Class 6

Practicing with SCERT Kerala Syllabus 6th Standard Hindi Textbook Solutions देशभक्ति गीत Kavita Summary in Malayalam & Hindi improves language skills.

Desh Bhakti Kavita Class 6 Summary

देशभक्ति गीत Kavita Summary in Malayalam & Hindi

Desh Bhakti Kavita Summary in Malayalam &Hindi

देश भक्ति गीत (ദേശദകതിഗാനം)
1962 के भारत – चीन युद्ध के दौरान शहीद हुए सैनिकों की श्रद्धांजलि में कवि प्रदीप ने ‘ए मेरे वतन के लोगों’ नामक देश भक्ति गीत लिखा है ।
1962 ലെ ഭാരത – ചൈന യുദ്ധത്തിൽ രക്തസാക്ഷിത്വം വരിച്ച വീരസൈനികർക്ക് ആദരാഞ്ജലി അർപ്പിച്ചു കൊണ്ട് കവി പ്രദീപ് എഴുതിയ ഒരു ദേശഭക്തിഗാനം ആണ് ‘എ മേരേ വതൻ കേ ലോഗോം’.

लता मंगेशकर द्वारा गाए इस गीत का तत्कालीन प्रधानमंत्री जवाहरलाल नेहरु की उपस्थिति में २६ जनवरी १९६३ को दिल्ली के रामलीला मैदान में सीधा प्रसारण किया गया ।
ലതാ മങ്കേഷ് കറിനാൽ ആലപിക്കപ്പെട്ട ഈ ഗീതം അന്നത്തെ പ്രധാന മന്ത്രി ആയിരുന്ന ജവഹർലാൽ നെഹ് റുവിൻ്റെ അദ്ധ്യക്ഷതയിൽ 26 ജനുവരി 1963 ൽ ഡൽഹിയിലെ രാംലീല മൈതാനത്തിൽ നിന്നാണ് പ്രക്ഷേപണം ചെയ്തത്.

ऐ मेरे वतन के लोगो,
तुम खूब लगा लो नारा ।
ये शुभदिन है, हम सब का
लहरा लो तिरंगा प्यारा ।
पर मत भूलो सीमा पर,
वीरों ने हैं प्राण गवाए,
कुछ याद उन्हें भी कर लो,
जो लौट के घर न आए।

എന്റെ രാജ്യത്തെ ജനങ്ങളെ
നിങ്ങൾ നന്നായി മുദ്രാവാക്യം മുഴക്കു
ഇത് നമ്മുടെ ശുഭദിനമാണ്
ത്രിവർണ്ണ പതാക വീശു
എന്നാൽ നമ്മുടെ അതിർത്തിയിൽ
വീരമൃത്യു വരിച്ച വീരന്മാരായ ജവാന്മാരെ മറക്കരുത്
വീട്ടിൽ തിരികെ എത്താത്ത അവരെയും കൂടി ഈ
സമയത്ത് നിങ്ങൾ ഓർമിക്കു

ऐ मेरे वतन के लोगो,
ज़रा आँख में भर लो पानी,
जो शहीद हुए हैं उनकी.
ज़रा याद करो कुर्बानी ।
जब घायल हुआ हिमालय,
खतरे में पडी आज़ादी
जब तक थी साँस लजे वो,
फिर अपनी लाश बिछा दी ।
संगी पे घर कर माथा,
सो गए अमर बलिदानी ।

എന്റെ രാജ്യത്തെ ജനങ്ങളെ
നിങ്ങൾ കുറച്ച് കണ്ണുനീർ പൊഴിക്കു
നമുക്ക് വേണ്ടി വീരമൃത്യു വരിച്ചവരെയും അവരുടെ
ത്യാഗത്തെപ്പറ്റിയും ചിന്തിക്കൂ.
ഹിമാലയത്തിൽ പരിക്കേറ്റിട്ടും സ്വാതന്ത്ര്യം അപകടത്തിലായപ്പോൾ അവസാന ശ്വാസം വര നമുക്ക് വേണ്ടി പോരാടിയതാണവർ, എന്നിട്ടും അവർ മരണം വരിച്ചു. തോക്കിൻ കുന്തത്തിൻ മേൽ നെറ്റി ചായിച്ച്, നിദ്രയിലാണ്ടവർ, അവരുടെ അനശ്വരമായ ത്യാഗത്തെപ്പറ്റി ഓർക്കൂ.

जब देश में भी दीवाली,
वो खेल रहे थे होली ।
जब हम बैठे थे घरों में,
वो झेल रहे थे गोली ।
थे धन्य जवान वो अपने,
भी धन्य वो उनकी जवानी ।

ദീപാവലിയും ഹോളിയുമൊക്ക നമ്മുടെ വീട്ടിൽ ഇരുന്ന് നമ്മൾ ആഘോഷിക്കുമ്പോൾ നമുക്കായി വെടിയുണ്ടകളെ ചെറുത്ത് പോരാടുകയായിരുന്നു അവർ, അനുഗ്രഹീതരായ ജവാൻമാർ, അവരുടെ യൗവനവും അനുഗ്രഹീതമാണ്.

