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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

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Panorama English Book Class 9 Solutions Chapter 2 Yayati

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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.

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Bihar Board Class 8 Maths Solutions Chapter 2 एक चर वाले रैखिक समीकरण Text Book Questions and Answers.

BSEB Bihar Board Class 8 Maths Solutions Chapter 2 एक चर वाले रैखिक समीकरण

Bihar Board Class 8 Maths एक चर वाले रैखिक समीकरण Ex 2.1

Bihar Board Class 8 Math Solution प्रश्न 1.
3(x – 3) = 15
उत्तर
3(x – 3) = 15
3x – 9 = 15
3x = 15 + 9
3x = 24
x = \(\frac{24}{3}\)
x = 8

Bihar Board Class 8 Math Solution Chapter 2 प्रश्न 2.
\(\frac{x}{2}\) – 7 = 15
उत्तर
\(\frac{x}{2}\) – 7 = 15
\(\frac{x-14}{2}\) = 15
x – 14 = 15 × 2 = 30
x – 14 = 30
x = 30 + 14
x = 44

Class 8 Bihar Board Math Solution प्रश्न 3.
\(\frac{-2 x}{7}\) + 2 = 8
उत्तर
\(\frac{-2 x+14}{7}\) = 8
-2x + 14 = 8 × 7 = 56
-2x + 14 = 56
-2x = 56 – 14
-2x = 42
-x = \(\frac{42}{2}\)
x = -21

Bihar Board Class 8 Math Solution In Hindi प्रश्न 4.
7 – 3x = 18
उत्तर
7 – 3x = 18
-3x = 18 – 7
-3x = 11
x = \(\frac{-11}{3}\)

Bihar Board Math Solution Class 8 प्रश्न 5.
18 = 40 – 3x
उत्तर
18 – 40 = -3x
-22 = -3x
x = \(\frac{22}{3}\)

Bihar Board 8th Class Math Solution प्रश्न 6.
\(\frac{25}{6}\) – 9y = 11
उत्तर

Class 8 Math Solution Bihar Board प्रश्न 7.
2.4 = \(\frac{x}{2.5}\) – 1
उत्तर
2.4 = \(\frac{x-2.5}{2.5}\)
2.4 × 2.5 = x – 2.5
6 = x – 2.5
6 + 2.5 = x
x = 8.5

Bihar Board Solution Class 8 Math प्रश्न 8.
3x + 10 – x = 0
उत्तर
3x + 10 – x = 0
3x – x = -10
2x = -10
x = -5

Bihar Board Class 8 Math प्रश्न 9.
2(x + \(\frac{11}{4}\)) = 13
उत्तर
2(\(\frac{4 x+11}{4}\)) = 13
2(4x + 11) = 13 × 4
8x + 22 = 52
8x = 52 – 22
8x = 30
x = \(\frac{30}{8}\)
x = \(\frac{15}{4}\)

Bihar Board Class 8 Maths Solutions प्रश्न 10.
\(\frac{x}{3}+\left(\frac{-14}{3}\right)=\frac{3}{7}\)
उत्तर

Bihar Board Class 8 Maths एक चर वाले रैखिक समीकरण Ex 2.2

Class 8 Math Bihar Board प्रश्न 1.
यदि किसी संख्या के आधे में से \(\frac{1}{4}\) घटाया जाए तो \(\frac{1}{8}\) प्राप्त होता है। संख्या ज्ञात कीजिए।
उत्तर
माना कि सं. = x
तो प्रश्नानुसार,

Bihar Board 8 Class Math Solution प्रश्न 2.
यदि किसी आयत की लम्बाई और चौड़ाई का अंतर 5 मी. हो और परिमिति 110 मी. हो तो लम्बाई एवं चौड़ाई ज्ञात करें।
उत्तर
माना कि आयत की लम्बाई = x
आयत की चौड़ाई = y
प्रश्न से,
(x – y) = 5 ……. (i)
आयत की परिमिति = 2(x + y) = 110
x + y = \(\frac{110}{2}\) = 55 …….. (ii)
अब, (i) तथा (ii) से,
x + y = 55
x – y = 5
—————-
2x = 60
x = 30
(i) से,
x + y = 55
3x + y = 55
y = 55 – 30 = 25
आयत की लम्बाई = x = 30 m.
आयत की चौड़ाई = y = 25 cm.

