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Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Text Book Questions and Answers.

BSEB Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

Bihar Board Class 10 Maths निर्देशांक ज्यामिति Ex 7.4

प्रश्न 1.
बिन्दुओं A (2, -2) और B(3, 7) को जोड़ने वाले रेखाखण्ड को रेखा 2x + y – 4 = 0 जिस अनुपात में विभाजित करती है, उसे ज्ञात कीजिए।
हल
दिया है, बिन्दु A = (2, -2) तथा B = (3, 7)
यहाँ x₁ = 2, y₁ = -2, x₂ = 3, y₂ = 7
माना दिए हुए बिन्दुओं से बना रेखाखण्ड रेखा 2x + y – 4 = 0 को m₁ : m₂ के अनुपात में विभाजित करता है जबकि प्रतिच्छेद बिन्दु (x, y) है।

बिन्दु (x, 3) रेखा 2x + y – 4 = 0 पर स्थित होगा;
अतः इसके निर्देशांक रेखा 2x + y – 4 = 0 को सन्तुष्ट करेंगे।

अत: अभीष्ट अनुपात = 2 : 9

प्रश्न 2.
x और y में एक सम्बन्ध ज्ञात कीजिए यदि बिन्दु (x, y), (1, 2) और (7, 0) संरेखी है।
हल
माना बिन्दु A = (x, y), B = (1, 2) तथा C = (7, 0)
यहाँ, x₁ = x, y₁ = y, x₂ = 1, y₂ = 2, x₃ = 7, y₃ = 0
∆ का क्षेत्रफल = \(\frac{1}{2}\) [{x₁y₂ + x₂y₃ + x₃y₁} – {y₁x₂ + y₂x₃ + y₃x₁}]
= \(\frac{1}{2}\) [{x × 2 + 1 × 0 + 7 × y} – {y × 1 + 2 × 7 + 0 × x}]
= \(\frac{1}{2}\) [{2x + 0 + 7y} – {y + 14 + 0}]
= \(\frac{1}{2}\) [2x + 7y – y – 14]
= \(\frac{1}{2}\) [2x + 6y – 14]
= \(\frac{2}{2}\) (x + 3y – 7)
= x + 3y – 7
परन्तु यदि बिन्दु A, B, C संरेख हों तो ΔABC का क्षेत्रफल शून्य होना चाहिए।
x + 3y – 7 = 0
अतः x और में सम्बन्ध : x + 3y – 7 = 0

प्रश्न 3.
बिन्दुओं (6, -6), (3, -7) और (3, 3) से होकर जाने वाले वृत्त का केन्द्र ज्ञात कीजिए।
हल
माना A(6, -6), B(3, -7) तथा C(3, 3) बिन्दु एक वृत्त की परिधि पर हैं और वृत्त का केन्द्र O(h, k) है।
तब, OA, OB तथा OC वृत्त की त्रिज्याएँ होंगी।
अतः OA = OB = OC
⇒ OA² = OB² = OC²
OA² = [ केन्द्र O(h, k) और बिन्दु A (6, -6) के बीच की दूरी]²
⇒ OA² = (h – 6)² + (k + 6)²
⇒ OA² = h² – 12h + 36 + k² + 12k + 36
⇒ OA² = h² + k² – 12h + 12k + 72 ……..(1)
OB² = [केन्द्र O (h, k) और बिन्दु B (3, -7) के बीच की दूरी]²
⇒ OB² = (h – 3)² + (k + 7)²
⇒ OB² = h² – 6h + 9 + k² + 14k + 49
⇒ OB² = h² + k² – 6h + 14k + 58 ………(2)
OC² = [केन्द्र O(h, k) और बिन्दु C(3, 3) की दूरी]²
⇒ OC² = (h – 3)² + (k – 3)²
⇒ OC² = h² – 6h + 9 + k² – 6k + 9
⇒ OC² = h² + k² – 6h – 6k + 18 ………(3)
समीकरण (2) में से समीकरण (3) को घटाने पर,
20k + 40 = OB² – OC² = 0
⇒ k = -2
समीकरण (1) में से समीकरण (2) को घटाने पर,
-6h – 2k + 14 = OA² – OB² = 0
⇒ 6h + 2k = 14
⇒ 6h + (2 × -2) = 14 (∵ k = -2)
⇒ 6h – 4 = 14
⇒ 6h = 14 + 4 = 18
⇒ h = 3
अत: वृत्त का केन्द्र = (3, -2)

प्रश्न 4.
किसी वर्ग के दो सम्मुख शीर्ष (-1, 2) और (3, 2) हैं। वर्ग के अन्य दोनों शीर्ष ज्ञात कीजिए।
हल
दिया है, वर्ग के दो सम्मुख शीर्ष (-1, 2) व (3, 2) हैं।
वर्ग के एक विकर्ण का मध्य-बिन्दु = \(\left(\frac{-1+3}{2}, \frac{2+2}{2}\right)\) = (1, 2)
वर्ग के विकर्ण की लम्बाई = \(\sqrt{(-1-3)^{2}+(2-2)^{2}}\)
= \(\sqrt{(-4)^{2}+0}\)
= √16
= 4 मात्रक
तब, विकर्णों के प्रतिच्छेद बिन्दु E(मध्य बिन्दु) से प्रत्येक शीर्ष विकर्ण × \(\frac {1}{2}\) = 2 मात्रक दूरी पर होगा।

चित्र से स्पष्ट है कि शेष दोनों बिन्दु विकर्ण BD पर होंगे जो AC पर लम्ब होगा। तब प्रत्येक बिन्दु का भुज +1 होगा। माना कोटि y है।
तब, बिन्दु (1, 2) की बिन्दु (+1, y) से दूरी = 2 मात्रक
\(\sqrt{(1-1)^{2}+(y-2)^{2}}=2\)
⇒ \(\sqrt{0+(y-2)^{2}}=2\)
⇒ ±(y – 2) = 2
⇒ y – 2 = ±2
⇒ y = ±2 + 2
⇒ y = 0 या 4
अत: वर्ग के शेष दोनों शीर्ष (1, 0) व (1, 4) हैं।

प्रश्न 5.
कृष्णानगर के एक सेकेण्डरी स्कूल के कक्षा X के विद्यार्थियों को उनके बागवानी क्रियाकलाप के लिए एक आयताकार भूखण्ड दिया गया है। गुलमोहर की पौध (sapling) को परस्पर 1 मीटर की दूरी पर इस भूखण्ड की परिसीमा (boundary) पर लगाया जाता है। इस भूखण्ड के अन्दर एक त्रिभुजाकार घास लगा हुआ लॉन (lawn) है, जैसा कि आकृति में दर्शाया गया है। विद्यार्थियों को भूखण्ड के शेष भाग में फूलों के पौधे के बीज बोने हैं।
(i) A को मूलबिन्दु मानते हुए, त्रिभुज के शीर्षों के निर्देशांक ज्ञात कीजिए।
(ii) यदि मूलबिन्दु C हो तो ∆PQR के शीर्षों के निर्देशांक क्या होंगे?
साथ ही उपर्युक्त दोनों स्थितियों में, त्रिभुजों के क्षेत्रफल ज्ञात कीजिए। आप क्या देखते हैं?

हल
बिन्दुओं P, Q व R से सम्मुख अक्षों पर लम्ब खींचे गए हैं। (चित्र देखिए)

(i) यदि A मूलबिन्दु हो तो
बिन्दु P = (4, 6),Q = (3, 2) तथा R = (6, 5)
यहाँ x₁ = 4, y₁ = 6, x₂ = 3, y₂ = 2, x₃ = 6, y₃ = 5
∆PQR का क्षेत्रफल = \(\frac{1}{2}\) [{x₁y₂ + x₂y₃ + x₃y₁} – {y₁x₂ + y₂x₃ + y₃x₁}]
= \(\frac{1}{2}\) {{4 × 2 + 3 × 5 + 6 × 6} – {6 × 3 + 2 × 6 + 5 × 4}]
= \(\frac{1}{2}\) [(8 + 15 + 36) – (18 + 12 + 20)]
= \(\frac{1}{2}\) [59 – 50]
= \(\frac{1}{2}\) × 9
= \(\frac{9}{2}\) वर्ग मात्रक

(ii) जब C मूलबिन्दु हो तो
बिन्दु P = (-12, -2), Q = (-13, -6) तथा R = (-10, -3)
यहाँ x₁ = -12, y₁ = -2, x₂ = -13, y₂ = -6, x₃ = -10, y₃ = -3
∆PQR का क्षेत्रफल = \(\frac{1}{2}\) [{x₁y₂ + x₂y₃ + x₃y₁} – {y₁x₂ + y₂x₃ + y₃x₁}]
= \(\frac{1}{2}\) [{(-12 × -6) + (-13 × -3) + (-10 × -2)} – {(-2 × -13) + (-6 × -10) + (-3 × -12)}]
= \(\frac{1}{2}\) (72 + 39 + 20) – (26 + 60 + 36)]
= \(\frac{1}{2}\) [(131) – (122)]
= \(\frac{1}{2}\) × 9
= \(\frac{9}{2}\) वर्ग मात्रक
अत: दोनों ही स्थितियों में त्रिभुज का क्षेत्रफल समान है।

प्रश्न 6.
एक त्रिभुज ABC के शीर्ष A (4, 6), B(1, 5) और C (7, 2) हैं। भुजाओं AB और AC को क्रमशः D और E पर प्रतिच्छेद करते हुए एक रेखा इस प्रकार खींची गई है कि \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) है। ΔADE का क्षेत्रफल परिकलित कीजिए और इसकी तुलना ΔABC के क्षेत्रफल से कीजिए।

हल
दिया है, ΔABC के शीर्ष A (4, 6), B(1, 5) और C (7, 2) हैं।
\(\frac{A D}{A B}=\frac{1}{4}\)
⇒ AB = 4AD
⇒ AD + DB = 4AD
⇒ DB = 3AD
⇒ \(\frac{A D}{D B}=\frac{1}{3}\)
माना D के निर्देशांक यदि (x, y) हों तो

