्भुज का क्षेः = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × विकर्ण₁ × विकर्ण₂

Bihar Board Class 8 Maths क्षेत्रमिति Ex 13.2

Class 8 Bihar Board Math Solution प्रश्न 1.
एक समलम्ब चतुर्भुज PORS के ∠P और ∠S समकोण है। इसकी भुजाओं की माप चित्र में दर्शाई गई है, समलम्ब चतुर्भुज का क्षेत्रफल ज्ञात कीजिए।
उत्तर
समलम्ब का क्षे० = \(\frac {1}{2}\) × h × (b₁ + b₂)
= \(\frac {1}{2}\) × 13 cm × (20 + 25)
= \(\frac {1}{2}\) × 12 × 45
= 270 m²

Bihar Board 8 Class Math Solution प्रश्न 2.
एक समलम्ब चतुर्भुज ABCD में AB, CD का समान्तर है AB = 30 सेमी., BC = 15 सेमी., DC = 44 सेमी. और AD = 13 सेमी.। समलम्ब चतुर्भुज का क्षेत्रफल ज्ञात कीजिए।
उत्तर

समलम्ब चतुर्भुज का क्षे = \(\frac {1}{2}\) × h × (b₁ + b₂)
= \(\frac {1}{2}\) × 13 × (30 + 44)
= \(\frac {1}{2}\) × 13 × 74
= 841 m²

Bihar Board Class 8 Math Solution In Hindi प्रश्न 3.
किसी समलम्ब चतुर्भुज की समानान्तर भुजाएं 52 सेमी. और 27 सेमी. है तथा अन्य दो भुजाएँ 25 सेमी. और 30 सेमी. की हैं। समलम्ब चतुर्भुज का क्षेत्रफल ज्ञात कीजिए।
उत्तर

ABCD का क्षे = \(\frac {1}{2}\) × 25 × (27 + 52)
= \(\frac {1}{2}\) × 25 × 79
= \(\frac{1975}{2}\)
= 987.5
= 988

Bharti Bhawan Class 8 Math Solution Hindi प्रश्न 4.
किसी समलम्ब चतुर्भुज का क्षेत्रफल 200 मी. है और इसकी ऊँचाई 8 मी. है। यदि समान्तर भुजाओं में एक भुजा दूसरी भुजा से 6 मी. अधिक है तो समान्तर भुजाओं की लम्बाई ज्ञात कीजिए।
उत्तर
माना एक भुजा = x
दूसरी भुजा = 6 + x
x = 8 m
समलंब का क्षे = \(\frac {1}{2}\) × h × (b₁ + b₂)
200 = \(\frac {1}{2}\) × 8 × (x + x + 6)
\(\frac{200 \times 2}{8}\) = 2x + 6
50 = 2x + 6
50 – 6 = 2x
2x = 44
x = 22
एक भुजा = 22 m
दूसरी भुजा = x + 6 = 22 + 6 = 28 m

Bihar Board Class 8 Ka Math Solution प्रश्न 5.
किसी समलम्ब चतुर्भुज की समान्तर भुजाएँ क्रमशः 24 सेमी. और 20 सेमी. हैं तथा दोनों भुजाओं के बीच की दूरी 15 सेमी. है, इसका क्षेत्रफल ज्ञात कीजिए।
उत्तर
h = 15 m
b₁ = 24 cm
b₂ = 20 m
समलम्ब चतुर्भुज का क्षे० = \(\frac {1}{2}\) × h × (b₁ + b₂)
= \(\frac {1}{2}\) × 15 × (24 + 20)
= \(\frac {1}{2}\) × 15 × 44
= 330 cm²

Bihar Board Class 8 Solutions Math प्रश्न 6.
किसी समलम्ब चतुर्भुज का क्षेत्रफल 384 सेमी. है। यदि समान्तर भुजाओं का अनुपात 3 : 5 हो और दोनों की लम्बात्मक दूरी 12 सेमी. हो तो प्रत्येक समान्तर भुजाओं की माप ज्ञात कीजिए।
उत्तर
h = 12 cm
b₁ = 3x
b₂ = 5x
समलम्ब चतुर्भुज का क्षे० = \(\frac {1}{2}\) × h × (b₁ + b₂)
384 = \(\frac {1}{2}\) × 12 × (3x + 5x)
8x = \(\frac{384 \times 2}{12}\)
8x = 64
x = 8
पहली भुजा = 3x = 8 × 3 = 24 m
दूसरी भुजा = 5x = 5 × 8 = 40 m

