Class 9

Punjab State Board PSEB 9th Class Physical Education Book Solutions Chapter 1 शारीरिक शिक्षा-इसके गुण एवं उद्देश्य Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Physical Education Chapter 1 शारीरिक शिक्षा-इसके गुण एवं उद्देश्य

PSEB 9th Class Physical Education Guide शारीरिक शिक्षा-इसके गुण एवं उद्देश्य Textbook Questions and Answers

बहुत छोटे उत्तरों वाले प्रश्न

प्रश्न 1.
शारीरिक शिक्षा का लक्ष्य क्या है ?
उत्तर-
शारीरिक शिक्षा का उद्देश्य व्यक्ति के लिए ऐसा वातावरण प्रदान करना है जो उसके शरीर, दिमाग तथा समाज के लिए लाभदायक हो।

प्रश्न 2.
शारीरिक शिक्षा क्या है ?
उत्तर-
शारीरिक शिक्षा बच्चे के सम्पूर्ण व्यक्तित्व के विकास के लिए शारीरिक कार्यक्रम द्वारा, शरीर, मन तथा आत्मा को पूर्णता की ओर ले जाती है।

प्रश्न 3.
शारीरिक शिक्षा के कोई दो उद्देश्यों के नाम लिखो।
उत्तर-
(1) शारीरिक विकास (2) मानसिक विकास।

प्रश्न 4.
व्यक्ति और समाज के विकास में शारीरिक शिक्षा के कोई दो योगदानों के नाम लिखें।
उत्तर-
(1) खाली समय का उचित प्रयोग, (2) जीवन के लक्ष्य की प्राप्ति में सहायक, 3. सामाजिक भावना।

प्रश्न 5.
खेल के मैदान में आप कौन-कौन से शारीरिक शिक्षा के उद्देश्य ग्रहण करते हो ? किसी दो के नाम लिखें।
उत्तर-
(1) सहनशीलता, (2) अनुशासन, (3) चरित्र विकास ।

प्रश्न 6.
खेलें व्यक्ति में कैसे गुण विकसित करती हैं ?
उत्तर-
नेतृत्व के गुण।

बड़े उत्तरों वाले प्रश्न

प्रश्न 1.
शारीरिक शिक्षा का क्या लक्ष्य है ?
(What is the aim of Physical Education ?)
उत्तर-
शारीरिक शिक्षा का लक्ष्य (Aim of Physical Education)-शारीरिक शिक्षा साधारण शिक्षा की भान्ति उच्च मंजिल पर पहुंचने के लिए शिक्षा प्रदान करती है।
सुप्रसिद्ध शारीरिक शिक्षा शास्त्री जे० एफ० विलिअम्ज (J.F. Williams) का कहना है .. कि यदि हमें शारीरिक शिक्षा की मंजिल प्राप्त करनी है तो यह हमारा उद्देश्य होना चाहिए। शारीरिक शिक्षा का उद्देश्य एक कुशल तथा योग्य नेतृत्व देना तथा ऐसी सुविधाएं प्रदान करना है जो किसी एक व्यक्ति या समुदाय को कार्य करने का अवसर दें तथा ये सभी क्रियाएं शारीरिक रूप से सम्पूर्ण, मानसिक रूप से उत्तेजक तथा सन्तोषजनक तथा सामाजिक रूप से निपुण हों। इसके अनुसार व्यक्ति के लिए केवल उन्हीं क्रियाओं का चयन करना चाहिए जो शारीरिक रूप से लाभदायक हों।

प्रत्येक व्यक्ति केवल वे क्रियाएँ करे जो शरीर को तेज़ करने वाली हों तथा उसकी चिन्तन शक्ति बढ़ाने वाली हों। खेलों में कुछ उलझनें तथा प्रतिबन्ध इस प्रकार लगाये जाते हैं कि व्यक्ति का दिमाग ताज़ा रहे और उसे मानसिक सन्तोष मिले। इन क्रियाओं को समाज का समर्थन भी प्राप्त होना चाहिए। इसके अतिरिक्त समाज के अन्य सदस्य भी इन्हें आदर की दृष्टि से देखें। . संक्षेप में, समूचे तौर पर शारीरिक शिक्षा का लक्ष्य व्यक्ति के लिए ऐसा वातावरण प्रदान करना है जो उसके शारीरिक, मानसिक तथा सामाजिक पक्ष के लिए उपयोगी हो। इस प्रकार व्यक्ति का सर्वपक्षीय विकास (All-round Development) ही शारीरिक शिक्षा का एकमात्र लक्ष्य है।

प्रश्न 2.
खाली समय का उचित प्रयोग किस प्रकार किया जा सकता है ? संक्षेप में लिखो।
(How can Leisure time be usefully spent ? Describe in brief.)
उत्तर-
किसी ने ठीक ही कहा है कि “खाली मन शैतान का घर होता है।” (“An idle brain is a devil’s workshop.”) यह प्रायः देखा भी जाता है कि बेकार आदमी को शरारतें ही सूझती हैं। कई बार तो वह ऐसे गलत काम करने लगता है जो सामाजिक तथा नैतिक दृष्टि से उचित नहीं ठहराये जा सकते। इसका कारण यह है कि उसके पास फालतू समय तो है परन्तु उसके पास इसे व्यतीत करने का ढंग नहीं है। फालतू समय का उचित प्रयोग न होने के कारण उसका दिमाग कुरीतियों में फंस जाता है और कई बार अनेक उलझनों में उलझ कर रह जाता है जिनसे बाहर निकलना उसके वश से बाहर होता है।

यदि व्यक्ति इस फालतू समय का सदुपयोग जानता हो तो वह जीवन में उच्च शिखरों को छू सकता है। संसार के अनेक आविष्कार उन व्यक्तियों ने किये जो फालतू समय को कुशल ढंग से व्यतीत करने की कला से परिचित थे। इस प्रकार संसार के अनेक आविष्कार फालतू समय की ही देन हैं।।

यदि बच्चों के फालतू समय व्यतीत करने के लिए कोई उचित प्रबन्ध न हो तो वे बुरी आदतों का शिकार हो जाएंगे। इस प्रकार वे समाज पर बोझ बनने के साथ-साथ इसके लिए कलंक भी बन जाएंगे। इसलिए उनके फालतू समय के उचित प्रयोग की अच्छी व्यवस्था करनी चाहिए। स्कूलों, कॉलेजों, पंचायतों या नगरपालिकाओं या सरकार को अच्छे खेल के मैदानों तथा खेलों के सामान का पूरा प्रबन्ध करना चाहिए ताकि बच्चे खेलों में भाग लेकर अपने खाली समय का उचित प्रयोग कर सकें। इस प्रकार हम देखते हैं कि फालतू समय के प्रभावशाली उपयोग का सर्वोत्तम साधन खेलें हैं।

