CBSE Class 10

These NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3

Question 1.
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
Solution:

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° tan (90° – 48°) tan (90° – 23°)
= tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° .\(\frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}\)
= 1 = RHS [Verified]

(ii) LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin 38° sin (90° – 38°)
= cos 38° sin 38°- sin 38° cos 38° = 0 = RHS [Verified]

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵cot (90° – θ) = tan θ]
⇒ 90° – 2A = A – 18° ⇒ 3A = 108° ⇒ A = \(\frac { 108° }{ 3 }\)
∴ ∠ A = 36°

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
tan A = cot B ⇒ tan A = tan (90° – B) [ ∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B ⇒ A + B = 90° Proved

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°) [cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20° ⇒ 5A = 110°
A = \(\frac { 110° }{ 5 }\)
A = 22°
∴ ∠ A = 22°

Question 6.
If A, Band Care interior angles of a triangle ABC, then show that: sin (\(\frac { B+C }{ 2 }\)) = cos \(\frac { A }{ 2 }\)
Solution:
We have given that,
A, B and C are the interior angles of a triangle ABC

Question 7.
Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75°
or sin (90° – 23°) + cos (90° – 15°)
or cos 23° + sin 15° (∵ sin (90° – θ) = cos θ
and cos (90° – θ) = sin θ)
Now, sin 67° + cos 75° in term if trigonometric ratios of angles between 0° and 45°.
(cos 90° – 67°) + (sin 90° – 75°)
= cos 23° + sin 15°

Detailed, Step-by-Step NCERT Solutions for Class 10 Sanskrit Shemushi Chapter 11 प्राणेभ्योऽपि प्रियः सुह्रद् Questions and Answers were solved by Expert Teachers as per NCERT (CBSE) Book guidelines covering each topic in chapter to ensure complete preparation.

Shemushi Sanskrit Class 10 Solutions Chapter 11 प्राणेभ्योऽपि प्रियः सुह्रद्

प्रश्न 1.
अधोलिखितप्रश्नानाम् उनराणि संस्कृतभाषया लिखत-

(क) चन्दनदासः कस्य गृहजनं स्वगृहे रक्षति स्म?
उत्तर:
चन्दनदासः अमात्यराक्षस्य गृहजनं स्वगृहे रक्षति

(ख) तृणानां केन सह विरोधः अस्ति?
उत्तर:
तृणानां अग्निना सह विरोधः अस्ति।

(ग) कः चन्दनदासं द्रष्टुमिच्छति?
उत्तर:
चाणक्यः चन्दनदासं द्रष्टुमिच्छति।

(घ) पाठेऽस्मिन् चन्दनदासस्य तुलना केन सह कृता?
उत्तर:
पाठेऽस्मिन् चन्दनदासस्य तुलना शिविना सह कृता?

(ङ) प्रीताभ्यः प्रकृतिभ्यः प्रतिप्रियं के इच्छन्ति?
उत्तर:
प्रीताभ्यः प्रकृतिभ्यः प्रतिप्रियं राजान इच्छन्ति?

(च) कस्य प्रसादेन चन्दनदासस्य वणिज्या अखण्डिता?
उत्तर:
चन्द्रगुप्तस्य प्रसादेन चन्द्रनदासस्य वाणिज्या अखण्डिता?

प्रश्न 2.
स्थूलाक्षरपदानि आधृत्य प्रश्ननिर्माणं कुरुत-

(क) शिविना विना इदं दुष्कर कार्य कः कुर्यात्।
उत्तर:
केन् विना इदं दुष्कर कार्य कः कुर्यात्?

(ख) प्राणेभ्योऽपि प्रियः सुहृत्।
उत्तर:
प्राणेभ्योऽपि प्रियः कः?

(ग) आर्यस्य प्रसादेन मे वणिज्या अखण्डिता।
उत्तर:
कस्य प्रसादेन मे वाणिज्या अखण्डिता?

(घ) प्रीताभ्यः प्रकृतिभ्यः राजानः प्रतिप्रियमिच्छन्ति।
उत्तर:
प्रीताभ्यः प्रकृतिभ्यः के प्रतिप्रियमिच्छन्ति?

(ङ) तृणानाम् अग्निना सह विरोधो भवति।
उत्तर:
केषाम् अग्निना सह विरोधो भवति?

प्रश्न 3.
यथानिर्देशमुनरत-

(क) ‘अखण्डिता में वणिज्या’-अस्मिन् वाक्ये व्यिापदं किम्?
उत्तर:
अखण्डिता

(ख) पूर्वम् ‘अनृतम्’ इदानीम् आसीत् इति परस्परविरुद्ध वचने-अस्मात् वाक्यात् ‘अधुना’ इति पदस्य समानार्थकपदं चित्वा लिखत।
उत्तर:
इदानीम्

(ग) ‘आर्य! किं मे भयं दर्शयसि’ अत्र ‘आर्य’ इति सम्बोधनपदं कस्मै प्रयुक्तम्?
उत्तर:
चाणक्यः

(घ) ‘प्रीताभ्यः प्रछतिभ्यः प्रतिप्रियमिच्छन्ति राजानः’ अस्मिन् वाक्ये कर्तृपदं किम्?
उत्तर:
राजानः

(ङ) तस्मिन् समये आसीदस्मद्गृहे’ अस्मिन् वाक्ये विशेष्यपदं किम्?
उत्तर:
गृहे

प्रश्न 4.
निर्देशानुसारं सन्धि/सन्धिविच्छेदं कुरुत-

(क) यथा – कः + अपि – कोऽपि
प्राणेभ्यः + अपि – _________
_________ + अस्मि – सज्जोऽस्मि।
आत्मनः + _________ – आत्मनोणधिकारसदृशम्
उत्तर:
यथा – कः + अपि – कोऽपि
प्राणेभ्यः + अपि – प्राणेभ्यऽपि
सज्जः + अस्मि – सज्जोऽस्मि।
आत्मनः + अधिकारसदृशम् – आत्मनोधिकारसदृशम्

(ख) यथा- सत् + चित् – सच्चित्
शरत् + चन्द्रः – __________
कदाचित् + च – __________
उत्तर:
यथा – सत् + चित् – सच्चित्
शरत् + चन्द्रः – शरच्चन्द्र
कदाचित् + च – कदाचित

प्रश्न 5.
अधोलिखितवाक्येषु निर्देशानुसार परिवर्तनं कुरुत-
यथा – प्रतिप्रियमिच्छन्ति राजानः। (एकवचने) प्रतिप्रियमिच्छति राजा।

(क) सः प्रकृतेः शोभा पश्यति (बहुवचने)
उत्तर:
ते प्रकृतेः शोभा पश्यन्ति।

(ख) अहं न जानामि। (मध्यमपुरुषैकवचने)
उत्तर:
त्वम् न जानासि।

(ग) त्वं कस्य गृहजन स्वगृहे रक्षसि? (उत्तमपुरुषैकवचने)
उत्तर:
अहं कस्य गृहजनं स्वगृहे रक्षामि?

(घ) कः इदं दुष्करं कुर्यात्? (प्रथमपुरुषबहुवचने)
उत्तर:
के इदं दुष्करं कुर्यु:?

(ङ) चन्दनदासं द्रष्टुमिच्छामि। (प्रथमपुरुषैकवचने)
उत्तर:
चन्दन दासः द्रष्टुम इच्छति।

(च) राजपुरुषाः देशान्तरं व्रजन्ति। (प्रथमपुरुषैकवचने)
उत्तर:
राजपुरुषः देशान्तरं व्रजति।

प्रश्न 6.
(अ) कोष्ठकेषु दनयोः पदयोः शुद्धं विकल्पं विचित्य रिक्तस्थानानि पूरयत-

(क) __________ विना इदं दुष्करं कः कुर्यात्। (चन्दनदासस्य / चन्दनदासेन)
उत्तर:
चन्दनदासेन्

(ख) _____________ इदं वृनान्तं निवेदयामि। (गुरवे / गुरोः)
उत्तर:
गुरवे

(ग) आर्यस्य ____________ अखण्डिता में वणिज्या। (प्रसादात् / प्रसादेन)
उत्तर:
प्रसादेन

(घ) अलम् ___________। (कलहेन / कलहात्)
उत्तर:
कलहेन्

(ङ) वीरः ______________ बालं रक्षति। (सिंहेन /ष्ट्वसिंहात्)
उत्तर:
सिहात्

(च) ______________ भीतः मम भ्राता सोपानात् अपतत्। (कुक्कुरेण / कुक्कुरात्)
उत्तर:
कुक्कुरात्

(छ) छात्रः ______________ प्रश्नं पृच्छति। (आचार्यम् / आचार्येण)
उत्तर:
आचार्यम्।

(आ) अधोदत्तमञ्जूषातः समुचितपदानि गृहीत्वा विलोमपदानि लिखत-

असत्यम् पश्चात् गुणः आदरः तदानीम् तत्र

(क) अनादरः ___________
उत्तर:
आदरः

(ख) दोषः ___________
उत्तर:
गुणः

(ग) पूर्वम् _____________
उत्तर:
पश्चात्

(घ) सत्यम् ___________
उत्तर:
असत्यम्

(ङ) इदानीम् __________
उत्तर:
तदानीय

(च) अत्र ___________
उत्तर:
तत्र

प्रश्न 7.
उदाहरणमनुसृत्य अधोलिखितानि पदानि प्रयुज्य पञ्चवाक्यानि रचयत-

यथा निष्क्रम्य-शिक्षिका पुस्तकालयात् निष्क्रम्य कक्षा प्रविशति।

(क) उपसृत्य
उत्तर:
आदित्यः मातरम् उपसृत्य प्रणमति।

(ख) प्रविश्य
उत्तर:
आचार्यः कक्षा प्रविश्य पाठयति।

(ग) द्रष्टुम्
उत्तर:
लता चलचित्रं द्रष्टम् इच्छति।

(घ) इदानीम्
उत्तर:
सः इदानी खेलति।

(ङ) अत्र
उत्तर:
अत्र मनोहरम् तडागः अस्ति।

अन्यपरीक्षोपयोगी प्रश्नाः

प्रश्न 1.
पाठात् समुचित-विलोमपदान चित्वा लिखत-

(पाठ से उचित विलोत शब्दों को चुनकर लिखिए)

(क) प्रविश्य
उत्तर:
निष्क्रम्य्

(ख) अप्रियम्
उत्तर:
प्रियम

(ग) सत्यम्
उत्तर:
असव्यम

(घ) आयान्ति, आगच्छन्ति
उत्तर:
गच्छन्ति

(च) सन्तम्
उत्तर:
असन्तम्

प्रश्न 2.
पाठात् समुचित पार्यय पदानि लिखा।

(क) मन्त्री
उत्तर:
अमात्यः

(ख) आच्छाद्य
उत्तर:
पिधाय

(ग) दु:खाभावः
उत्तर:
अपरिक्लेशः

(घ) स्वगतम्
उत्तर:
आत्मगतम्

(च) उपगम्य
उत्तर:
उपसृत्य।

प्रश्न 3.
प्रस्तुत पाठ पठित्वा अधोलिखित प्रश्नानां उत्तरणि लिखत-

(प्रस्तुत पाठ को पढ़कर निम्नलिखित प्रश्नों के उत्तर लिखिए-)

1. एकपदेन उत्तरत-

(क) चन्दनदासः कः आसीत्स?
उत्तर:
मणिकारश्रेष्ठी

(ख) राजप्रसादेन तस्य वणिज्या की दुशी?
उत्तर:
अखण्डिता

(ग) राजनि कीदृशः भव?
उत्तर:
अवरुद्धवृतिः

(घ) अमात्यः कीदृशः आसीत्?
उत्तर:
राजापथ्यकिारी

(च) अस्मद् गृहे कस्य अहजनः पूर्वम् आसीत?
उत्तर:
अमात्यसक्षसस्य।

योग्यताविस्तारः

कविपरिचयः
‘मुद्राराक्षसम्’ इति नाटकस्य प्रणेता विशाखदत्तः आसीत्। सः राजवंशे उत्पन्नः आसीत्। तस्य पिता भास्करदत्तः महाराजस्य पदवीं प्राप्नोत्। विशाखदत्तः राजनीते: न्यायस्य ज्योतिषविषयस्य च विद्वान् आसीत्। वैदिकधर्मावलम्बी भूत्वाऽपि सः बौद्धध र्मस्य अपि आदरमकरोत्।

ग्रन्थपरिचयः
‘मुद्राराक्षसम्’ एकम् ऐतिहासिक नाटकम् अस्ति। दशाङ्केषु विरचिते अस्मिन्नाटके चाणक्यस्य राजनीतिककौशलस्य बुद्धिवैभवस्य राष्ट्रसञ्चालनार्थम् कूटनीतीनाम् निदर्शनमस्ति। __अस्मिन्नाटके चाणक्यस्यामात्यराक्षसस्य च कूटनीत्योः संघर्षः।