Bharti Bhawan Class 8 Math Solution In Hindi Pdf Download प्रश्न 7.
एक समचतुर्भुज की प्रत्येक भुजा 8 सेमी. है और इसका क्षेत्रफल 11.2 सेमी² है तो इस चतुर्भुज का शीर्ष लम्ब ज्ञात करें।
उत्तर
समलम्ब चतुर्भुज का क्षे० = \(\frac {1}{2}\) × d₁ × d₂, or, b × h
h = 6 cm
b = 10 cm
समलम्ब चतुर्भुज का क्षे० = 6 cm × 10 cm = 60 m²

Bihar Board 8th Class Math Solution प्रश्न 8.
एक समचतुर्भुज की प्रत्येक भुजा 8 सेमी. है और इसका क्षेत्रफल 11.2 सेमी² है तो इस चतुर्भुज का शीर्ष लम्ब ज्ञात करें।
उत्तर
समचतुर्भुज का क्षे० = b × h
11.2 = 8 × h
\(\frac{11.2}{8}\) = h
h = 1.4 cm

Class 8 Math Solution Bihar Board प्रश्न 9.
किसी समचतुर्भुज का क्षेत्रफल 64 सेमी² है और इसकी परिमाप 64 सेमी. है। समचतुर्भुज का शीर्ष लम्ब ज्ञात कीजिए।
उत्तर
समचतुर्भुज की परिमाप = 4 × भुजा = 4 × x
\(\frac{64}{4}\) = 16 = x
x = 16 cm
समचतुर्भुज की क्षे० = b × h
64 = 16 × h
\(\frac{64}{16}\) = h
h = 4 cm

प्रश्न 10.
एक समचतुर्भुजाकार पार्क की प्रत्येक भुजा की लम्बाई 72 मीटर तथा शीर्ष लम्ब 18 मीटर है। उस वर्गाकार खेल के मैदान का भुजा क्या होगी जिसका क्षेत्रफल इस समचतुर्भुज के क्षेत्रफल के बराबर है?
उत्तर
समचतुर्भुज का क्षे० = b × h = 72 × 18 = 1296 m²
समचतुर्भुज का क्षे० = वर्ग का क्षे० = 1296 m²
भुजा² = 1296 m²
x = √1296 = 36 m

प्रश्न 11.
किसी चतुर्भुज का एक विकर्ण 30 मीटर और सम्मुख शीर्षों से डाले गए लम्ब 10 मी. और 8 मी. हैं तो चतुर्भुज का क्षेत्रफल निकालिए।
उत्तर
चतुर्भुज का क्षे० = ∆ACB + ∆CDB
= \(\frac {1}{2}\) × 30 × 10 + \(\frac {1}{2}\) × 30 × 8
= 150 + 120
= 270 cm²

प्रश्न 12.
निम्न आकृति का क्षेत्रफल तथा शीर्ष लम्ब ज्ञात कीजिए।
उत्तर
∆PCB का क्षे० = \(\frac {1}{2}\) × आ० × ॐ
= \(\frac {1}{2}\) × 12 × 10
= 60 cm²
DCPA का क्षे० = भुजा² = (10)² = 100 cm²
आकृति का क्षे० = 100 + 60 = 160 cm²
∆PCB में,
शीर्ष = \(\sqrt{(10)^{2}-\left(\frac{12}{2}\right)^{2}}\) (पाइथागोरस प्रमेय)
= \(\sqrt{100-6^{2}}\)
= \(\sqrt{100-36}\)
= \(\sqrt{64}\)
= 8 m

Bihar Board Class 8 Maths क्षेत्रमिति Ex 13.3

प्रश्न 1.
दिए गए दोनों घनों को जोड़कर एक घनाभ बनाया गया, तो घनाभ के सम्पूर्ण पृष्ठ का क्षेत्रफल ज्ञात कीजिए।
उत्तर
पहले घन का क्षे० = 6 × भुजा²
= 6 × 8²
= 6 × 64 cm²
= 384 cm²
दूसरे घन का क्षे० = 6 × 8² cm
= 6 × 64 cm²
= 384 cm²
घनाभ का क्षे० = (384 + 38.4) cm² = 768 cm²

प्रश्न 2.
एक घन की एक भुजा 12 सेन्टीमीटर है तो धन का सम्पूर्ण. पृष्ठ क्षेत्रफल ज्ञात कीजिए।
उत्तर
घन की भुजा = 12 cm
घन का सम्पूर्ण पृष्ठ क्षे० = 6 × भुजा²
= 6 × (12)²
= 6 × 144
= 864 cm²