एक चर वाले रैखिक समीकरण क्लास 8 प्रश्न 3.
चीनी के मूल्य में 25 प्रतिशत की वृद्धि होने पर अब 1 किग्रा. चीनी का मूल्य 32 रु. है तो प्रारम्भ में चीनी का मूल्य प्रति किग्रा. क्या था?
उत्तर
चीनी का वर्तमान मू० = 32
माना की पहले चीनी का मूल्य = 100
वर्तमान मू० = \(\frac{32 \times 100}{125}=\frac{3200}{125}\) = 25.60

Bihar Board Class 8 Math Solution In Hindi Pdf Download प्रश्न 4.
दो विभिन्न मूल्यवाली 35 कलमों का कुल मूल्य 60 रु. है। यदि 1 सस्ती कलम का मूल्य 1.50 रु. एवं 1 महँगी कलम का मूल्य 2 रु. है तो कितनी महँगी कलमें खरीदी गईं?
उत्तर
माना कि महंगी कलमों की सं० = x
ता सस्ती कलम का मू = 60 – 2x
प्रश्नानुसार,
1.5 (60 – 2x) + 2x = 60
90 – 3x + 2x = 60
-x = 60 – 90
-x = -30
x = 30
महँगी कलमों की सं० = \(\frac{30}{2}\) = 15
तथा सस्ती कलमों की सं. = \(\frac{300}{15}\) = 20

Bihar Board Class 8 Math Book Solution प्रश्न 5.
एक त्रिभुज के तीनों कोण 2 : 3 : 5 के अनुपात में है तो उनके तीनों कोण ज्ञात कीजिए।
उत्तर
त्रिभुज के तीनों कोणों का योग = 180°
प्रश्नानुसार,
2x + 3x + 5x = 180
10x = 180
x = 18
त्रिभुज का प्रथम कोण = 2x = 18 × 2 = 36°
त्रिभुज का दूसरा कोण = 3x = 18 × 3 = 54°
त्रिभुज का तीसरा कोण = 5x = 18 × 5 = 90°

Bihar Board Class 8th Math Solution प्रश्न 6.
बिल्लू के पास 1 रु. एवं 5 रु. के कुल 160 सिक्के हैं जिनका कुल मूल्य 310 रु. है। यदि 2 रु. के सिक्कों की संख्या 5 रु. के सिक्कों की संख्या की तिगुनी हो तथा 1 रु. के सिक्कों की संख्या = 40 है तो उसके पास प्रत्येक प्रकार के कितने सिक्के हैं?
उत्तर
माना कि एक रु. के सिक्कों की सं० = 40
5 रु. के सिक्कों की सं० = x
2 रु. के सिक्कों की संख्या = 3x
तो प्रश्न से,
40 + x + 3x = 160
40 + 4x = 160
4x = 160 – 40 = 120
x = 30
तो 5 रु. के सिक्कों की सं. = x = 30
2 रु. के सिक्कों की सं = 3x = 90
1 रु. के सिक्कों की सं० = 4x = 40
कुल सिक्कों की सं० = 160
कुल मु० = 30 × 5 = 150
90 × 3 = 120
40 × 1 = 40
कुल मु० = 310