प्रश्न 7.
मान लीजिए A(4, 2), B(6, 5) और C (1, 4)एक त्रिभुज ABC के शीर्ष हैं।
(i) A से होकर जाने वाली माध्यिका BC से D पर मिलती है। बिन्दु D के निर्देशांक ज्ञात कीजिए।
(ii) AD पर स्थित ऐसे बिन्दु P के निर्देशांक ज्ञात कीजिए कि AP : PD = 2 : 1 हो।
(iii) माध्यिकाओं BE और CF पर ऐसे बिन्दुओं Q और R के निर्देशांक ज्ञात कीजिए कि BQ : QE = 2 : 1 हो और CR : RF = 2 : 1 हो।
(iv) आप क्या देखते हैं?
[नोट – वह बिन्दु जो तीनों माध्यिकाओं में सार्वनिष्ठ हो, उस त्रिभुज का केन्द्रक (centroid) कहलाता है और यह प्रत्येक माध्यिका को 2 : 1 के अनुपात में विभाजित करता है।]
(v) यदि A(x₁, y₁), B(x₂, y₂) और C(x₃, y₃) त्रिभुज ABC के शीर्ष हैं तो इस त्रिभुज के केन्द्रक के निर्देशांक ज्ञात कीजिए।
हल

प्रश्न 8.
बिन्दुओं A(-1, -1), B(-1, 4), C(5, 4) और D(5, -1) से एक आयत ABCD बनता है। P, Q, R और S क्रमश: भुजाओं AB, BC, CD और DA के मध्य-बिन्दु हैं। क्या चतुर्भुज PQRS एक वर्ग है? क्या यह एक आयत है? क्या यह एक समचतुर्भुज है? सकारण उत्तर दीजिए।
हल
दिए हुए बिन्दु A = (-1, -1), B = (-1, 4), C = (5, 4) और D = (5, -1)


∵ चतुर्भुज PQRS में, PQ = QR = RS = SP और विकर्ण PR ≠ विकर्ण QS
अत: चतुर्भुज PQRS एक समचतुर्भुज है।

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Panorama English Book Class 9 Solutions Chapter 7 Kathmandu

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Bihar Board Class 9 English Kathmandu Text Book Questions and Answers

A. Answer the following questions orally:

Write any eight-digit number with 6 in ten lakhs place and 9 in ten-thousandth place.

Question 1.
Have you ever visited a sacred place? Share your experiences with your friends.
Answer:
Yes, I have visited a sacred place. It is Deoghar or Baba Dham in Jharkhand. It is a very fine place. The whole place was charmed. One can get, people worshipping Lord Shiva there, I also entered the temple and worshipped the Lord. It is a worth visiting place. It was the month of Shrawan (July) in this month Lakhs of people come here bringing the Ganga water in Karwar. The whole city turns in Kawarias colour. That is ochrous (Gerua). The whole environment looks beautiful.

Question 2.
Name some of the holy places of your state.
Answer:
Some holy places of our state are Bodh Gaya, Rajgir, Pawapuri, many places in Darbhanga, Pumiya and Patna.

Question 3.
Describe the surroundings of a holy place you have visited.
Answer:
Once I got a chance to visit Gurudawara the temple of Patna Saheb. It is famous all over India and abroad. It is a place of pilgrimage for the Sikhs. It is the holy place of Guru Gobind Singhji. Thousands of devotees and worshippers visit it daily. Recitations from Holy Granth go on in the temple at all times. It is very ennobling and to sit in the temple for an hour. Surroundings of the holy place are full of shops. Only Sikhs are seen here as shopkeepers. A visit to this religious place is not without merit.

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

1. At Pashupatinath, there is an atmosphere of ‘febrile confusion’.
2. By the main gate an Indian struggles for permission to enter.
3. I consider what route should take back home.
4. From a balcony a basket of flowers and leaves, old offerings now wilted, is dropped into the lake.
5. I enter a Nepal Airport office and buy a ticket for the day after tomorrow flight.

Answer:

1. — T
2. — F
3. — T
4. — T
5. — F

B.1.2. Answer the following questions very briefly:

Question 1.
With whom does Mr Vikram Seth visit the two temples in Kathmandu?
Answer:
Mr Vikram Seth visited the two temples in Kathmandu with his son and nephew.

Question 2.
Why does a party of saffron-clad Westerner struggle?
Answer:
A party of saffron-clad Westerner struggles for permission to enter the temple.

Question 3.
Briefly describe Baudhnath Stupa?
Answer:
At the Boudhanath Stupa there is a sense of stillness. It has a white dome small shops are there. That is a heaven of quietness in the busy streets around.

Question 4.
What does the author buy at Nepal Airlines?
Answer:
At Nepal Airlines, the author buys a ticket for the next day’s flight.

Question 5.
When will the Kaliyug end on earth?
Answer:
The Kaliyug will end on earth when the small shrine half protruding from the stone platform on the river bank at the Pashupatinath emerges fully. The goddess inside will escape, and the evil period of the Kaliyug will end on earth.

B.2. Answer the following questions very briefly:

Question 1.
Where does the anther look at the flute seller?
Answer:
The author looks at the flute seller standing in a comer of the square near the hotel.

Question 2.
Name three kinds of the flute.
Answer:
The three kinds of the flute are the reed neh, the recorders and the deep baosuri of Hindustani classical music.

Question 3.
What does the flute seller have in his hand?
Answer:
In the flute seller’s hand, there is a pole with an attachment at the top from which fifty or sixty bansuri protrude.

Question 4.
Why does the author find it difficult to go away from the square?
Answer:
Tire author attracted by the flute music so he finds it difficult to go away from the square.

C. 1. Long Answer Type Questions

Question 1.
Why is Kathmandu famous? Describe briefly.
Answer:
Kathmandu is the capital of Nepal. It is famous for its two temples that are most sacred to Hindu and Buddhists. The first is die Pashupatinath temple. The second is the Baudhnath stupa that is a heaven of the compass in the busy streets around. Besides these, Kathmandu is a very lovely wealthy and religious place.

Question 2.
Describe Baudhnath Stupa and its surroundings.
Answer:
At Baudhnath stupa there is calm and quiet just outside and in the shrine. There are no crowds. Outside the shrine there are small shops of Tibetan immigrants selling felt bags. Tibetan prints and silver jewellery. The Buddhist shrine is a haven of quietness. It is a complete contrast of the Pashupatinath shrine.

Question 3.
Describe daily happenings at Pashupatinath.
Answer:
At Pashupatinath, there is an atmosphere of utter confusion mixed with excitement in and outside the shrine. Priests, hawkers, devotees, tourists, cows, monkeys, dogs and pigeons – all roam through the grounds. The crowd of worshippers push each other to go to the front to get “darshan”. There is noise confusion and disorder. There is a fight of monkeys inside the temple and quarrel for permission to enter by the saffron-clad foreigners.

Question 4.
What, according to the author, has been the pattern of the flute seller’s life?
Answer:
According to the author die flute seller stands in a comer of the square near. In his hand is a pole with an attachment at the top from which fifty or sixty bansuris protrude in all directions like die quills of a porcupine. From time to time he stands the pole on the ground selects a flute and plays for a few minutes. The sound rises clearly above the noise of the traffic and the hawker’ cries. He plays slowly meditatively, without excessive display. He makes sale also. Sometimes he breaks off playing to talk to the fruit seller. This has been the pattern of his life for years.

Question 5.
The author was moved by the music of the flute. Describe a similar experience of your own.
Answer:
Once there was a cultural programme at Ravindra Kala Bhawan in Patna. I had been there. I was watching it. A man appeared at the stage, at first sight, he looked like an ordinary man. He played the guitar and it was so good that I listened patiently. I was so charmed that I felt to be in a world of pleasure.

C. 2. Group Discussion

Discuss the following in groups or pairs

Question 1.
Religious tolerance is inbuilt in Indian society.
Answer:
India is a secularism country. Here people of all religions, irrespective of caste or creed enjoy equal rights. There is no religion of the state. This thought is nothing new for us, as it is embedded in the cultural ethos that makes us tolerant, magnanimous and receptive to ail religions like Sikhism, Jainism and Buddhism and of course Hinduism. People over the centuries lived in peace, except for the last two centuries when foreign rules visited this harmony by their much-maligned policy of divide and rule. Yet we are tolerant Benjamin Franklin said, “We must indeed hang together or, most assuredly we shall all hang separately” The true success of this belief will be when we regard ourselves first as Indians and then as Hindu. Sikhs and Muslims.

Question 2.
Music has overwhelming power.
Answer:
It is true to say that music haš overwhelming power. Music is such an art that can soothe and relax our heart. It is an art of making pleasing. It is combinations of sounds in rhythm and harmony. It finds people with a còmmon bond. It is also a source of living to hear music it is to be drawn into the commonality of all mankind to be moved by music. Its motive force too is living breath, It clearly differs from noise.

Comprehension Based Questions with Answers

1. I get a cheap room in the centre of town and sleep for hours. The next morning, with Mr Shah’s son and nephew, I visit the two temples in Kathmandu that are most sacred to Hindus and Buddhists.
At Pashupatinath (outside which a sign proclaims ‘Entrance for the Hindus only’) there is an atmosphere of ‘febrile confusion’. Priests, hawkers, devotees, tourists, cows, monkeys, pigeons and dogs roam through the grounds. We offer a few flowers. There are so many worshippers that some people trying to get the priest’s attention are elbowed aside by others pushing their way to the front. A princess of
the Nepalese royal house appears; everyone bows and makes way. By the main gate, a party of saffron-clad Westerners J struggle for permission to enter. The policeman is not convinced that they are ‘The Hindus’ (only Hindus are allowed to enter the temple). A light breaks out between two monkeys. One chases the other, who jumps onto a shiva linga, then runs screaming around the temples and down, to the river, the holy Bagmati, which flows below. A corpse is being cremated on its banks; washerwomen are at their work and children bathe. From a balcony a basket of flowers and leaves, old offerings now wilted, is dropped into the river. A small shrine half protrudes from the stone platform on the river bank. When it emerges fully, the goddess inside will escape, and the evil period of the Kaliyug will end on earth.