प्रश्न 3.
एक धनाभाकार पिंड की लम्बाई 15 सेमी., चौड़ाई 14 सेमी. एवं ऊँचाई 13 सेमी. है, पिंड का पृष्ठ क्षेत्रफल ज्ञात कीजिए।
उत्तर
ल. (l) = 15 cm
चौ० (b) = 14 cm
ॐ (h) = 13 cm क्षे० = 2(lb + bh + lh)
= 2(15 × 14 + 14 × 13 + 13 × 15)
= 2 (210 + 182 + 195)
= 2(587)
= 1174 m²

प्रश्न 4.
ऐसे घनाभाकार पिंड की भुजा ज्ञात कीजिए जिसका पृष्ठीय क्षेत्रफल 2400 वर्ग मीटर है।
उत्तर
कुल पृष्ठीय क्षे० = 6 × भुजा²
2400 = 6 × x²
\(\frac{2400}{6}\) = x²
x = √400 = 20 m

प्रश्न 5.
एक घनाभाकार साबुन की लम्बाई 6 सेमी., चौड़ाई 5 सेमी. एवं सम्पूर्ण पृष्ठ का क्षेत्रफल 148 वर्ग सेमी. है तो उसकी ऊँचाई ज्ञात कीजिए।
उत्तर
घनाभाकार साबुन की लं० = 2 (lb + bh + lh)
l = 6 cm
b = 5 cm
h = x cm
148 m² = 2(6 × 5 + 5 × x + 6 × x)
148 = 2(30 + 5x + 6x)
148 = 2(30 + 11x)
148 = 60 + 22x
148 – 60 = 22x
88 = 22x
x = 4 cm (ऊँचाई)

प्रश्न 6.
एक घनाकार लकड़ी के टुकड़े की एक किनारे की लम्बाई 10 सेमी. है। उसमें से 3 सेमी. × 2 सेमी. × 1 सेमी. आकार का घनाभ एक कोने से काटकर निकाल दिया गया तो शेष क्षेत्रफल कितना होगा?
उत्तर
घनाभाकार लकड़ी के टुकड़े का क्षे० = 6 × l²
= 6 × 10²
= 6 × 100
= 600 cm
घनाभ का क्षे० = 2(lb + bh + lh)
= 2(3 × 2 + 2 × 1 + 1 × 3)
= 2(6 + 2 + 3)
= 2 × 11
= 22 cm²
शेष क्षे = (600 – 22) cm² = 578 cm²

प्रश्न 7.
एक बेलन की ऊँचाई 25 सेमी. है और आधार का क्षेत्रफल 154 वर्ग सेमी. है तो बेलन के वक्रपृष्ठ का क्षेत्रफल ज्ञात करें।
उत्तर
h= 25 cm
बेलन का आधार वृत्त का क्षे० = 154 m²
πr² = 154
r² = 154 × \(\frac{7}{22}\)
r² = \(\sqrt{7 \times 7}\)
r = 7 cm
बेलन का सम्पूर्ण पृष्ठीय क्षे० = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 7 (7 + 25)
= 2 × \(\frac{22}{7}\) × 7 × 32
= 1408 m²

प्रश्न 8.
एक बेलनाकार लकड़ी की लम्बाई 50 सेमी. है तथा आधार की त्रिज्या 14 सेमी. है। इसके सम्पूर्ण पृष्ठ का क्षेत्रफल ज्ञात करें।
उत्तर
h = 50 cm
r = 14 cm
सम्पूर्ण पृष्ठ का क्षे० = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 14 (14 + 50)
= 2 × \(\frac{22}{7}\) × 14 × 64
= 5632 cm²

प्रश्न 9.
यदि आपको इन आकृतियों को कागज से पूरा-पूरा ढंकना हो तो कम से कम कितने कागज की आवश्यकता होगी?
उत्तर
(i) बेलन का क्षे० = 2πr(r + h)
h = 15 cm
r = 7 cm
क्षे० = 2 × \(\frac{22}{7}\) × 7 (15 + 7)
= 2 × \(\frac{22}{7}\) × 7 × 22
= 968 m²
(ii) घनाभ का क्षे० = 2(lb + bh + lh)
= 2(4 × 6 + 6 × 3 + 3 × 4)
= 2(24 + 18 + 12)
= 2 × 54
= 108 m²