कोई सिख कोई जाट मराठा
कोई गुरखा कोई मद्रासी,
सरहद पर मरनेवाला,
हर वीर था भारतवासी ।
जो खून गिरा पर्वत पर,
वो खून था हिंदुस्तानी

ചിലർ സിക്കുകാരാണ്, ചിലർ ജാറ്, ചിലർ മറാഠികളാണ്. ചിലർ ഗൂർഖകൾ, ചിലർ മദ്രാസികൾ, വിഭിന്ന സമുദായത്തിലുള്ളവർ, അതിർത്തിയിൽ മരിക്കുന്ന ഓരോ ജവാനും ഇവരെല്ലാം ഭാരതവാസികളാണ്. പർവ്വതങ്ങളിൽ ചൊരിഞ്ഞ രക്തം ഭാരതത്തിന്റേതാണ്.

भी खून से लथपथ काया
फिर भी बंदूक उठाके,
दस दस को एक ने मारा,
फिर गिर गए होरा गँवा के
जब अंत समय आया तो,
कहा गए कि अब मरते हैं
खूब कहना देश के प्यारो
अब हम तो सफर करते हैं।
क्या लोग थे वो दीवाने,
क्या लोग थे वो अभिमानी
जय हिंद, जय हिंद की सेना

പരിക്കേറ്റ് രക്തത്തിൽ കുതിർന്ന ശരീരമാണെങ്കിലും തോക്കെടുത്ത് ഒരാൾ തന്നെ പത്ത് വീതം ആൾക്കാരെ കൊന്നൊടുക്കുന്നു. പിന്നീട് അവർ അന്ത്യശ്വാസമെടുക്കുന്നു. അന്ത്യശ്വാസമെടുക്കുന്ന സമയം വന്നപ്പോൾ അവർ പറഞ്ഞു. ഇനി നമുക്ക് മരിക്കാം. ദേശത്തിലെ പ്രിയപ്പെട്ടവരെ സന്തോഷമായിരിക്കു, ഞങ്ങൾ യാത്രയാകുകയാണ്. ദേശസ്നേഹികളും അഭിമാനികളുമായ വീരജവാൻമാർ, ജയ്ഹിന്ദ്

देशभक्ति गीत शब्दार्ध

• नारा – മുദ്രാവാക്യം slogan
• लहराना – വീശുക fluctuate
• लौटना – മടങ്ങിവരിക return
• ടി – രക്തസാക്ഷി martyr
• कुर्बानी – ത്രാഗം sacrifice
• घायल – പരിക്കേൽക്കുക wound
• लाश – മൃതദേഹം – dead body
• झेलना – ചെറുക്കുക – to bear
• खतरा – അപകടം – disaster
• खून – രക്തം – blood
• गँवाई – ത്വജിച്ചു – testimony
• माथा – നെറ്റി – forehead
• संगीन – ബയോണറ്റ് – bayonet
• प्यारा – പ്രിയപ്പെട്ട – lovely
• धन्य – അനുഗ്രഹീതർ blessed
• जवानी – യൗവനം – youth
• काया – ശരീരം body

Practicing with SCERT Kerala Syllabus 6th Standard Hindi Textbook Solutions Unit 2 Chapter 2 हिंदुस्तान हमारा Question Answer Notes Summary in Malayalam & Hindi improves language skills.

Hindustan Hamara Class 6 Question Answer Notes Summary

SCERT Class 6 Hindi Unit 2 Chapter 2 Question Answer Kerala Syllabus हिंदुस्तान हमारा

Hindustan Hamara Question Answer

प्रश्न 1.
लेख पढ लिया है न? अब ये वाक्य पढें।
ലേഖനം വായിച്ചല്ലോ; ഇനി ഈ വാക്വം വായിക്കാം

‘घने पेडों से सजे पर्वत ।’
‘घने पेडों से सजे’ ऐसे प्रयोग वाक्य की शोभा बढाती है।
വൃക്ഷങ്ങൾ തിങ്ങി നിൽക്കുന്ന പർവ്വതം/തിങ്ങി നിറഞ്ഞ
വൃക്ഷങ്ങളാൽ അലങ്കരിക്കപ്പെട്ട പർവ്വതം.
തിങ്ങി നിറഞ്ഞ വൃക്ഷങ്ങളാൽ അലങ്കരിക്കപ്പെട്ട – ഈ പ്രയോഗം വാക്യത്തിന്റെ ശോഭ വർദ്ധിപ്പിക്കുന്നു.

लेख से ऐसे वाक्यों को चुनकर लिखें।
ലേഖനത്തിൽ നിന്ന് ഇങ്ങനെയുള്ള വാക്യങ്ങൾ തിരഞ്ഞെടുത്ത് എഴുതാം.
उत्तर :
खिल-खिलाकर बहती नदियाँ ।
काली घटा से अपने बालों को खोलकर झूमती बारिश |
हम घाँघरा पहनकर नाचते लहलहाते खेत ।

प्रश्न 2.
तालिका की पूर्ती करें (താഴെ കൊടുത്തിരിക്കുന്ന കോളം പൂർത്തിയാക്കുക)

उत्तर :
खिल-खिलाकर बहती नदियाँ – सुंदर नदियाँ
हरा घाँघरा पहनकर नाचते – हरा खेत
लह-लहाते खेत
घने पेडों से सजे पर्वत – घने जंगल