Questions:

1. Name the lesson and the author of the above passage.
2. What does ‘febrile confusion’ apply here?
3. What made the ‘febrile confusion’?
4. Why can some people not get the priest’s attention?
5. Which actions show that the members of the royal of Nepal are respected by common people?
6. Which word/words in the passage mean the following:
   (a) pushed to one side
   (b) bends with respect.

Answers:

1. The name of the lesson is ‘Kathmandu’. The author is Vikram Seth.
2. ‘Febrile confusion’ here implies excited, disorderly nervous movements and noises.
3. The febrile confusion was created by a crowd of devotees, hawkers, priests tourists, cows, monkeys, pigeons and dogs roaming through the ground.
4. because they are pushed by others who want to come to the front.
5. When a Nepalese princess came for worshipping in the temple every one bowed and made way for her.
6. (a) elbowed
   (b) bows

2. At the Boudhanath stupa, the Buddhist shrine of Kathmandu, there is, in contrast, a sense of stillness. Its immense white dome is ringed by a road. Small shops stand on its outer edge: many of these are owned by Tibetan immigrants; felt bags, Tibetan prints and silver jewellery can be bought here. There are no crowds: this is a haven of quietness in the busy streets around.
Kathmandu is vivid, mercenary, religious, with small shrines to flower-adorned deities along with the narrowest and busiest streets; with fruit sellers, flute sellers, hawkers of postcards; shops selling Western cosmetics, film rolls and chocolate; or copper utensils and Nepalese antiques. Film songs blare out from the radios, 6ar horns sound, bicycle bells ring, stray cows low questioningly at motorcycles, vendors shout out their wares. I indulge myself mindlessly: buy a bar of ’ marzipan, a com-on-the-cob roasted in a charcoal brazier on the pavement rubbed with salt, chilli powder and lemon); a couple of love story comics, and even a Reader’s Digest. All this I wash down with Coca Cola and a nauseating orange drink and feel much the better for it.

Questions:

1. What does the author want to contrast the Buddhist shrine with?
2. What does ring with a road imply?
3. Whose shops are there outside the Buddhist Shrine? What do they sell?
4. What contrast is there between the inside of the Buddhist shrine and its surroundings?
5. What type of city is Kathmandu?
6. What does the writer do to pass his time?
7. What does ‘wash down’ imply here?
8. Which words in the passage mean the following.
   (a) to satisfy one’s desire
   (b) safe place.

Answers:

1. It is contrasted with the Pashupatinath temple in Kathmandu.
2. It implies that there is a road all around the dome.
3. There are shops of Tibetan immigrants. They sell felt bags, Tibetan prints and silver jewellery is located in the busy streets of Kathmandu.
4. Inside the shrine, there is peace and quiet whereas the surroundings are noisy as the temple in located in the busy streets of Kathmandu.
5. Kathmandu is a religious city with narrow streets. These have shrines dedicated to different gods. There is noise and disorder but Kathmandu is full of life.
6. He buys some light reading and eating materials.
7. It implies that he reads and eats while sipping a Coca Cola and orange drink.
8. (a) Indulge
   (b) haven

3. I consider what route I should take back home. If I were -propelled by enthusiasm for travel purse, I would go by bus and train to Patna, then sail up the Ganges past Benaras to Allahabad, then up the Yamuna, past Agra to Delhi. But I am too exhausted and homesick; today is the last day of August. Go home, I tell myself: move directly towards home. I enter a Nepal Airlines office and buy a ticket for tomorrow’s flight.
I look at the flute seller standing in a comer of the square near the hotel. In his hand is a pole with an attachment at the top from which fifty or sixty bansuris protrude in all directions, like the quills of a porcupine. They are of bamboo: there are cross-flutes and recorders. From time to time he stands the pole on the ground, selects a flute and plays for a few minutes. The sound rises clearly above the noise of the traffic and the hawkers’ cries. He plays slowly, meditatively without excessive display. He does not shout out his wares. Occasionally he makes a sale, but in a curiously offhanded way as if this were incidental to his enterprise. Sometimes he breaks off playing to talk to the fruit seller. I imagine that this has been the pattern of his life for years.

Questions:

1. Why does the speaker not travel by bus or train?
2. How does the author describe the flute seller in such a manner?
3. Why does the flute seller play on the flute for just a few minutes?
4. While other’s shout out their, wares how does the flute seller attracts the attention of the customers?
5. How does he play the flute?
6. Which words in the passage mean the following:
   (a) without any preparation
   (b) not much show off.

Answers:

1. The speaker is feeling homesick. So to reach his home early he decides to go by air instead of by bus or train.
2. The author is attracted by the flutes and also by the flute seller. The flute reminds him of the commonality of all mankind.
3. The flute seller does not seem to bother much about the sale of his flutes.
4. He attracts the attention of his customers by playing on different flutes.
5. He plays the flute slowly, meditatively and without much show-off.
6. (a) offhanded way
   (b) excessive display.

4. I find it difficult to tear myself away from the square. Flute music always does this to me: it is at once the most universal and most particular of sounds. There is no culture that does not have its flute the reed neh, the recorder, the Japanese shakuhachi, the deep bansuri of Hindustani classical music, the clear or breathy flutes of South America, the high-pitched Chinese flutes. Each has its specific fingering and compass. It weaves its own associations. Yet to hear any flute is,, it seems to me, to be drawn into the commonality of all mankind, to be moved by music closest in its phrases and sentences to the human voice. Its motive force too is living breath: it too needs to pause and breathe before it can go on. That I can be so affected by a few familiar phrases on the bansuri, surprises me at first, for on the previous occasions that I have returned home after a long absence abroad, ! have hardly noticed such details, and certainly have not invested them with the significance I now do.

Questions:

1. What does ‘tear me away’ imply?
2. What fact shows that the writer is very fond of the flute?
3. How is flute music universal?
4. What does it mean to hear any flute?
5. How can the author be affected by the flute?
6. Which words in the passage mean the following
   (a) used by all
   (b) special.

Answers:

1. It implies that the author could not leave the place.
2. The fact that is difficult for him to move away from it shows that he likes it.
3. There is no culture in the world which does not have its own flute. This shows that the flute is universal.
4. To hear any flute means to be drawn into the commonality of all mankind.
5. The author can be so affected by a few familiar phrases on the bansuri, surprises him that he had hardly noticed such details.
6. (a) universal
   (b) particular.

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Bihar Board Class 12th Intermediate Entrepreneurship Objective Questions and Answers Key 2019, 2020, 2021 Pdf free download in Hindi Medium and English Medium.

Bihar Board 12th Objective Questions and Answers

Bihar Board 12th Intermediate Entrepreneurship Objective Questions and Answers Key 2020-2021 Pdf Download

Bihar Board 12th Entrepreneurship Objective Questions and Answers in Hindi

• Chapter 1 उद्यमितीय / साहसिक अवसरों की अनुभूति एवं पहचान
• Chapter 2 पर्यावरण अथवा वातावरण का सूक्ष्म अध्ययन
• Chapter 3 विपणन / बाजार मूल्यांकन
• Chapter 4 उद्यमितीय / साहसिक की पहचान एवं व्यवहार्थता अध्ययन
• Chapter 5 उपक्रम का चुनाव एवं स्थापना
• Chapter 6 व्यावसायिक नियोजन
• Chapter 7 परियोजना की अवधारणा एवं नियोजन
• Chapter 8 परियोजना प्रतिवेदन का निर्माण एवं परियोजना मूल्यांकन
• Chapter 9 संसाधन निर्धारण : वित्तीय एवं गौर-वित्तीय
• Chapter 10 स्थायी एवं कार्यशील पूँजी की आवश्यकताएँ
• Chapter 11 कोष प्रवाह विवरण
• Chapter 12 लेखांकन अनुपात
• Chapter 13 सम-विच्छेद विश्लेषण
• Chapter 14 उद्यमी पूँजी : कोषों के स्रोत एवं साधन
• Chapter 15 प्रौद्योगिकी का चयन
• Chapter 16 प्रबंध के बुनियादी आधार
• Chapter 17 उत्पादन प्रबंध एवं किस्म नियन्त्रण
• Chapter 18 विपणन प्रबंध
• Chapter 19 वित्त प्रबंधन
• Chapter 20 लागत एवं लाभ का निर्धारण
• Chapter 21 व्यवसाय में संवृध्दि एवं विकास की सम्भावनाएँ एवं रणनीति
• Chapter 22 उद्यमितीय अनुशासन एवं सामाजिक उत्तरदायित्व
• Bihar Board 12th Entrepreneurship Objective Model Set 1
• Bihar Board 12th Entrepreneurship Objective Model Set 2
• Bihar Board 12th Entrepreneurship Objective Model Set 3
• Bihar Board 12th Entrepreneurship Objective Model Set 4
• Bihar Board 12th Entrepreneurship Objective Model Set 5
• Bihar Board 12th Entrepreneurship Objective Model Set 6
• Bihar Board 12th Entrepreneurship Objective Model Set 7

Bihar Board 12th Entrepreneurship Objective Questions and Answers in English

• Chapter 1 Sensing and Identification of Entrepreneurial Opportunities
• Chapter 2 Environment Scanning
• Chapter 3 Market Assessment
• Chapter 4 Identification of Entrepreneurial Opportunities and Feasibility Study
• Chapter 5 Selection and Setting Up of An Enterprise
• Chapter 6 Business Planning
• Chapter 7 Concept of Project and Planning
• Chapter 8 Formulation of Project Report and Project Appraisal
• Chapter 9 Resource Assessment – Financial and Non-Financial
• Chapter 10 Fixed and Working Capital Requirements
• Chapter 11 Fund Flow Statement
• Chapter 12 Accounting Ratios
• Chapter 13 Break-Even Analysis
• Chapter 14 Venture Capital: Sources and Means of Funds
• Chapter 15 Selection of Technology
• Chapter 16 Fundamentals of Management
• Chapter 17 Production Management and Quality Control
• Chapter 18 Marketing Management
• Chapter 19 Financial Management
• Chapter 20 Determination of Cost and Profit
• Chapter 21 Possibilities and Strategies for Growth and Development in Business
• Chapter 22 Entrepreneurial Discipline and Social Responsibility
• Bihar Board 12th Entrepreneurship Objective Model Set 1
• Bihar Board 12th Entrepreneurship Objective Model Set 2
• Bihar Board 12th Entrepreneurship Objective Model Set 3
• Bihar Board 12th Entrepreneurship Objective Model Set 4
• Bihar Board 12th Entrepreneurship Objective Model Set 5
• Bihar Board 12th Entrepreneurship Objective Model Set 6
• Bihar Board 12th Entrepreneurship Objective Model Set 7

Bihar Board Class 10 Maths Solutions Chapter 2 बहुपद Ex 2.2 Text Book Questions and Answers.