प्रश्न 10.
एक भवन में 20 बेलनाकार खंभे लगे हैं जिसकी ऊँचाई 4 मीटर है तथा त्रिज्या 14 सेमी. है। 4 रुपये प्रति वर्गमीटर की दर से वक्रपृष्ठीय क्षेत्रफल में रँगाई करने का खर्च ज्ञात कीजिए।
उत्तर
h = 4 m
r = 14 cm
एक बेलन का क्षे० = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 14 (14 + 4)
= 2 × \(\frac{22}{7}\) × 14 × 18
= 1584
20 बेलनों का क्षे० = 20 × 1584 = 31680
4 रु. वर्गमीटर की दर से रंगाई का खर्च = \(\frac{31680}{100}\) × 4 = 1267.20

Bihar Board Class 8 Maths क्षेत्रमिति Ex 13.4

प्रश्न 1.
(अ) एक घन में कितनी सतहें होती हैं?
(ब) किसी घनाभ में किनारों की कुल संख्या कितनी है?
(स) घन और घनाभ के सतहों में क्या अंतर है?
(द) घन में कितने शीर्ष होते हैं?
उत्तर
(अ) 6
(ब) 12
(स) घन की सतहें सर्वांगसम होती हैं जबकि, घनाभ की नहीं होती।
(द) 8

प्रश्न 2.
नीचे घनाभ के किनारों की लम्बाइयाँ दी हुई हैं, उनके
अ. कुल पृष्ठ का क्षेत्रफल एवं
ब. आयतन निकालिए।
(i) 10 मी., 5 मी., 6 मी.
(ii) 17 सेमी., 12 सेमी., 10 सेमी.
उत्तर
(i) कुल पृष्ठ का आयतन = l × b × h = 10 m × 5 m × 6 m = 300 m³
(ii) कुल पृष्ठ का क्षे० = 2(lb + bh + hl)
= 2(10 × 5 + 5 × 6 + 6 × 10)
= 2 (50 + 30 + 60)
= 2(140)
= 280 m³
(iii) कुल पृष्ठ का क्षे० = 2(lb + bh + hl)
= 2(17 × 12 + 12 × 10 + 10 × 17)
= 2(204 + 120 + 170)
= 2(494)
= 988 m²
आयतन = l × b × h
= 17 × 12 × 10
= 2040 cm³

प्रश्न 3.
5 सेमी. किनारेवाले एक घन से 1 सेमी. किनारेवाले कितने घन काटे जा सकते हैं?
उत्तर

प्रश्न 4.
एक घनाभ का आयतन 576 घनमीटर है और आधार वर्गाकार है जिसकी एक भुजा 6 मीटर है तो घनाभ की ऊँचाई ज्ञात कीजिए।
उत्तर
घनाभ का आयतन = lbh = 576
b = 6 m
l = 6 m
(∵ आधार वर्गाकार है)
lbh = 576
6 × 6 × h = 576
h = \(\frac{576}{6 \times 6}\) = 16 m

प्रश्न 5.
12 सेमी. किनारेवाले दो घन बराबर से जोड़ दिए जाएँ तो नए घनाभ का पृष्ठ क्षेत्रफल ज्ञात कीजिए।
उत्तर

नए घनाभ में
h = 12 cm
b = 12 cm
l = 24 cm
पृष्ठ क्षे० = 2(lb + bh + lh)
= 2 (24 × 12 + 12 × 12 + 24 × 12)
= 2 (288 + 144 + 288)
= 2 × 720
= 1440 m³

प्रश्न 6.
एक लड़का 2 लीटर दूध खरीदने गया। दुकानदार ने उसे एक आयताकार आधार वाले बरतन से जो 20 सेमी. लम्बा, 15 सेमी. चौड़ा और 5 सेमी. गहरा था एक बार मापकर दे दिया । बताइए उस लड़के को कितना कम या अधिक दूध मिला। (यदि 1 लीटर = 1000 घन सेमी.)।
उत्तर
बरतन का आयतन = l × b × h
l = 20 cm
b = 15 cm
h = 5 cm
आयतन = 20 cm × 15 cm × 5 cm = 1500 cm³
1 लीटर = 1000 cm³
दूध दिया = 1500 cm³
2 लीटर = 2000 cm³
दूध कम दिया = 2000 – 1500 = 500 cm³ (आधा लीटर)