भावविस्तारः
चाणक्य-चाणक्यः एकः विद्वान् ब्राह्मणः आसीत्। तस्य पितृप्रदत्तं नाम विष्णुगुप्तः आसीत्। अयमेव ‘कौटिल्य’ इति नाम्ना प्रसिद्धः। केषाञ्चित् विदुषाम् इदमपि मतमस्ति यत् राजनीतिशास्त्रे कुटिलनीतेः प्रतिष्ठापनाय तस्याः स्व-जीवने उपयोगाय च अयं ‘कौटिल्यः’ इत्यपि कथ्यते। चणकनामकस्य कस्यचित् आचार्यस्य पुत्रत्वात् ‘चाणक्यः’ इति नाम्ना स प्रसिद्धः जातः। नन्दानां राज्यकालः शतवर्षाणि पर्यन्तम् आसीत्। तेषु अन्तिमेषु द्वादशवर्षेषु एतेन सुमाल्यादीनाम् अष्टनन्दानां संहार: कारितः तथा च चन्द्रगुप्तमौर्यः नृपत्वेन राजसिंहासने स्थापितः। अयमेक: महान् राजनीतिज्ञः आसीत्। एतेन भारतीयशासनव्यवस्थायाः प्रामाणिकतत्त्वानां वर्णनेन युक्तं “अर्थशास्त्रम्” इति अतिमहत्त्वपूर्णः ग्रन्थः रचितः।

चन्द्रगुप्तमौर्य:- चन्द्रगुप्तः महापद्मनन्दस्य मुरायाः च पुत्रः आसीत्। चाणक्यस्य मार्गदर्शने अनेन चतुर्विशतिवर्षपर्यन्तं राज्यं कृतम्।
राक्षसः-नन्दराज्ञः स्वामिभक्तः चतुरः प्रधानामात्यः आसीत्।
चन्दनदासः-कुसुमपुर नाम्नि नगरे महामात्यस्य राक्षसस्य प्रियतम पात्रं मित्रञ्च आसीत्। स मणिकारः श्रेष्ठी च आसीत्। अस्यैव गृहात् राक्षसः सपरिवारः नगरात् बहिरगच्छत्।

भाषिकविस्तारः

1. पृथक् और विना शब्दों के योग में द्वितीया तृतीया और पंचमी तीनों विभक्तियों का प्रयोग-
यथा – जलं विना जीवनं न सम्भवति। – द्वितीया
जलेन विना जीवनं न सम्भवति। – तृतीया
जलात् विना जीवनं न सम्भवति। – पंचमी
परिश्रमं पृथक् नास्ति सुखम्। – द्वितीया
परिश्रमेण पृथक् नास्ति सुखम्। – तृतीया
परिश्रमात् पृथक् नास्ति सुखम्। – पंचमी

2. अनीयर् प्रत्ययप्रयोगः
अत्यादरः शङ्कनीयः
जन्तुशाला दर्शनीया
याचकेभ्यः दानं दानीयम्
वेदमन्त्राः स्मरणीयाः

पुस्तकमेलापके पुस्तकानि क्रयणीयानि।
(क) अनीयर् प्रत्ययस्य प्रयोगः योग्यार्थे भवति।
(ख) अनीयर् प्रत्यये ‘अनीय’ इति अवशिष्यते।
(ग) अस्य रूपाणि त्रिषु लिगेषु चलन्ति।

यथा-
पुंल्लिने – स्त्रीलिङ्गे – नपुंसकलिङ्गे
पठनीयः – पठनीया – पठनीयम्
इनके रूप क्रमशः देववत्, लतावत् तथा फलवत् चलेगें।

3. उभ सर्वनामपदम्
पुंल्लिङ्गे – स्त्रीलिङ्गे – नपुंसकलिड़े
उभे – उभे – उभे
उभौ – उभे – उभे
उभाभ्याम् – उभाभ्याम् – उभाभ्याम्
उभाभ्याम् – उभाभ्याम् – उभाभ्याम्
उभाभ्याम् – उभाभ्याम् – उभाभ्याम्
उभयोः – उभयोः – उभयोः
उभयोः – उभयोः – उभयोः

Class 10 Sanskrit Shemushi Chapter 11 प्राणेभ्योऽपि प्रियः सुह्रद् Summary Translation in Hindi and English

पाठपरिचय-
प्रस्तुत नाट्यांश महाकवि विशाखदत्त द्वारा रचित ‘मुद्राराक्षसम्’ नामक नाटक के प्रथम अङ्क से उद्ध त किया गया है। नन्दवंश का विनाश करने के बाद उसके हितैषियों को खोज-खोजकर पकड़वाने के क्रम में चाणक्य, अमात्य राक्षस एवं उसके कुटुम्बियों की जानकारी प्राप्त करने के लिए चन्दनदास से वार्तालाप करता है किन्तु चाणक्य को अमात्य राक्षस के विषय में कोई सुराग न देता हुआ चन्दनदास अपनी मित्रता पर दृढ़ रहता है। उसके मैत्री भाव से प्रसन्न होता हुआ भी चाणक्य जब उसे राजदण्ड का भय दिखाता है, तब चन्दनदास राजदण्ड भोगने के लिये भी सहर्ष प्रस्तुत हो जाता है। इस प्रकार अपने सुहृद् के लिए प्राणों का भी उत्सर्ग करने के लिये तत्पर चन्दनदास अपनी सुहृद्-निष्ठा का एक ज्वलन्त उदाहरण प्रस्तुत करता है।

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चाणक्यः – वत्स! मणिकार श्रेष्ठिनं चन्दनदासनिदानों द्रष्टुमिच्छामि।
शिष्यः – तथेति (निष्क्रम्य चन्दनदासेन सह प्रविश्य) इत: इतः श्रेष्ठिन्! (उभौ परिक्रामत:)
शिष्यः – (उपसृत्य) उपाध्याय! अयं श्रेष्ठी चन्दनदासः।
चन्दनदासः – जयत्वार्यः
चाणक्यः – श्रेष्ठिन्! स्वागत ते। अपि प्रचीयन्ते संव्यवहाराणां वृद्धिलाभाः?
चन्दनदासः – (आत्मगतम्) अत्यादरः शटनीयः। (प्रकाशम्) अथ किम्। आर्यस्य प्रसादेन अखण्डिता में वणिज्या।
चाणक्य: – भो श्रेष्ठिन्! प्रीताभ्यः प्रकृतिभ्यः प्रतिप्रियमिच्छन्ति राजानः।
चन्दनदास: – आज्ञापयतु आर्य:, किं कियत् च अस्मज्जनादिष्यते इति।
चाणक्यः – भो श्रेष्ठिन्! चन्द्रगुप्तराज्यमिदं न नन्दराज्यम्। नन्दस्यैव अर्थसम्बन्धः प्रीतिमुत्पादयति। चन्द्रगुप्तस्य तु भवतामपरिक्लेश एव।
चन्दनदासः – (सहर्षम्) आर्य! अनुगृहीतोऽस्मि।
चाणक्यः – भो श्रेष्ठिन्! स चापरिक्लेशः कथमाविर्भवति इति ननु भवता प्रष्टव्याः स्मः।
चन्दनदास: – आज्ञापयतु आर्यः। चाणक्यः-राजनि अविरुद्धवृत्तिर्भव।
चन्दनदास: – आर्य! कः पुनरधन्यो राज्ञो विरुद्ध इति आर्यणावगम्यते?
चाणक्यः – भवानेव तावत् प्रथमम्।
चन्दनदासः – (कर्णी पिधाय) शान्तं पापम्, शान्तं पापम्। कीदृशस्तृणानामग्निना सह विरोधः?
चाणक्य: – अयमीदृशो विरोधः यत् त्वमद्यापि राजापथ्यकारिणोऽमात्यराक्षसस्य गृहजनं स्वगृहे रक्षसि।

शब्दार्थ:
मणिकारश्रेष्ठिनम् – रत्नकारं वणिज (मणियों का व्यापारी)
निष्क्रम्य – बहिर्गत्वा (निकलकर)
उपसृत्य – समीपं गत्वा (पास जाकर)
परिक्रामतः – परिभ्रमणं कुरुतः (दोनों) परिभ्रमण करते हैं
प्रचीयन्ते – वृद्धिं प्राप्नुवन्ति (बढ़ते हैं)
संव्यवहाराणाम् – व्यापाराणाम् (व्यापारों का)
आत्मगतम् – स्वगतम् (मन ही मन)
शटनीयः – सन्देहास्पदम् (शंका करने योग्य)
अखण्डिता – निर्बाधा (बाधारहित)
वणिज्या – वाणिज्यम् (व्यापार)
प्रीताभ्यः – प्रसन्नाभ्यः (प्रसन्न जनों के प्रति)
प्रतिप्रियम् – प्रत्युपकारम् (उपकार के बदले किया गया उपकार)
अपरिक्लेशः – दु:खाभावः (दुख का अभाव)
आज्ञापयतु – आदिशतु (आदेश दें)
अर्थसम्बन्धः – धनस्य सम्बन्धः (धन का सम्बन्ध)
परिक्लेशः – दुःखम् (दु:ख)
प्रष्टव्याः – प्रष्टुं योग्याः (पूछने योग्य)
अवगम्यते – ज्ञायते (जाना जाता है)
अविरुद्धवृत्तिः – अविरुद्धस्वभावः (विरोधरहित स्वभाव वाला)
पिधाय – आच्छाद्य (बन्द कर)
राजापथ्यकारिणः – नृपापकारकारिणः (राजाओं का अहित करने वाले)

सरलार्थ
चाणक्य – हे वत्स रत्नाकर (सुनार) चन्दनदास को अभी देखना चाहता हूँ।
शिष्य – ठीक है (निकलकर चन्दनदास के साथ प्रवेश करक) इधर से सेठ जी इधर से (दोनों घूमते हैं)
शिष्य – (पास जाकर) हे आचार्य! यह है सेठ चन्दनदास।
चन्दनदास – आर्य की जय हो। चाणक्य-हे सेठ! स्वागत है तुम्हारा। क्या व्यापार की बढ़ोतरी हो रही है?
चन्दनदास – (मन-ही-मन) अधिक आदर शंका के योग्य होता है। (स्पष्ट शब्दों में और क्या। आर्य की कृपा से निर्बाध है मेरा व्यापार।
चाणक्य – हे सेठ! प्रसन्न जनों से उपकार का बदला चाहा करते हैं राजा लोग।
चन्दनदास – आज्ञा दीजिए आर्य! क्या और कितना हमें आदेश किया जा रहा है? ऐसा।
चाणक्य – हे सेठ यह चन्द्रगुप्त का राज्य है नंद का नहीं। नन्द के ही धन का सम्बन्ध प्रसन्नता उत्पन्न करताहै। चन्द्रगुप्त के(धन का सम्बन्ध तो) आपके लिए दु:ख का अभाव ही है।
चन्दनदास – (हर्ष के साथ) आर्य! मैं अनुगृहीत हूँ।
चाणक्य – हे सेठ! और क्लेश दू:ख का अभाव कैस उत्पन्न होता है? हमें आपसे वही पूछना है।
चन्दनदास – आज्ञा दीजिये आर्य। चाणक्य-राजा के प्रति अनुकूल हो जाओं।
चन्दनदास – है आर्य! कौन अभागा राजा के विरुद्ध है? यह आर्य को ज्ञात है?
चाणक्य – पहले तो आप ही हैं।
चन्दनदास – (दोनों कान बन्द करके) पाप शांत हो, पाप शांत हो। घास-फूस का अग्नि के साथ विरोध कैसा है?
चाणक्य – यह ऐसा विरोध है कि तुम आज भी राजा का अहित करने वाले अमात्य राक्षस के घर वालों को अपने घर में रखते हो।

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चन्दनदास: – आर्य! अलीकमेतत्। केनाप्यनार्येण आर्याय निवेदितम्।
चाणक्यः – भो श्रेष्ठिन्! अलमाशटया। भीताः पूर्वराजपुरुषाः पौराणामिच्छतामपि गृहेषु गृहजनं निक्षिप्य देशान्तरं व्रजन्ति। ततस्तत्प्रच्छादनं दोषमुत्पादयति।
चन्दनदासः – एवं नु इदम्। तस्मिन् समये आसीदस्मद्गृहे अमात्यराक्षसस्य गृहजन इति।
चाणक्यः – पूर्वम् ‘अनृतम्’, इदानीम् “आसीत्” इति परस्परविरुद्ध वचने।
चन्दनदासः – आर्य! तस्मिन् समये आसीदस्मद्गृहे अमात्यराक्षसस्य गृहजन इति।
चाणक्यः – अथेदानी क्व गतः?
चन्दनदास: – न जानामि।
चाणक्यः – कथं न ज्ञायते नाम? भो श्रेष्ठिन्! शिरसि भयम्, अतिदूरं तत्प्रतिकारः।
चन्दनदासः – आर्य! किं मे भयं दर्शयसि? सन्तमपि गेहे अमात्यराक्षसस्य गृहजनं न समर्पयामि, किं पुनरसन्तम्?
चाणक्य: – चन्दनदास! एष एव ते निश्चयः?
चन्दनदासः – बाढम्, एष एव मे निश्चयः।
चाणक्यः – (स्वगतम्) साधु! चन्दनदास साधु।