BSEB Bihar Board Class 10 Maths Solutions Chapter 2 बहुपद Ex 2.2

Bihar Board Class 10 Maths बहुपद Ex 2.2

Bihar Board Class 10 Math Book Solution In Hindi प्रश्न 1.
निम्न द्विघात बहुपदों के शून्यक ज्ञात कीजिए और शून्यकों तथा गुणांकों के बीच के सम्बन्ध की सत्यता की जाँच कीजिए-
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4
हल
(i) दिया गया बहुपद = x² – 2x – 8
= x² – (4 – 2)x – 8
= x² – 4x + 2x – 8
= (x² – 4x) + (2x – 8)
= x(x – 4) + 2 (x – 4)
= (x – 4) (x + 2)
x² – 2x – 8 = (x – 4) (x + 2)
जब बहुपद x² – 2x – 8 = 0 हो तो (x – 4) (x + 2) भी शून्य होगा जिसका अर्थ है कि या तो x – 4 = 0 या फिर x + 2 = 0
यदि हो x – 4 = 0 हो तो x = 4 और यदि x + 2 = 0 हो तो x = -2
अत: बहुपद x² – 2x – 8 के शून्यक = 4 व -2
बहुपद x² – 2x – 8 की तुलना बहुपद ax² + bx + c से करने पर,
a = 1, b = -2 तथा c = -8
तब, बहुपद के गुणांकों और शून्यकों में सम्बन्ध
शून्यकों का योगफल = \(-\frac{b}{a}=-\left(\frac{-2}{1}\right)=2\)
तथा शून्यकों का गुणनफल: = \(\frac{c}{a}=\frac{-8}{1}=-8\)
और जो शून्यक हमने ज्ञात किए थे उनका योगफल भी 2 तथा गुणनफल (-8) है।
अत: बहुपद के गुणांकों और शून्यकों के बीच के उपर्युक्त सम्बन्ध सत्य हैं।
इति सिद्धम्

(ii) दिया गया बहुपद = 4s² – 4s + 1
= (2s)² – 2(2s) . 1 + (1)² [∵ (a – b)² = a² – 2ab + b²]
= (2s – 1)²
4s² – 4s + 1 = (2s – 1)²
जब बहुपद 4s² – 4s + 1 = 0 हो तो (2s – 1)² भी शून्य होगा जिसका अर्थ है कि
(2s – 1)² = 0
⇒ (2s – 1) = 0
⇒ 2s = 1
⇒ s = \(\frac{1}{2}\)
यहाँ बहुपद के दोनों शून्यक समान हैं,
अत: बहुपद 4s² – 4s + 1 के शून्यक = \(\frac{1}{2}\) व \(\frac{1}{2}\),
बहुपद 4s² – 4s + 1 की तुलना बहुपद as² + bs + c से करने पर,
a = 4, b = -4 तथा c = 1
तब, बहुपद के शून्यकों और गुणांकों में सम्बन्ध
शून्यकों का योगफल = \(-\frac{b}{a}=-\left(\frac{-4}{4}\right)=1\)
तथा शून्यकों का गुणनफल = \(\frac{c}{a}=\frac{1}{4}\)
और जो शून्यक हमने ज्ञात किए हैं उनका भी योगफल (\(\frac{1}{2}+\frac{1}{2}\) = 1) तथा गुणनफल (\(\frac{1}{2} \times \frac{1}{2}=\) \(\frac{1}{4}\)) है।
अत: बहुपद के शून्यकों और गुणांकों के बीच उपर्युक्त सम्बन्ध सत्य हैं।
इति सिद्धम्

(iii) दिया गया बहुपद = 6x² – 3 – 7x
= 6x² – 7x – 3
= 6x² – (9 – 2)x – 3
= 6x² – 9x + 2x – 3
= (6x² – 9x) + (2x – 3)
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
बहुपद 6x² – 3 – 7x = (2x – 3) (3x + 1)
जब बहुपद 6x² – 3 – 7x = 0 हो तो (2x – 3) (3x + 1) भी शून्य होगा जिसका अर्थ है कि या तो 2x – 3 = 0 या फिर 3x + 1 = 0
यदि 2x – 3 = 0 हो तो 2x = 3 ⇒ x = \(\frac{3}{2}\)
और यदि 3x + 1 = 0 हो तो 3x = -1 ⇒ x = \(-\frac{1}{3}\)
अत: बहुपद 6x² – 3 – 7x के शून्यक \(\frac{3}{2}\) व \(-\frac{1}{3}\)
अब, बहुपद 6x² – 3 – 7x की तुलना मानक द्विघात बहुपद ax² + bx + c से करने पर,
a = 6, b = -7 तथा c = -3
तब, बहुपद के शून्यकों और गुणांकों में सम्बन्ध

अत: बहुपद के गुणांकों और शून्यकों के बीच के उपर्युक्त सम्बन्ध सत्य हैं।
इति सिद्धम्

(iv) दिया गया बहुपद = 4u² + 8u = 4u(u + 2)
यदि उक्त बहुपद 4u² + 8u = 0 हो तो 4u(u + 2) = 0 जिसका अर्थ है कि
4u = 0 ⇒ u = 0 या फिर u + 2 = 0 ⇒ u = -2
अत: बहुपद 4u² + 8u के शून्यक = 0 व -2
अब, बहुपद 4u² + 8u की तुलना बहुपद au² + bu + c से करने पर,
a = 4, b = 8 तथा c = 0
तब, बहुपद के गुणांकों और शून्यकों में सम्बन्ध
शून्यकों का योगफल = \(-\frac{b}{a}=-\frac{8}{4}=-2\)
और शून्यकों का गुणनफल \(\frac{c}{a}=\frac{0}{4}=0\)
और हमने जो शून्यक ज्ञात किए हैं, उनका योगफल (-2 + 0) = -2 तथा गुणनफल {(-2) × 0} = 0 है
अत: बहुपद के गुणांकों और शून्यकों के बीच के उपर्युक्त सम्बन्ध सत्य हैं।
इति सिद्धम्

(v) दिया गया बहुपद = t² – 15
जब बहुपद t² – 15 = 0 हो तो t² = 15 या t = ±√15
अत: बहुपद t² – 15 के शून्यक = +√15 व -√15
यहाँ शून्यकों का योगफल (+√15 – √15) = 0 तथा गुणनफल {√15 × (-√15)} = -15 है।
दिए गए बहुपद t² – 15 = 0 की तुलना मानक द्विघात बहुपद at² + bt + c से करने पर,
a = 1, b = 0 तथा c = -15
तब, बहुपद के गुणांकों और शून्यकों के मध्य सम्बन्ध
शून्यकों का योगफल = \(-\frac{b}{a}=-\frac{0}{1}=0\)
और शून्यकों का गुणनफल = \(\frac{c}{a}=\frac{-15}{1}=-15\)
जो कि उपर्युक्त फलन से मेल खाता है।
अत: बहुपद के गुणांकों और शून्यकों के मध्य उपर्युक्त सम्बन्ध सत्य हैं।
इति सिद्धम्

(vi) दिया गया बहुपद = 3x² – x – 4
= 3x² – (4 – 3)x – 4
= 3x² – 4x + 3x – 4
= (3x² – 4x) + (2x – 4)
= x(3x – 4) + 1(3x – 4)
= (3x – 4) (x + 1)
= (3x – 4) (x + 1)
जब बहुपद 3x² – x – 4 = 0 हो तो (3x – 4) (x + 1) = 0
जिसका अर्थ है कि या तो 3x – 4 = 0 या फिर x + 1 = 0 है।
यदि 3x – 4 = 0 हो तो 3x = 4 ⇒ x = \(\frac {4}{3}\)
और यदि x + 1 = 0 हो तो x = -1
अत: बहुपद के शून्यक = \(\frac {4}{3}\) व -1
अब दिए हुए बहुपद 3x² – x – 4 की तुलना मानक द्विघात बहुपद ax² + bx + c से करने पर,
a = 3, b = -1 तथा c = -4
तब, बहुपद के गुणांकों a, b, c और बहुपद के शून्यकों के बीच सम्बन्ध
शून्यकों का योगफल = \(-\frac{b}{a}=-\frac{(-1)}{3}=\frac{1}{3}\)
और शून्यकों का गुणनफल = \(\frac{c}{a}=\frac{-4}{3}\)
और जो शून्यक हमने ज्ञात किए हैं उनका भी योगफल = \(\left(-1+\frac{4}{3}\right)=\frac{1}{3}\) और गुणनफल \(\left(\frac{4}{3} \times-1\right)=-\frac{4}{3}\) है।
अत: बहुपद के गुणांकों और शून्यकों के बीच उपर्युक्त सम्बन्ध सत्य हैं।
इति सिद्धम्