प्रश्न 7.
पिता ने अपने तीन संतानों के बीच अपनी संपत्ति का बँटवारा 1 : 2 : 3 के अनुपात में करता है और अपने लिए 100000 रु. रखता है। यदि उसकी कुल संपत्ति 2.5 लाख रु. की हो तो प्रत्येक संतान को हिस्से के रूप में क्या मिला?
उत्तर
पिता की कुल संपत्ति = 2.5 लाख
पिता अपने लिए 1 लाख रु० रखता है।
पिता ने अपने बच्चों में बाँटने लिए रखे 2.5 – 1.0 = 1.5
अब प्रश्नानुसार,
x + 2x + 3x = 1.5 लाख
6x = 1500000
x = 25000
पिता ने अपनी प्रथम संतान को दिए x = 25000
पिता ने अपनी द्वितीय संतान को दिए 2x = 2 × 25000 = 50000
पिता ने अपनी द्वितीय संतान को दिए 2x = 2 × 25000 = 50000
पिता ने अपने तृतीय संतान को दिए = 3x = 25000 × 3 = 75000

प्रश्न 8.
11 के लगातार तीन गुणजों का योग 231 है तो उन्हें ज्ञात कीजिए।
उत्तर
माना कि 11 का पहला गुणज = x
द्वितीय गुणज = x + 1
तृतीय गुणज = x + 2
प्रश्नानुसार,
11 × x + 11 (x + 1) + 11 (x + 2) = 231
11x + 11x + 11 + 11x + 22 = 231
33x + 33 = 231
33x = 231 – 33
33x = 198
x = 6
पहला गुणज = 11x = 11 × 6 = 66
गज = 11(x + 1) = 11(6 + 1) = 11 × 7 = 77
तृतीय गुणज = 11(x + 2) = 11(6 + 2) = 11 × 8 = 88

प्रश्न 9.
संकुल संसाधन केन्द्र म.वि. फरना में आयोजित बाल मेले में प्रत्येक विजेता छात्र को 2 कलम एवं विजेता को छोड़कर शेष सभी प्रतिभागियों को 1 कलम दिया गया। यदि 100 छात्रों के बीच 120 कलम दिये गये तो विजेताओं की संख्या ज्ञात कीजिए।
उत्तर
विजेताओं को दिए गए कलमों की सं० = 2x
दूसरे विद्यार्थियों को दिए गए कलमों की सं० = x
2x + x = 120
3x = 120
x = 40
विजेताओं की सं० = 2x = 40
x = 20

प्रश्न 10.
रवि के पिता का वर्तमान उम्र रवि के वर्तमान उम्र के तिगुने से 5 वर्ष अधिक है। 5 वर्ष बाद उनके उम्रों का योग 47 वर्ष होगा। दोनों की वर्तमान उम्र ज्ञात कीजिए।
उत्तर
माना कि रवि की उम्र = x
पिता की उम्र = 3x + 5
प्रश्न से,
x + 5 + 3x + 5 + 5 = 47
4x + 15 = 47
4x = 47 – 15 = 32
x = 8
रवि की उम्र = 8
पिता की वर्तमान उम्र = 3x + 5 = 3 × 8 + 5 = 24 + 5 = 29

Bihar Board Class 8 Maths एक चर वाले रैखिक समीकरण Ex 2.3

निम्नलिखित समीकरणों का हल ज्ञात कीजिए

प्रश्न 1.
\(\frac{7-6 x}{9 x}=\frac{1}{15}\)
उत्तर
\(\frac{7-6 x}{9 x}=\frac{1}{15}\)
15 (7 – 6x) = 9x
105 – 90x = 9x
105 = 9x + 90x
105 = 99x
x = \(\frac{35}{33}\)

प्रश्न 2.
\(\frac{z}{4}=\frac{z+15}{9}\)
उत्तर
\(\frac{z}{4}=\frac{z+15}{9}\)
9z = 4(z + 15)
9z = 4z + 60
9z – 4z = 60
5z = 60
z = 12

प्रश्न 3.
x² – (x – 2)² = 32
उत्तर
x² – (x² – 4x + 4) = 32
x² – x² + 4x – 4 = 32
4x – 4 = 32
4x = 32 + 4
4x = 36
x = 9