सुलभेष्वर्थलाभेषु परसंवेदने जने।
क इदं दुष्करं कुर्यादिदानीं शिविना विना॥

शब्दार्थ:
अलीकम् – असत्यम् (झूठ)
अनार्येण – दुष्टेन (दुष्ट के द्वारा)
पौराणाम् – नगरवासिनाम् (नगर के लोगों के)
निक्षिप्य – स्थापयित्वा (रखकर)
व्रजन्ति – गच्छन्ति (जाते हैं)
प्रच्छादनम् – आच्छादनम् (छिपाना)
अमात्यः – मन्त्री (मन्त्री)
असन्तम् – न निवसन्तम् (न रहने वाले)
बाढम् – आम् (हाँ)
संवेदने – समर्पणे कृते सति (समर्पण पर)
जने – लोके (संसार में)

सरलार्थ-
चन्दनदास – आर्य! यह झूठ है। यह किसी दुष्ट ने आपको बताया है।
चाणक्य – हे सेठ! आशंका मत करो। डरे हुए पूर्वराजकर्मचारी नगरवासियों के चाहने पर भी घरों में घरवालों को रखकर दूसरे देश चले जाते हैं फिर उन्हें छिपाना ही
दोष को उत्पन्न करता है।
चन्दनदास – अच्छा ऐसा है। उस समय हमारे घर में अमात्य राक्षस के घर के लोग थे। ऐसा।
चाणक्य – पहले ‘झूठ’, अब ‘थे’ ये परस्पर विरोधी बातें
चन्वनदास – आर्य! क्यामुझे भय दिखा रहे हैं? अमात्य राक्षस के परिजनों के मेरे घर में होने पर भी नहीं देता तो (उनके) न होने पर क्या (दे सकता हूँ)?
चाणक्य – चन्दनदास! क्या यही तुम्हारा निश्चय है?
चन्दनदास – हाँ, यही मेरा निश्चय है।
चाणक्य – (मन-ही-मन)अच्छा है। चन्दनदास, अच्छा है।

अंतिम श्लोक का अन्वय तथा भावार्थ

अन्वयः – परस्य संवेदने अर्थलाभेषु सुलभेषु इदं दुष्कर कर्म जने (लोके) शिविना विना कः कुर्यात्।।

भावः – परस्य परकीयस्य अर्थस्य संवेदने समर्पणे छते सति अर्थलाभेषु सुलभेषु सत्सु स्वार्थ तृणीकृत्य परसंरक्षणरूपमेवं दुष्करं कर्म जने (लोके) एकेन शिविना विना त्वदन्यः कः कुर्यात्। शिविरपि कृते युगे छतवान् त्वं तु इदानीं कलौ युगे करोषि इति ततोऽप्यतिशयित-सुचरितत्वमिति भावः।

अर्थ – दूसरों की वस्तु को समर्पित करने पर बहुत धन प्राप्त होने की स्थिति में भी दूसरों की वस्तु की सुरक्षा रूपी कठिन कार्य को एक शिवि को छोड़कर तुम्हारे अलावा दूसरा कौन कर सकता है?

आशय – इस श्लोक के द्वारा महाकवि विशाखदत्त ने बड़े ही संक्षिप्त शब्दों में चन्दनदास के गुणों का वर्णन किया है। इसमें कवि ने कहा है कि दूसरों की वस्तु की रक्षा करनी कठिन होती है। यहा! चन्दनदास के द्वारा अमात्य राक्षस के परिवार की रक्षा का कठिन काम किया गया है। न्यासरक्षण को महाकवि भास ने भी दुष्कर कार्य मानते हुए स्वप्नवासवदनम् में कहा है-दुष्करं न्यासरक्षणम्।

चन्दनदास अगर अमात्य राक्षस के परिवार को राजा को समर्पित कर देता, तो राजा उससे प्रसन्न भी होता और बहुत-सा धन पारितोषिक के रूप में देता, पर उसने भौतिक लाभ व लोभ को दरकिनार करते हुए अपने प्राणप्रिय मित्र के परिवार की रक्षा को अपना कर्नव्य माना और इसे निभाया भी। कवि ने चन्दनदास के इस कार्य की तुलना राजा शिवि के कार्यों से की है, जिन्होंने अपने शरणागत कपोत की रक्षा के लिए अपने शरीर के अंगों को काटकर दे दिया था। राजा शिवि ने तो सतयुग में ऐसा किया था, पर चन्दनदास ने ऐसा कार्य इस कलियुग में किया है, इसलिए वे और अधिक प्रशंसा के पात्र हैं।

भावार्थ-
सुलभेष्वर्थलाभेषु परसंवेदने जन।
क इदं दुष्कर कुयादिदानी शिविना विना॥

अन्वयः – परस्य संवेदने अर्थलाभेषु सुलभेषु इंद दुष्कर कर्म जने (लोक) शिविना विना कः कुर्यात्।

शब्दार्थ: – सुल भेषु – सरलता से मिलने पर। परसंवेदने – परसमर्पणे (दूसरों को देने पर)। दुष्करम् – कठिनम् (कठि)। जने – संसारे, लोके (संसार में)।

व्याख्या – दूसरों की वस्तु को समर्पित करने पर बहुत धन प्राप्त होने की स्थिति में भी दूसरों की वस्तु की सुरक्षा रूपी कठिन कार्य को एक शिवि को छोड़कर तुम्हारे अलावा दूसरा कौन कर सकता है? भावार्थ इस श्लोक के द्वारा महाकवि विशाखदत्त ने बड़े ही संक्षिप्त शब्दों में चन्दनदास के गुणों का वर्णन किया है इसमें कवि ने कहा है कि दूसरों कीवस्भ्तु की रक्षा करनी कठिन होती है। यहां चन्दनदास के द्वारा अमात्य राक्षस के परिवार की रक्षा का कठिन काम किया गया है। न्यासरक्षण को महाकवि भास ने भी दुष्कर कार्य मानतेहुए स्वपवासवदत्तम् में कहा है-दुष्कर न्यासरक्षणम्।

चन्दनदास अगर अमात्य राक्षस के परिवार को राजा को समर्पित कर देता, तो राजा उससे प्रसन्न भी होता और बहुत-सा धन पारितोषिक के करूकप में देता, पर उसने भौतिक लाभ व लोभ को दरकिनार करते हुए अपने प्राणप्रिय मित्र के परिवार की रक्षा को अपना कर्त्तव्य माना और इसे निभाया भी कवि ने चन्दनदास के इस कार्य की तुलना राजा शिवि के कार्यों से की है। जिन्होंने अपने शरणागत कपोत की रक्षा के लिए अपने शरीर के अंगों को काटकर दे दिया था। राजा शिवि ने तो सत्ययुग में ऐसा किया था, पर चन्दनदास ने ऐसा कार्य इस कलियुग में किया है, इसलिए वे और अधिक प्रशंसा के पात्र हैं।

Here you will find Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

Extra Questions for Class 10 Maths Quadratic Equations with Answers Solutions

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers

Quadratic Equations Class 10 Extra Questions Objective Type

Question 1.
The solution of quadratic equation x² – x – 2 = 0 are:
(a) 1, -2
(b) -1, -2
(c) -1, 2
(d) 1, 2.
Answer:
(c) -1, 2

Question 2.
The zero of the polynomial x² + 2x – 3 are:
(a) 1, -3
(b) -1, 3
(c) -1, -3
(d) 1, 3
Answer:
(a) 1, -3

Question 3.
The degree of the polynomial x³ – x + 7 is:
(a) 1
(b) 2
(c) 3
(d) none of these.
Answer:
(c) 3

Question 4.
The zero of the polynomial p(x) = x² + 1 are:
(a) real
(b) not real
(c) (a) and (b) both
(d) none of these.
Answer:
(b) not real

Question 5.
(i) Every quadratic polynomial can have at the most:
(a) one zero
(b) two zeroes
(c) three zeroes
(d) none of these.
Answer:
(b) two zeroes

(ii) From the equation 4√5 x² + 7x – 3√5 = 0, the value of x will be:
(a)
(b)
(c)
(d)
Answer:
(a)

Question 6.
If one root of quadratic equation x² + kx + 3 = 0 is 1, then the value of k will be:
(a) 1
(b) -3
(c) -4
(d) -5.
Answer:
(c) -4

Question 7.
If one root of quadratic equation 2x² + px – 4 = 0 is 2, then the value of p will be:
(a) -3
(b) -2
(c) 2
(d) 3.
Answer:
(b) -2

Question 8.
If one root of quadratic equation ax² + bx + c = 0 is 1, then:
(a) a = 1
(b) b = 1
(c) c = 1
(d) a + b + c = 0
Answer:
(d) a + b + c = 0

Question 9.
If 2x² + 1 = 33, then the value of x will be:
(a) ±2
(b) ±3
(c) ±4
(d) ±1.
Answer:
(c) ±4

Question 10.
One root of quadratic equation x² + 3x – 10 = 0:
(a) -2
(b) +2
(c) 0
(d) 1
Answer:
(b) +2

Question 11.
If = \(\frac {x}{6}\) = \(\frac {6}{x}\), then the value of x will be:
(a) 6
(b) -6
(c) 6
(d) none of these.
Answer:
(c) 6

Question 12.
Discriminant of ax² + bx + c = 0) is:
(a)
(b)
(c)
(d) none of these.
Answer:
(d) none of these.

Question 13.
A quadratic equation ax² + bx + c = 0, a ≠ 0 has equal roots if b² – 4ac is:
(a) equal to 0
(b) ≥ 0
(c) ≤ 0
(d) none of these.
Answer:
(a) equal to 0

Question 14.
The roots of ax² + bx + c = 0, a 50 are real and unequal, if b² – 4ac is:
(a) equal to 0
(b) ≥ 0
(c) ≤ 0
(d) none of these.
Answer:
(b) ≥ 0

Question 15.
The roots of the equation 2x² – 8x + c = 0 are eaual, the value of c is:
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 16.
If the discriminant (D) of a quadratic equation ax² + bx + c = 0, a ≠ 0 is greater than zero, the roots are:
(a) real and unequal
(b) real and equal
(c) not real
(d) none of these.
Answer:
(a) real and unequal

Question 17.
An expression in α and β is called symmetrical expression if by interchanging α and β, the expression is:
(a) changed
(b) not changed
(c) may be (a) and (b)
(d) none of these.
Answer:
(b) not changed

Question 18.
If x² + 5bx + 16 = 0 has no real roots, then:
(a) b > \(\frac {8}{5}\)
(b) b < \(\frac {-8}{5}\)
(c) \(\frac {-8}{5}\) < b < \(\frac {8}{5}\)
(d) none of these.
Answer:
(c) \(\frac {-8}{5}\) < b < \(\frac {8}{5}\)

Question 19.
If the roots of 5x² – px + 1 = 0 are real and distinct, then:
(a) p > √15
(b) p > – 2√5
(c) -2√5 < p < 2√5
(d) p > 2√5 or p < – 2√5.
Answer:
(d) p > 2√5 or p < – 2√5

Question 20.
The sum of the roots of the quadratic equation 3x² + 4x = 0 is:
(a) 0
(b) –\(\frac {3}{4}\)
(c) –\(\frac {4}{3}\)
(d) \(\frac {4}{3}\)
Answer:
(c) –\(\frac {4}{3}\)

Question 21.
The product of the roots of quadratic equation 3x² – 4x = 0) is:
(a) 0
(b) –\(\frac {3}{4}\)
(c) –\(\frac {4}{3}\)
(d) –\(\frac {4}{3}\)
Answer:
(a) 0

Question 22.
If one root of x² + kx + 3 = 0 is 1, then the value of k will be:
(a) -4
(b) -3
(c) 1
(d) 5.
Answer:
(b) -3

Question 23.
If the sum of the roots of the equation 3x² + (2k + 1)x – (k + 5) = 0 be equal to their product, the value of k is:
(a) 5
(b) 4
(c) 3
(d) 2
Answer:
(b) 4

Question 24.
The sum of the roots of quadratic equation 5 – 7x + 3x² = 0 is:
(a) +\(\frac {7}{5}\)
(b) –\(\frac {7}{5}\)
(c) –\(\frac {7}{3}\)
(d) +\(\frac {7}{3}\)
Answer:
(d) +\(\frac {7}{3}\)

Question 25.
If one root of quadratic equation x³ – 3x + 2 = 0 is 2, then the second root is:
(a) 3
(b) -1
(c) 1
(d) 2
Answer:
(c) 1

Quadratic Equations Class 10 Extra Questions Very Short Answer Type

Question 1.
If one root quadratic equation x² + 2x – p = 0 is -2, then find the value of p.
Solution.
Given that – 2 is a root of given quadratic equation x² + 2x – p = 0
∴ (-2)² + 2 (-2) – p = 0
⇒ 4 – 4 – p = 0
∴ p = 0
Hence, p = 0

Question 2.
If x² – 5 = 0, then find the value of x:
Solution.
x² – \(\frac {1}{9}\) = 0
x = ±\(\frac {1}{3}\)

Question 3.
If x = 12 and y = 5, then find the value of 9x² + 49y² – 42xy.
Solution.
We have:
9x² + 49y² – 42xy
= (3x)² + (7y)² – 2.3x.7y
= (3x – 7y)²
= (3.12 – 7.5)²
= (36 – 35)² = 1

Question 4.
Solve the following quadratic equation for x:
4x² + 46x – (a² – 12) = 0
Solution.
4x² + 4bx – a² + b² = 0
⇒ (2x)² + 2.2xb + (b)²-(a)² = 0
⇒ (2x + b)²-(a)² = 0
⇒ (2x + b + a) (2x + b – a) = 0

Question 5.
Find the roots of the quadratic equation:
3x² – 2 √6 x + 2 = 0
Solution.
Given quadratic equation
3x² – 2√6 x + 2 = 0
= (√3x)² – 2√3x × √2 +(√2)² = 0
(√3x – √2)² = 0
∴ roots are \(\frac{\sqrt{2}}{\sqrt{3}}\) or \(\frac{\sqrt{2}}{\sqrt{3}}\)

Question 6.
Find the equation whose roots are b – 2a and b + 2a.
Solution.
α = b – 2a
β = b + 2a
Sum of roots = α + β = 6-2a +b + 2a = 2b.
Product of roots = αß = (b – 2a)(b + 2a)
= b² – (2a)²
= b² – 4a²
The required quadratic equation is
∴ x² – (sum of roots) x + product of roots = 0
∴ x² – (2b)x + (b² – 4a²) = 0
∴ x² – 2bx + b² – 4a² = 0

Question 7.
If α and β are the roots of the equation x² – x – 2 = 0, find that quadratic equation whose roots are (2α + 1) and (2β + 1).
Solution.
We have
x² – x – 2 = 0

The required equation is is

Question 8.
If – 5 is a root of quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots. Find the value of k.
Solution.
Given – 5 is a root of quadratic equation 2x² + px – 15 = 0
∴ 2(-5)² + p(-5) – 15 = 0
50 – 5p – 15 = 0
5p = 35 ⇒ p = 7
Putting the value of p = 7 in given (ii) quadratic equation.
7x² + 7x + k = 0
as the roots of this equation are equal
∴ b² = 4ac
⇒ (7)² = 4 × 7 × k
∴ k = \(\frac {7}{4}\)

Question 9.
If aand Bare the roots of the equation ax² + bx + b = 0, then prove that:

Solution.
Since, a and ß are the roots of the equation
ax² + bx + b = 0

Proved.