Bihar Board Class 10 Math Book Solution In Hindi Pdf Download प्रश्न 2.
एक द्विघात बहुपद ज्ञात कीजिए, जिसके शून्यकों के योगफल तथा गुणनफल क्रमशः दी गई संख्याएँ हैं
(i) \(\frac{1}{4}\), -1
(ii) √2, \(\frac{1}{3}\)
(iii) 0, √5
(iv) 1, 1
(v) \(\frac{-1}{4}\), \(\frac{1}{4}\)
(vi) 4, 1
हल
(i) माना द्विघात बहुपद के शून्यक α तथा β हैं।
तब, शून्यकों का योगफल = α + β तथा शून्यकों का गुणनफल = αβ
दिया गया है कि शून्यकों का योगफल \(\frac{1}{4}\) तथा गुणनफल -1 है।
α + β = \(\frac{1}{4}\) और αβ = -1
तब, द्विघात बहुपद = (x – α) (x – β)
= x² – (α + β) x + αβ
= x² – \(\frac{1}{4}\) x + (-1)
= \(\frac{4 x^{2}-x-4}{4}\)
= k(4x² – x – 4)
अत: अभीष्ट बहुपद 4x² – x – 4 या k(4x² – x – 4) है, जहाँ k = \(\frac{1}{4}\) एक वास्तविक संख्या है।

(ii) माना द्विघात बहुपद के शून्यक α तथा β हैं।
तब, शून्यकों का योगफल = α + β तथा गुणनफल = αβ
दिया गया है कि बहुपद के शून्यकों का योगफल √2 तथा गुणनफल \(\frac{1}{3}\) है।
α + β = √2 तथा αβ = \(\frac{1}{3}\)
तब, द्विघात बहुपद = (x – α) (x – β)
= x² – (α + β) + αβ
= x² – √2x + \(\frac{1}{3}\)
= \(\frac{1}{3}\)(3x² – 3√2x + 1)
= k(3x² – 3√2x + 1)
अत: अभीष्ट बहुपद k(3x² – 3√2x + 1) है, जहाँ k = \(\frac{1}{3}\) एक वास्तविक संख्या है।

(iii) माना द्विघात बहुपद के शून्यक α तथा β हैं।
तब, शून्यकों का योगफल = (α + β) और शून्यकों का गुणनफल = αβ
दिया गया है कि शून्यकों का योगफल 0 तथा गुणनफल √5 है।
तब, α + β = 0 तथा αβ = 15
द्विघात बहुपद = (x – α)(x – β)
= x² – (α + β)x + αβ
= x² – 0 . x + √5
= x² + √5
अतः अभीष्ट बहुपद = x² + √5

(iv) माना द्विघात बहुपद के शून्यक α तथा β हैं।
तब, शून्यकों का योगफल = α + β तथा शून्यकों का गुणनफल = αβ
दिया गया है कि शून्यकों का योगफल 1 तथा गुणनफल 1 है।
तब, α + β = 1 तथा αβ = 1
द्विघात बहुपद = (x – α) (x – β)
= x² – (α + β)x + αβ
= x² – (1) . x + 1
= x² – x + 1
अतः अभीष्ट बहुपद = x² – x + 1

(v) माना द्विघात बहुपद के शून्यक α व β हैं।
तब, शून्यकों का योगफल = α + β तथा शून्यकों का गुणनफल = αβ
दिया गया है कि शून्यकों का योगफल \(-\frac{1}{4}\) तथा गुणनफल \(\frac{1}{4}\) है।

(जहाँ k एक वास्तविक संख्या है)
अत: अभीष्ट बहुपद = 4x² + x + 1 अथवा k(4x² + x + 1) जहाँ k = \(\frac{1}{4}\) एक वास्तविक संख्या है।

(vi) माना द्विघात बहुपद के शून्यक α व β हैं।
तब, शून्यकों का योगफल = (α + β) तथा गुणनफल = αβ
दिया गया है कि शून्यकों का योगफल 4 तथा गुणनफल 1 है।
α + β = 4 तथा αβ = 1
द्विघात बहुपद = (x – α) (x – β)
= x² – (α + β)x + αβ
= x² – 4x + 1
अत: अभीष्ट बहुपद = x² – 4x + 1

Bihar Board Class 10 Maths Solutions Chapter 5 समांतर श्रेढ़ियाँ Ex 5.2 Text Book Questions and Answers.

BSEB Bihar Board Class 10 Maths Solutions Chapter 5 समांतर श्रेढ़ियाँ Ex 5.2

Bihar Board Class 10 Maths समांतर श्रेढ़ियाँ Ex 5.2

Bihar Board Class 10 Math प्रश्न 1.
निम्नलिखित सारणी में, रिक्त स्थानों को भरिए, जहाँ A.P. का प्रथम पद a, सार्वान्तर d और n वाँ पद a_n है:

हल
(i) दिया है, a = 7, d = 3, n = 8, a_n = ?
n वाँ पद (a_n) = a + (n – 1)d
= 7 + (8 – 1) × 3
= 7 + (7 × 3)
= 7 + 21
= 28
अत: an = 28

(ii) दिया है, a = -18, n = 10, a_n = 0, d = ?
n वाँ पद (a_n) = a + (n – 1)d
⇒ 0 = -18 + (10 – 1)d
⇒ -18 + 9d = 0
⇒ 9d = 18
⇒ d = 2
अतः d = 2

(iii) दिया है, d = -3, n = 18, a_n = -5, a = ?
n वाँ पद (a_n) = a + (n – 1)d
⇒ -5 = a + (18 – 1) × (-3)
⇒ -5 = a + (-51)
⇒ a = -5 + 51 = 46
अत: a = 46

(iv) दिया है, a = -18.9, d = 2.5, a_n = 3.6, n = ?
n वाँ पद (a_n) = a + (n – 1)d
⇒ 3.6 = -18.9 + (n – 1) (2.5)
⇒ 18.9 + 3.6 = (n – 1) (2.5)
⇒ (n – 1)(2.5) = 22.5
⇒ n – 1 = 9
⇒ n = 1 + 9 = 10
अतः n = 10

(v) दिया है, a = 3.5, d = 0, n = 10.5, a_n = ?
n वाँ पद (a_n) = a + (n – 1)d
= 3.5 + (10.5 – 1) (0)
= 3.5 + 0
= 3.5
अत: a_n = 3.5

Exercise 5.2 Class 10 Solutions In Hindi Bihar Board प्रश्न 2.
निम्नलिखित में सही उत्तर चुनिए और उसका औचित्य दीजिए :
(i) A.P.: 10, 7, 4,……, का 30 वाँ पद है :
(A) 97
(B) 77
(C) -77
(D) -87
(ii) A.P.: -3, \(-\frac{1}{2}\), 2,….. का 11 वाँ पद है :
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
हल
(i) दी हुई A.P. : 10, 7, 4, …….
यहाँ a = 10 तथा d = 7 – 10 = -3
A.P. का 30 वाँ पद (a₃₀) = a + (n – 1)d
= 10 + (30 – 1) × (-3)
= 10 + (-87)
= -77
अत: विकल्प (C) सही है।

(ii) दी हुई A.P. : -3, \(-\frac{1}{2}\), 2,…….
यहाँ a = -3 तथा d = \(-\frac{1}{2}\) – (-3) = \(\frac{5}{2}\)
A.P.का 11वाँ पद (a₁₁) = a + (n – 1)d
= -3 + (11 – 1) × \(\frac{5}{2}\)
= -3 + 10 × 5
= -3 + 25
= 22
अतः विकल्प (B) सही है।

Bihar Board Class 10th Math Solution प्रश्न 3.
निम्नलिखित समान्तर श्रेढ़ियों में रिक्त खानों (boxes) के पदों को ज्ञात कीजिए :

हल
(i) पहला पद (a) = 2, तीसरा पद = 26, दूसरा पद = ?
माना सार्वान्तर (d) है,
तब, तीसरा पद (a₃) = a + 2d = 26
⇒ 2 + 2d = 26
⇒ 2d = 24
⇒ d = 12
दूसरा पद = a + d = 2 + 12 = 14
अत: रिक्त बॉक्स का पद (a₂) = 14

(ii) पहला पद = ?, दूसरा पद = 13, तीसरा पद = ?, चौथा पद = 3
माना पहला पद (a) तथा सार्वान्तर (d) है।
तब, दूसरा पद = a + d
प्रश्नानुसार, a + d = 13 …….(1)
और चौथा पद = a + 3d
प्रश्नानुसार, a + 3d = 3 ………(2)
समीकरण (2) में से समीकरण (1) को घटाने पर,
2d = -10 ⇒ d = -5
समीकरण (1) में d का मान रखने पर,
a + d = 13
⇒ a + (-5) = 13
⇒ a = 13 + 5 = 18
और तीसरा पद = a + 2d = 18 + 2(-5) =18 – 10 = 8
अत: रिक्त बॉक्सों के पद क्रमशः 18 व 8 हैं।

(iv) -4, a₂, a₃ , a₄, a₅, 6
पहला पद (a) = -4
माना सार्वान्तर d है।
तब, छठा पद = a + 5d
परन्तु छठा पद = 6
a + 5d = 6
⇒ -4 + 5d = 6
⇒ 5d = 10
⇒ d = 2
दूसरा पद (a₂) = a + d = -4 + 2 = -2
तीसरा पद (a₃) = a + 2d = -4 + 2 × 2 = -4 + 4 = 0
चौथा पद (a₄) = a + 3d = -4 + 3 × 2 = -4 + 6 = 2
पाँचवाँ पद (a₅) = a + 4d = -4 + 4 × 2 = -4 + 8 = 4
अत: बॉक्सों के रिक्त पद क्रमशः -2, 0, 2, 4 हैं।