प्रश्न 3.
वर्णों से खेलें (അകയരങ്ങൾ കൊണ്ട് കളിക്കുക

यहाँ ‘ताल’ शब्द से शुरू करके केवल एक अक्षर को बदलते हुए हम ‘गीत’ शब्द तक पहूँचे हैं। इसी तरह ‘केरल’ शब्द से शुरू करके ‘भारत’ शब्द तक पहूँचना है। शर्त यह है कि एक बार केवल एक ही अक्षर बदलें और प्रत्येक शब्द सार्थक हो ।

ഇവിടെ ‘ताल’ എന്ന് വാക്കിൽ തുടങ്ങി കേവലം ഒരു അക്ഷരം മാത്രം മാറ്റി നമ്മൾ ‘गीत’ എന്ന വാക്കിലാണ് എത്തിച്ചേർന്നിരിക്കുന്നത്. ഇതുപോലെ ‘केरल’ എന്ന വാക്കിൽ തുടങ്ങി ‘भारत’ എന്ന വാക്കിൽ എത്തിച്ചേരുക. നിയമം ഇതാണ്, ഒരു പ്രാവശ്യം കേവലം ഒരു അക്ഷരം മാത്രം മാറ്റുക. അതുപോലെ തന്നെ ഓരോ വാക്കും അർത്ഥമുള്ളതായിരിക്കണം.

उत्तर :
केरल → सरल → सरस → नीरस → विरस → विरह → विरल → तरल → तरह → शस्त → भारत
अथवा / or
केरल → लाल → लड़का → काल → रंग → गाना → नदी → दर्द → दीवार → रेल → लाभ → भारत

प्रश्न 4.
ये नारे गीत पढें ।
ഈ മുദ്രാവാക്വം ഗീതം വായിക്കാം

“सत्य अहिंसा का रखवारा “भारत हे देश हमारा भारत मेरा प्यारा – प्यारा ।” फहराएंगे तिरंगा प्यारा।”
और भी है भारत की खूबियाँ
ഇനിയുമുണ്ട് ഭാരതത്തിന്റെ ഗുണമേന്മകൾ

आप भी लिखें नारागीत
നിങ്ങളും എഴുതു ഒരു മുദ്രാവാകഗീതം

मेरा नारा गीत (എന്റെ മുദ്രാവാക്യ ഗീതം)

“चमके जैसे जगमग तारा
भारत मेरा जान से प्यारा । ”
ഒരു നക്ഷത്രം പോലെ മിന്നി തിളങ്ങി ശോഭിക്കുന്ന ഭാരതം എന്റെ ജിവനേക്കാൾ പ്രിയപ്പെട്ടതാണ്.

हिंदुस्तान हमारा अन्य महत्वपूर्ण परीक्षा केंद्रित प्रश्नोत्तर

प्रश्न 1.
भारत की खूबसूरत क्या-क्या है ?
ഭാരതത്തിന്റെ മനോഹാരിത എന്തെല്ലാമാണ് ?
उत्तर :
भौगोलिक झलक
മനോഹരമായ ഭൂപ്രകൃതി

समय-समय पर रंग बदलते अनोखे मौसम
സമയാസമയങ്ങളിൽ മാറി വരുന്ന അതുല്യമായ കാലാവസ്ഥ

शान बनी संपन्न संस्कृति
അഭിമാനം തരുന്ന സമ്പന്നത

प्रश्न 2.
भारत की अपनी विशेषताएँ क्या-क्या है?
ഭാരതത്തിന്റെ വിശേഷതകൾ എന്തൊക്കെയാണ്?
उत्तर :
घने पेडों से सजे पर्वत
വൃക്ഷങ്ങൾ തിങ്ങി നിൽക്കുന്ന പർവ്വതങ്ങൾ

खिल – खिलाकर बहती नदियाँ
കളകളാരവം പാടി ഒഴുകുന്ന നദികൾ

काली घटा से से अपने बालों को खोलकर झूमती बारिश
കറുത്ത മേഘങ്ങളിൽ നിന്ന് ഇടതൂർന്ന് മുടിയഴിച്ചിട്ടതുപോലെ ആടിയുലഞ്ഞു പെയ്യുന്ന മഴ

हरा घाँघरा पहनकर नाचते लह लहाते खेत
പച്ചപ്പാവാടയണിഞ്ഞതു പോലെ പച്ചപ്പ് നിറഞ്ഞു നിൽക്കുന്ന വയലുകൾ

प्रश्न 3.
भारत के छह ऋतु कौन-कौन है ?
ഭാരതത്തിന്റെ ആറ് ഋതുക്കൾ ഏതെല്ലാമാണ്?
उत्तर :
वसंत ऋतु (വസന്തകാലം, Spring Season)
ग्रीष्म ऋतु (വേനൽക്കാാലം, Summer Season)
वर्ष ऋतु (മഴയകാലം, rainy season)
शरद ऋतु (ശരത്കാലം, autumn season)
हेमंत ऋतु (ഹോന്തകാലം, pre-winter season)
शिशिर ऋतु (തണുപ്പുകാലം, winter season)

प्रश्न 4.
विख्यात गीत ‘ सारे जहाँ से अच्छा ………….. ‘ का गीतकार कौन है ?
പ്രസിദ്ധമായ പാട്ട് ‘സാരേ ജഹാം സേ അച്ചാ’ എഴുതിയത് ആരാണ് ?
उत्तर :
मशहूर शायर मुहम्मद इकबाल

प्रश्न 5.
उत्तर चुनकर लिखो ഉത്തരം തിരഞ്ഞെടുത്തെഴുതുക
i) कवि के अनुसार हमारा देश कैसे चमके ?
കവിയുടെ അഭിപ്രായമനുസരിച്ച് നമ്മുടെ ദേശം എങ്ങനെ ആണ് ശോഭിക്കുന്നത്?