Question 10.
The length and breadth of a room are 15 m and 12 m. There is a verandah surrounding the room and area of this is 90m². Find the width of verandah.
Solution.
Suppose the width of Verandah is x metre.
Length of verandah = (2x + 15) m.
Breadth of verandah = (2x + 12) m.
Area of verandah = (2x + 15) (2x + 12) – 15 × 12 = 90

∴ 4x² + 24x + 30x + 180 – 180 – 90 = 0
∴ 4x² + 54x – 90 = 0
∴ 2x² + 27x – 45 = 0
∴ 2x² + 30x – 3x – 45 = 0
∴ 2x(x + 15) – 3(x + 15) = 0
(x + 15) (2x – 3) = 0
x + 15 = 0
or 2x – 3 = 0
2x = 3
x = 3/2
The width of verandah = 3/2 m.

Question 11.
The square of a positive integer is greater than 11 times the integer by 26. Find the positive integer.
Solution.
Let the positive integer be x then according to the given condition,
x² = 11x + 26
⇒ x² – 11x – 26 = 0
⇒ x² – 13x + 2x – 26 = 0
⇒ x(x – 13) + 2(x – 13) = 0
⇒ (x – 13)(x + 2) = 0
∴ x = -2 or x = 13
Neglecting the negative value of x, we have x = 13.
Hence, the integer is 13.

Question 12.
A dealer sells an article for ₹ 24 and gains as much per cent as the cost price of the article. Find the cost price of the article.
Solution.
Let cost price of article be ₹ x. Then, gain = x%.

⇒ x² + 100x – 2400 = 0
⇒ (x + 120)(x – 20) = 0
⇒ x = -120 or x = 20
⇒  x = 20
[∴ Cost Price can never be negative]
Hence, the cost price of the article is ₹ 20.

Quadratic Equations Class 10 Extra Questions Short Answer Type

Question 1.
The area of a right angle triangle is 30cm². Find the lengths of the base if its height is 7 cm more then base.
Solution.
Let the length of base of triangle = x cm
∴ length of its height = (x + 7) cm

Area of right angle ∆ = \(\frac {1}{2}\) × base × height
⇒ 30 = \(\frac {1}{2}\)x.(x + 7)
⇒ x² + 7x – 60 = 0
⇒ x² + (12 – 5)x – 60 = 0
⇒ x² + 12x – 5x – 60 = 0
⇒ x(x +12) – 5(x + 12) = 0
⇒ (x + 12) (x – 5) = 0
∴ x = 5 or – 12
as length is never negative
∴ length of base = 5 cm.

Question 2.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution.
Given that, hypotenuse of right triangle = 13 cm
Let the base of the right triangle = x
According to the question,
Altitude of the triangle = 7 cm
Less than its base = (x – 7) cm

By Pythagoras theorem, we have
In ∆ABC AC² = BC² + Ab²
⇒ (13)² = x² + (x – 7)²
⇒ 169 = x² + x² – 14x + 49
[∵ (a – b)² = a² – 2ab + b²]
⇒ 2x² – 14x – 120 = 0
⇒ x² – 7x – 60 = 0
x² – (12x – 5x) – 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12) (x + 5) = 0
Now, x – 12 = 0 x = 12
and x + 5 = 0 = x = -5
Since, altitude of the triangle cannot be negative, hence x ≠ -5
Hence, base of the triangle = 12 cm
and altitude of the triangle = 12 – 7 = 5 cm.

Question 3.
Solve the equation

Solution.
Given,

⇒ (3x + 4)(x + 4) = 4(x + 1)(x + 2)
⇒ 3x² + 16x + 16 = 4(x² + 3x + 2)
⇒ 3x² + 16x + 16 = 4x² + 12x + 8
⇒ x² – 4x – 8 = 0
⇒ Here a =1, b = -4, and c = -8.

Question 4.
A motor boat whose speed in still water is 24 km/h. takes 1 hour more to go 32 km up stream than to return down stream to the same spot. Find the speed of stream.
Solution.
Let the speed of stream be x km/h.
Given that the speed of boat in still water is 24 km/h
∴ Speed of boat in down stream = (24 + x) km/h
and speed of boat in up stream = (24 – x) km/h
Time taken by boat to travel 32 km in down

⇒ 32 × 2x = (24 – x) (24 + x)
⇒ 64x = 576 – x²
⇒ x² + 64x – 576 = 0
⇒ x² + (72 – 8)x – 576 = 0
⇒ x² + 72x – 8x – 576 = 0
⇒ x(x + 72) – 8(x + 72) = 0 (x – 8)(x + 72) = 0
∴ x = 8 or -72
as speed is never negative
∴ x = 8
i.e., speed of stream = 8 km/h.

Question 5.
Solve the quadratic equation

by factorization method
Solution.

⇒ x(a + b + x) = – ab
⇒ ax + bx + x² + ab = 0
⇒ x² + ax + bx + ab = 0
⇒ x(x + a) + b (x + a) = 0
⇒ (x + a) (x + b) = 0
∴ x = -a or -6.

Question 6.
If the equations x² + kx + 64 = 0 and x² – 8x + k = 0 have real roots. Find the positive value of k.
Solution.
Given quadratic equation are
x² + kx + 64 = 0. …(i)
and x² – 8x + k = 0…(ii)
If the equations have real roots, then D ≥ 0 Discriminant of (i) k² – 256 and equation

(ii) 64 – 4k
= k² – 256 ≥ 0 and 64 – 4k ≥ 0
⇒ k² ≥ 256 and 64 ≥ 4k
⇒ k ≥ 16 and 16 ≥ k.
Hence, k = 16.

Question 7.
If the speed of a train is increased by 5 km/H then train takes 1 hour less time to cover a distance of 360 km. Find the speed of the train.
Solution.
Let the speed of train be x km/ Hr.
∴ Time taken by the train to cover 360
km = \(\frac {360}{x}\) Hours
New speed of the train = (x + 5) km/Hr.
∴ New time taken to cover 360 km

⇒ 360 × 5 = x² + 5x
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45) (x – 40) = 0
∴ x = 40 or -45
as speed is never negative
∴ speed of train = 40 km/hour.

Question 8.
Solve the following equation

Solution.
Let = \(\frac{x-2}{x+2}\) = y
∴ Equation reduced to
y + \(\frac {3}{y}\) – 4 = 0
⇒ y² + 3 – 4y = 0
⇒ y² – 4y + 3 = 0
⇒ y² – 3y – y + 3 = 0
⇒ y(y – 3)-1 (y – 3) = 0
⇒ (y – 3)(y – 1) = 0
∴ y = 3 or 1.
Where
y = 3 = \(\frac{x-2}{x+2}\)
3x + 6 = x – 2
⇒ 2x = -8
∴ x = -4
Where
y = 1 = \(\frac{x-2}{x+2}\)
x + 2 x + 2 = x – 2 which is not true
∴ x = -4.

Question 9.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is a. Find his present age.
Solution.
Let the present age of Rehman – x year
∴ Rehman’s age, 3 yr ago = (x – 3) yr
Rehman’s age, after 5 yr = (x + 5) yr
According to the question,

∴ 6x + 6 = x² + 2x – 15
⇒ x² + 2x – 6x – 15 – 6 = 0
x² – 4x − 21 = 0
x² – (7x – 3x) – 21 = 0
⇒ x² – 7x + 3x – 21 = 0
(By factorization method)
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7)(x + 3) = 0
∴ x – 7 = 0 ⇒ x = 7 or x + 3 = 0 ⇒ x = -3
But x +-3 age cannot be negative.
So, present age of Rehman = 7 yr.

Question 10.
The difference of two numbers is 2 and the sum of their square is 34. Find the number.
Solution.
Let Numbers be x and y, such x > y
∴ x – y = 2 …..(i)
and x² + y² = 34 …(ii)
from (i) x = 2 + y
⇒ (2 + y)² + y² = 34
⇒ 2y² + 4y – 30 = 0
⇒ y² + 2y – 15 = 0
⇒ (y + 5) (y – 3) = 0
∴ y = 3 or – 5
where y = 3, x = 2 + 3 = 5
∴ Numbers are 5 and 3.

Question 11.
The speed of boat in still water is 15km/H. The boat completes 30 km distance in going downstream and 30 km up stream in total 4 Hr 30 minute. Find the speed of stream.
Solution.
Let the speed of stream = x km/ Hr.
∴ speed of boat in down stream = (15 + x) km/H
and speed of boat in up stream = (15 – x) km/H
Time taken 30 km in down stream = \(\frac{30}{(15+x)}\) Hrs

⇒ 2 × 30 × 30 = 9(225 – x²)
⇒ 200 = 225 – x²
⇒ x² = 225 – 200
= 25
x = √25 = ± 5
as speed is never negative:. speed to stream = 5 km/Hr.

Question 12.
Determine whether the quadratic equation 9x² + 7x – 2 = 0 has real roots. If so, find roots of quadratic equation.
Solution.
Given, quadratic equation
9x² + 7x – 2 = 0
Here a = 9, b = 7 and c= -2
D = b² – 4ac .
= (7)² – 4 × 9 × -2
= 49 + 72 = 121
∴ roots of quadratic equation are real.
∴ roots are

Question 13.
By solving the equation

get a quadratic equation. Find the nature of roots. Using formula, solve the quadratic equation.
Solution.
Given,

Nature of root, D = b² – 4ac = (4)² – 4 × 1 × 0 = 16 > 0
∴ root of the equation are real.

Question 14.
Some vessels are manufactured in a day in a small industry. On a particular day it is observed that the cost of manufacturing each item was 3 more than twice the number of vessels manufactured on that day. If the total manufacturing cost on that day was 90, then find the number of manufactured vessels and manufacturing cost of each item.
Solution.
Let Number of vessel manufacture in a day = x
∴ Cost of manufacturing = (2x + 3).
According to Question,
x(2x +3) = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x( 2x + 15)-6 (2x + 15) = 0
⇒ (x – 6) (2x + 15) = 0
∴ x = 6 or – \(\frac {15}{2}\)
as number of vessels are not negative
∴ x = 6
Also the manufacturing cost = 2 × 6 + 3 = ₹ 15.

Question 15.
Is the following situation possible ? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution.
Let age of one of the two friends = x yr
Then, age of other friend = (20 – x) yr
(∵ The sum of the ages of two friends is 20 yr)
4 yr ago age of one of the two friends = (x – 4) yr
4 yr ago age of the other friend
= (20 – x – 4) yr = (16 – x) yr
According to the question,
(x – 4)(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112=0
On comparing the above equation with ax² + bx + c = 0, we get
a = 1, b = -20 and c = 112
∴ Discriminant, D = b² – 4ac
= (-20)² – 4 × 1 × 112
= 400 – 448 = 48 < 0
which implies that the real roots are not possible because this condition represents imaginary roots. So, the solution does not exist.

Question 16.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth.
Solution.
Let the breadth of the park = x metre
Then, according to the question, Perimeter of a rectangular park = 80 m
= 2(Length + Breadth)=80 m
= Length + Breadth = 40 m
Length = (40 – x) m
∴ Area of a rectangular park = Length × Breadth = (40 – x)x m²
But according to the question,
Area of the rectangular park is 400 m².
∴ (40 – x)x = 400
⇒ x² – 40x + 400 = 0
⇒ x² – 2x × 20 + (20)² = 0
⇒ (x – 20)² = 0
[∵ a² – 2ab + b² = (a – b)²]
⇒ x = 20
Thus, breadth of the park = 20 m
and length of the park = (40 – 20) m = 20 m
So, it is possible to design the rectangular park of perimeter 80 m and area 400 m². But this rectangle will be a square of side 20 m.