प्रश्न 4.
(x + 4)² – (x – 5)2 = 9
उत्तर
(x + 4)² – (x – 5)2 = 9
x² + 8x + 16 – (x² – 10x + 25) = 9
18x – 9 = 9
18x = 9 + 9
18x = 18
x = 1

प्रश्न 5.
(y + 3) (y – 3) – y (y + 5) = 6
उत्तर
(y + 3)(y – 3) – y(y + 5) = 6
y2 – 6y + 9 – y2 – 5y = 6
-11y = 6 – 9
-11y = -3
y = \(\frac{3}{11}\)

प्रश्न 6.

उत्तर

प्रश्न 7.

उत्तर

प्रश्न 8.
\(\frac{0.3+0.7 x}{x}\) = 0.95
उत्तर
\(\frac{0.3+0.7 x}{x}\) = 0.95
0.3 + 0.7x = 0.95 × x
0.3 + 0.7x = 0.95x
0.3 = 0.95x – 0.7x
0.3 = 0.25x
x = 1.2

प्रश्न 9.
\(\frac{15(2-x)-5(x+6)}{1-3 x}=6\)
उत्तर
\(\frac{15(2-x)-5(x+6)}{1-3 x}=6\)
15(2 – x) – 5(x + 6) = 6(1 – 3x)
30 – 15x – 5x – 30 = 6 – 18x
-20x = 6 – 18x
-20x + 18x = 6
-2x = 6
x = -3

प्रश्न 10.
दो अंकों की संख्या का दहाई अंक, इकाई अंक का तिगुना है। यदि अंक बदल दिये जाएँ तो प्राप्त संख्या मूल संख्या से 36 कम हो जाती है। वह संख्या ज्ञात कीजिए।
उत्तर
माना कि इकाई का अंक = x
दहाई का अंक = 10 × 3x
अंक बदल दिए जाने पर इकाई का अंक = 3x
दहाई का अंक = x
प्रश्नानुसार,
3x × 10 + x = 10x × x + 3x + 36
30x + x = 10x + 3x + 36
31x = 13x + 36
31x – 13x = 36
18x = 36
x = 2
इकाई का अंक x = 2
दहाई का अंक = 3x = 3 × 2 = 6
इसलिए अंक = 10 × 3x + x = 31x = 31 × 2 = 62

प्रश्न 11.
एक नाव धारा की दिशा में एक घाट से दूसरे घाट तक जाने में 9 घंटे लगाती है । धारा की विपरीत दिशा में यही दूरी 10 घंटे में पूरा करती है। यदि धारा की चाल 1 किमी./प्रति घंटा हो तो शांत जल में नाव की चाल एवं दोनों घाटों के बीच की दूरी ज्ञात कीजिए।
उत्तर
धारा की दिशा में नाव की चाल = 9 km/h (u)
धारा की विपरीत दिशा में चाल = 10 km/h (v)
शांत जल में नाव की चाल = (u + v) km/h = (10 + 9) km/h = 19 km/h
दोनों घाटों के बीच की दूरी = 2 × v × u = 2 × 10 × 9 = 20 × 9 = 180 km.

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Panorama English Book Class 9 Solutions Chapter 1 Dharam Juddha

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Bihar Board Class 9 English Dharam Juddha Text Book Questions and Answers

A. Answer the following questions very briefly.

Dharam Juddha Class 9 Question Answer Bihar Board Question 1.
What was the question that the Padma wanted to know from her parents?
Answer:
The Padma wanted to know from her parents what the identity of a woman was and how she was different from others.

Dharam Juddha Question Answer Bihar Board Class 9 English Question 2.
How many brother and sister did the Padma have?
Answer:
The Padma was the only child of her parents. She had no brother and sister.

Dharam Juddha In Hindi Bihar Board Class 9 English Question 3.
Why was the Padma unhappy with her father?
Answer:
The Padma was unhappy with her father as she thought that he cared more for money than for her.