(तारा जैसे / सूरज जैसे )
(നക്ഷത്രം പോലെ / സൂര്യൻ പോലെ)

ii) हमारा देश किसका रखवारा है ?
നമ്മുടെ ദേശം എന്തിന്റെ സൂക്ഷിപ്പുകാരൻ ആണ്?

(सत्य, अहिंसा का / झूठ – कपट का )
(സത്യത്തിന്റെയും അഹിംസയുടെയും കള്ളത്തിന്റെയും കാപട്യത്തിന്റെയും)

iii) हमारे देश को किसने संवारा?
നമ്മുടെ രാജ്യത്തെ അലങ്കരിക്കുന്നതാര്?
(बलिदानों ने / फूलदानों ने)
(യാഗങ്ങൾ / പുഷ്പപാത്രങ്ങൾ വഴി)
उत्तर :
i) तारा जैसे
ii) सत्य, अहिंसा का
iii) बलिदानों ने

प्रश्न 6.
भारत की खूबियों पर पदसूर्य पढकर एर लेख तैयार कीजिए:
ഭാരതത്തിന്റെ ഗുണമേന്മകളെപ്പറ്റി പദസൂര്യൻ വായിച്ച് ഒരു ലേഖനം തയ്യാറാക്കുക.

उत्तर :
भारत में कई राज्य है । यहाँ लोगों की वेश-भूषाएँ भिन्न-भिन्न है। भारत में अनेक त्योहार है । यहाँ अनेक नदियाँ और हरे भरे जंगल है। यहाँ भाषाएँ भिन्न- भिन्न है। सचमुच में भारत एक विशाल देश है

ഭാരതത്തിൽ അനേകം സംസ്ഥാനങ്ങളുണ്ട്. ഇവിടെ ആളുകളുടെ വസ്ത്രധാരണം വ്യത്യസ്തമാണ്. ഭാരതത്തിൽ ഒരുപാട് ഉത്സവങ്ങൾ ഉണ്ട്. ഇവിടെ അനേകം നദികളും പച്ചപ്പ് നിറഞ്ഞ് നിൽക്കുന്ന വനങ്ങളും ഉണ്ട്. ഇവിടെ ഭാഷകൾ വ്യത്യസ്തമാണ്. വാസ്തവത്തിൽ ഭാരതം ഒരു വിശാലമായ രാജ്യമാണ്.

प्रश्न 7.
कोष्ठक से उचित शब्द चुनकर पूरा कीजिए
(ബ്രായ്ക്കറ്റിൽ നിന്ന് ഉചിതമായ വാക്ക് തിരഞ്ഞെടുത്ത് പുരിപ്പിക്കുക.
(कला, केरल, कृषि, खेल-कूद, जगमगाते तारे, साहित्य, संगीत, भारत, जगमगाता : तारा)
__________ जगमगाता तारा है। दुनिया के देशों के बीच __________ के समान भारत चमकता है। विज्ञान, __________, __________, __________, __________ आदि क्षेत्रों में भारत दुनिया के देशों की तुलना में आगे हैं । इसलिए भारत को __________ कहा है।
उत्तर :
भारत जगमगाता तारा है । दुनिया के देशों के बीच जगमगाते तारे के समान भारत चमकता है। विज्ञान, कला, कृषि, खेलकूद, साहित्य, आदि क्षेत्रों में भारत दुनिया के देशों की तुलना में आगे हैं। इसलिए- भारत को जगमगाता तारा : कहा है ।
ഭാരതം ഒരു തിളങ്ങുന്ന നക്ഷത്രമാണ്. ലോകരാജ്യങ്ങൾക്കിടയിൽ ഒരു തിളങ്ങുന്ന നക്ഷത്രം പോലെ ഭാരതം ശോഭിക്കുന്നു. ശാസ്ത്രം, കലാരംഗം, കൃഷി, കായികരംഗം, സാഹിത്യം തുടങ്ങിയ മേഖലകളിലെല്ലാം ഭാരതം ലോകത്തിലെ മറ്റു രാജ്യങ്ങളുമായി താരതമ്യം ചെയ്യുമ്പോൾ ഏറെ മുന്നിലാണ്. അതുകൊണ്ട് ഭാരതത്തെ തിളങ്ങുന്ന നക്ഷത്രം എന്ന് വിളിക്കുന്നു.

प्रश्न 8.
लिखें सही या गलत (ശരിയോ തെറ്റോ എഴുതുക)
क) हमारा देश हमें अपनी जान से प्यारा है।
നമ്മുടെ രാജ്യം നമുക്ക് സ്വന്തം ജീവനേക്കാളേറെ പ്രിയപ്പെട്ടതാണ്.

ख) सागर ने भारत का पाँव पसारा ।
സമുദ്രം ഭാരതത്തിലേക്ക് കാല് നീട്ടി വച്ചിരിക്കുന്നു.