Quadratic Equations Class 10 Extra Questions Long Answer Type

Question 1.
Solve:

x ≠ 0.
Solution.
The given equation is:

Therefore, the given equation can be written as

Now, substituting y = x + \(\frac {1}{x}\), we get
4y² + 8y – 45 = 0
⇒ 4y² + 18y – 10y – 45 = 0
⇒ 2y (2y + 9) – 5(2y +9) = 0
⇒ (2y + 9) (2y – 5) = 0

Now, from (ii), we get,
2(x² + 1) = 5x
⇒ 2x² – 5x + 2 = 0
⇒ 2x² – 4x – x + 2 = 0
⇒ 2x(x – 2) – 1(x – 2) = 0
⇒ (x – 2)(2x – 1) = 0
x = 2 and x = \(\frac {1}{2}\)
Hence, the solutions of the given equation are

Question 2.
Solve: 2x⁴ – x³ – 11x² + x + 2 = 0.
Solution.
Given equation is
2x⁴ – x³ – 11x² – x + 2 = 0
Divide both sides by x²,

Now substitute x + \(\frac {1}{x}\) = y
∴ 2y² – y – 15 = 0
⇒ 2y² – 6y + 5y – 15 = 0
⇒ 2y(y – 3) + 5(y – 3) = 0
⇒ (y – 3)(2y + 5) = 0
∴ y = 3 or y = –\(\frac{5}{2}\)
y = 3 = \(\frac{x^{2}+1}{x}\)
When
⇒ x² + 1 = 3x
⇒ x² – 3x + 1 = 0
Here, a = 1, b = -3 and c = 1

⇒ 2x² + 5x + 2 = 0
⇒ 2x² + 4x + x + 2 = 0
⇒ 2x(x + 2) + 1(x + 2) = 0
⇒ (x + 2)(2x + 1) = 0
x = -2, – \(\frac{1}{2}\)
Hence, the solutions are -2, –\(\frac{1}{2}\), \(\frac{3 \pm \sqrt{5}}{2}\).

Question 3.
Solve \(\sqrt{3 x+10}\) + \(\sqrt{6-x}\) = 6.
Solution.
Given equation is:
\(\sqrt{3 x+10}\) + \(\sqrt{6-x}\) = 6.
We must look for solutions which satisfy
3x + 10 ≥ 0 and 6 – x ≥ 0
x ≥ –\(\frac {10}{3}\) and x ≤ 6
i.e., –\(\frac {10}{3}\) ≤ x ≤ 6.
Now, we first transform one of the radicals co the R.H.S.
We have,
\(\sqrt{3 x+10}\) = 6 – \(\sqrt{6-x}\)
Squaring both sides, we get
3x + 10 = 36 + 6 – x – 12\(\sqrt{6-x}\)
or 4x – 32 = -127\(\sqrt{6-x}\)
or 8 – x = 3\(\sqrt{6-x}\)
Again, squaring both sides, we get
64 + x² – 16x = 9 (6 – x)
or 64 + x² – 16x – 54 + 9x = 0
or x² – 7x + 10 = 0
or (x − 2)(x – 5) = 0
∴ x = 2 and x = 5.
which satisfy \(\frac{-10}{3}\) ≤ x ≤ 6.
Hence, the solutions are x = 2 and x = 5.

Question 4.
One person purchased some cloth for ₹ 2250. Had the price of cloth been ₹ 50 per metre higher, then he would have been able to purchase the cloth less by 1.5 metre. Find the measure of the cloth purchased and price per metre.
Solution.
Let the price per metre of cloth be ₹ x
Amount spend = ₹ 2250
∴ Length of cloth purchased
\(\frac{2250}{x}\) = metre.
Now, new price of cloth per metre become = ₹ (x + 50)
∴ Length of cloth purchase
= \(\frac{2250}{x + 50}\) metre.
According to the question,

⇒ 2250 × 50 × 2 = 3(x² + 50x)
⇒ x² + 50x – 75000 = 0
⇒ x² + 300x – 250x – 75000 = 0
⇒ x(x + 300) – 250(x + 300) = 0
⇒ (x + 300)(x – 250) = 0
∴ x = -300 or x = 250
Neglecting the -ve sign value i.e.,
x = -300
Hence, cost of cloth per metre = ₹ 250.
and length of cloth \(\frac{2250}{250}\) = 9 metre.

Question 5.
A two-digit number is such that the product of the digits is 12, when 36 is added to the number, the digit interchange their places. Find the number.
Solution.
Suppose two digit number is 10x + y.
Whose y = unit’s place and x = ten’s place digit.
According two question
xy = 12 …..(i)
and 10x + y + 36 = 10y + x.
∴ 9x – 9y = -36
∴ x – y = -4
x = y – 4
Put x = y – 4 in (i)
(y – 4)y = 12
∴ y² – 4y – 12 = 0
∴ y² – 6y + 2y – 12 = 0
∴ y(y – 6) + 2(y – 6) = 0
∴ (y – 6) (y + 2) = 0
y – 6 = 0 or y + 2 = 0
y = 6 or y = -2
(not acceptable).
∴ Put y = 6 in (i) we get
x (6) = 12
x = 2
The two digit number is
10x + y = 10(2) + 6 = 26.

Question 6.
The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides of the field.
Solution.
Let PQRS be the rectangular field. Let the shorter side QR of the rectangle = x m.

According to the question,
Diagonal of the rectangle PR = 60 m more than the shorter side = (x + 60) m
Side of the rectangle PQ = 30 m more than the shorter side = (x + 30) m
∴ By Pythagoras theorem,
In ∆PQR, PR² = PQ² + QR²
(∵ In rectangle every adjacent side makes an angle 90° to each other)
⇒ (x + 60)² = (x + 30)² + x²
⇒ x² + 120x + 3600 = x² + 60x + 900 + x²
[∵ (a + b)² = a² + 2ab + b²]
⇒ (2x² – x²) + (60x – 120x) + (900 – 3600) = 0
x² – 60x – 2700 = 0
x² – (90x – 30x) – 2700 = 0
x² – 90x + 30x – 2700 = 0
(By factorization method)
⇒ x(x – 90) + 30(x – 90) = 0
⇒ (x – 90)(x + 30) = 0
∴ Either x – 90 = 0 ⇒ x = 90 or x + 30 = 0
⇒ x = – 30 which is not possible because side cannot be negative.
∴ x = 90
So, breadth of the rectangle = 90 m and length of the rectangle = 90 + 30 = 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution.
Let the required numbers be x and y, where x > y
Difference of squares of two numbers = 180
Given, x² – y² = 180 …(i)
Square of smaller number = 8 × Larger number
⇒ y² = 8x …..(ii)
From Eqs. (i) and (ii), we have
x² – 8x = 180
⇒ x² – 8x – 180 = 0
x² – (18x – 10x) – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
(By factorization method)
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x – 18)(x + 10) = 0
∴ x – 18 = 0 or x + 10 = 0
∴ x = 18 or x = – 10
Now, x = 18
⇒ y² = 8 × 18 = 144
⇒ y = ± 12
⇒ y = 12 or -12 [From EQuestion (ii)]
Again, x = -10
⇒ y² = [8 (-10)] = -80
which is not possible i.e., imaginary value.
Hence, the numbers are (18 and 12) or (18 and -12).

Question 8.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 kmh^-1 more than that of the passenger train, find the average speed of the two trains.
Solution.
Let the average speed of the passenger train = x kmh^-1.
Then, the average speed of the express train = (x + 11) kmh^-1
Time taken by passenger train to cover

⇒ x (x + 11) = 132 × 11
⇒ x² + 11x – 1452 = 0
x² + (44 – 33)x – 1452 = 0
⇒ x² + 44x – 33x – 1452 = 0
(By fatorization method)
= x(x + 44) – 33(x + 44) = 0
⇒ (x + 44)(x – 33) = 0
∴ x + 44 = 0
⇒ x = -44 or x – 33 = 0
⇒ x = 33
Since, x ≠ -44 because speed can’t be negative.
Hence, the average speed of the passenger train = 33 km/h
and the average speed of the express train = (33 + 11) km/h = 44 km/h.

Question 9.
Out of a certain number of Saras birds, one-fourth of the number are moving about in lotus plants, \(\frac {1}{9}\)th coupled with \(\frac {1}{4}\)th as well as 7 times the square root of the number move on a hill, 56 birds remain in vakula tree. What is the total number of birds ?
Solution.
Let the total number of birds be x.
∴ Number of birds moving about in lotus plants = \(\frac {x}{4}\)
and the number of birds moving on a hill
⇒ \(\frac {x}{9}\) + \(\frac {x}{4}\) + 7√x
Also, the number of birds in vakula tree = 56
∴ According to the given information, we have

9x + 2x + 126 √x – 18x + 1008 = 0
⇒ -7x + 126 √x + 1008 = 0
⇒ x – 18√x – 144 = 0
Putting √x = y, we get
y² – 18y – 144 = 0
⇒ y² – 24y + 6y – 144 = 0
⇒ y(y – 24) + 6(y – 24) = 0
⇒ (y – 24)(y + 6) = 0
⇒ y = 24 or y = -6.
But y = -6, since √x = y is positive
∴ y = 24
⇒ √x = 24
∴ x = 576.
Hence, the total number of birds is 576.

Question 10.
Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m² ? If so, find its length and breadth.
Solution.
Let breadth of a rectangular mango grove = x metre
∴ Length of a rectangular mango grove = 2x metre(By given condition)
∴ According to the question,
Area of rectangular mango grove = 800 m²
⇒ Length × Breadth = 2x (x) = 800
⇒ 2x² = 800
⇒ x² = 400
⇒ x = ± 20
But (∵ breadth never be negative)
∴ Length 2x = 40 m and Breadth = 20 m.

Here you will find Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

Extra Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers Solutions

Extra Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Solutions Answers

Pair of Linear Equations in Two Variables Class 10 Extra Questions Objective Type

Question 1.
If the pair of linear equations x – y – k = 0 and 6x – 2y – 3 = 0 represent an infinite solution, then the value of k is:
(a) k = 1
(b) k = 2
(c) k = 0
(d) No value of k
Answer:
(d) No value of k

Question 2.
If the pair of equations 6x + 5y = 4 and 12x + py = -8 has no solution, then the value of p is:
(a) 9
(b) 10
(c) 7
(d) 6
Answer:
(b) 10

Question 3.
The solution of the equation 2x + 3y = 18 and x – 2y = 2 is
(a) x = 2, y = 6
(b) x = 6, y = 2
(c) x = – 6, y = -2
(d) None of these
Answer:
(b) x = 6, y = 2

Question 4.
Find the values of x and y in the following equations:
x – 3y = 8 and 5x + 3y = 10
(a) x = 3, y = – \(\frac {5}{3}\)
(b) x = -3, y = \(\frac {5}{3}\)
(c) x = -3, y = – \(\frac {5}{3}\)
(d) None of these
Answer:
(a) x = 3, y = – \(\frac {5}{3}\)

Pair of Linear Equations in Two Variables Class 10 Extra Questions Very Short Answer Type

Question 1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14, x – y = 4
(ii) 3x – y = 3, 9x – 3y = 9
(iii) √2x + √3y = 0, √3x – √8y = 0
Solution:
(i) We have,
x + y = 14 …..(i)
and x – y = 4 ……(ii)
From Eq. (ii), y = x – 4 …(iii)
Substituting y from Eq. (iii) in Eq. (i), we get
x + x – 4 = 14
⇒ 2x = 18.
⇒ x = 9
On substituting x = 9 in Eq. (iii), we get
y = 9 – 4 = 5
⇒ y = 5
x = 9, y = 5

(ii) We have,
3x – y = 3 …(i)
and 9x – 3y = 9 …(ii)
From Eq. (i)y = 3x – 3 …. (iii)
On substituting y from Eq. (iii) in Eq. (ii), we get
9x – 3(3x – 3) = 9
⇒ 9 = 9
It is a true statement. Hence, every solution of Eq. (i) is a solution of Eq. (ii) and vice-versa.
On putting x = k in Eq. (i), we get
3k – y = 3 ⇒ y = 3k – 3
∴ x = k, y = 3k – 3 is a solution for every real k.
Hence, infinitely many solutions exist.