प्रश्न 3.
‘खेलें अच्छे नेता बनाती हैं।’ कैसे ?
(Games make good Leader. How ?)
अथवा
खेलें एक अच्छे नेता के गुण कैसे पैदा करती हैं?
(How sports produce the qualities of a good Leader ?)
उत्तर-
खेलें व्यक्ति में अच्छे नेतृत्व के गुण विकसित करती हैं। शारीरिक शिक्षा का क्षेत्र अत्यधिक विशाल है। खेलों में एक खिलाड़ी को अनेक ऐसे अवसर मिलते हैं जब उसे टीम के कप्तान, सैक्रेटरी, रैफरी या अम्पायर की भूमिका निभानी पड़ती है। इन परिस्थितियों में वह अपनी योग्यता के अनुसार आचरण करता है। एक कप्तान के रूप में वह अपनी टीम के खिलाड़ियों को समय-समय पर निर्देश देकर उन्हें उचित ढंग से तथा पूर्ण विश्वास के साथ खेलने की प्रेरणा देता है। एक रैफरी या अम्पायर के रूप में वह निष्पक्ष रूप से उचित निर्णय देता है। सैक्रेटरी के रूप में वह टीम का ठीक ढंग से गठन एवं संचालन करता है। इस प्रकार उसमें एक सफल और अच्छे नेता के गुण विकसित हो जाते हैं।

एक नेता में अच्छे चारित्रिक गुणों का होना आवश्यक है। उसमें आज्ञा-पालन, समय की पाबन्दी, सभी के साथ समान व्यवहार, प्रेम, सहानुभूति, सहनशीलता आदि गुण प्रचुर मात्रा में होने चाहिएं। ये सभी गुण वह खेल के मैदान से ग्रहण कर सकता है। एक नेता को अपने आस-पास के लोगों के साथ सद्भावना से रहना चाहिए। खेलें उसमें यह गुण विकसित करती हैं। जब एक खिलाड़ी अन्य स्थानों के खिलाड़ियों के साथ मिल-जुल कर खेलता है, वह उनके स्वभाव तथा सभ्यता के साथ भली-भान्ति परिचित हो जाता है। उसमें उनके प्रति सद्भावना उत्पन्न हो जाती है। यह सद्भावना ही एक नेता का महत्त्वपूर्ण गुण है।

नेता को चुस्त और फुर्तीला होना चाहिए। खेलों में भाग लेने से व्यक्ति में चुस्ती और स्फूर्ति आती है। इस प्रकार खेलें अच्छे नेताओं के निर्माण में विशेष योगदान देती हैं।

बड़े स्तरों वाले प्रश्न

प्रश्न 1.
शारीरिक शिक्षा के मुख्य उद्देश्य कौन-कौन से हैं ?
(Describe the main objectives of Physical Education.)
उत्तर-
शारीरिक शिक्षा के उद्देश्य (Objectives of Physical Education)किसी भी कार्य को आरम्भ करने से पहले उसके उद्देश्य निर्धारित कर लेना आवश्यक है। बिना उद्देश्य के किया गया काम छाछ को मथने के समान है। उद्देश्य निश्चित कर लेने से
हमारे यत्नों को प्रोत्साहन मिलता है और उस काम को करने में लगाई गई शक्ति व्यर्थ नहीं जाती है। आज तो शारीरिक शिक्षा के उद्देश्य की जानकारी प्राप्त करना और भी ज़रूरी हो गया है क्योंकि अब तो स्कूलों में एक विषय (Subject) के रूप में इसका अध्ययन किया जाता है।

साधारणतया शारीरिक शिक्षा के निम्नलिखित उद्देश्य हैं-

• शारीरिक वृद्धि एवं विकास (Physical Growth and Development)
• मानसिक विकास (Mental Development) (3) सामाजिक विकास (Social Development)
• चरित्र निर्माण या नैतिक विकास (Formation of Character or Moral Development)
• नाड़ियों और मांसपेशियों में समन्वय (Neuro-muscular Co-ordination )
• बीमारियों से बचाव (Prevention of Diseases)

1. शारीरिक वृद्धि एवं विकास (Physical Growth and Development)अच्छा, सफल तथा सुखद जीवन व्यतीत करने के लिए सुदृढ़, सुडौल तथा स्वस्थ शरीर का होना परमावश्यक है। हमारे शरीर का निर्माण मज़बूत हड्डियों से हुआ है। इसमें काम करने वाले सभी अंग उचित रूप से अपने कर्तव्य का पालन कर रहे हैं। शरीर के अंगों के सुचारु रूप से कार्य करते रहने से शरीर का निरन्तर विकास होता है। इसके विपरीत इनके भलीभान्ति काम न करने से शारीरिक विकास भी रुक जाता है। इस प्रकार शारीरिक प्रफुल्लता के उद्देश्य की प्राप्ति के लिए शारीरिक शिक्षा सुखद वातावरण जुटाती है।

2. मानसिक विकास (Mental Development)- शारीरिक विकास के साथसाथ मानसिक विकास की भी आवश्यकता है। शारीरिक शिक्षा ऐसी क्रियाएं प्रदान करती हैं जो व्यक्ति के मस्तिष्क को उत्तेजित करती हैं। उदाहरणस्वरूप, बास्केटबाल के खेल में एक टीम के खिलाड़ियों को विरोधी टीम के खिलाड़ियों से गेंद (बॉल) को बचा कर रखना होता है तथा इसके साथ-साथ अपना लक्ष्य भी देखना पड़ता है । अपनी शक्ति का अनुमान लगा कर बॉल को ऊपर लगी बास्केट में भी डालना होता है। कोई भी खिलाड़ी जो शारीरिक रूप से हृष्ट-पुष्ट है, परन्तु मानसिक रूप में विकसित नहीं है, कभी भी अच्छा खिलाड़ी नहीं बन सकता है। शारीरिक शिक्षा मानसिक विकास के लिए उचित वातावरण प्रदान करती है। इसलिए खेलों में भाग लेने वाले व्यक्ति का शारीरिक विकास के साथ-साथ मानसिक विकास भी हो जाता है।

प्रायः शारीरिक रूप से अस्वस्थ तथा मानसिक रूप से सुस्त व्यक्ति बहुत ही भावुक हो जाते हैं। वे जीवन की साधारण समस्याओं को हंसी-हंसी में सुलझा लेने के स्थान पर उनमें उलझ कर रह जाते हैं। वे अपनी खुशी, गम, पसन्द तथा नफ़रत को आवश्यकता से कहीं अधिक महत्त्व देने लगते हैं और वे अपना कीमती समय तथा शक्ति व्यर्थ ही गंवा देते हैं। इस प्रकार कोई महान् सफलता प्राप्त करने से वंचित रह जाते हैं। शारीरिक शिक्षा इन भावनाओं पर नियन्त्रण पाने की कला सिखाती है।