Dharam Juddha Class 9 In Hindi Bihar Board Question 4.
For whom is life hell according to Padma’s mother?
Answer:
According to Padma’s mother, life is hell for the women whose husband is dead.

Question 5.
What, according to the Padma, is the identity of the human race?
Answer:
According to the Padma, the identity of the human race is the identity of women.

C. Long Answer Questions

Question 1.
‘Father cares more for money than for me’. Why does the Padma say so? Have you felt like the Padma? Explain.
Answer:
The Padma says so because she feels her father lacking in emotion towards her. He is always behind earning money. She feels unhappy with him for her neglect. According to Padma her father gives more importance to money than to her. Occasionally I feel like the Padma when my parents do not listen to my talk and they do not give proper care and attention. I find that my demand is ignored and I feel that I am no more than a commodity to my parents and an object to gratify their sense of achievement. They lack personal feelings and emotions for me.

Question 2.
What, according to Padma’s father, is the identity of a woman? Do you agree with him?
Answer:
According to Padma’s father marriage lends identity to a woman. A woman’s identity depends on the identity of her husband. I do not agree with this point. Women play a complementary and important role in the life of men. Nature has gifted them with special attributes so that the human race would continue to grow to prosper. Thus while men have been endowed with virile strength, women are endowed with human sensibilities and emotion like love. Compassion and caring. These are extremely vital for the upbringing of the future generation. Who is the hope of tomorrow? Their contribution to the field of education, doctors, nurses or a social worker. They provide health care facilities to the community. Their contributions are indeed their identity.

Question 3.
What question did the Padma ask that her mother calls senseless? Why did Padma’s mother call it senseless? Explain.
Answer:
The question that the Padma asked her mother, “what if one does not get married?” Padma’s mother was a devotee lade. She knew’ that for a woman husband is everything without a husband a woman life is like a hell so when the Padma asked the question repeatedly she called it senseless.

Question 4.
What is the injustice that the Padma talked about?
Answer:
The Padma calls it an injustice when her mother tells that life is hell for the woman whose husband is dead. She hates such idea of woman’s identity.

Question 5.
Sketch the character of Padma.
Answer:
The Padma is a young girl with a modem outlook. She is observed of all observers. She is the centre of the play. She is educated and advanced she is independent in thinking and has a strong personality. She does not suffer from any sense of inferiority or fear complex that is why she feels aggrieved and agitated by the uncharitably orthodox attitude of the society towards women. She repeatedly questions her parents and her teacher about the right and the identity of a woman and asks them whether her identity is subject to her marriage. She wants equality and justice for the woman.

Question 6.
Padma’s mother is the true image of a typical traditional woman. Elaborate,
Answer:
The word tradition, however, means, unwritten beliefs and customs handed down from generation to generation, which we all knowingly or unknowingly adhere to in our daily lives. It is the emotional fabric, which binds us to our fore-lather and makes us distinct from one another.
Padma’s mother is the true image of a typical traditional woman. She loves her daughter and her husband truly. She does not believe in the independent identity of a woman without her husband she does not think herself different from her husband, she thinks, that her identity is linked with her husband and she cherishes the bond. She strongly believes that a woman without a husband has no existence and for a widow life is a hell. She also believes that a woman has no right to ask odd questions from her husband. Thus, Padma’s mother proves herself a complete Indian traditional woman.

Question 7.
Discuss the main idea contained in the text?
Answer:
The main idea contained in the text is an enquiry into the identity of the woman and search for justice and the right of equality for her. The Padma is a young girl and a feminist with the modern outlook. She opposes traditional beliefs and questions to her parents about the identity of a woman. She is progressive in her thinking. She wants to enjoy equal rights and should have identity of her own. She does not want to be dependent on marriage or her husband for her identity.