(iii) We have,
√2x + √3y = 0 …..(i)
and √3x – √8y = 0 …(ii)
From Eq. (ii),
√8y = √3x
⇒ y = \(\frac{\sqrt{3} x}{\sqrt{8}}\) …..(iii)
On substituting y from Eq. (iii) in Eq. (i), we get

⇒ √2 x √8x + 3x = 0
⇒ √16x + 3x = 0
⇒ 4x + 3x = 0
⇒ 7x = 0
⇒ x = 0
Putting x = 0 in Eq. (iii), y = 0

Question 2.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac {1}{2}\), if we only add 1 to the denominator. What is the fraction ?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
Solution:
(i) Let the fraction be \(\frac {x}{y}\)
According to the given conditions,

⇒ x + 1 = y – 1; 2x = y + 1 ……..(i)
⇒ x – y = -2 and 2x – y = 1 ……..(ii)
On subtracting Eq. (i) from Eq. (ii),
(2x – y) – (x – y) = 1 + 2
⇒ x = 3
On substituting x = 3 in Eq. (i),
3 – y = -2 ⇒ y = 5
Hence, the fraction is \(\frac {3}{5}\)

(ii) Let present age of Nuri = x years. Present age of Sonu = y years
According to the given conditions,
Five years ago,
x – 5 = 3(y – 5)
⇒ x – 3y = -10 …(i)
Ten years later,
x + 10 = 2(y + 10)
⇒ x – 2y = 10 …(ii)
On subtracting Eq. (i) from Eq. (ii),
(x – 2y) – (x – 3y) = 10 + 10
⇒ – 2y + 3y = 20
⇒ y = 20
From Eq. (ii), substituting y = 20, we get
⇒ x = 2y + 10 = 2 × 20 + 10
⇒ x = 50
Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years

Question 3.
For What value of k, will the pair of linear equations kx + y = k² and x + ky = 1 have infinitely many solutions ?
Solution:
Given equation are
kx + y = k² ……(i)
and x + ky = 1 ……(ii)
have infinitely many solution

∴ k² = 1 = k ⇒ ± 1
and k³ = 1 ⇒ k = 1
Hence k = 1

Pair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type

Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) \(\frac {4}{x}\) + 3y =14 and \(\frac {3}{x}\) – 4y = 23
(ii)

Solution:
(i) We have,
\(\frac {4}{x}\) + 3y =14 and \(\frac {3}{x}\) – 4y = 23
On putting \(\frac {1}{x}\) = X, we get
4x + 3y = 14 …..(i)
and 3x – 4y = 23 … (ii)
On multiplying Eq. (i) by 4 and Eq. (ii) by 3 and then adding, we get
16X + 9X = 4 × 14 + 3 × 23
⇒ 25X = 56 + 69
⇒ 25X = 125 = X = 5
Then, \(\frac {1}{x}\) = 5
⇒ x = \(\frac {1}{5}\) From Eq. (i), substituting x = 5, we get
4 × 5 + 3y = 14 = 3y = 14 – 20
⇒ 3y = -6 = y = -2
Hence, x = and y = -2


⇒ x – 1 = 3 and y – 2 = 3
⇒ x = 3 + 1
and y = 3 + 2
⇒ x = 4 and y = 5

Question 2.
Prove that the pair of linear equation \(\frac {22}{7}\)x + \(\frac {22}{7}\)y = 7 and y = 7 and 9x – 10y = 14 is coinsistent find its solution by method cross-multiplication.
Solution:
Given equation are
\(\frac {22}{7}\)x + \(\frac {22}{7}\)y = 7
⇒ 3x + 5y = 14 ….(i)
and 9x – 10y = 14 …(ii)
If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) then equation are coinsistance.
Here a₁ = 3, b₁ = 5, c₁ = – 14
and a₂ = 9, b₂ = -10, c₂ = -14

Question 3.
Solve the following pair of linear equations.

x ≠ 0, y ≠ 0
Solution:
Suppose \(\frac{1}{x-y}\) = a
and \(\frac{1}{x+y}\) = b
The reduced equation will be
15a + 22b = 5 ….(i)
and 40a + 55b = 13 (ii)
Multiply equation (i) by 8 and equation (ii) by 3, we get
120a + 176b = 40
and 120a + 165b = 39
on substracting 11b = 1

∴ x – y = 5 … (iv)
Solving Eq. (iii) and Eq. (iv) we get.
x = 8 and y = 3

Question 4.
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2. Find the number.
Solution:
Let unit digit of the number bex and ten’s digit of the number be y
∴ Number = 10y + x
and Number obtain by reversing the digits is 10x + y
∴ By Ist condition,
(10y + x) + (10x + y) = 66
⇒ 11x + 11y = 66
∴ x + y = 6 ……(i)
By IInd condition, x – y = 2 (ii)
On adding (i) & (ii)
2x = 8
∴ x = 4
from equation (1), y = 6 – 4 = 2
Hence, required Number be 2 × 10 + 4 = 24

Pair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type 2

Question 1.
7 chairs and 4 tables for a classroom cost ₹ 7000 while 5 chairs and 6 tables cost ₹ 5080. Find the cost of each chair and that of each table.
Solution:
Let the cost of each chair be ₹ x and that of each table be ₹ y.
Then,
7x + 4y = 7000 …(i)
5x + 3y = 5080 …(ii)
On multiplying (i) by 3, (ii) by 4 and subtracting, we get:
(21x – 20x) = (21000 – 20320)
⇒ x = 680.
On substituting x = 680 in (i), we get:
(7000 – 4760) 4y = 7000
⇒ 4y = (7000 – 4760)
⇒ y = 560.
∴ cost of each chair = ₹ 680 and cost of each table = ₹ 560.

Question 2.
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.
Solution:
Let the tens and units digits of the required number be x and y respectively.
Then, the number = (10x + y).
The number obtained on reversing the digits = (10y + x).
∴ (10y + x) + (10x + y) = 99
= 11(x + y) = 99
⇒ x + y = 9.
Also, (x – y) = ± 3.
∴ x + y = 9 …..(i)
x – y = 3 …..(ii)
x + y = 9 ….(iii)
x – y = -3 …..(iv)
From (i) and (ii), we get: x = 6, y = 3.
From (iii) and (iv), we get: x = 3, y = 6.
Hence, the required number is 63 or 36.

Question 3.
The monthly incomes of A and B are in the ratio 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves ₹ 5000 per month, find the monthly income of each.
Solution:
Let the monthly incomes of A and B be ₹ 8x and ₹ 7x respectively, and let their expenditures be ₹ 19y and ₹ 16y respectively.
Then, A’s monthly savings = ₹ (8x – 19y).
And, B’s monthly savings = ₹ (7x – 16y).
But, the monthly saving of each is ₹ 5000.
∴ 8x – 19y = 5000 …..(i)
7x – 16y = 5000…..(ii)
Multiplying (ii) by 19, (i) by 16 and subtracting the results, we get:
(19 × 7 – 16 × 8)x = (19 × 5000 – 16 × 5000)
⇒ (133 – 128)x = (19 – 16) 5000
⇒ 5x = 15000
⇒ x = 3000.
∴ A’s monthly income = ₹ (8x) = ₹ (8 × 3000) = ₹ 24000.
And, B’s monthly income = ₹ (7x) = ₹(7 × 3000) = ₹ 21000.

Question 4.
A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.
Solution:
Let the tens and units digits of the required number be x and y respectively.
Then, xy = 14.
Required number= (10x + y).
Number obtained on reversing its digits = (10y + x).
∴ (10x + y) + 45 = (10y + x)
⇒ 9(y – x) = 45
⇒ y – x = 5 …(i)
Now, (y + x)² – (y – x)² = 4xy

= y + x = 9 …(ii)
[∴ digits are never negative)
2y = 14 ⇒ y = 7.
Putting y = 7 in (ii), we get:
7 + x = 9 ⇒ x = (9 – 7) = 2.
x = 2 and y = 7.
Hence, the required number is 27.

Question 5.
The sum of the numerator and the denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \(\frac {1}{2}\). Find the fraction.
Solution:
Let the required fraction be \(\frac {x}{y}\).
Then,
∴ x + y = 12 ……(i)
and \(\frac{x}{y+3}\) = \(\frac{1}{2}\)
⇒ 2x = y + 3
⇒ 2x – y = 3 …(ii)
Adding (i) and (ii), we get:
3x = 15 x = 5.
Putting x = 5 in (i), we get:
5 + y = 12 y = (12 – 5) = 7.
Thus, x = 5 and y = 7.
Hence, the required fraction is \(\frac {5}{7}\).

Question 6.
Five year ago, A was thrice as old as B and ten years later A shall be twice as old as B. What are the present ages of A and B ?
Solution:
Let the present ages of B and A be x years and y years respectively. Then,
B’s age 5 years ago = (x – 5 ) years and A’s age 5 years ago = (y – 5) years.
∴ (y – 5) = 3(x – 5) = 3x – y = 10 …(i)
B’s age 10 years hence = (x + 10) years.
A’s age 10 years hence = (y + 10) years.
∴ (y + 10) = 2(x + 10) = 2x – y = -10 (ü)
On subtracting (ii) from (i), we get:
x = 20.
Putting x = 20 in (i), we get:
(3 x 20) – y = 10 = y = (60 – 10) = 50.
∴ x = 20 and y = 50.
Hence, B’s present age = 20 years and A’s present age = 50 years.

Question 7.
If the length of a rectangle is reduced by 5 units and its breadth is increased by 2 units, then the area of the rectangle is reduced by 80 sq units. However, if we increase its length by 10 units and decrease the breadth by 5 units, its area is increased by 50 sq units. Find the length and breadth of the rectangle.
Solution:
Let the length and breadth of the rectangle be x units and y units respectively.
Then, area of the rectangle = xy sq units
Case I: When the length is reduced by 5 units and the breadth is increased by 2 units.
Then,new length= (x – 5) units
andnew breadth = (y + 2) units.
∴ new area = (x – 5)(y + 2) sq units.
∴ xy – (x – 5)(y + 2) = 80
⇒ 5y – 2x = 70…(i)
Case II: When the length is increased by 10 units and the breadth is decreased by 5 units.
Then, new length= (x + 10) units
andnew breadth = (y – 5) units.
∴ new area = (x + 10)(y – 5) sq units.
∴ (x + 10)(y – 5) – xy = 50
⇒ 10y – 5x = 100
⇒ 2y – x = 20 …….(ii)
On multiplying (ii) by 2 and subtracting esult from (i), we get:
y = 30.
Putting y = 30 in (ii), we get:
(2 × 30) – x = 20 ⇒ 60 – x = 20
⇒ x = (60 – 20) = 40.
∴ x = 40 and y = 30.
Hence, length = 40 units and breadth = 30 units.

Question 8.
8 men and 12 boys can finish a piece of work in 5 days, while 6 men and 8 boys can finish it in 7 days. Find the time taken by 1 man alone and that by 1 boy alone to finish the work…
Solution:
Suppose 1 man alone can finish the work in x days and 1 boy alone can finish it in y days.
Then, 1 man’s 1 day’s work = \(\frac {1}{x}\)
And, 1 boy’s 1 day’s work = \(\frac {1}{y}\)
8 men and 12 boys can finish the work in 5 days
⇒ (8 men’s 1 day’s work) + (12 boys’ 1 day’s work) = \(\frac {1}{5}\)
⇒ \(\frac {8}{x}\) + \(\frac {12}{y}\) = \(\frac {1}{5}\),
⇒ 8υ + 12υ = \(\frac {1}{5}\),
[Where \(\frac {1}{x}\) = υ and \(\frac {1}{y}\) = υ] …..(i)
Again, 6 men and 8 boys can finish the work in 7 days
⇒ (6 men’s 1 day’s work) + (8 boys’ 1 day’s work) = \(\frac {1}{7}\)

∴ one man alone can finish the work in 70 days,
and one boy alone can finish the work in 140 days.

Question 9.
A boat goes 16 km upstream and 24 km downstream in 6 hours. Also, it covers 12 km upstream and 36 km downstream at the same time. Find the speed of the boat in still water and that of the stream.
Solution:
Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr.
Then, speed upstream = (x – y) km/hr
and speed downstream = (x + y) km/hr.
Time taken to cover 16 km upstream


On adding (v) and (vi), we get: 2x = 16 ⇒ x = 8.
On subtracting (vi) from (v), we get:
2y = 8 y = 4.
∴ speed of the boat in still water
= 8 km/hr.
And, speed of the stream = 4 km/hr.

Question 10.
90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid Solution: Find the quantity of each type of acid to be mixed to form the mixture.
Solution:
Let the given solutions be labelled as A and B respectively.
Let x litres of A be mixed with y litres of B. Then,
x + y = 21
Quantity of acid in x litres of A = (90% of x)


⇒ 90x + 97y – 1995 …….(ii)
Multiplying (i) by 90 and subtracting the result from (ii), we get:
7y = 105 ⇒ y = 15.
Putting y = 15 in (i), we get:
x + 15 = 21 = x = 6.
∴ x = 6 and y = 15.
So, 6 litres of 90% solution is mixed with 15 litres of 97% Solution.

Question 11.
On selling a tea-set at 5% loss and a lemon-set at 15% gain, a shopkeeper gains ₹ 84. However, if he sells the tea-set at 5% gain and the lemon-set at 10% gain, he gains ₹ 104. Find the price of the tea-set and that of the lemon-set paid by the shopkeeper.
Solution:
Let the CP of the tea-set and the lemon-set be ₹ x and ₹ y respectively.

On adding (i) and (ii), we get:
5y = 3760 ⇒ y = 752.
Putting y = 752 in (ii), we get:
x + (2 × 752) = 2080
x = (2080 – 1504) = 576.
∴ x = 576 and y = ₹ 752.
Hence, CP of the tea-set = ₹ 576
and CP of the lemon-set = ₹ 752.