प्रश्न 7.
एक तालाब की लम्बाई 20 मीटर, चौड़ाई 12 मीटर और गहराई 8 मीटर है तथा एक दूसरे तालाब की लम्बाई और चौड़ाई 20 मीटर के बराबर है तथा गहराई पहले तालाब के बराबर है। किस तालाब में अधिक पानी अँटेगा?
उत्तर
पहले तालाब का आयतन = 20m × 12m × 8m = 1920 m³
दूसरे तालाब का आयतन = 20m² × 8m = 160 m³
पहले तालाब में अधिक पानी अँटेगा।

प्रश्न 8.
एक खाली डिब्बा जिसमें साबुन रखा जाना है, डिब्बों की लम्बाई 0.40 मीटर, चौड़ाई 0.25 मीटर तथा ऊँचाई 0.25 मीटर है। साबुन 5 सेमी. × 4 सेमी. × 2 सेमी. साइज का है। डिब्बा में कितने साबून रखे जा सकते हैं?
उत्तर
डब्बों की l = 0.40 m, b = 0.25 m, h = 0.25 m
साबुन की l = 5 cm, b = 4 cm, h = 2 cm

प्रश्न 9.
30 मीटर लम्बा, 20 सेमी. चौड़ा तथा 4 मीटर ऊँची दीवार बनवानी है। यदि एक ईंट की लम्बाई 25 सेमी., चौड़ाई 12.5 सेमी. तथा ऊँचाई 7.5 सेमी. हो तो उस दीवार के बनवाने में कितने ईंट लगेंगी। (सीमेंट व बालू का आयतन नगण्य माना गया है।)
उत्तर
दीवार का क्षे० = 30 m × 0.20 cm × 4 m
ईंट की क्षे० = 25 cm × 12.5 cm × 7.5 cm

प्रश्न 10.
एक कमरे की लम्बाई 15 मीटर, चौड़ाई 10 मीटर तथा ऊँचाई 8 मीटर है। उस घर में कितनी हवा भरेगा?
उत्तर
कमरे की लं० = 15 m
चौ० = 10 m
ॐ = 8 m
कमरे का आयतन = l × b × h
= 15 m × 10 m × 8 m
= 1200 m³

Get Updated Bihar Board Class 9th English Book Solutions in PDF Format and download them free of cost. Bihar Board Class 9 English Book Solutions Prose Chapter 2 Yayati Questins and Answers provided are as per the latest exam pattern and syllabus. Access the topics of Panorama English Book Class 9 Solutions Chapter 2 Yayati through the direct links available depending on the need. Clear all your queries on the Class 9 English Subject by using the Bihar Board Solutions for Chapter 2 Yayati existing.

Panorama English Book Class 9 Solutions Chapter 2 Yayati

If you are eager to know about the Bihar Board Solutions of Class 9 English Chapter 2 Yayati Questins and Answers you will find all of them here. You can identify the knowledge gap using these Bihar Board Class 9 English Solutions PDF and plan accordingly. Don’t worry about the accuracy as they are given after extensive research by people having subject knowledge alongside from the latest English Textbooks.

Bihar Board Class 9 English Yayati Text Book Questions and Answers

A. Work in small groups and discuss the following:

Yayati Class 9 Questions And Answers Bihar Board Question 1.
Which is the golden period of life – childhood, youth or old age? Give reasons for your choice?
Answer:
I consider youth as the golden period of life. This is the period when a man works for himself or for the country. In this stage, a person assumes the role of a lover. He is passionately in love with his beloved. He is sad when he is separated from his darling. He heaves sighs like the bellows of a furnace. He sings sorrowful songs composed in the praise of the beauty of his beloved. At this stage, he acts like a soldier. He has learnt many swearing terms. His beard is like that of a leopard. He saves the honour of his country at the risk of his life. He fights in the article. He tries to earn fame how so ever short-lived by jumping before the mouth of a firing cannon. So this is the age for doing something. For example, Alexander the great who died at the age of 32 only. Napolian, J. C. Bose, they all did at his youth.

Yayati Question Answer Bihar Board Question 2.
Why does the man want to remain always young?
Answer:
A man always wants to be young, because this is the age when one can enjoy this life in various ways. I consider youth as the golden period of life. This is the period when a man works for himself or for the country. In this stage, a person assumes the role of a lover. He is passionately in love with his beloved. He is sad when he is separated from his darling. He heaves sighs like the bellows of a furnace. He sings sorrowful songs composed in the praise of the beauty of his beloved. At this stage, he acts like a soldier. He has learnt many swearing terms. His beard is like that of a leopard. He saves the honour of his country at the risk of his life. He fights in the article. He tries to earn fame how so ever short-lived by jumping before the mouth of a firing cannon. So this is the age for doing something. For example, Alexander the great who died at the age of 32 only. Napolian, J.C. Bose, they all did at his youth.