ग) भारत हमेशा असत्य की जय का संदेश देता है।
ഭാരതം എപ്പോഴും അസത്വം ജയിക്കട്ടെ എന്ന സന്ദേശമാണ്

घ) जगमग तारा चैना है।
തിളങ്ങുന്ന നക്ഷത്രം ചൈനയാണ്

ङ) भारत सत्य और अहिंसा का रखवारा है ।
ഭാരതം സത്വത്തിന്റെയും അഹിംസയുടെയും സൂക്ഷിപ്പുകാരൻ ആണ്.
उत्तर :
(क) सही
(ख) सही
(ग) गलत
(घ) गलत
(ङ) सही

प्रश्न 9.
सही मिलान करके लिखें (ശരിയായ പൊരുത്തം എഴുതുക)

1) भारत	भारत चमकता है।
2) जगमगाते तारे के समान	रखवारा है।
3) सत्य और अहिंसा का	अपनी जान से अधिक प्यार करते हैं ।
4) बलिदानों ने	हमारा देश है।
5) भारतीय भारत को	भारत के पाँव के समान फैला पडा है।
6) दक्षिण में हिन्द महासागर	भारत को संवारा है।

उत्तर :

1) भारत	हमारा देश है।
2) जगमगाते तारे के समान	भारत चमकता है।
3) सत्य और अहिंसा का	रखवारा है।
4) बलिदानों ने	भारत को संवारा है।
5) भारतीय भारत को	अपनी जान से अधिक प्यार करते हैं।
6) दक्षिण में हिन्द महासागर	भारत के पाँव के समान फैला पडा है ।

प्रश्न 10.

i) यह चित्र देखो, इसमें क्या-क्या हैं? लिखें
ഈ ചിത്രം നോക്കൂ, ഇതിൽ നാം എന്തെല്ലാം കാണുന്നു?

ii) रात के समय आकाश में कौन-कौन है ?
രാത്രി സമയം ആകാശത്ത് ആരൊക്കെയാണുള്ളത്?

iii) कौन हमें दिन में प्रकाश देता है ?
ആരാണ് പകൽ നമുക്ക് പ്രകാശം നൽകുന്നത്?
उत्तर :
(i) चित्र में सूरज, चाँद, तारे और मेघ हैं ।
ചിത്രത്തിൽ സൂര്യനും, ചന്ദ്രനും, നക്ഷത്രങ്ങളും മേഘങ്ങളും ഉണ്ട്.

ii) रात के समय आकाश में चाँद और तारे हैं।
രാത്രി സമയം ആകാശത്ത് ചന്ദ്രനും നക്ഷത്രങ്ങളും ഉണ്ട്

iii) सूरज हमें दिन में प्रकाश देता है।
സൂര്യൻ നമുക്ക് പകൽ പ്രകാശം നൽകുന്നു.

प्रश्न 11.
वाक्यांशों को मिलाएँ और नारागीत बनाएँ ।
വാക്യങ്ങൾ കൂട്ടി യോജിപ്പിച്ച് മുദ്രാവാക്യഗീതം ഉണ്ടാക്കാം

1) प्यारा भारत देश हमारा	भारत देश मेरा प्यारा
2) हिंद देश का प्यारा झंडा	हमें हैं अपनी जान से प्यारा
3) दुनिया में है सबसे सुंदर	ऊँचा सदा रहेगा झंडा

उत्तर :

1) प्यारा भारत देश हमारा	ऊँचा सदा रहेगा झंडा
2) हिंद देश का प्यारा झंडा	भारत देश मेरा प्यारा
3) दुनिया में है सबसे सुंदर	भारत देश मेरा प्यारा

प्रश्न 12.
नमूना के अनुसार विशेषण शब्द जोड़कर लिखिए
മാതൃക അനുസരിച്ച് വിശേഷണ വാക്കുകൾ ചേർത്തെഴുതുക.
खेत
लहलहाते खेत
नाचते लह
लहाते खेत
घाँघरा पहनकर नाचते लह
लहाते खेत
हरा घाँघरा पहनकर नाचते लह- लहाते खेत

(i) समय – समय पर रंग बदलते अनोखे मौसम
उत्तर :
मौसम
अनोखे मौसम
रंग बदलते अनोखे मौसम
समय-समय पर रंग बदलते अनोखे मैसम

(ii) घने पेडों से सजे पर्वत
उत्तर :
पर्वत
सजे पर्वत
घने पेडों से सजे पर्वत

प्रश्न 13.
निम्नलिखित गद्यांश पढकर नीचे दिए गए प्रश्नों को उत्तर दीजिए ।
താഴെ തന്നിരിക്കുന്ന ഗദ്യഭാഗം വായിച്ചിട്ട് താഴെ തന്നിരിക്കുന്ന ചോദ്യങ്ങൾക്ക് ഉത്തരം എഴുതുക. मैं कबूतर हूँ। मैं शांति का प्रतीक हूँ। मुझे कपोत भी कहते हैं। पुराने ज़माने में राजमहलों में भी मेरा स्थान था। मैं संदेश लेकर दूर देशों में जाता था । मैं बिल्ली से डरता हूँ। मुझे गेहूँ का दाना बहुत पसंद है। मैं मानव का दोस्त हूँ ।
ഞാനൊരു പ്രാവ് ആണ്. ഞാൻ ശാന്തിയുടെ പ്രതീകമാണ്. എന്നെ കപോത് എന്നും വിളിക്കുന്നു. പണ്ടു കാലത്ത് രാജകൊട്ടാരങ്ങളിലും എനിക്ക് സ്ഥാനമുണ്ടായിരുന്നു. ഞാൻ സന്ദേശവുമായി ദൂരെ ദേശങ്ങളിൽ പോകാറുണ്ട്. ഞാൻ പൂച്ചയെ പേടിക്കുന്നു. എനിക്ക് ഗോതമ്പു മണി