Question 12.
A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 litres of a 40% acid solution ?
Solution:
Let x liters of 50% solution be mixed with y liters of 25% Solution. Then,
x + y = 10 and 50% of x + 25% of y = 40% of 10
⇒ x + y 10 and \(\frac {50}{100}\) × x + \(\frac {25}{100}\) × y = \(\frac {40}{100}\) × 10
⇒ x + y 10 and \(\frac {x}{2}\) + \(\frac {y}{4}\) = 4
⇒ x + y = 10 …(i)
and 2x + y = 16 …(ii)
On solving (i) and (ii), we get: x = 6 and y = 4.
∴ 6 litres of 50% solution is to be mixed with 4 litres of 25% Solution.

Question 13.
In a ∆ABC, ∠C = 3∠B = 2(∠A + ∠B). Find the angles.
Solution:
Let ∠A = x° and ∠B = y°. Then,
∠C = 3∠B = (3y)°.
Now, ∠A + ∠B + ∠C = 180°
⇒ x + y + 3y = 180°
x + 4y = 180 ……(i)
Also, ∠C = 2(∠A + ∠B)
⇒ 3y = 2(x + y)
⇒ 2x – y = 0 ……(ii)
Multiplying (ii) by 4 and adding the result to (i), we get:
9x = 180 ⇒ x = 20.
Putting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = 160
⇒ y = \(\frac {160}{4}\) = 40.
∴ x = 20 and y = 40.
∴ ∠A = 20°, ∠B = 40°
and ∠C = (3 x 40)° = 120°.

Here you will find Polynomials Class 10 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

Extra Questions for Class 10 Maths Polynomials with Answers Solutions

Extra Questions for Class 10 Maths Chapter 2 Polynomials with Solutions Answers

Polynomials Class 10 Extra Questions Objective Type

Question 1.
The number of polynomials having zeroes -2 and 5 is:
(a) 1
(b) 2
(c) 3
(d) More than 3
Answer:
(d) More than 3

Question 2.
If 1 is zero of the polynomial p(x) = ax² – 3(a – 1)x – 1, then the value of ‘a’ is:
(a) 1
(b) -1
(c) 2
(d) – 2
Answer:
(a) 1

Question 3.
If a, ß are zeroes of x² – 6x + k. What is the value of k if 3a + 2B = 20.
(a) – 16
(b) 8
(c) – 2
(d) -8
Answer:
(a) – 16

Question 4.
If one zero of 2x² – 3x + k is reciprocal to the other, then the value of k is:
(a) 2
(b) \(\frac {-2}{3}\)
(c) \(\frac {-3}{2}\)
(d) -3
Answer:
(a) 2

Question 5.
If p(x) = x² + 6x + 9 and q(x) = x + 3 then remainder will be when p(x) is divided by q(x):
(a) – 1
(b) 0
(c) 11
(d) 2
Answer:
(b) 0

Question 6.
Dividing (x² + 1) by (x + 1), the remainder will be:
(a) -1
(b) 11
(c) 0
(d) – 2
Answer:
(c) 0

Question 7.
Dividing x³ + 3x + 3 by (x + 2), the remainder will be:
(a) -2
(b) -1
(c) 0
(d) 1
Answer:
(d) 1

Question 8.
The zero’s of the polynomial (x² – 2x – 3) will be:
(a) – 3, 1
(b) -3, -1
(c) 3, -1
(d) 3, 1
Answer:
(c) 3, -1
Solution:
x² – 2x – 3 = x² – 3x + x – 3
= x(x – 3) + 1 (x – 3 ) = (x – 3) (x + 1)
∴ Zeros of the polynomial are 3, -1
Hence, choice (c) is correct.

Question 9.
Dividing x³ – 3x² – x + 3 by x – 4x + 3 the remainder will be:
(a) – 3
(b) 3
(c) 1
(d) 0
Answer:
(d) 0

Polynomials Class 10 Extra Questions Very Short Answer Type

Question 1.
Find a quadratic polynomial each with the given numbers as the sum and product of the zeroes respectively.
(i) \(\frac {1}{4}\), -1
(ii) √2, \(\frac {1}{3}\)
(iii) 0, √5
(iv) 1, 1
(v) – \(\frac {1}{4}\), \(\frac {1}{4}\)
(vi) 4, 1
Solution:
Let the polynomial be ax² + bc + c and its zeroes be α and β.
(i) Here, α + β = \(\frac {1}{4}\) and αβ = -1
Thus, the polynomial formed = x² – (Sum of the zeroes)x + Product of the zeroes

If k = 4 then the polynomial is 4x² – 3 – 4.

(ii) Here, α + β = √2 and αβ = \(\frac {1}{3}\)
Thus, the polynomial formed = x² – (Sum of the zeroes) x + Product of the zeroes

If k = 3, then, the polynomial is 3x² – 3√2x + 1.

(iii) Here, α + β = 0 and aß = √5
Thus, the polynomial formed = x² – (Sum of the zeroes) x + Product of zeroes
= x² – (0)x + √5
= x² + √5

(iv) Let the polynomial be ax² + bx + c and its zeroes α and β. Then

If a = 1 then b = -1 and c = 1
∴ One quadratic polynomial which satisfy the given conditions is x² – x + 1.

(v) Let the polynomial be ax² + bx +c and its zeroes be a and B. Then,

if a = 4, then b = 1 and c = 1.
∴ One quadratic polynomial which satisfy the given conditions is 4x² + x + 1.

(vi) Let the polynomial be ax² + bx + c and its zeroes be a and ß. then,

and αβ = 1 if a = 1 then b = -4 and c = 1
∴ One quadratic polynomial which satisfy the given conditions is x² – 4x + 1.

Question 2.
Divide p(y) by g(y) if p(y) = y³ – 3y² – y + 3 and g(y) = y² – 4y + 3.
Answer:
We write
y³ – 3y² – y + 3 = (y + 1) (y² – 4y + 3).
Solution:

Here, the quotient is y + 1 and the remainder is zero.

Question 3.
Examine if x – 1 is a factor of 2x³ – 5x + 3.
Solution:

Here, remainder is 0, hence, x – 1 is one factor of 2x³ – 5x + 3.

Polynomials Class 10 Extra Questions Short Answer Type

Question 1.
Divide the polynomial f(x) = 3x² – x³ – 3x + 5 by the polynomial g(x) = x – 1 – x² and verify the division alogrithm.
Solution:
f(x) = – x³ + 3x² – 3x + 5 and g(x) = – x² + x- 1

∴ Quotient = x – 2 and remainder = 3.
as dividend = Q x divisior + remainder
– x³ + 3x² – 3x + 5 = (x – 2)(-x² + x – 1) +3
= – x³ + x² – x + 2x² – 2x + 2 + 3
= – x³ + 3x² – 3x + 5
Verify

Question 2.
Divide 3 – x + 2x² by (2 – x) and verify alogrithm.
Solution:
First we write the terms of dividend and divisor in decreasing order of their degrees and then perform the division as shown below:

Clearly, degree (9) = 0 < degree (- x + 2).
∴ quotient = (- 2x – 3) and remainder = 9.
= (quotient x divisor) + remainder
= (- 2x – 3)(- x + 2) + 9
= 2x² – 4x + 3x – 6 + 9
= 2x² – x + 3
= dividend
Thus, (quotient x divisor) + remainder = dividend. Hence, the division algorithm is verified.

Question 3.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectvely. Find g(x).
Solution:
Given, p(x) = x² – 3x² + x + 2, q(x) = x – 2 and r(x) = – 2x + 4. By division algorithm, we know that
Dividend = Divisor × Quotient + Remainder
p(x) = q(x) × g(x) + r(x).
Therefore,
x³ + 3x² + x + 2 = (x – 2) × g(x) + (-2x + 4)
⇒ x³ – 3x² + x + 2 + 2x – 4 = (x – 2) × g(x)
⇒
On dividing x³ – 3x² + 3x – 2 by x – 2, we get

Hence, g(x) = x² – x + 1.

Question 4.
Divide 6x⁵ + 5x⁴ + 11x³ – 5x² + 2x + 7 by 3x² – 2x + 4.
Solution:

Question 5.
If – 3 is one of the zeros of the quadratic polynomial (k – 1) x² + kx + 1. Find the value of other zeros.
Solution:
Given that – 3 is one zeros of the polynomial
(k – 1) x² + kx + 1
∴ (k – 1)(-3)² + k (-3) + 1 = 0 .
⇒ 9k – 9 – 3k + 1 = 0
⇒ 6k = 8
⇒ Putting the value of k = \(\frac {4}{3}\) in given polynomial,
we get (\(\frac {4}{3}\)– 1) x² + \(\frac {4}{3}\) x + 1

⇒ x² + 4x + 3 = x² + 3x + x + 3
= x (x + 3) + 1 (x + 3)
⇒ (x + 3) (x + 1)
∴ zeros are – 3 and – 1
Hence their zeros of the quadratic polynomial is – 1.

Question 6.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg g(x)
(ii) deg g(x) = deg r(x)
(iii) deg q(x) = 0
Solution:
Let
q(x) = 3x² + 2x + 6,
degree of g(x) = 2
p(x) = 12x² + 8x + 24,
degree of p(x) = 2

(i) Using division algorithm,
We have,’ p(x) = (x) × g(x) + r(x)
On dividing 12x² + 8x + 24 by 3x² + 2x + 6.
we get

Since, the remainder is zero, therefore 3x² + 2x + 6 is a factor of 12x² + 8x + 24.
∴ g(x) = 4 and r(x) = 0.

(ii) p(x) = x⁵ + 2x⁴ + 3x³ + 5x² + 2
q(x) = x² + x + 1, degree of g(x) = 2
g(x) = x³ + x² + x + 1
r(x) = 2x² – 2x + 1, degree of r(x) = 2
Here, deg q(x) = deg r(x)
On dividing x⁵ + 2x⁴ + 3x³ + 5x² + 2 by x² + x + 1, we get

Here, g(x) = x³ + x² + x + 1
and r(x) = 2x² – 2x + 1

(iii) Let p(x) = 2x⁴ + 8x³ + 6x² + 4x + 12, r(x) = 2, degree of r(x) = 0
g(x) = x⁴ + 4x³ + 3x² + 2x + 1
⇒ 9(x) = 10
Here deg r(x) = 0
On dividing 2x⁴ + 8x³ + 6x² + 4x + 12 by 2,
we get

Question 7.
Find the zero’s of quadratic polynomial f(x) = 3x² – 3 – 4. Verify the relationship between the zeros and its coefficients.
Solution:
∴ f(x) = 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)
∴ zero’s are \(\frac {4}{3}\) and -1.
sum of the zeros =

Question 8.
Solve the pair of linear simultaneous equations
2x – y = 1 and x + 2y = 13
By drawing their graphs. Find the coordinates of vertices of a triangle formed by these lines and y-axis.
Solution:
2x – y = 1 ⇒ x + 2y = 13
2x = 1 + y ⇒ x = 13 – 2y

Now plot the points on the graph paper.

The coordinate of required ∆ are
A(3, 5), B (0, \(\frac {13}{2}\)) and C(0, -1) respectively
Solution is x = 3, y = 5

Question 9.
Find all the zeros of polynomial 2x⁴ – 3x² – 3x² + 6x – 2. If two of its zeros are √2 and √2.
Solution.
quad. polynomial form by the given zeros is
(X – √2) (x + √2)
⇒ x² – 2
Now divided the given polynomial by x² – 2

∴ x⁴ – 3x³ – 3x² + 6x – 2
= (x² – 2) (2x² – 3x + 1)
= (x² – 2) (2x² – 2x – x + 1)
= (x² – 2) [2x(x -1) – 1(x-1)]
= (x √2 )(x + √2)(x – 1) (2x – 1)
Hence all zeros are
1/2, 1, √2 and – √2

Here you will find Real Numbers Class 10 Extra Questions Maths Chapter 1 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

Extra Questions for Class 10 Maths Real Numbers with Answers Solutions

Extra Questions for Class 10 Maths Chapter 1 Real Numbers with Solutions Answers

Real Numbers Class 10 Extra Questions Objective Type

Question 1.
The graph of y = f(x) is given. The no. of zeros of f(x) will be:

(a) 1
(b) 3
(c) 2
(d) 4
Answer:
(b) 3
Solution.
As graph of/(x) cut the x and at 3 points
∴ Zeros of fix) are 3.
Choice (b) is correct.

Question 2.
For some integer m, the form of every positive even integer will be:
(a) m
(b) m + 1
(c) 2m
(d) 2m + 1
Answer:
(c) 2m
Solution.
2m is even integer
Hence, Choice (c) is correct.

Question 3.
The rational number which can be expressed as a terminating decimal number will be:
(a) \(\frac {77}{210}\)
(b)
(c) \(\frac {13}{3125}\)
(d) \(\frac {8}{17}\)
Answer:
(c) \(\frac {13}{3125}\)
Solution.
As the prime factorization of denominators of \(\frac {13}{3125}\) is only at the ninth place of 2 and 5.
Hence, choice (c) is correct.

Question 4.
For some integer n, the form of every positive odd integer will be:
(a) n
(b) n + 1
(c) 2n
(d) 2n + 1
Answer:
(d) 2n + 1
Solution.
For some integer n, the odd positive integer will be (2n + 1)
Hence, choice (d) is correct.