3. सामाजिक विकास (Social Development) शारीरिक शिक्षा व्यक्ति को अपने इर्द-गिर्द के लोगों के साथ सद्भावना से रहने की शिक्षा प्रदान करती है। जब एक व्यक्ति विभिन्न स्थानों के खिलाड़ियों के साथ मिल कर खेलता है तो सामाजिक विकास का अच्छा वातावरण पनपता रहता है। प्रत्येक खिलाड़ी अन्य खिलाड़ियों के स्वभाव, रीति-रिवाज, पहनावा, सभ्यता तथा संस्कृति से भली-भान्ति परिचित हो जाता है। कई बार दूसरों की अच्छी बातें तथा गुण ग्रहण कर लिए जाते हैं। देखने में आता है कि यूनिवर्सिटी, राज्यीय, राष्ट्रीय तथा अन्तर्राष्ट्रीय स्तरों पर खेल मुकाबलों का आयोजन किया जाता है। इनका मुख्य उद्देश्य लोगों में प्रेम तथा आदर की भावनाएं विकसित करना होता है।

4. चरित्र-निर्माण अथवा नैतिक विकास (Character Formation or Moral Development)-खेल का मैदान चरित्र-निर्माण की पाठशाला है। इसका कारण यह है कि खेल के मैदान में ही व्यक्ति खेल के नियमों को निभाते हैं। यहीं से वे अच्छा जीवन व्यतीत करने की कला सीखते हैं तथा सुलझे हुए इन्सान बन जाते हैं। खेल खेलते समय यदि रैफरी कोई ऐसा निर्णय दे देता है जो उन्हें पसन्द नहीं तो भी वे खेल जारी रखते हैं और कोई अभद्र व्यवहार नहीं करते। खेल के मैदान में ही आज्ञा-पालन, अनुशासन, प्रेम तथा दूसरों से सहयोग आदि के गुण सीखे जाते हैं। इस प्रकार प्रत्येक व्यक्ति का चरित्र-निर्माण और नैतिक विकास होता है।

5. नाड़ियों और मांसपेशियों में समन्वय (Neuro-muscular Coordination)-हमारी प्रतिदिन की क्रियाओं को समुचित ढंग से पूर्ण करना अनिवार्य है ताकि नाड़ियों और मांसपेशियों में समन्वय पैदा हो। शारीरिक शिक्षा इनमें समन्वय पैदा करने में सहायता देती है।

6. बीमारियों से बचाव (Prevention from Diseases) शारीरिक शिक्षा का उद्देश्य विद्यार्थियों को बीमारी से बचाना भी है। बहुत-सी बीमारियां अज्ञानता के कारण लग जाती हैं। शारीरिक शिक्षा का उद्देश्य बच्चों की बीमारियों के कारणों का ज्ञान देना है। वे इन कारणों से बच कर स्वयं भी बीमारियों से बच सकते हैं।

अन्त में, हम कह सकते हैं कि शारीरिक शिक्षा मनुष्य के सर्वपक्षीय विकास के लिए, नागरिकता के लिए, मानवीय भावनाओं के निर्माण तथा राष्ट्रीय एकता के लिए बहुत ही उपयोगी है।

प्रश्न 2.
हॉकी के खेल के मैदान में आप कौन-कौन से शारीरिक शिक्षा के उद्देश्य ग्रहण करते हो ?
(What are the objectives of Physical Education that one acquires in the game of Hockey ?)
उत्तर-
हॉकी का मैदान भी एक तरह की पाठशाला है जहां से विद्यार्थी शारीरिक शिक्षा के अनेक गुण ग्रहण करता है जिनसे वह जीवन में उन्नति के उच्च शिखर को छूता है और जीवन का हर पक्ष से भरपूर आनन्द उठाता है। हॉकी के क्रीड़ा-क्षेत्र में हम शारीरिक शिक्षा के निम्नलिखित गुणों को ग्रहण करते हैं –

1. सहनशीलता (Toleration) खेल के मैदान में हम सहनशीलता का पाठ पढ़ते हैं। वैसे तो सभी खिलाड़ी चाहते हैं कि जीत उनकी टीम की ही हो। परन्तु कई बार लाख चाहने पर भी विरोधी टीम विजयी हो जाती है। ऐसी स्थिति में पराजित टीम के खिलाड़ी दिल छोड़ कर नहीं बैठ जाते बल्कि अपना मनोबल ऊंचा रखते हैं। वे हार-जीत को एक ही समान समझते हैं। इस प्रकार खेल के मैदान से विद्यार्थियों को सहनशीलता की व्यावहारिक ट्रेनिंग मिलती है।

2. अनुशासन (Discipline)-खेल के मैदान में खिलाड़ी अनुशासन में रहने की कला सीखते हैं। उन्हें पता चलता है कि अनुशासन ही सफलता की कुंजी है। वे खेल में भाग लेते समय अनुशासन का पालन करते हैं। वे अपने कप्तान की आज्ञा मानते हैं तथा रैफरी के निर्णयों को सहर्ष स्वीकार करते हैं। खेल में पराजय को सामने स्पष्ट शारीरिक शिक्षा-इसके गुण एवं उद्देश्य देखते हुए भी वे कोई ऐसा अभद्र व्यवहार नहीं करते जिससे कोई उन्हें अनुशासनहीन – कह सके।

3. चरित्र विकास (Character Development) हॉकी के खेल में भाग लेने से विद्यार्थियों में सहयोग, प्रेम, सहनशीलता, अनुशासन आदि गुणों का विकास होता है जिनसे उनके चरित्र का विकास होता है। इस खेल में भाग लेने से उनमें सहयोग की भावना विकसित होती है। वे निजी हितों को समूचे हितों पर न्योछावर कर देते हैं।

4. व्यक्तित्व का विकास (Development of Personality) हॉकी के खेल में भाग लेने से विद्यार्थियों में कुछ ऐसे गुण विकसित हो जाते हैं, जिनसे उनके व्यक्तित्व का विकास हो जाता है। उनमें सहयोग तथा सहनशीलता आदि गुण विकसित होते हैं तथा उनका शरीर सुन्दर एवं आकर्षक बन जाता है। ये सभी अच्छे व्यक्तित्व के चिन्ह हैं।

5. अच्छे नागरिक बनाना (Creation of Good Citizens)-हॉकी के मैदान में खिलाड़ी में कर्त्तव्य-पालन, आज्ञा पालन, सहयोग, सहनशीलता आदि गुण विकसित हो जाते हैं जो उन्हें एक अच्छा नागरिक बनने में पर्याप्त सहायता पहुंचाते हैं। वे नागरिकता के सभी कर्तव्यों का भली-भान्ति पालन करते हैं। इस प्रकार हॉकी का मैदान अच्छे नागरिकों के निर्माण में महत्त्वपूर्ण भूमिका निभाता है।