Question 8.
Why is the lesson culled “Dharam Juddha”? Give reasons.
Answer:
The lesson has been titled Dharam Juddha which reminds the Mahabharata war between the Kauravas and the Pandavas. Lord Krishna encouraged Arjuna to fight against his Guru and grandfather for his rights. It was called Dharam Juddha a war for righteousness. Here is also a war of justice and rights of equality of a woman. The Padma fights for a woman in modern ways whereas her parents are of on traditional path. The Padma wants woman’s independent… identity in society. So It is just like a war of right and equality. So the title is ‘Dharam Juddha’ is correct.

A. Work in small groups and discuss the following:

Question 1.
What forms the identity of a woman in society?
Answer:
I think the woman in free India enjoys perfect equality with man. She is no longer a slave. She is no longer at the mercy of a man. She has a voice in shaping the destiny of the nation. We have women in different roles as administrators, ambassador, governors, leaders, minters, doctors teachers, officers, educationists and even in defence services. They are proving good in every field of work. They have on their side brain beauty and personality. They are sweet, soft and peace-loving. You may see them in the forms, said above.

Question 2.
How is marriage related to the identity of a woman?
Answer:
The present social, customs and practices also determinate against her I thus after marriage, she has to adopt the surname of her spouse and move in with him. What title she had of her identity, is also lost after marriage. She is also divided her share in parental property. But she only was known as the housewife of a particular house and the wife of a particular person of a family, she is considered the builder of the nation. She enriches society. and help in the growth and development of the country. Cheat men are known to their mothers. All great men and their autobiographies have acknowledged the fact that behind every successful man there is a woman. Thus it is clear that marriage is related to the succession of men. It is her great identity.

Question 3.
An unmarried man is venerated and called a saint, while an unmarried woman is called immoral and wanton. Is it justified?
Answer:
If a man decides to remain unmarried, he is considered a saint and is also regarded as a sacred man, his character is also considered sacred. He now becomes a holy man. On the other hand, if a woman remains unmarried she is called immortal and wanton. Her character is suspected in society. She is not respected Gdike a holy man. It is not justified according to the author. So he pleads for equal treatment.

Question 4.
Both men and women are equally important for human society. Do you agree?
Answer:
Yes, I wholeheartedly agree that both men and women are equally important for human society. Behind every successful! man, there is a woman. All great men in their autobiographies have acknowledged this fact. Woman today play a complementary and important role in the life of men and thus in human society. Nature has gifted them with special attributes so that the human race would continue to grow and prosper. Thus while men have been endowed with virile strength, women are endowed with human sensibilities and emotion like love, compassion and caring. These extremely vital for the upbringing of the future generation, who are the hope of tomorrow. They are the fulcrums around which the family and our lives revolve. Thus a wife, mother, sister they play an indispensable part in the life of men. So like men, they are equally important for human society.

Question 5.
‘Man and women should have equal rights’. Discuss.
Answer:
Undoubtedly, men and women should have equal right in all respects. The need of the hour is not to question this reality, but to devise ways and means to ameliorate their lot. The Government has taken many steps to give women a fair deal. Marriage customs are rapidly changing. A social atmosphere is being created to abolish dowry. It must, however, be admitted that freedom has distributed its blessing mainly among the woman belonging to rich classes in society. But the majority of women are still backward and superstitious. By the government, the women have been given the same rights as are enjoyed by men. Some of them are bringing honour and glory to India by serving the country wholeheartedly.

Question 6.
“The hand that rocks the cradle rules the world”. Elaborate.
Answer:
This prover runs regarding women. As a mother, she inculcated spiritual and moral values in the child. Outside home, she moulds the destiny of the children as teacher, counsellor and guide. As doctors, nurses or a social worker, she provides health care facilities to the community. Their contributions are indeed invaluable for the well-being, growth and economic development of society and the country. She played a crucial role in men’s life no doubt, as a mother. Thus in all crucial matters pertaining to property, or ascendance to the throne, it was the son who inherited the property or the throne.