Question 5.
Which of the following graph is not of a quadratic polynomial?
(a)
(b)
(c)
(d)
Answer:
(d)

Solution.
Choice (d) is correct, (d)

Question 6.
Decimal expansion of rational number \(\frac {14587}{1250}\) will end after the following decimal places:
(a) one
(b) two
(c) three
(d) four
Answer:
(d) four
Solution.
Given Rational Number \(\frac {14587}{1250}\)
The factors of the denominator 1250 is 5 × 2
So the Number is terminating decimal expansion

∴ Number will end after 4 is decimal places.
Hence choice (d) is correct. Answer:

Question 7.
The decimal expansion of \(\frac {131}{120}\) will terminate after how many places of decimals?
(a) 3
(b) 4
(c) 1
(d) 2
Answer:
(a) 3

Question 8.
The decimal expansion of the rational number \(\frac{11}{2^{3} \cdot 5^{2}}\) will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) more than 3 decimal places.
Answer:
(c) three decimal places

Question 9.
The decimal expansion of the rational number \(\frac{43}{2^{4} \times 5^{3}}\)will terminate after:
(a) 3 places
(b) 4 places
(c) 5 places
(d) 1 place.
Answer:
(b) 4 places

Question 10.
The decimal expansion of the rational numbers \(\frac{23}{2^{2} \cdot 5}\) will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) more than three decimal places.
Answer:
(b) two decimal places

Real Numbers Class 10 Extra Questions Very Short Answer Type

Question 1.
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) We start with the larger number 225.
By Euclid’s division algorithm, we have Dividend = (Divisor × Quotient + Remainder)
225 = 135 × 1 + 90

We apply Euclid’s division algorithm on divisor 135 and the remainder 90.
Dividend = (Divisor × Quotient + Remainder)
135 = 90 × 1 + 45 1

Again, we apply Euclid’s division algorithm on divisor 90 and remainder 45.
90 = 45 × 2 + 0
∴ HCF (135,225) =45

(ii) We have,
Dividend = 38220 and Divisor = 196
Dividend = (Divisor × Quotient + Remainder)
38220 = 196 × 195 + 0
Hence, HCF (196, 38220) = 196

(iii) By Euclid’s division algorithm,
we have
Dividend = (Divisor × Quotient + Remainder)
867 = 255 + 3 + 102
We apply Euclid’s division algorithm on the divisor 255 and the remainder 102.
Dividend = (Divisor × Quotient + Remainder)
255 = 102 × 2 + 51
Again, we apply Euclid’s division algorithm on the divisor 102 and the remainder 51.
Dividend = (Divisor × Quotient + Remainder)
102 = 51 × 2 + 0
∴ HCF (867, 255) = 51

Question 2.
Using Euclid’s division algorithm prove that: 847, 2160 are co-primes/relatively prime.
Solution.
Definition of co-primes or relatively primes : Two numbers are said to be co-prime or relatively prime. If their HCF is 1. Hence to prove 847 and 2160 as co-prime numbers we will find their HCF and which should be 1.
Now steps to find HCF will be as under
2160 = 847 × 2 + 466
847 = 466 × 1 + 381
466 = 381 × 1 + 85
381 = 85 × 4 + 41
85 = 41 × 2 + 3
41 = 3 × 13 + 2
3 = 2 × 1 + 1 .
2 = 1 × 2 + 0
Therefore, the HCF = 1.
Hence, the numbers are co-prime relatively prime.

Question 3.
Find the largest number that divides 628, 3129 and 15630 to leave remainders 3, 4 and 5 respectively.
Solution.
Required number = HCF of (628 – 3), (3129 – 4) and (15630 – 6) i.e., required number
= HCF of 625, 3125,15625
To find HCF of 625, 3125 and 15625 first we will find HCF of 625 and 3125 and then we will find HCF of HCF of (625 & 3125) and 15625.
Step (1): HCF of 625 and 3125 by using Euclid’s division algorithm is
3125 = 625 × 5 + 0
⇒ HCF of 625 and 3125 is 625.
Step (2): Now HCF of625 and 15625 will be : 15625 = 625 × 25 + 0
⇒ HCF of 625 and 15625 is 625.
Hence HCF of 625, 3125 and 15625 is 625.
Hence required number is 625.

Question 4.
An Army contingent of 616 members is to march behind an army bond of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution.
To find the maximum number of columns, we have to find the HCF of 616 and 32. By Euclid’s division algorithm 19 4

i.e., 616 = 32 × 19 + 8 i.e., 32 – 8 × 4 + 0
∴ The HCF of 616 and 32 is 8.
Hence, maximum number of columns is 8.

Question 5.
Show that the square of an odd positive integer is of the form 8m + 1, for some whole number m.
Solution.
Any positive odd integer is of the form 2q + 1, where q is a whole number.
Therefore,
(2q + 1² = 4q² + 4q + 1
= 4 q(q + 1) + 1 …(1)
q(Question + 1) is either 0 or even because out of q and q + 1 one of the number is even. So, it is 2m, where m is a whole number.
Therefore, (2q + 1)²
= 4.2 m + 1 = 8m + 1. [From (1)]

Question 6.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution.
We have,
HCF (306, 657) = 9
We know that,
Product of LCM and HCF
= Product of two numbers
⇒ LCM × 9 = 306 × 657
⇒
Hence, LCM (306, 657) = 22338

Real Numbers Class 10 Extra Questions Short Answer Type

Question 1.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solution:
By Euclid’s division algorithm, we have
a – bq + r …(i)
On putting b = 3 in Eq (i), we get a = 3q + r, [0 ≤ r < 3, i.e. , r = 0, 1, 2]
If r = 0 ⇒ a – 3q
⇒ a² = 9q² …(ii)
If r = 1 ⇒ a = 3q + 1
⇒ a² = 9q² + 6q + 1 …(iii)
If r = 2 ⇒ a = 3q + 2
⇒ a² = 9q² + 12g + 4 …(iv)
From Eq (ii), 9q² is a square of the form 3m, where m 3q²
From Eq (iii), 9q² + 6q + 1 i.e., 3(3q² + 2q) + 1 is a square which is of the form
3m + 1, where m = 3q² + 2q
From Eq (iv) 9q² + 12q + 4 i.e.,3(3q² + 4q + 1) + 1 is a square which is of the form 3m + 1, where m – 3q² + 4q + 1.
∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Question 2.
Show that that square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
Solution:
We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.
Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5.
Thus we have :
(6m + 1)² – 36m² + 12m + 1
= 6(6m² + 2m) + 1 = 6q + 1,
q = 6m² + 2m is an integer.
(6m + 3)² – 36m² + 36m + 9
6(6m² + 6m + 1) + 3
= 6q + 3,
q = 6m² + 6m + 1 is an integer.
(6m + 5)² 36m² + 60m + 25
= 6(6m² + 10m + 4) + 1 = 6q+ 1, q
= 6m² + 10m + 4 is an integer.
Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

Question 3.
Find the LCM and HCF of the fo11owing pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91 (ii) 510 and 92
Sol:
(i) 26 and 91
26 = 2 × 13 and 91 = 7 × 13
LCM of 26 and 91
= 2 × 7 × 13 = 182
and HCF of 26 and 91 = 13

Now, 182 × 13 = 2366 and 26 × 91 = 2366
Hence, 182 × 13 = 26 × 91

(ii) 510 and 92
510 = 2 × 3 × 5 × 17
and 92 = 2 × 2 × 23

∴ LCM of 510 and 92
= 2 × 2 × 3 × 5 × 17 × 23 = 23460
and HCF of 510 and 92 = 2
Now, 23460 × 2 = 46920
and 510 × 92 = 46920
Hence, 23460 × 2 = 510 × 92

Question 4.
Prove that 5 + √3 is an irrational number.
Solution:
Let us assume, to the contrary, that 5 + √3 is rational.
So that we can find integers p and q(q ≠ 0) such that
5 + √3 = P/q,
Where p and q are coprime.

Since, p and q are integers, we get \(\frac {p}{q}\) – \(\frac {5q}{q}\) is a rational. So √3 is rational.
But this contradict the fact that √3 is irrational.
So, we conc1ude that 5 + √3 is irrational.
Proved

Question 5.
With out performing 1ong division, show whether the rational number is terminating decimal or non-terminating decimal. Find a1so its decimal expansion without performing actual division.
Solution:
Given Number is \(\frac {17}{8}\) in this 17 and 8 are coprime and the denominator is on1y the mu1tip1e of 2.
∴ \(\frac {17}{8}\) is a terminating decimal.
Now,

Question 6.
Without performing 1ong division procidure, show that the rational Number \(\frac {17}{90}\) is terminating decimal or non terminating repeating decimal.
Solution:
Given, Number

In denominator it contains 32 therefore this number is non-terminating repeating decimal.

Question 7.
Exp1ain, Why
(17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.
Solution:
17 × 5 × 11 × 3 × 2 + 2 × 11
= 2 × 11 [17 × 5 × 3 + 1]
= 22 × 256 = 2 × 11 × 4 × 4 × 4 × 4
Which is a product of more than two prime numbers i.e., 2 and 11. Hence it is a composite number.

Question 8.
Without performing the 1ong division process, find whether the rational Number \(\frac {987}{10500}\) is terminating or non terminating decimal. Give reason of your answer.
Solution:

∵ The prime factor of denominator is of the form 53 × 22
∴ The given rational number is terminating decimal.

Real Numbers Class 10 Extra Questions Long Answer Type

Question 1.
Prove that √2 is irrational.
Solution:
If possible, let √2 be rational and let its simp1est form be \(\frac {a}{b}\) .
Then, a and b are integers having no common factor other than 1, and b ≠ 0.
Now,

[on squaring both sides]
⇒ 2 b² = a² …(i)
⇒ 2 divides a² [∵ 2 divides 2b²]
⇒ 2 divides a
[ ∵ 2 is prime and divides c² ⇒ 2 divides a]
Let a = 2c for some integer c.
Putting a = 2c in (i), we get
2b² = 4c²
⇒ b² = 2c²
⇒ 2 divides b² [ ∵ 2 divides 2c²]
⇒ 2 divides b[ ∵ 2 is prime and 2 divides b² ⇒ 2 divides b]
Thus, 2 is a common factor of a and b.
But, this contradicts the fact that a and b have no common factor other than 1.
The contradiction arises by assuming that √2 is rational.
Hence, √2 is irrational.

Question 2.
Why n is an irrational number.
Solution:
We know that the ratio of circumference of a circle to 1ength of its corresponding diameter is known as ‘π’ (pie).
i.e., π = \(\frac {c}{d}\)
Now, read this to understand why ‘π’ is an irrational number.
If we take circumference of a circle as an integer (natural number) then we find length of its corresponding diameter comes in decimal not an integer (natura1 number)] so ratio of \(\frac {c}{d}\) is not in the form of \(\frac {p}{q}\), i.e., p and q bothare not integers.

Simi1arly if we take diameter of circle as an integer (natura1 number) then circumference of its corresponding circle comes out to be in the decimal form so again ratio \(\frac {c}{d}\) = π does not comes in the form of \(\frac {p}{q}\) because diameter, i.e., ‘q’ is an integer and ‘p’ is in decimal form.

So it is clear that for any circle if length of diameter is natura1 number then circumference is the decimal form or if circumference is a natura1 number then diameter is a decimal number. So we can say that the ratio \(\frac{\pi}{d}\) can never come in the form of \(\frac {p}{q}\) where p and q both are integers and q ≠ 0. So it is clear that the ratio \(\frac{c}{d}\), i.e., π can never be rational hence ‘π’ is a1ways an irrational number.

Question 3.
Prove that 3 + 2√5 is irrational.
Solution:
let us assume, to the contrary, that 3 + 2√5 is a rational number.
Now, let 3 + 2√5 = x, where a and b are coprime and b ≠ 0.

Since, a and b are integers, therefore
\(\frac {a}{2b}\) – \(\frac {3}{2}\) is a rational number.
∴ √5 is a rational number.
But √5 is an irrational number.
This shows that our assumption is wrong.
So, 3 + 2 √5 is an irrational number.
Hence proved.
Example 4. Prove that √3 is irrational.
Solution:
If possible, let √3 be rational and let its simplest form be \(\frac{a}{b}\)
Then, a and b are integers having no common factor other than 1, and b ≠ 0.
[on squaring both sides]
⇒ 3b² = a² …(i)
⇒ 3 divides a² [ ∵ 3 divides 3b²]
⇒ 3 divides a [∵ 3 is prime and divides a² ⇒ 3 divides a]
Let a = 3c for some integer c.
Putting a = 3c in (i), we get
3b² = 9c²
⇒ b² = 3c²
⇒ 3 divides b² [ ∵ 3 divides 3c²]
⇒ 3 divides b [ ∵ 3 is prime and 3 divides b² ⇒ 3 divides b]
Thus, 3 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1.
The contradiction arises by assuming that √3 is rational.
Hence, √3 is irrational.

Question 3.
The fo11owing rea1 numbers have decimal expansions as given be1ow. In each case, decide whether they are rational or not. If they are rational and of the form \(\frac{p}{q}\), what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000….
(iii) \(43 . \overline{123456789}\)
Solution:
(i) 43.123456789 is terminating. So, it represents a rational number.
Thus,

Thus, q = 10⁹
(ii) 0.120120012000120000…. is non-terminating and non-repeating. So, it is an irrational number.
(iii) \(43 . \overline{123456789}\) is non-terminating but repeating. So it is a rational.

Thus q = 999999999