6. सहयोग (Co-operation) हॉकी के खेल में भाग लेने वाला खिलाड़ी प्रत्येक खिलाड़ी का कहना मानता है। वह अपना विचार दूसरों पर बलात् लागू नहीं करवाता है, अपितु अपने विचार विनिमय के द्वारा खेल के मैदान में संयुक्त विचारधारा बनाता है। इस प्रकार सहयोग की भावना उत्पन्न होती है।

7. राष्ट्रीय भावना (National Spirit)-हॉकी का मैदान एक ऐसा स्थान है जहां हम बिना धर्म और वर्ग के आधार पर भाग ले सकते हैं। कोई भी खिलाड़ी खेल के मैदान में से किसी खिलाड़ी को धर्म के आधार पर टीम से बाहर नहीं निकाल सकता। इस प्रकार . खेल के मैदान में समानता और राष्ट्रीय एकता की भावना पैदा होती है।

8. आत्म-विश्वास की भावना (Self-Confidence) हॉकी के खेल के मैदान में खिलाड़ियों में आत्म-विश्वास की भावना पैदा होती है। जैसे, वह विजय-पराजय को एक समान समझता है। वही खिलाड़ी खेल के मैदान में सफल होता है जो धैर्य
और विश्वास के साथ खेले। इससे सिद्ध होता है कि हॉकी के खेल के द्वारा खिलाड़ियों में आत्म-विश्वास की भावना पैदा होती है।

9. विजय-पराजय को समान समझने की भावना (Spirit of giving equal importance to Victory of Defeat) हॉकी के खेल के द्वारा खिलाड़ियों में विजय-पराजय को एक समान समझने की भावना पैदा होती है।

हमें कभी भी विरोधी टीम का मज़ाक नहीं उड़ाना चाहिए या विजय की प्रसन्नता में पागल नहीं होना चाहिए। पराजित टीम को सदैव प्रोत्साहन देना चाहिए। यदि उसकी पराजय होती है तो उसे निराश और उत्साहहीन नहीं होने देना चाहिए अपितु उसका हौसला बढ़ाना चाहिए।

10. त्याग की भावना (Spirit of Sacrifice)-हॉकी के खेल के मैदान में त्याग की भावना अत्यावश्यक है। जब हम खेल में भाग लेते हैं तो हम अपने स्कूल, प्रान्त, क्षेत्र और सारे राष्ट्र के लिए अपने हित का त्याग करके उसकी विजय का श्रेय राष्ट्र को देते हैं। अत: यह सिद्ध होता है कि खेलें सदैव त्याग चाहती हैं। – ड्यूक ऑफ़ विलिंग्टन ने नेपोलियन को वाटरलू (Waterloo) के युद्ध में पराजित करने के पश्चात् कहा था, “वाटरलू का युद्ध एटन और हैरो के खेल के मैदानों में जीता गया।” (“The Battle of Waterloo was won at the playing-fields of Eton and Harrow.”)
इससे यह सिद्ध होता है कि खेलें अच्छे नेता पैदा करने में सहायक होती हैं।

शारीरिक शिक्षा-इसके गुण एवं उद्देश्य PSEB 9th Class Physical Education Notes

• .शारीरिक शिक्षा-शारीरिक क्रियाओं से हमें जो अनुभव प्राप्त होता है, उसे शारीरिक शिक्षा कहते हैं।
• शारीरिक शिक्षा का लक्ष्य-मनुष्य का सर्वोन्मुखी विकास ही शारीरिक शिक्षा का लक्ष्य है।
• शारीरिक शिक्षा के उद्देश्य-शारीरिक, मानसिक और नैतिक विकास ही शारीरिक शिक्षा के उद्देश्य हैं।
• खेल के मैदान में शारीरिक शिक्षा का उद्देश्य-खेल के मैदान से सहनशीलता, अनुशासन और चरित्र का विकास होता है।
• खाली समय का उचित प्रयोग-खेलों में भाग लेने से बच्चे खाली समय का उचित प्रयोग करते हैं जिससे उनमें बुरी आदतें नहीं पनपती हैं।
• खेलों के नेतृत्व का गुण-एक नेता में अच्छे चारित्रिक गुणों का होना आवश्यक है और ये खेलें मनुष्य में नेतृत्व के गुण पैदा करती हैं।

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 15 Probability MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 15 Probability MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
When a balanced die is thrown, the probability of getting 3 is …………….. .
A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{6}\)
Answer:
D. \(\frac{1}{6}\)

Question 2.
A card is drawn at random from a well shuffled pack of cards. The probability of that card being a king is …………………. .
A. \(\frac{1}{52}\)
B. \(\frac{1}{26}\)
C. \(\frac{1}{13}\)
D. 1
Answer:
C. \(\frac{1}{13}\)

Question 3.
A card is drawn at random from a well shuffled pack of cards. The probability of that card being a card other than picture cards is ……………….. .
A. \(\frac{4}{13}\)
B. \(\frac{10}{13}\)
C. \(\frac{3}{13}\)
D. \(\frac{1}{13}\)
Answer:
B. \(\frac{10}{13}\)

Question 4.
When an unbiased coin is tossed thrice, the probability of receiving three heads is ………………… .
A. \(\frac{1}{8}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{8}\)
Answer:
A. \(\frac{1}{8}\)

Question 5.
When three unbiased coins are tossed simultaneously, the probability of receiving exactly one tail is ………………… .
A. \(\frac{1}{8}\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{8}\)
Answer:
D. \(\frac{3}{8}\)

Question 6.
When a balanced die is thrown, the probability of receiving an even number is ………………… .
A. \(\frac{1}{6}\)
B. \(\frac{5}{6}\)
C. \(\frac{1}{2}\)
D. \(\frac{1}{4}\)
Answer:
C. \(\frac{1}{2}\)

Question 7.
When a balanced die is thrown, the probability of receiving a prime number is ……………….. .
A. \(\frac{2}{3}\)
B. \(\frac{3}{4}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{2}\)
Answer:
D. \(\frac{1}{2}\)

Question 8.
When two balanced dice are thrown simultaneously, the probability of getting the total of numbers on dice as 9 is ………………. .
A. \(\frac{1}{9}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{12}\)
Answer:
A. \(\frac{1}{9}\)

Question 9.
Out of 100 days, the forecast predicted by the wheather department proved to be true on 20 days. Chosen any one day from these 100 days, the probability that the forecast proved to be false is ………………… .
A. \(\frac{1}{3}\)
B. \(\frac{1}{4}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
Answer:
D. \(\frac{4}{5}\)

Question 10.
The probability of a month of January having 5 Sundays is ………………….. .
A. \(\frac{2}{7}\)
B. \(\frac{3}{7}\)
C. \(\frac{5}{7}\)
D. \(\frac{1}{7}\)
Answer:
B. \(\frac{3}{7}\)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Answer:
The batswoman played 30 balls. Hence, the total number of trials = 30. If the event that she did not hit a boundary is denoted by A, then. the number of trials when event A occured is 30 – 6 = 24.
∴ p(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{24}{30}\)
= \(\frac{4}{5}\)
Thus, the probability that she did not hit a boundary is \(\frac{4}{5}\).