വളരെ ഇഷ്ടമാണ്. ഞാൻ മനുഷ്യന്റെ അടുത്ത സുഹൃത്താണ്.
i) कबूतर का दूसरा शब्द क्या है ?
പ്രാവിന്റെ മറ്റൊരു പേരെന്താണ്?
(कोयल, कपोत, कौआ)

ii) सही प्रस्ताव चुनकर लिखिए ।
ശരിയായ വാക്വം തിരഞ്ഞെടുത്തെഴുതുക.

(क) कबूतर मानव से डरता है।
പ്രാവ് മനുഷ്യനെ പേടിക്കുന്നു

(ख) कबूतर बिल्ली से डरता है ।
പ്രാവ് പൂച്ചയെ പേടിക്കുന്നു.

iii) कबूतर किसका प्रतीक है?
പ്രാവ് എന്തിന്റെ പ്രതീകമാണ്.
(क्रोध, दुख, शांति)

iv) कबूतर को कौन-सा दाना बहुत पसंद है?
പ്രാവിന് ഏത് ധാന്യമാണ് ഒരുപാടിഷ്ടം ഉള്ളത്?
(बाजरा, गेहूँ, चावल)

v) पुराने ज़माने में राजमहलों में भी मेरा स्थान था । क्यों ?
പണ്ടുകാലത്ത് രാജകൊട്ടാരങ്ങളിലും എനിക്ക് സ്ഥാനം ഉണ്ടയിരുന്നു എന്തുകൊണ്ട്
(कबूतर दूर देशों तक संदेश लेकर जाता था, राजा लोग कबूतर को पसंद करता है)
उत्तर :
i) कपोत
ii) कबूतर बिल्ली से डरता है।
iii) शांति
iv) गेहूँ
v) कबूतर दूर देशों तक संदेश लेकर जाता था ।

प्रश्न 14.
चित्र देखकर और कविता पढ़कर उत्तर दीजिए
ചിത്രം കണ്ടിട്ടും കവിത വായിച്ചും ഉത്തരമെഴുതുക
(i) वह जंगल की शान होता है
അവൻ കാടിന്റെ അഭിമാനമാണ്.
देश का राष्ट्र पशु होता है
രാജ്യത്തിന്റെ രാഷ്ട്രീയ മൃഗമാണ്
सबसे तेज वह पशु होता है
ഏറ്റവും വേഗത്തിലോടുന്ന മൃഗമാണ്
कौन है वह ?
അതാരാണ് ?

ii) फूलों में सबसे सुंदर
देश का राष्ट्र फूल
पानी में तैरती
हवा के साथ झूमती
പൂക്കളിൽ ഏറ്റവും സുന്ദരം
രാജ്യത്തിന്റെ രാഷ്ട്രീയ പൂവ്
വെള്ളത്തിൽ നീന്തി
വായുവിൽ ആടിക്കളിച്ച്
कौन है वह ?
ആരാണ് അവൾ?

iii) हमें हैं इसके फलों का राजा
പഴങ്ങളുടെ രാജാവെന്ന് പറയപ്പെടുന്നു
प्यारा – प्यारा मीठा फल
പ്രിയപ്പെട്ട മധുരമുള്ള ഫലം
जन जन को मन भाता फल
ഓരോ ജനത്തിന്റെയും മനസ്സ് കീഴടക്കിയ ഫലം
देश का राष्ट्र फल
ദേശത്തിന്റെ രാഷ്ട്രീയ ഫലം
कौन है वह ?
അതാരാണ് ?

उत्तर :
i) बाघ (Tiger)
ii) कमल (Lotus)
(iii) आम (Mango)

प्रश्न 15.
अपनी असीम खूबसूरती से घिरे हुए भारत के सात अजूबे का चित्र एकत्रित करो और उनके नाम भी लिखो ।
തന്റെ അതിരില്ലാത്ത സൗന്ദര്യത്താൽ ചുറ്റപ്പെട്ട ഭാരതത്തിന്റെ ഏഴ് അത്ഭുതങ്ങളുടെ ചിത്രങ്ങൾ കണ്ടെത്തുക, കൂടെ അതിന്റെ പേരുകൾ കൂടി എഴുതു.
उत्तर :
(i) ताजमहल (Tajmahal)
(ii) स्वर्ण मंदिर (Golden temple)
(iii) हम्पी मन्दिर (Hampi temple
(iv) खजुराहो (Khajuraho)
(v) विक्टोरिया स्मारक (Victoria Memorial)
(vi) हवा महल (Hawa Mahal)
(vii) नालंदा विश्वविद्दालय (Nalanda University)
(बच्चे स्वयं चित्र एकत्रित करो)

प्रश्न 16.
कवितांश पढ़ें और उत्तर लिखें
കവിത വായിച്ച് ഉത്തരമെഴുതാം.