(v) a, 38, a₃, a₄, a₅, -22
माना पहला पद (a) तथा सार्वान्तर (d) है।
तब, दूसरा पद = a + d
परन्तु a + d = 38 ……..(1)
और छठा पद = a + (6 – 1)d = a + 5d
परन्तु a + 5d = -22 ………(2)
समीकरण (2) में से समीकरण (1) को घटाने पर,
(a + 5d) – (a + d) = -22 – 38
⇒ 4d = -60
⇒ d = -15
समीकरण (1) में d का मान रखने पर,
a + (-15) = 38
⇒ a = 38 + 15
⇒ a = 53
तीसरा पद (a₃) = a + 2d = 53 + 2(-15) = 53 – 30 = 23
चौथा पद (a₄) = a + 3d = 53 + 3(-15) = 53 – 45 = 8
पाँचवाँ पद (a₅) = a + 4d = 53 + 4(-15) = 53 – 60 = -7
अत: बॉक्सों के रिक्त पद क्रमशः 53, 23, 8, -7 हैं।

Bihar Board Class 10 Math Book Solution In Hindi प्रश्न 4.
A.P.: 3, 8, 13, 18, …… का कौन-सा पद 78 है?
हल
दी गई A.P. : 3, 8, 13, 18, ……..
पहला पद (a) = 3 तथा सार्वान्तर (d) = 8 – 3 = 5
माना n वा पद (a_n) 78 है।
n वाँ पद (a_n) = 78
⇒ a + (n – 1)d = 78
⇒ 3 + (n – 1)5 = 78
⇒ 3 + 5n – 5 = 78
⇒ 5n = 78 + 5 – 3 = 80
⇒ n = 16
अत: 16 वाँ पद 78 है।

Bihar Board Class 10 Math Solution In Hindi प्रश्न 5.
निम्नलिखित समान्तर श्रेढ़ियों में से प्रत्येक श्रेढ़ी में कितने पद हैं?
(i) 7, 13, 19, ……, 205
(ii) 18, 15\(\frac{1}{2}\), 13, ….., -47
हल
(i) दी गई समान्तर श्रेढ़ी (A.P.) : 7, 13, 19, …… , 205
पहला पद (a) = 7 तथा सार्वान्तर (d) = 13 – 7 = 6
माना दी गई A.P. में n पद हैं जिसमें n वाँ पद (a_n) = 205
n वाँ पद (a_n) = 205
⇒ a + (n – 1)d = 205
⇒ 7 + (n – 1)6 = 205
⇒ 7 + 6n – 6 = 205
⇒ 6n = 205 + 6 – 7 = 204
⇒ n = 34
अतः दी गई श्रेढी (A.P.) में 34 पद हैं।

(ii) दी गई समान्तर श्रेढ़ी (A.P.) : 18, 15\(\frac{1}{2}\), 13, ….., -47
पहला पद (a) = 18
तथा सार्वान्तर (d) = 15\(\frac{1}{2}\) – 18
= \(\frac{31}{2}\) – 18
= \(\frac{31-36}{2}\)
= \(\frac{-5}{2}\)
माना दी गई श्रेढ़ी में n पद हैं।
n वाँ पद 4 (a_n) = -47
⇒ a + (n – 1)d = -47
⇒ 18 + (n – 1)(\(\frac{-5}{2}\)) = -47
⇒ \(-\frac{5(n-1)}{2}\) = -47 – 18 = -65
⇒ (n – 1) = \(\frac{65 \times 2}{5}\) = 26
⇒ n = +1 + 26 = 27
अतः दी गई श्रेढी (A.P.) में 27 पद हैं।

Bihar Board Class 10 Math Book Solution In Hindi Pdf Download प्रश्न 6.
क्या A.P. : 11, 8, 5, 2 का एक पद -150 है? क्यों?
हल
दी गई A.P. : 11, 8, 5, 2
पहला पद (a) = 11 तथा सार्वान्तर (d) = 8 – 11 = -3
माना n वाँ पद (a_n) = -150 है।
n वाँ पद (a_n) = -150
⇒ a + (n – 1)d = -150
⇒ 11 + (n – 1) × -3 = -150
⇒ -3(n – 1) = -150 – 11 = -161
⇒ (n – 1) = 53.6 (लगभग)
⇒ n = 53.6 + 1 = 54.6
n का मान एक पूर्ण संख्या नहीं है।
अतः दी गई A.P. का कोई पद -150 नहीं है।

Bihar Board 10th Class Math Solution प्रश्न 7.
उस A.P. का 31 वाँ पद ज्ञात कीजिए, जिसका 11 वाँ पद 38 है और 16 वाँ पद 73 है।
हल
माना A.P. का पहला पद (a) तथा सार्वान्तर (d) है।
दिया है, A.P. का 11 वाँ पद (a₁₁) = 38
⇒ a + (n – 1)d = 38
⇒ a + (11 – 1)d = 38
⇒ a + 10d = 38 ………(1)
पुनः दिया है, A.P. का 16 वाँ पद (a₁₆) = 73
⇒ a + (16 – 1)d = 73
⇒ a + 15d = 73 ……..(2)
समीकरण (2) में से समीकरण (1) को घटाने पर,
(a + 15d) – (a + 10d) = 73 – 38
⇒ 5d = 35
⇒ d = 7
समीकरण (1) में d का मान रखने पर,
a + 10 × 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70 = -32
श्रेढ़ी का 31 वाँ पद (a₃₁) = a + (31 – 1)d
= -32 + 30 × 7
= -32 + 210
= 178
अतः A.P. का 31 वाँ पद = 178

10 Class Ka Math Bihar Board प्रश्न 8.
एक A.P. में 50 पद हैं, जिसका तीसरा पद 12 है और अन्तिम पद 106 है। इसका 29 वाँ पद ज्ञात कीजिए।
हल
माना A.P. का प्रथम पद (a) तथा सार्वान्तर (d) है।
तब, तीसरा पद = a + (3 – 1)d = a + 2d
और अन्तिम 50 वाँ पद = a + (50 – 1)d = a + 49d
तब प्रश्नानुसार,
a + 2d = 12 ………(1)
a + 49d = 106 ……(2)
समीकरण (2) में से समीकरण (1) को घटाने पर,
(a + 49d) – (a + 2d) = 106 – 12
⇒ 47d = 94
⇒ d = 2
तब समीकरण (1) में d का मान रखने पर,
a + 2 × 2 = 12
⇒ a + 4 = 12
⇒ a = 8
A.P. का 29 वाँ पद = a + (29 – 1)d
= 8 + 28 × 2
= 8 + 56
= 64
अतः दी गई A.P. का 29 वाँ पद 64 है।

Bihar Board Math Solution प्रश्न 9.
यदि किसी A.P. के तीसरे और नौवें पद क्रमशः 4 और -8 हैं तो इसका कौन-सा पद शून्य होगा?
हल
माना श्रेढ़ी का पहला पद (a) तथा सार्वान्तर (d) है।
A.P. का तीसरा पद (a₃) = a + 2d
तथा नौवाँ पद (a₉) = a + (9 – 1) d = a + 8d
तब प्रश्नानुसार,
a + 2d = 4 ……(1)
a + 8d = -8 ……(2)
समीकण (2) में से समीकरण (1) को घटाने पर,
(a + 8d) – (a + 2d) = -8 – 4
या 6d = -12
या d = -2
समीकरण (1) में d का मान रखने पर,
a + 2 x (-2) = 4
⇒ a – 4 = 4
⇒ a = 8
माना श्रेढ़ी का n वाँ पद शून्य होगा, अर्थात्
a_n = 0
n वाँ पद (a_n) = 0
⇒ a + (n – 1)d = 0
⇒ 8 + (n – 1) × (-2) = 0
⇒ -2(n – 1) = -8
⇒ (n – 1) = 4
⇒ n = 5
अतः दी गई A.P. का 5 वाँ पद शून्य होगा।

Bihar Board Class 5 Math Solution In Hindi प्रश्न 10.
किसी A.P. का 17 वाँ पद उसके 10 वें पद से 7 अधिक है। इसका सार्वान्तर ज्ञात कीजिए।
हल
माना A.P. का पहला पद (a) तथा सार्वान्तर (d) है।
तब, 17 वाँ पद (a₁₇) = a + (17 – 1)d = a + 16d
10 वा पद (a₁₀) = a + (10 – 1)d = a + 9d
प्रश्नानुसार, 17 वाँ पद, 10 वें पद से 7 अधिक है।
17 वाँ पद (a₁₇) – 10 वाँ पद (a₁₀) = 7
⇒ (a + 16d) – (a + 9d) = 7
⇒ 7d = 7
⇒ d = 1
अत: श्रेढ़ी का सार्वान्तर (d) = 1

Bihar Board Math Solution Class 10 प्रश्न 11.
A.P. : 3, 15, 27, 39, ….. का कौन-सा पद उसके 54 वें पद से 132 अधिक होगा?
हल
माना अभीष्ट पद n वाँ पद है।
दी गई A.P. : 3, 15, 27, 39, …..
प्रथम पद (a) = 3 तथा सार्वान्तर (d) = 15 – 3 = 12
तब, श्रेढ़ी का 54 वाँ पद (a₅₄) = a + (54 – 1)d
= 3 + (53 × 12)
= 3 + 636
= 639
n वॉ पद (a_n) = 54 वें पद से 132 अधिक
= 639 + 132
= 771
n वाँ पद (a_n) = 771
a + (n – 1)d = 771
3 + (n – 1) 12 = 771
(n – 1)12 = 771 – 3 = 768
n – 1 = 64
n = 64 + 1 = 65
अतः श्रेढ़ी का 65 वाँ पद 54 वें पद से 132 अधिक है।