Question 2.
1500 families with 2 children were selected randomly, and the following data were s recorded:

Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) No girl. Also check whether the sum of these probabilities is 1.
Answer:
Here, the total number of families is 1500.
Hence, the total number of trials = 1500

(i) Let event A denote the event that the family chosen at random is having 2 girls.
Then, the number of trials when event A occured is 475.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{475}{1500}\)
= \(\frac{19}{60}\)

(ii) Let event B denote the event that the family chosen at random is having 1 girl.
Then, the number of trials when event B occured is 814.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{814}{1500}\)
= \(\frac{407}{750}\)

(iii) Let event C denote the event that the family chosen at random Is having no girl.
Then, the number of trials when event C occured is 211.
∴ p(C) = \(\frac{\text { No. of trials in which event } \mathrm{C} \text { occured }}{\text { The total number of trials }}\)
= \(\frac{211}{1500}\)
Now,
P(A) + P(B) + P(C) = \(\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}\)
= \(\frac{475+814+211}{1500}\)
= \(\frac{1500}{1500}\)
= 1

Question 3.
Refer to sum no. 5 of “Sums to Enrich ‘Remember’” in chapter 14. Find the probability that a student of the class was born in August.
Answer:
From the Bar graph in the sum which is referred here, we get the following information:

Total number of students = 40 and the number of students born in August = 6.
Hence, if event A denotes the event that a student of the class is born in August, then the number of trials when event A occured is 6 and the total number of trials is 40.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{6}{40}\)
= \(\frac{3}{20}\)

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Answer:
Here, the total number of trials = 200. If event A denotes the event that 2 heads come up, then the number of trials when event A occured is 72.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{72}{200}\)
= \(\frac{9}{25}\)

Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below :

Suppose a family is chosen. Find the probability that the family chosen is ( i ) earning ? 10000- ? 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not .own any vehicle.
(iv) earning ₹ 13000 – ₹ 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Answer:
Here, the total number of families is 2400. Hence, the total number of trials = 2400

(i) Let event A denote the event that the family is earning ₹ 10000 – ₹ 13000 per month and owning exactly 2 vehicles.
Then, the number of trials when event A occured = 29.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{29}{2400}\)

(ii) Let event B denote the event that the family is earning ₹ 16000 or more per month and owning exactly 1 vehicle.
Then, the number of trials when even B occured = 579.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{579}{2400}\)
= = \(\frac{193}{800}\)

(iii) Let event C denote the event that the family is earning less than ₹ 7000 per month and does not own any vehicle.
Then, the number of trials when event C occured = 10.
∴ P(C) = \(\frac{\text { No. of trials in which event C occured }}{\text { The total number of trials }}\)
= \(\frac{10}{2400}\)
= = \(\frac{1}{240}\)

(iv) Let event D denote the event that the family is earning ? 13000 -? 16000 per month and is owning more than 2 vehicles. Then, the number of trials when event D occured = 25.
∴ P(D) = \(\frac{\text { No. of trials in which event D occured }}{\text { The total number of trials }}\)
= \(\frac{25}{2400}\)
= \(\frac{1}{96}\)

(v) Let event E denote the event that the family is owning not more than 1 vehicle, i.e., 1 vehicle or no vehicle.
Then, the number of trials when event E occured.
= 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062
∴ P(E) = \(\frac{\text { No. of trials in which event E occured }}{\text { The total number of trials }}\)
= \(\frac{2062}{2400}\)
= \(\frac{1031}{1200}\)

Question 6.
Refer to table 7 of sum no. 7 in “Sums to Enrich ‘Remember’” in chapter 14.
(i) Find the probability that a student obtained less than 20 marks in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Answer:
According to the table referred here, the total number of students = 90.
Hence, the total number of trials = 90.
(i) According to the same table, the number of students who obtained less than 20 marks in the mathematics test is 7. So, if the event that a student obtained less than 20 marks in mathematics test is called event A, then the number of trials when event A occured is 7.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{7}{90}\)

(ii) Let event B denote the event that a student obtained 60 or more marks. Then, , according to the same table, the number of trials when event B occured = 15 + 8 = 23.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{23}{90}\)

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table:

Opinion	Number of students
Like	135
Dislike	65

Find the probability that a student chosen at random
(i) Likes statistics,
(ii) Does not like it.
Answer:
Here, the total number of students = 200.
Hence, the total number of trials = 200.

(i) Let event A denote the event that a student likes statistics.
Then, the number of trials when event A occured = 135
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{135}{200}\)
= \(\frac{27}{40}\)

(ii) Let event B denote the event that a student does not like statistics. Then, the number of trials when event B occured = 65.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{65}{200}\)
= \(\frac{13}{40}\)

Question 8.
Refer to sum no. 2, Exercise 14.2. What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work ?
(ii) more than or equal to 7km from her place of work ?
(iii) within \(\frac{1}{2}\)km from her place to work?
Answer:
The total number of observations in the question referred here is 40.
Hence, the total number of trials = 40.

(i) Let event A denote the event that the distance between her residence and the place of work is less than 7 km. Then there are 9 such observations, viz., 5, 3, 2, 3, 6, 5, 6, 2, 3.
Hence, the number of trials when event A occured = 9.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{9}{40}\)

(ii) Let event B denote the event that the said distance is 7 km or more than 7 km. Then, all the remaining 31(40-9) observations refer to event B.
Hence, the number of trials when event B occured = 31
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{31}{40}\)

(iii) Let event C denote the event that the engineer lives within \(\frac{1}{2}\) km from her place of work. There is no observation which is \(\frac{1}{2}\) or less than \(\frac{1}{2}\).
Hence, the number of trials when event C occured = 0.
∴ P(C) = \(\frac{\text { No. of trials in which event C occured }}{\text { The total number of trials }}\)
= \(\frac{0}{40}\)
= 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Answer:
Note: Students should do this Activity themselves.

Question 10.
Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her / him is divisible by 3 ? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Answer:
Note: Students should do this Activity themselves.

Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg) :
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer:
The total number of bags = 11.
Hence, the total number of trials = 11.
Let event A denote the event that a bag contains more than 5 kg of flour.
There are 7 bags weighing more than 5 kg.
Their weights (in kg) are 5.05, 5.08, 5.03, 5.06, 5.08, 5.04 and 5.07. Hence, the number of trials when event A occured = 7.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{7}{11}\)

Question 12.
In sum no. 5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Answer:
In sum no. 5, Exercise 14.2, total number of days is 30.
Hence, the total number of trials = 30.
In the table prepared there, we see that the frequency of class 0.12 – 0.16 is 2.
Hence, during 2 days the concentration of sulphur dioxide (in ppm) was in the interval 0.12 – 0.16.
Let event A denote the event that the concentration of sulphur dioxide (in ppm) is in the interval 0.12 – 0.16.
Hence, the number of trials when event A occured = 2.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{2}{30}\)
= \(\frac{1}{15}\)

Question 13.
In sum no. 1, Exercise 14.2, you were asked) to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Answer:
In sum no. 1, Exercise 14.2, the total number of students is 30.
Hence, the total number of trials = 30.
Let event A denote the event that a student has blood group AB. The number of students having blood group AB is 3.
Hence, the number of trials when event A occured = 3.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{3}{30}\)
= \(\frac{1}{10}\)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 14 Statistics MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The marks scored by Kavya in 10 tests of Mathematics are 35, 18, 41, 24, 45, 10, 28, 32, 40, 15. Then, the range of the data is …………….. .
A. 45
B. 10
C. 35
D. 28.8
Answer:
C. 35

Question 2.
The average of the observations 3, 4, 5, 8, 12, 10, 13, 16, 18, 11 is …………………. .
A. 100
B. 10
C. 18
D. 3
Answer:
B. 10

Question 3.
The mean of first five odd natural numbers is ……………….. .
A. 3
B. 5
C. 4
D. 25
Answer:
B. 5

Question 4.
The mean of first four even natural numbers is ……………….. .
A. 5
B. 10
C. 20
D. 4
Answer:
A. 5

Question 5.
The mean of first five prime numbers is
A. 28
B. 2.8
C. 5.6
D. 1.4
Answer:
C. 5.6

Question 6.
If the mean of 2x, 5, 3x, 12, 5x, 17 and 6 is 20, then x = ………………….. .
A. 10
B. 20
C. 15
D. 40
Answer:
A. 10

Question 7.
The mean of the following distribution is ………………. .

A. 3.9
B. 7.8
C. 78
D. 39
Answer:
A. 3.9

Question 8.
If the mean of 12, 13, x, 17, 18 and 20 is 16, then x = ………………. .
A. 8
B. 4
C. 16
D. 32
Answer:
C. 16

Question 9.
For a given frequency distribution, n = 20 and Σf_ix_i = 140, then X̄ = ………………… .
A. 20
B. 14
C. 7
D. 28
Answer:
C. 7

Question 10.
The mean of \(\frac{2}{5},\), \(\frac{5}{7},\), \(\frac{3}{5},\) and \(\frac{2}{7},\) is ……………… .
A. \(\frac{1}{2},\)
B. \(\frac{3}{5},\)
C. \(\frac{5}{7},\)
D. 2
Answer:
A. \(\frac{1}{2},\)

Question 11.
The median of 14, 6, 2, 13, 9, 15 and 12 is …………………. .
A. 12
B. 10
C. 2
D. 15
Answer:
A. 12

Question 12.
The median of 21, 17, 13, 33, 19, 23 is ………………… .
A. 21
B. 20
C. 33
D. 19
Answer:
B. 20

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches:
2, 3. 4, 5, 0. 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Answer:
Here, n = 10.
Mean X̄ = \(\frac{\Sigma x_{i}}{n}\)
= \(\frac{2+3+4+5+0+1+3+3+4+3}{10}\)
= \(\frac{28}{10}\)
= 2.8
Thus, the mean of the given scores is 2.8 goals.

Arranging the observations in the ascending order, we get:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Since n = 10 is an even number, \(\frac{n}{2}\) = 5 and \(\frac{n}{2}\) + 1 = 6.

Median M
= \(\frac{\left(\frac{n}{2}\right) \text { th observation }+\left(\frac{n}{2}+1\right) \text { th observation }}{2}\)
= \(\frac{5 \text { th } \text { observation }+6 \text { th } \text { observation }}{2}\)
= \(\frac{3+3}{2}\) = 3
Thus, the median of the given scores is 3 goals.
In the given data, observation 3 occurs most frequently (4 times). Hence, the mode of the data is 3 goals.

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded :
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Answer:
Here, n = 15.
Mean X̄ = \(\frac{\Sigma x_{i}}{n}\)
= \(\begin{gathered}
41+39+48+52+46+62+54+40 \\
+96+52+98+40+42+52+60 \\
\hline 15
\end{gathered}\)
= \(\frac{822}{15}\) = 54.8
Thus, the mean of the data is 54.8 marks.
Arranging the observations in the ascending order, we get:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Here, n = 15 is an odd number.
Median M = \(\left(\frac{n+1}{2}\right)\)th observation
= \(\left(\frac{15+1}{2}\right)\)th observation
= 8 th observation
= 52
Thus, the median of the data is 52 marks.
In the given data, observation 52 occurs most frequently (3 times). Hence, the mode of the data is 52 marks.

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Answer:
Here, the median = 63 and n = 10.
∴ \(\frac{n}{2}\) = 5 and \(\frac{n}{2}\) + 1 = 6

Median M
= \(\frac{\left(\frac{n}{2}\right) \text { th observation }+\left(\frac{n}{2}+1\right) \text { th observation }}{2}\)
∴ 63 = \(\frac{5 \text { th } \text { observation }+6 \text { th } \text { observation }}{2}\)
∴ 63 = \(\frac{(x)+(x+2)}{2}\)
∴63 × 2 = x + x + 12
∴126 = 2x + 2
∴ 2x = 124
∴ x = 62

Question 4.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Answer:
Here, just by simple observation, it is clearly seen that observation 14 occurs most frequently, i.e., 4 times.
Hence, the mode of the data is 14.

Question 5.
Find the mean salary of 60 workers of a factory from the following table:

Answer:

Mean X̄ = \(\frac{\Sigma f_{i} x_{i}}{n}\)
= \(\) = \(\frac{3,05,000}{60}\) = 5083.33
Thus, the mean salary is ₹ 5083.33.

Question 6.
Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
Answer:
For the students studying in the same class, usually their level of knowledge and understanding would be more or less equal. There would be a few student having this level low and there would be a few students having this level high. Their level of knowledge and understanding would be reflected in the marks scored by them at an examination. Hence, the mean of marks scored by them at an examination is an appropriate measure of central tendency.

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
If we consider the monthly income of the people of certain region, the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %):

Causes	Female fatality rate (%)
1. Reproductive health conditions	31.8
2. Neuropsychiatric conditions	25.4
3. Injuries	12.4
4. Cardiovascular conditions	4.3
5. Respiratory conditions	4.1
6. Other causes	22.0

(i) Represent the information given above graphically.
Answer:

(ii) Which condition is the major cause of women’s ill health and death worldwide?
Answer:
‘Reproductive health conditions’ is the major cause of womens ill health and death worldwide.