B. 1.1. Answer the following questions briefly:

Yayati Story In Hindi Class 9 Bihar Board Question 1.
Who was Emperor Yayati?
Answer:
Emperor Yayati was one of the ancestors of the Pandavas.

Yayati Class 9 In Hindi Bihar Board Question 2.
How did Yayati become old? Who cursed him?
Answer:
Yayati became prematurely old for having wronged his wife Devayani, Sukracharya cursed him.

Question 3.
Was Yayati devoid of sensual desires?
Answer:
No, Yayati was not devoid of sensual desires.

Question 4.
What was the reply of his eldest son?
Answer:
Yayati, eldest son replied that if he took upon himself his old age women and servants would mock at him. So he could not do so.

Question 5.
Why did he become angry?
Answer:
He became angry as his three sons had declined to do as he wished.

B.1.2. Answer the following questions briefly:

Question 1.
What are the symptoms of an old man?
Answer:
In the old age strength and beauty are destroyed. His face wrinkles, his hair becomes grey. He can’t ride on a horse or an elephant. His speech falters. He has to seek the help of others even to keep his body clean. These are the symptoms of an old man.

Question 2.
Who was Puru? Did he accept his father’s proposal?
Answer:
Puru was the youngest son of Yayati. Yes, he gladly accepted his father’s proposal.

Question 3.
Sensual desire is everlasting. Whose thought is this?
Answer:
This is Indian ancient thought. This is Yayati’s thought here.

Question 4.
Why did Yayati resume his old age?
Answer:
When Yayati realised that sensual desire cannot be quenched he took back his old age and went to the forest to live a life of austere.

Question 5.
Do corn, gold, cattle and woman satisfy the desire of a man?
Answer:
No, corn, gold, cattle and woman do not satisfy the desire of a man.

B. 1.3. Answer the following questions briefly:

Question 1.
What was Yayati famous for?
Answer:
Yayati was famous as a ruler devoted to the welfare of his subjects.

Question 2.
Why did Yayati call his sons?
Answer:
Yayati found himself suddenly an old man. But was still haunted by the desire for sensual enjoyment. So he called his five sons hoping that one of them would bear the burden of his old age and give his youth in return so that he might enjoy his life.

Question 3.
What did he say to them?
Answer:
He piteously appealed to the affection of his sons and asked one of them to exchange his youth with his old age so that he might enjoy his life in the full vigour of youth.

Question 4.
Which son agreed to give Yayati his youth and take his old age?
Answer:
Puru, the youngest son agreed to give Yayati his youth and take his old age.

Question 5.
Why did Yayati go to the garden of Kubera?
Answer:
Yayati went to Kubera’s garden to get more sensual satisfaction with an apsara maiden because his desire could not be launched on the earth.

Long Answer Type Questions

C.1. Answer the following questions briefly:

Question 1.
Why did Yayati become prematurely old? Why did he dis¬like it?
Answer:
Yayati became prematurely old by the curse of Sukracharya. his father-in-law, having wronged his wife, Devayani. Yayati disliked it because old age destroyed beauty and brought on miseries. He wanted to enjoy the desire of sensual enjoyment in the full vigour of youth. That was not possible in old age so he disliked it.

Question 2.
In order to enjoy the pleasures of youth, Yayati wished to take the youth of one of his sons. Was he right in doing so? Explain with the argument of your own.
Answer:
Yayati became old due to curse. This was not good, that brought him miseries. Everyone has the right to enjoy his life vigorously. But wished to take the youth of one of his sons for his sake and send the son in miseries was not good in any way. As an old man, Yayati should accept as truth because no pleasure can quench the desire.

Question 3.
Write in your own words the responses of the first three sons to their father’s request.
Answer:
When Yayati asked his sons to exchange their youth with his old age the eldest son said that the women and servants would mock at him. The second son declined the proposal by saying that old age destroys not only strength and beauty but also wisdom and he was not strong enough to do that. The third son opined that an old man can not ride a horse or an elephant. His speech is faltered his plight is helpless.