Question 3.
‘Women in India are underprivileged’. Elaborate.
Answer:
Women in India are underprivileged. We like it or not and no matter what facts and figures we may quote, the facts remain that the fruits of development have not reached them in the measure that it should have. They constitute nearly one half of the population of the country and therefore, no real development can take place without their involvement and participation. It is indeed a matter of concern that after more than sixty years of independence, they continue to live an underprivileged existence more like a second class citizen. We may have done away with Sati Pratha and the Pardah system. But evils like dowry and female infanticide continue to thrive in our society even in age of enlightenment. This is why a girl child is looked upon as a burden for her parents. This discrimination does not anger well for the growth and development of the country. Immediate steps are called for ameliorating their lot, by providing them with free education and ensuring their economic independence.

Comprehension Based Questions with Answers

1. Padma: Mother, what is the identity of a woman?
Maa: Why do you ask that?
Padma: How am I different from others?
Maa: You are my daughter, our only child and your . father’s darling.
Padma: Father cares more for money than for me. It takes a heart to bestow love.
Maa: For whom does he save? He does all this only for your sake. You shouldn’t think that I’ve no heart.
Padma: But I didn’t say it of you.
Maa: I am not in any way different from him. My identity is linked with his and I cherish the bond.

Questions:

1. What is the relation between the talkers?
2. Who is the daughter?
3. What does she ask her mother?
4. Who is darling and whose?
5. What does the daughter feel about her father?
6. What is the reply of the mother?

Answers:

1. The talkers are mother and daughter.
2. The Padma is the daughter.
3. She asks from her mother what the identity of a woman is.
4. The Padma is darling of her father.
5. The daughter feels that her father cares more for money than for her which takes a heart to bestow love.
6. The mother replies that her father saves money for her only daughter, Padma.

2. Father: What is this talk about the bond, Padma’s mother?
Maa: Here is your father. Ask him whatever you want to know.
Father: What’s it. my child?
Maa: She wants to know what is the identity of a woman.
Father: You will know it, dear, when you are married.
Padma: Does marriage lend identity to a woman?
Maa: What is a woman without a husband?
Padma: What if one does not get married?
Maa: Why do you always ask such senseless questions?
Padma: What about a woman whose husband is dead?
Maa: Life is hell for her.
Padma: That’s injustice!

Questions:

1. Who enters?
2. What does he ask and from whom?
3. What does the daughter want to know from her father?
4. What do her fathers reply?
5. Find out from the lesson which means ‘unjust’.

Answers:

1. Padma’s fathers enter.
2. He asks from Padma’s mother what that talk is about the bond.
3. The daughter wants to know from her father what the identity of a woman is.
4. Her father replies that she would know the fact when she is married.
5. Injustice.

3. Maa: You needn’t talk of justice and injustice here.
Padma: Why?
Maa: Over here women don’t have the right to ask questions.
Padma: What right do they have then?
Father: Rights are for equals, dear.
Padma: Why is a home considered a place for bargain¬ing? One shouldn’t live in such a place.
Father: You will get the same replies to these questions wherever you go.
Padma: But the identity of a woman is the identity of the human race. A society sans women…
Father: That’s why they say that marriage lends this identity.
Padma: So if a girl does not get married she has no identity! Why is it that an unmarried man is venerated and called a saint while an unmarried woman is called immoral and wanton?

Questions:

1. ‘A home is like a’… What type of place is, according to the Padma?
2. Is she considering the house a living place?
3. What is the identity of the human race?
4. What is the difference between an unmarried man and an unmarried woman?
5. Find a word in the passage which means ‘unchaste’.

Answers:

1. According to Padma a home is considered a place for bargaining!
2. No, such a home is not a living place for the Padma.
3. The identity of a woman is the identity of the human race.
4. An unmarried man is venerated and called a saint but an unmarried woman is called immoral and wanton.
5. Wanton.

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