Bihar Board Class 5 Math Solution प्रश्न 12.
दो समान्तर श्रेढ़ियों का सार्वान्तर समान है। यदि इनके 100 वें पदों का अन्तर 100 है, तो इनके 1000 वें पदों का अन्तर क्या होगा?
हल
माना पहली A.P. का पहला पद a तथा सार्वान्तर d है और दूसरी A.P. का पहला पद A तथा सार्वान्तर d है क्योंकि सार्वान्तर समान है।
तब, पहली श्रेढ़ी का 100 वाँ पद = a + (100 – 1)d = a + 99d
दूसरी श्रेढ़ी का 100 वा पद = A + (100 – 1) d = A + 99d
दोनों श्रेढ़ियों के 100 वें पदों का अन्तर = (A + 99d) – (a + 99d) = A – a
तब, प्रश्नानुसार, A – a = 100 ……(1)
अब, पहली श्रेढ़ी का 1000 वाँ पद = a + (1000 – 1)d = a + 999d
दूसरी श्रेढ़ी का 1000 वाँ पद = A + (1000 – 1)d = A + 999d
दोनों श्रेढ़ियों के 1000 वें पदों का अन्तर = (A + 999d) – (a + 999d) = A – a
दोनों श्रेढ़ियों के 1000 वें पदों का अन्तर = A – a = 100 [समीकरण (1) से]
अत: 1000 वें पदों का अन्तर = 100

Bihar Board Class 5 Math Book Solution प्रश्न 13.
तीन अंकों वाली कितनी संख्याएँ 7 से विभाज्य हैं?
हल
तीन अंकों की संख्याओं की सूची : 100, 101, 102, ……., 999
3 अंकों की 7 से विभाज्य पहली संख्या = 105
और अन्तिम संख्या = 994
तब, 7 से विभाज्य 3 अंकीय संख्याओं की सूची :
105, (105 + 7), (105 + 7 + 7), ……….., 994
= 105, 112, 119,…….,994
माना कुल संख्याएँ n हैं।
पहली संख्या (a) = 105, सार्वान्तर (d) = 7, n वाँ पद (a_n) = 994
n वा पद (a_n) = 994
a + (n – 1)d = 994
105 + (n – 1) × 7 = 994
(n – 1) × 7 = 994 – 105 = 889
(n – 1) = \(\frac{889}{7}\) = 127
n = 127 + 1 = 128
अतः 7 से विभाज्य तीन अंकीय संख्याएँ 128 हैं।

Bihar Board Class 5th Math Solution In Hindi प्रश्न 14.
10 और 250 के बीच में 4 के कितने गणज हैं?
हल
10 से बड़ा 4 का पहला गुणज = 12
250 से छोटा 4 का पहला गुणज = 248
10 और 250 के बीच 4 के गुणजों की सूची :
12, 12 + 4, (12 + 4 + 4),……., 248
12, 16, 20, 24, …….., 248.
माना गुणजों की संख्या n है।
यहाँ, पहला पद (a) = 12, सार्वान्तर (d) = 16 – 12 = 4
n वा पद (a_n) = 248
⇒ a + (n – 1)d = 248
⇒ 12 + (n -1) 4 = 248
⇒ 12 + 4n – 4 = 248
⇒ 4n = 248 + 4 – 12 = 240
⇒ n = 60
अत: 10 और 250 के बीच 4 के गुणजों की संख्या = 60

Bihar Board Class 10th Math Solution In Hindi प्रश्न 15.
n के किस मान के लिए, दोनों समान्तर श्रेढ़ियों 63, 65, 67, ….. और 3, 10, 17,…..के n वें पद बराबर होंगे?
हल
पहली समान्तर श्रेढ़ी : 63, 65, 67,……
पहला पद (a) = 63, सार्वान्तर (d) = 65 – 63 = 2
श्रेढ़ी का n वा पद = a + (n – 1)d
= 63 + (n – 1)2
= 63 + 2n – 2
= 61 + 2n
दूसरी समान्तर श्रेढ़ी : 3, 10, 17, ……
पहला पद (A) = 3 सार्वान्तर (D) = 10 – 3 = 7
श्रेढ़ी का n वाँ पद = A + (n – 1)D
= 3 + (n – 1)7
= 3 + 7 n – 7
= 7n – 4
दोनों श्रेढ़ियों के n वें पद बराबर हैं।
7n – 4 = 61 + 2n
⇒ 7n – 2n = 61 + 4
⇒ 5n = 65
⇒ n = 13
अतः दी गई दोनों श्रेढ़ियों के 13 वें पद समान हैं।

Bihar Board Class 10 Math Solution प्रश्न 16.
वह A.P. ज्ञात कीजिए जिसका तीसरा पद 16 है और 7 वाँ पद 5 वें पद से 12 अधिक है।
हल
माना पहला पद a है तथा सार्वान्तर d है।
दिया है, A.P. का तीसरा पद = 16
a + 2d = 16 ……(1)
श्रेढ़ी का 7 वाँ पद (a₇) = a + (7 – 1)d = a + 6d
तथा 5 वाँ पद (a₅) = a + (5 – 1)d = a + 4d
7 वाँ पद 5 वें पद से 12 अधिक है
(a + 6d) – (a + 4d) = 12
⇒ a + 6d – a – 4d = 12
⇒ 2d = 12
⇒ d = 6
d का मान समीकरण (1) में रखने पर,
a + 2 × 6 = 16
⇒ a = 4
श्रेढ़ी का पहला पद (a) = 4
दूसरा पद (a₂) = a + d = 4 + 6 = 10
तीसरा पद (a₃) = a + 2d = 4 + 2 × 6 = 4 + 12 = 16
चौथा पद (a₄) = a + 3d = 4 + 3 × 6 = 4 + 18 = 22
अत: अभीष्ट A.P. : 4, 10, 16, 22, …….. है।

Bihar Board 10th Math Solution प्रश्न 17.
A.P. : 3, 8, 13, ….., 253 में अन्तिम पद से 20 वाँ पद ज्ञात कीजिए।
हल
दी गई A.P. : 3, 8, 13, ………, 253
पहला पद (a) = 3, सार्वान्तर (d) = 8 – 3 = 5
यदि श्रेढ़ी को अवरोही क्रम में लिखें तो यह निम्नवत् होगी
253, (253 – 5), (253 – 10), (253 – 15), ……., 3
या 253, 248, 243, 238, ………, 3
पहला पद (a) = 253 तथा सार्वान्तर (d) = 248 – 253 = -5
श्रेढ़ी का 20 वाँ पद = a + (20 – 1)d
= 253 + 19 × (-5)
= 253 – 95
= 158
अतः दी गई A.P. के अन्तिम पद से 20 वाँ पद = 158
वैकल्पिक विधिः A.P. का अन्त से n वाँ पद = l – (n – 1)d
जहाँ, पर l = अन्तिम पद यहाँ, l = 253, d = 8 – 3 = 5, n = 20
A.P. का अन्त से 20वाँ पद = 253 – (20 – 1) (5)
= 253 – 19 × 5
= 253 – 95
= 158

Bihar Board Class 5th Math Solution प्रश्न 18.
किसी A.P. के चौथे और 8 वें पदों का योग 24 है तथा छठे और 10 वें पदों का योग 44 है। इस A.P. के प्रथम तीन पद ज्ञात कीजिए।
हल
माना A.P. का पहला पद a तथा सार्वान्तर d है।
प्रश्नानुसार, 4वाँ पद (a₄) + 8 वाँ पद (a₈) = 24
⇒ a + (4 – 1)d + a + (8 – 1)d = 24
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ……..(1)
तथा 6वाँ पद (a₆) + 10वाँ पद (a₁₀) = 44
⇒ a + (6 – 1)d + a + (10 – 1)d = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 ……..(2)
समीकरण (2) में से समीकरण (1) को घटाने पर,
(a +7d) – (a + 5d) = 22 – 12
⇒ 2d = 10
⇒ d = 5
d का यह मान समीकरण (1) में रखने पर,
a + 5 × 5 = 12
⇒ a + 25 = 12
⇒ a = -13
तब, श्रेढ़ी का पहला पद (a₁) = -13
दूसरा पद (a₂) = a + d = -13 + 5 = -8
तीसरा पद (a₃) = a + 2d = -13 + 2 × 5 = -13 + 10 = -3
अतः दी गई A.P. के प्रथम तीन पद = -13, -8, -3

प्रश्न 19.
सुब्बाराव ने 1995 में ₹ 5000 के मासिक वेतन पर कार्य आरम्भ किया और प्रत्येक वर्ष ₹ 200 की वेतन वृद्धि प्राप्त की। किस वर्ष में उसका वेतन ₹ 7000 हो गया?
हल
पहले वर्ष में प्रारम्भिक वेतन = ₹ 5000 प्रतिमास
दूसरे वर्ष में वेतन = ₹ (5000 + 200) = ₹ 5200
प्रतिमास तीसरे वर्ष में वेतन = ₹ (5200 + 200) = ₹ 5400 प्रतिमास
इस प्रकार प्रत्येक वर्ष के वेतन (₹)
5000, 5200, 5400, …….. एक समान्तर श्रेढ़ी बनाते हैं।
जिसका पहला पद (a) = 5000 तथा सार्वान्तर (d) = 200
माना n वर्ष बाद वेतन ₹ 7000 होगा।
तब, n वाँ पद = 7000
a + (n – 1)d = 7000
⇒ 5000 + (n – 1) 200 = 7000
⇒ (n – 1) × 200 = 7000 – 5000
⇒ (n – 1) × 200 = 2000
⇒ (n – 1) = 10
⇒ n = 10 + 1 = 11
अत: 11 वें वर्ष में सुब्बाराव का वेतन ₹ 7000 हो जायेगा।

प्रश्न 20.
रामकली ने किसी वर्ष के प्रथम सप्ताह में ₹ 5 की बचत की और फिर अपनी साप्ताहिक बचत ₹ 1.75 बढ़ाती गई। यदि n वें सप्ताह में उसकी साप्ताहिक बचत ₹ 20.75 हो जाती है, तो n ज्ञात कीजिए।
हल
प्रथम सप्ताह की बचत = ₹ 5
प्रत्येक सप्ताह की बचत में उत्तरोत्तर ₹ 1.75 की वृद्धि होती है।
प्रत्येक सप्ताह की बचतें एक A.P. का निर्माण करती हैं जिसका पहला पद (a) = 5 तथा सार्वान्तर (d) = ₹ 1.75
n वें सप्ताह में बचत = 20.75
a + (n – 1)d = 20.75
⇒ 5 + (n – 1) 1.75 = 20.75
⇒ (n – 1) × 1.75 = 20.75 – 5
⇒ (n – 1) × 1.75 = 15.75
⇒ n – 1 = 9
⇒ n = 9 + 1 = 10
अत: n = 10

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Bihar Board Class 9 English Abraham Lincoln’s Letter to His Son’s Teacher Text Book Questions and Answers

A. Answer the following questions orally:

Abraham Lincoln’s Letter To His Son’s Teacher Solutions Bihar Board Question 1.
What do you know about Abraham Lincoln, the famous President of America?
Answer:
Abraham Lincoln was the sixteenth President of the United States of America. He was the man who abolished slavery by Proclamation in 1863. He was shot-dead in 1865.