Question 4.
How did the fourth son respond to his father’s appeal? How would you have responded if you were the fourth son?
Answer:
The fourth son begged to be forgiven, as this was a thing he could by no means consent to. An old man has to seek the help of others even to keep his body clean a most pitiful plight. No, much as he loved his father he could not do it. If I were the fourth son. I took pity on him. I would speak to him politely and did not use hard words at least…

Question 5.
Why did Puru agree to give his youth to his father and take his father’s old age in the bargain? Did he do the right thing?
Answer:
Puru was fifth and last son of Yayati who had never yet opposed his father’s wishes. He moved by Filial love and agreed to give his youth to his father. He took his father’s old age in the bargain. He could not see his mighty father begging for anything and relieved him of the sorrow of old age and the cares of state, to be happy. Yes, as a son he did the right thing. It was his duty to protect his father.

Question 6.
Is it right for a father to make such a request to his sons as Yayati did?
Answer:
No, it is not morally right for a father to make such a request to his sons as Yayati did. He should have accepted his old age gracefully and ruled his kingdom more and more wisely.

Question 7.
Were the four sons justified in refusing their father’s request? If yes, give reasons.
Answer:
Yes, the four sons were fully justified in refusing their father’s request. I think every individual has the right. As enjoy his share of pleasure, who wants to become an old man in his full youth. None, the old age destroys beauty and brings on miseries and it is needless to describe the misery of vigorous youth sud¬denly plighted into old ages so the four sons rightly to refuse their father’s desires.

Question 8.
What lesson do you learn from this story?
Answer:
The story Yayati tells us a very good lesson that youth is the best time of life. It is full of beauty, strength and sensual pleasures. This is the period when a young man can do for himself or for the welfare of the country. But it also tells us it is in vain efforts to quench desire by indulgence. Sensual desire is never quenched by indulgence, any more than fire is by putting ghee in it. No, the object of desire corn, gold, cattle and women nothing can ever satisfy the desires of man. So mental peace and balance are the only remedy which beyond likes and dislikes. Such is the state of Brahman.

Comprehension Based Questions with Answers

1. Emperor Yayati was one of the ancestors of the Pandavas. He had never known defeat. He followed the dictates of the Sastras, adored the gods and venerated his ancestors with intense devotion. He became famous as a ruler devoted to the welfare of his subjects. He became prematurely old by the curse of Sukracharya for having wronged his wife Devayani. In the words of the poet of the Mahabharata, “Yayati attained that old age which destroys beauty and brings on miseries.” It is needless to describe the misery of vigorous youth suddenly blighted into age, where the horrors of loss are accentuated by pangs of recollection.

Questions:

1. Name the lesson and its writer.
2. Who was Yayati? Why was he famous for?
3. How did he become old prematurely?
4. According to the poet of the Mahabharata, what does the old age do?
5. Why is it needless to describe the misery of vigorous youth?

Answers:

1. The name of the lesson is Yayati and the writer is C. Raj Gopala Chari.
2. Yayati was an emperor and was one of the ancestors of the Pandavas. He became famous as a ruler devoted to the welfare of his subjects.
3. He became prematurely old by the curse of Sukracharya . for having wronged his wife Devayani.
4. According to the poet of the Mahabharata old age destroys beauty and brings on miseries.
5. It is needless to describe the misery of vigorous youth suddenly blighted into age, where the horrors of loss are accentuated by pangs of recollection.

2. Yayati, who found himself suddenly an old man. was still haunted by the desire for sensual enjoyment. He had five beautiful sons, all virtuous and accomplished. Yayati called them and appealed piteously to their affection. ’’The curse of your grandfather Sukracharya has made me unexpectedly and prematurely old. I have not had my feel of the joys of life; for not knowing what was in store for me. I lived a life of restraint, denying myself even lawful pleasures. One of you ought to bear the burden of my old age and give his youth in return. He who agrees to this and bestows his youth on me will be the ruler of my kingdom. I desire to enjoy life in the full vigour of youth”.

Questions:

1. What was the desire of Yayati after becoming an old man?
2. How many sons had Yayati?
3. What did Yayati tell his sons?
4. Find out the word from the passage which is an antonym to ‘ugly’.

Answers:

1. Yayati found himself suddenly an old man, was still haunted by the desire for sensual enjoyment.
2. Yayati had five sons. They were beautiful, virtuous and accomplished.
3. Yayati told his sons that curse of their grandfather Sukracharya had made him unexpectedly and prematurely old. He had not his feel of the joys of life and further said that one of them ought to bear the burden of his old age and give his youth in return. He who agreed to that would be the ruler of his kingdom.
4. Beautiful.