(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Answer:
‘Malnutrition’ and ‘Lack of necessary medical facilities’ can be considered as two other factors which play a major role in female fatality.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:

Section	Number of girls per thousand bays
Scheduled Caste (SC)	940
Scheduled Tribe (ST)	970
Non-SC/ST	920
Backward districts	950
Non-backward districts	920
Rural	930
Urban	910

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Answer:

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

(i) Draw a bar graph to represent the polling results.
Answer:
Seats won by different political parties

(ii) Which political party won the maximum number of seats?
Answer:
Political party: A won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	.5
163-171	4
172-180	2

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous.]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
Answer:
Making the class intervals continuous, we get the following table:

Length (in mm)	Number of leaves
117.5-126.5	3
126.5- 135.5	5
135.5-144.5	9
144.5-153.5	12
153.5- 162.5	5
162.5-171.5	4
171.5-180.5	2

(i) Length of leaves in millimetre

(ii) Yes. The given data can also be represented by ‘Frequency polygon’.

(iii) It is not correct to conclude that the maximum number of leaves are 153 mm long, because even if the frequency of class 145-153 is 12, we do not have the information about the length of each of those 12 leaves individually.

Question 5.
The following table gives the life times of 400 neon lamps:

Life time (in hours)	Number of lamps
300 – 400	14
400 – 500	56
500 – 600	60
600 – 700	86
700 – 800	74
800 – 900	62
900 – 1000	48

(i) Represent the given information with the help of a histogram.
Answer:

(ii) How many lamps have a life time of 700 hours or more than 700 hours ?
Answer:
The-frequencies of classes 700-800, 800-900 and 900-1000 are 74, 62 and 48 respectively.
Hence, the life time of 184 (74 + 62 + 48) lamps is 700 hours or more than 700 hours.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Answer:
To draw the frequency polygons of both the sections, we find the class marks of each class and prepare the following tables:

Section A

Marks	Class mark	Frequency
0-10	5	3
10-20	15	9
20-30	25	17
30-40	35	12
40-50	45	9

Section B

Marks	Class mark	Frequency
0-10	5	5
10-20	15	19
20-30	25	15
30-40	35	10
40-50	45	1

Comparing the performance of both the sections from the frequency polygons, we observe that the performance of students of section A is better than the performance of students of section B.

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Number of balls	Team A	Team B
1-6	2	5
7-12	1	6
13-18	8	2
19-24	9	10
25-30	4	5
31-36	5	6
37-42	6	3
43-48	10	4
49-54	6	8
55-60	2	10

Represent the data of both the teams on the same graph by frequency polygons.
[Hint: First make the class intervals continuous.]
Answer:

Number of runs made by Team A and Team B in first 60 balls.

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows:

Age (in years)	Number of children
1-2	5
2-3	3
3-5	6
5-7	12
7-10	9
10-15	10
15-17	4

Draw a histogram to represent the data above.
Answer:

Children of various age groups playing in a park

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of letters	Number of surnames
1-4	6
4-6	30
6-8	44
8-12	16
12-20	4

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Answer:

(i) Information regarding the number of surnames having given number of letters
Answer:

(ii) Write the class interval in which the maximum number of surnames lie.
Answer:
The maximum number of surnames lie in the class interval 6-8.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows : Height =110 cm, Depth = 25 cm, Breadth = 85 cm (see the given figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.

Answer:
Outer faces to be polished:

• One face on back side of the bookshelf, measuring 110 cm × 85 cm.
• Two faces on the sides, each of those measuring 110 cm × 25 cm.
• The top and the base, each of those measuring 85 cm × 25 cm.
• Two vertical strips on the front side, each of those measuring 110 cm × 5 cm.
• Four horizontal strips on the front side, each of those measuring 75 cm × 5 cm.

Thus, total area of region to be polished
= [(110 × 85) + 2(110 × 25) + 2 (85 × 25) + 2(110 × 5) + 4(75 × 5)] cm²
= (9350 + 5500 + 4250 + 1100+ 1500) cm²
= 21700 cm²
20 paise per cm² = ₹ 0.20 per cm²
Cost of polishing 1 cm2 region = ₹ 0.20
∴ Cost of polishing 21700 cm² region
= ₹ (21700 × 0.20)
= ₹ 4340

Inner faces to be painted:

• Two faces on the sides each of those measuring 90 cm × 20 cm.
• Two faces each of two shelves, the top face and the bottom face, in all six face, each of those measuring 75 cm × 20 cm.
• Face on the back side, measuring 90 cm × 75 cm.

Thus, total area of the region to be painted
= [2 (90 × 20) + 6 (75 × 20) + (90 × 75)] cm²
= (3600 + 9000 + 6750) cm²
= 19350 cm²
10 paise per cm² = ₹0.10 per cm²
Cost of painting 1 cm² region = ₹ 0.10
∴ Cost of painting 19350 cm² region = ₹ (19350 × 0.10) = ₹ 1935
Then, the total expense of polishing and painting = ₹ 4340 + ₹ 1935 = ₹ 6275

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be ‘ painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Answer:
For each wooden sphere,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{21}{2}\) cm
Curved surface area of 1 sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm²
= 1386 cm²
For each cylindrical support, radius r = 1.5 cm and height h = 7 cm.
Area of top of cylindrical support
= πr²
= \(\frac{22}{7}\) × 1.5 × 1.5 cm²
= 7.07 cm² (approx.)
Hence, the area of each sphere to be painted silver = 1386 cm² – 7.07 cm² = 1378.93 cm²
∴ Total area of eight spheres to be painted silver = 1378.93 cm² × 8 = 11031.44 cm²
25 paise per cm² = ₹ 0.25 per cm²
Cost of painting silver in 1 cm² region = ₹ 0.25
∴ Cost of painting silver in 11031.44 cm² region
= ₹ (11031.44 x 0.25)
= ₹ 2757.86 (approx.)
Curved surface area of 1 cylindrical support
= 2πrh
= 2 × \(\frac{22}{7}\) × 1.5 × 7 cm
= 66 cm²
∴ Total area of eight cylindrical supports to be painted black = 66 cm² × 8 = 528 cm²
5 paise per cm² = ₹ 0.05 per cm²
Cost of painting black in 1 cm² region = ₹ 0.05
∴ Cost of painting black in 528 cm² region = ₹ (528 × 0.05)
= ₹ 26.40
Thus, the total cost of painting = ₹ 2757.86 + ₹ 26.40
= ₹ 2784.26 (approx.)

Question 3.
The diameter of a sphere is decreased by 25 %. By what per cent does its curved surface area decrease?
Answer:
Suppose, the initial diameter of the sphere is d units and radius is r units.
∴ d = 2r
Original curved surface area of the sphere
= 4πr²
= π (4r²)
= π (2r)²
= πd² unit²
Now, the diameter of the sphere is reduced by 25 %. Hence, the new diameter of the sphere is 0.75d units.
New curved surface area of the sphere
= π (diameter)
= π (0.75d)² unit²
= 0.5625 πd² unit²
∴ The decrease in the curved surface area of the sphere = πd² – 0.5625 πd²
= 0.4375 πd² unit²
∴Percentage decrease in the curved surface area of the sphere = \(\frac{0.4375 \pi d^{2}}{\pi d^{2}}\) × 100 = 43.75 %
Thus, when the diameter of a sphere is decreased by 25 %, its curved surface area decreases by 43.75 %.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students ?
Answer:
Frequency distribution table

From the frequency distribution table, it is very clear that the most common blood group is O and the rarest blood group is AB.

Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Answer:
Grouped frequency distribution table

From the frequency distribution table, we can conclude that for the majority of engineers, s i.e., 31 engineers, the distance from their residence to their place to work is 5 km or more than 5 km but less than 20 km. For some engineers, i.e., 5 engineers, this distance is less than 5 km. Still, for some engineers, i.e., 4 engineers, this distance is 20 km or more than 20 km but less than 35 km.

Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows:

(i) Construct a grouped frequency distribution table with classes 84-86, 86 – 88, etc.
Answer:

(ii) Which month or season do you think this data is about ?
Answer:
During 24 days out of 30 days, the relative humidity is 92 % or more than 92 %. This suggests that the data must have been collected during Monsoon.

(iii) What is the range of this data ?
Answer:
Range of the data
= The greatest observation – The least observation
= 99.2 – 84.9
= 14.3

Question 4.
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:

(i) Represent the data given above by grouped frequency distribution table, taking the class intervals as 160 – 165, 165-170, etc.
Answer:
Grouped frequency distribution table

(ii) What can you conclude about their heights from the table?
Answer:
From the above frequency distribution, we can conclude that the height of 70 % students (35 students) is less than 165 cm while the height of only 10 % students (5 students) is 170 cm or more than that.

Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows :

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00-0.04, 0.04-0.08, and so on.
Answer:

(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
Answer:
The concentration of sulphur dioxide was more than 0.11 ppm for 8 days (2 + 4 + 2).

Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:

Prepare a frequency distribution table for the data given above.
Answer:

Question 7.
The value of π up to 50 decimal places is given below:
3.1415926535897932384626433832795028 8419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
Answer:
Frequency distribution table

(ii) Which are the most and the least frequently occurring digits?
Answer:
The most frequently occurring digits are 3 and 9 (8 times each) and the least occurring digit is 0 (2 times).

Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
Answer:

(ii) How many children watched television for 15 or more hours a week?
Answer:
Two children watched television for 15 or more hours a week.

Question 9.
A company manufactures car batteries of a Grouped frequency distribution table particular type. The lives (in years) of 40 such batteries were recorded as follows

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Answer:

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1

Question 1.
Give five examples of data that you can collect from your day-to-day life.
Answer:
Five examples of data that can be collected from day-to-day life can be given as below:

1. Election results obtained from newspapers or TV
2. The number of different kinds of trees grown in our school.
3. Amounts of invoices of electricity for last one year at our home.
4. The number of students studying in different standards of our school.
5. Percentage of marks scored at last examination by the students in our class.

Question 2.
Classify the data in Q. 1 above as primary or secondary data.
Answer:
Among the five data given as the answer to Q. 1, data no.

Primary data:

• The number of different kinds of trees grown in our school.
• Amounts of invoices of electricity for last one year at our home.
• Percentage of marks scored at last examination by the students in our class.
• primary data which we can collect ourselves.

Secondary data:

• Election results obtained from newspapers or TV
• The number of students studying in different standards of our school.
• Secondary data as they are received from the sources of the newspapers or TV or the office of our school.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 13 Surface Areas and Volumes MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The total surface area of a cuboid with length 20 cm, breadth 15 cm and height 10 cm is
A. 1300
B. 650
C. 3000
D. 1500
Answer:
A. 1300

Question 2.
The lateral surface area of a cuboid with length 15 cm, breadth 8 cm and height 5 cm is ………………. cm².
A. 115
B. 230
C. 600
D. 300
Answer:
B. 230

Question 3.
The diameter of a cylinder is 7 cm and its curved surface area is 220 cm². Then, its height is ……………….. cm.
A. 35
B. 10
C. 44
D. 20
Answer:
B. 10

Question 4.
The total surface area of a closed cylinder with radius 3.5 cm and height 6.5 cm is ………………… cm².
A. 110
B. 220
C. 330
D. 440
Answer:
B. 220

Question 5.
The curved surface area of a cone is 880 cm². If its slant height is 20 cm, then its diameter is …………………… cm.
A. 14
B. 7
C. 3.5
D. 28
Answer:
D. 28

Question 6.
The curved surface area of a cone with diameter 14 cm and slant height 10 cm is ………………. cm².
A. 220
B. 1540
C. 110
D. 440
Answer:
A. 220

Question 7.
The height of a cone is 24 cm and its slant height is 25 cm. Then, its diameter is ………………. cm.
A. 14
B. 7
C. 4
D. 49
Answer:
A. 14

Question 8.
The circumference of the base of a cone is 44 cm and its slant height is 15 cm. Then, its curved surface area is ……………….. cm².
A. 14
B. 154
C. 330
D. 115
Answer:
C. 330

Question 9.
The diameter of a cone is 7 cm and its slant ! height is 16.5 cm. Then, its total surface area is …………………. cm².
A. 110
B. 220
C. 105
D. 154
Answer:
B. 220

Question 10.
Total surface area of a hemisphere with radius 7 cm is ………………….. cm².
A. 231
B. 115.5
C. 462
D. 154
Answer:
C. 462

Question 11.
Total surface area of a hemisphere is 72 cm².
Then, its curved surface area is ………………….. cm².
A. 24
B. 36
C. 48
D. 72
Answer:
C. 48

Question 12.
The surface area of a sphere is 616 cm².
Then, its radius is ……………. cm.
A. 6
B. 8
C. 7
D. 14
Answer:
C. 7

Question 13.
In a cuboid, the area of the face with sides length and breadth is 120 cm². If the height of the cuboid is 5 cm, then its volume is …………………… cm³
A. 120
B. 240
C. 600
D. 300
Answer:
C. 600

Question 14.
The volume of a cylinder is 2200 cm³ and its height is 7 cm. Then, the radius of the cylinder is ……………….. cm.
A. 5
B. 15
C. 10
D. 20
Answer:
C. 10

Question 15.
The radius and height of a cone are 7 cm and 3 cm respectively. Then, the volume of the cone is ……………… cm³.
A. 154
B. 168
C. 148
D. 462
Answer:
A. 154

Question 16.
The volume of a sphere is 4.5 π cm³. Then, its diameter is …………………. cm.
A. 3
B. 2
C. 1.5
D. 4
Answer:
A. 3

Question 17.
The ratio of radii of two cones is 2 : 3 and the ratio of their heights is 9:4. Then, the ratio of their volumes is
A. 1 : 1
B. 3 : 2
C. 1 : 3
D. 2 : 3
Answer:
A. 1 : 1

Question 18.
The circumference of the base of a cone is 44 cm and its height is 6 cm. Then, its volume is ………………… cm³.
A. 49
B. 98
C. 308
D. 154
Answer:
C. 308