भारत मेरा प्यारा देश,
सब देशों से न्यारा देश ।
हिन्दू – मुस्लिम भाई – भाई,
मिलकर रहते सिख – ईसाई ।
धरती इसकी सबसे सुंदर,
उत्तर में हैं खडा हिमालय,
दक्षिण में लहराता सागर ।
भारत मेरा प्यारा देश |

क) कवितांश के अनुसार सही प्रस्ताव चुनकर लिखें।
കവിതാഭാഗം അനുസരിച്ച് ശരിയായ പ്രസ്താവന തിരഞ്ഞെടുത്ത് എഴുതുക.
1) भारत के उत्तर में सागर लहराता है।
ഭാരതത്തിന്റെ വടക്ക് സമുദ്രം അലയടിക്കുന്നു

2) भारत की धरती सुंदर नहीं है।
ഭാരതത്തിന്റെ ഭൂമി സുന്ദരമല്ല

3) भारत में सब लोग मिल जुलकर रहते हैं ।
ഭാരതത്തിലെ എല്ലാ ആളുകളും ഒരുമിച്ചാണ് താമസിക്കുന്നത്.

4) भारत के दक्षिण में हिमालय खडा है।
ഭാരതത്തിന്റെ തെക്ക് ഭാഗത്ത് ഹിമാലയമാണ് നിലനിൽക്കുന്നത്.
उत्तर :
भारत में सब लोग मिलजुलकर रहते हैं ।

ख) सही मिलान करके लिखें। (ശരിയായത് ചേർത്തെഴുതാം)

i) भारत	(क) लहराता सागर
ii) दक्षिण में	(ख) मिलकर रहते
iii) उत्तर में	(ग) प्यारा देश
iv) सिख- ईसाई	(घ) खडा हिमालय

उत्तर :

i) भारत	(ग) प्यारा देश
ii) दक्षिण में	(क) लहराता सागर
iii) उत्तर में	(घ) खडा हिमालय
iv) सिख- ईसाई	(ख) मिलकर रहते

हिंदुस्तान हमारा Summary in Hindi

Hindustan Hamara Summary in Hindi

भारत की खूबियों पर लिखा हुआ एक वर्णनात्मक लेख है – हिन्दुस्तान हमारा

भारत संसार का ऐसा ताज है, जिसकी चमक विश्व भर फैली है। इसकी खूबसूरत भौगोलिक झलक, समय – समय पर रंग बदलते अनोखे मौसम, शान बनी संपन्न संस्कृति सब इस ताज में जडे हीरे हैं । घने पेडों से सजे पर्वत, खिल-खिलाकर बहती नदियाँ, काली घटा से अपने बालों को खोलकर झूमती बारिश, हरा घाँघरा पहनकर नाचते लह-लहाते खेत सब भारत की अपनी विशेषताएँ हैं

छह ऋतुएँ मेहमान बनकर मुस्कराती हुई इस देश से गुज़र जाती है। गर्मी, सर्दी और बारिश सभी यहाँ हर साल अतिथि बनकर आते हैं। ये फल, धूप, पानी जैसे कई नज़ारे लाते हैं। जिससे सज धजता भारत देखने लायक है। रहन-सहन, वेश- भूषा, खान-पान, भाषा तथा धर्म अलग-अलग होने पर भी भारतवासी एकता के साथ रहते हैं। सब हिंदुस्तानी कहलाते हैं, इनकी पहचान एक हे। इसीलिए ही मशहूर शायर मुहम्मद इकबाल ने फरमाया- “सारे जहाँ से अच्छा हिंदोसिताँ हमारा!”

Hindustan Hamara Summary in Malayalam

हिंदुस्तान हमारा Summary in Malayalam

ഭാരതത്തിന്റെ ഗുണമേന്മകൾ വർണ്ണിക്കുന്ന ഒരു ലേഖനമാണ് ഹിന്ദുസ്ഥാൻ ഹമാരാ.

ഭാരതം ലോകം മുഴുവൻ പ്രകാശം പരത്തുന്ന ഒരു കിരീടം പോലെയാണ്. ഭാരതത്തിന്റെ മനോഹരമായ ഭൂപ്രകൃതിയും, മാറി മാറി വരുന്ന കാലാവസ്ഥയും സംസ്കാര സമ്പന്നതയുമെല്ലാം ഭാരതമാകുന്ന കിരീടത്തിൽ പതിച്ചിരിക്കുന്ന രത്നങ്ങളാണ്. വൃക്ഷങ്ങൾ തിങ്ങി നിൽക്കുന്ന പർവ്വതങ്ങൾ, കളകളാരവം പാടി ഒഴുകുന്ന നദികൾ, കറുത്ത മേഘങ്ങളിൽ നിന്ന് ഇടതൂർന്ന മുടിയഴിച്ചിട്ടതു പോലെ ആടി ഉലഞ്ഞു പെയ്യുന്ന മഴയും, പച്ച പാവാടയണിഞ്ഞതു പോലെ പച്ചപ്പ് നിറഞ്ഞ് വയലുകളുമെല്ലാം നിൽക്കുന്ന ഭാരതത്തിന്റെ സവിശേഷതകളാണ്.