Abraham Lincoln’s Letter To His Son’s Teacher Questions Answers Pdf Bihar Board Question 2.
Has your father ever written any letter to your teacher?
Answer:
No, he has never written any letter to my teacher.

Abraham Lincoln Letter To His Son’s Teacher Question Answer Bihar Board Question 3.
Can you imagine how much your father is worried about your future?
Answer:
Yes, like every father he is very much worried about my future. He wants it to be good at every step. So he gives proper care of my studies.

B.1. Write ‘T’ for true and ‘F’ for false statement:

1. It is a letter written by a father to his son.
2. All men are not just and true.
3. A dollar earned is of far more value than five pounds.
4. One should be taught to mourn over losing.
5. One should be taught to fail rather than to cheat.

Answer:

1. — F
2. — T
3. — T
4. — F
5. — T

B.2. Complete the sentences on the basis of your reading of the poem:

1. Teach him to sell his ________ and brain to the highest _________
2. Teach him to be _________ with gentle people.
3. Teach him to ________ at cynics.
4. Teach him it is far _________ to fail than to ________
5. Let him learn early that the bullies are the easiest to _________
6. Teach him the _______ of books.
7. He should be given quiet time to ponder the _________ mystery of birds in the sky.
8. One should have sublime _________ in himself to have faith in mankind.

Answer:

1. brawn, bidder
2. gentle
3. scoff
4. honourable, cheat
5. lick
6. wonder
7. eternal
8. faith.

C.1. Long Answer Type Questions

Abraham Lincoln Letter To His Son’s Teacher Poem Questions And Answers Bihar Board Question 1.
Why did Abrahafti Lincoln write a letter to his son’s teacher?
Answer:
Abraham Lincoln wrote a letter to his son’s teacher suggesting to give such an education which can make him strong with body, mind and spirit.

Abraham Lincoln’s Letter To His Son’s Teacher Poem Bihar Board Question 2.
What does Lincoln mean by saying that “for every scoundrel, there is a hero; that for every selfish Politician, there is a dedicated leader…”
Answer:
The saying is true. He means to say that there is always every scoundrel there is a hero and for every selfish Politician, there is a dedicated leader. It depends upon thinker as what he .thinks. We should consider it within our heart about the concerning man.

Letter To A Teacher Prose Questions And Answers Bihar Board Question 3.
A child should be treated gently but not cuddled. Do you agree? Give your opinion.
Answer:
Too much of love spoil the child. A child definitely needs well discipline and manner. Emersion says, “Secret of Education lies in respecting pupils.” At the time of giving education, he should be treated kindly but at his fault he must be punished. He.should be treated kindly but a not at his fault.

Abraham Lincoln Letter To His Son’s Teacher Question And Answer Bihar Board Question 4.
“All men are not just, all men are not true.” Comment on this statement.
Answer:
The poet is right to say that all men are not just and are not true. It’s human nature differs individually. Some are kind and some are rude. So Lincoln advises the teacher to teach the boy about human nature. So that he may beware of the people in Future.

Lincoln’s Son Will Learn That Bihar Board Question 5.
Why do you think Lincoln wants his son to steer away from envy and learn the secret of quiet laughter?
Answer:
Lincoln wants his son to steer away from envy because envy is such a person who believes that people do not do things for good. He is not sincere. Such type of man is not social, he does every work for his sake only. Rather he wants his son that a man who remains amused all the. can tackle any situation and should enjoy laughing quietly.

Abraham Lincoln Letter To His Son’s Teacher Poem Bihar Board Question 6.
Why does Lincoln not want his son to follow the crowd when everyone is getting on the bandwagon?
Answer:
Lincoln does not want his son to follow the crowd when everyone is getting on the bandwagon because he won’t be able to develop has personal qualities. He would not be a leader but a follower as a puppet.

Abraham Lincoln’s Letter To His Son’s Teacher Questions And Answers Bihar Board Question 7.
What qualities did Lincoln want his son’s teacher to teach him?
Answer:
Lincoln wants his son’s teacher to teach him to know about different types of people and able to face different situations. He wants that he should not have ill thoughts rather he should be always in a happy mood. He wished for his son a higher and higher post. He must be self-dependent.

C.2. Group Discussion

Lincoln’s Son Will Learn That Answer Bihar Board Question 1.
The present system of education is at variance with the learner’s experience.
Answer:
The multiplicity of subjects that we have to study bewilders me. 1 do some times wonder the utility of studying Shakespeare in English literature, History, Geography, Social studies etc. There are a host of other subjects, which have probably no use in our’ later lives. One could understand if we were to study the history and geography of our country. What is perplexing, however, is that we have to learn the history and geography of different countries and continents that we shall probably never even see in our lives. So present system of education is at variance with the learners is useless.

Comprehension Based Questions with Answers

1. He will have to learn, I know,
that all men are not just,
all men are not true.
But teach him also that
for every scoundrel there is a hero;
that for every selfish Politician,
there is a dedicated leader
Teach him for every enemy there is a friend,
It will take time, I now;
but teach him if you can,
that a dollar earned is of far more value than five-pound….

Questions:

1. Name the poem and its poet.
2. What will the pupil have to learn?
3. What does the poet advise the teacher about foe and friend,
4. Which earning is preferable to?

Answers:

1. The name of the poem is “Abraham Lincoln’s letter to his son’s teacher” and the poet is Abraham Lincoln.
2. The pupil will have to learn that all men are not just and true.
3. The poet advises the teacher to teach his pupil that for every enemy there is a friend.
4. A dollar earned honestly is better than five-pound earned otherwise.

Question 2.
Teach him to learn to lose…
and also to enjoy winning.
Steer him away from envy,
if you can,
teach him the secret of
quiet laughter.
Let him learn early that
the bullies are the easiest to lick…
Teach him, if you can,
the wonder of books…
But also give him quiet time
to ponder the eternal mystery of birds in the sky,
bees in the sun,
and the flowers on a green hillside.

Questions:

1. What are the poet’s suggestions for losing and winning?
2. What do you mean by envy?
3. What is an early lesson?
4. What does the poet say about books?
5. What does the poet advise about the mystery?

Answers:

1. The poet suggests the teacher to teach to learn to lose and to enjoy winning.
2. Envy means a feeling of discontent caused by’some else’s good fortune. The poet here asks the teacher to steer away from his son from envy.
3. The poet advises the teacher to give his son an early lesson that the bullies are easiest to be defeated.
4. The poet thinks that there is ‘wonder in books.
5. The poet advises the teacher to teach his son so he may able to think over birds in the sky. bees in the sun and flowers on the green hillside.

3. In the school teach him
it is far honourable to fail
than to cheat…
Teach him to have faith
in his own ideas,
even if everyone tells hint
they are wrong…
Teach him to be gentle
with gentle people,
and tough with the tough.
Try to give my son
the strength not to follow the crowd when everyone is getting oil the bandwagon…

Questions:

1. What does the poet advise the teacher to teach the boy at school?
2. What should he have faith in?
3. How should he behave with people?
4. Find the word from the given stanza which means, “cart”.

Answers:

1. It is far honourable to fail than to cheat.
2. He should have faith in his own ideas.
3. He should be gentle with the gentle and tough with the tough.
4. Wagon.

Question 4.
I each him to listen to all men…
but teach him also to filter
all he hears on a screen of truth,
and take only the good
that comes through.
Teach him if you can.
how to laugh when he is sad…
Teach him there is no shame in tears.
Teach him to scoff at cynics
and to beware of too much sweetness…
Teach him to sell his brawn
and brain to the highest bidders
but never to put a price-tag
on his heart and soul.
Teach him to close his ears
to a howling mob
and to stand and fight
if he thinks he’s right.

Questions:

1. What does the poet advise in the first line?
2. What should he do when he is sad?
3. Whom should he scoff at?
4. Whom should he sell his brawn and brain?
5. What should be done with the howling mob?

Answers:

1. The poet advised in the first line that he should be taught to listen to all men.
2. When hs is said he should learn to laugh.
3. He should be taught to scoff at cynics.
4. He should sell his brawn and brain to the highest bidders.
5. He should not hear it but should stand and fight if he thinks right.

5. Treat him gently,
but do not cuddle him,
because only the test
of fire makes fine steel.
Let him have the courage
to be impatient…
let him have the patience to be brave.
Teach him always
to have sublime faith in himself,
because then he will have
sublime faith in mankind.
This is a big order,
but see what you can do…
He is such a fine fellow, my son!

Questions:

1. What is the suggestion of the poet?
2. How is steel made?
3. What should he always think?
4. How is the son?

Answers:

1. The poet suggests the teacher treat the boy gently but forbade to caddie him.
2. When iron is made hot in fire it becomes fine steel.
3. He should always think about the sublime.
4. His son is very nice.

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