3. He first asked his eldest son to do his bidding. That son replied: “O great king, women and servants will mock at me if 1 were to take upon myself your old age. I cannot do so. Ask of my younger brothers who are dearer to you than myself.”
When the second son was asked, he gently refused with the word; “Father, you ask me to take up old age which destroys not only strength and beauty but also as I see wisdom. I am not strong enough to do so.”
The third son replied: “An old man cannot ride a horse or an elephant. His speech will falter. What can 1 do in such a helpless plight? I cannot agree.”
The king grew angry when he saw that his three sons had declined to do as he wished. He hoped for better from his fourth son, to whom he said: “You should take up my old age. If you exchange your youth with me, I shall give it back to you after some time and take back the old age with which I have been cursed.”

Questions:

1. What was the reply of the eldest son? When he asked him his bidding.
2. What did the second son reply?
3. What did the third son reply?
4. Why did the king grow angry with his sons?

Answers:

1. When the king asked his eldest son to do his bidding, he replied that women and servants would mock at him if he had been to take upon himself his old age. So he could not do and advised him to ask him youngest brothers who were dearer to him.
2. The second Yon gently refused to take up old age because it destroys strength, beauty and wisdom.
3. The third son replied that old age destroys strength to ride a horse or an elephant and to speak clearly.
4. The king saw that his three sons had declined to do as he wished, he grew angry.

4. The fourth son begged to be forgiven, as this was a thing he could by no means consent to. An old man has to seek the help of others even to keep his body clean, a most pitiful plight. No, much as he loved his father, he could not do it. Yayati was struck with sorrow’ at the refusal of the four sons. He paused for some time and then supplicated his last son,’ who had never yet opposed his wishes; “You must save me. I have got this old age with its wrinkles, debility and grey hairs as a result of the curse of Sukaracharya. I cannot bear it. If you take upon yourself these infirmities, I shall enjoy life for just a while more and then give you back your youth and resume my old age and all its sorrows. Puru. do not refuse as your elder brothers have done.” Puru. the youngest son, moved by filial love, said: “Father, I gladly give you my youth and relieve you of the sorrow’s of old age and the cares of State. Be happy.” Hearing these words Yayati became a youth. Puru, who accepted the old age of his father, ruled the kingdom and acquired great renown.

Questions:

1. Why was Yayati struck with sorrow?
2. What was the reply of the fourth.son?
3. Why did the king get his old age?
4. Who was Puru? Why did he accept his father’s proposal?
5. What happened to Puru?

Answers:

1. Yayati was struck with sorrow at the refusal of the four sons.
2. The fourth son replied that an old man had to seek the help of others even to keep his body clean. It looked a most pitiful plight so he begged not to accept the proposal.
3. The king got this old age as a result of the curse of Sukracharya, his father-in-law.
4. Puru was the fifth and last son of Yayati. who moved by filial love of his father he gladly accepted his father’s proposal.
5. Puru became old but ruled the kingdom and acquired great renown.

5. Yayati enjoyed life for long and, not satisfied, went later to the garden of Kubera and spent many years with an apsara ‘maiden. After long years spent in vain efforts to quench desire by indulgence, the truth dawned on him. Returning to furu, he said: “Dear son, sensual desire is never quenched by indulgence, any more than fire is by pouring ghee in it. I had heard and read this, but till now I had not realised it. No object of desire corn, gold, cattle and women nothing can ever satisfy the desires of man. We can reach peace only by a mental pose beyond likes and dislikes. Such is the state of Brahman. Take back your youth and rule kingdom wisely and well.” With these words, Yayati took back his old age. Puru, who regained his youth, was made king by Yayati who retired to the forest. He spent his time there in austerities and in due course attained heaven.

Questions:

1. How long did Yayati enjoy his life?
2. Why did he go to the garden of Kubera?
3. When did the truth dawn on him?
4. Why did he return to Puru?
5. What did he say to his son Puru after returning to him?
6. How can he get mental peace?
7. Find out the word in the passage which means ‘get again’.

Answers:

1. Yayati enjoyed life for a long time.
2. Yayati went to the garden of Kubera and spent many years with an apsara maiden.
3. The truth dawned on him-after long years spent in vain efforts to quench by indulgence in an apsara maiden.
4. After long years spent in vain efforts to quench desire by indulgence, he returned to Puru.
5. He said to Puru. “Sensual desire is never quenched so take your youth back and rule the kingdom wisely and well.”
6. He can get peace only by a mental pose beyond likes and dislikes. Such is the state of Brahman.
7. Regained.

Hope you liked the article on Bihar Board Solutions of Class 9 English Chapter 2 Yayati Questions and Answers and share it among your friends to make them aware. Keep in touch to avail latest information regarding different state board solutions instantly.