വിരുന്നുകാരെ പോലെ ഈ രാജ്യത്തിൽ ചിരിച്ചുകൊണ്ട് വന്നു പോകുന്ന ആറ് ഋതുക്കൾ. ചൂടും മഴയും തണുപ്പുമെല്ലാം എല്ലാ വർഷവും അതിഥിയെപ്പോലെ എത്തുന്നു. പഴങ്ങളും, പൂക്കളും, വെയിലും, വെള്ളവും തുടങ്ങിയ കാഴ്ചകളെല്ലാം വിലമതിക്കാനാവാത്തതാണ്. ഇവയെല്ലാമാണ് ഭാരതത്ത അലങ്കരിക്കുന്നത്. ജീവിതനിലവാരവും, വസ്ത്രധാരണവും, ഭക്ഷണക്രമവും, ഭാഷയും മതവും എല്ലാം വ്യത്യസ്തമാണെങ്കിലും ഭാരതീയർ ഐക്യത്തോടു കൂടിയാണ് കഴിയുന്നത്. എല്ലാവരെയും തിരിച്ചറിയുന്നത് ഒരുപോലെ ആണ്. അതുകൊണ്ടാണ് പ്രശസ്ത കവി മുഹമ്മദ് ഇക്ബാൽ ഇങ്ങനെ എഴുതിയത് . എല്ലാ സ്ഥലങ്ങളെയും കാൾ മികച്ചത് നമ്മുടെ ഭാരതമാണ്.

हिंदुस्तान हमारा शब्दार्ध

• संसार – ലോകം world
• ताज – കിരീടം crown
• मौसम – കാലാവസ്ഥ climat
• शान – അഭിമാനം pride
• पर्वत – പർവ്വതം mountain
• बारिश -മഴ rain
• ऋतु – ഋതുക്കൾ season
• घाँघरा – പാവാട skirt
• गर्मी – ചൂട് hot
• सर्दी – തണുപ്പ് cold
• धूप – വെയിൽ sunlight
• सज – धजता – – ധരിക്കുക dressed
• मशहूर – പ്രസിദ്ധനായ popular
• हमारा – നമ്മുടെ our
• अहिंसा – അഹിംസ non-violence
• जान – ജീവൻ life
• प्यारा – പ്രിയപ്പെട്ട dear
• सागर – സമുദ്രം sea
• वतन – രാജ്യം country
• ज़वान – പട്ടാളക്കാരൻ soldier
• तिरंगा – ത്രിവർണ്ണ Tricolour
• याद – ഓർമ്മ memory
• आँख – കണ്ണ് eye
• बंदूक – തോക്ക് gun
• आज़ादी – സ്വാതന്ത്ര്യം independence
• साँस – ശ്വാസം breath
• सफर – യാത്ര journey
• दीवाना – ഭ്രാന്തൻ crazy

Practicing with SCERT Kerala Syllabus 6th Standard Hindi Textbook Solutions Unit 1 Chapter 2 बारिश कैसे मिली Question Answer Notes Summary in Malayalam & Hindi improves language skills.

Barish Kaise Mili Class 6 Question Answer Notes Summary

SCERT Class 6 Hindi Unit 1 Chapter 2 Question Answer Kerala Syllabus बादल दानी

Barish Kaise Mili Question Answer

प्रश्न 1.
‘बादल काँटों में फँसा हुआ है।’ इसका मतलब क्या होगा? आजकल हमारे बादल की हालत क्या है ? चर्चा करें |
മേഘം മുള്ളുകളിൽ കുഴങ്ങി കിടങ്ങുകയാണ്. ഇതിന്റെ അർത്ഥം എന്തായിരിക്കും ? ഇന്നത്തെ നമ്മുടെ മേഘത്തിന്റെ അവസ്ഥ എന്താണ്? ചർച്ച ചെയ്യാം
उत्तर :
इसका मतलब यह है कि बादल के बरसने में कई बाधाएँ हैं। मनुष्य ने पेडों को काटकर प्रकृति को बरबाद करते रहते हैं । इसलिए आजकल हमारे बादल प्रदूषित है और बारिश भी कम देता है ।

ഇതിന്റെ അർത്ഥം ഇതാണ്, മഴ പെയ്യുന്നതിന് ഒരുപാട് തടസ്സങ്ങൾ ഉണ്ട്. മനുഷ്യൻ മരങ്ങൾ വെട്ടി പ്രകൃതിയെ നശിപ്പിച്ചുകൊണ്ടിരിക്കുന്നു. അതുകൊണ്ട് ഇന്ന് നമ്മുടെ മേഘങ്ങളും മലിനമാക്കപ്പെട്ടിരിക്കുന്നു. കൂടാതെ മഴ ലഭിക്കുന്നതും ഇന്ന് കുറവാണ്.

प्रश्न 2.
बारिश के आने पर प्रकृति में क्या-क्या बदलाव आता है ? विवरण तैयार करें।
മഴ വരുമ്പോൾ പ്രകൃതിയിലുണ്ടാകുന്ന മാറ്റങ്ങൾ എന്തൊക്കെയാണ്? വിവരണം തയ്യാറാക്കാം.
उत्तर :
आकाश में बादल उठता है
बिजली चमकती है
पानी बरसता है
कुओं में, नदी में और तालाबों में पानी भर जाता है।
पेड-पौधे हरे-भरे हो जाते हैं
खेतों में फसल होती है
सब जीव-जंतु खुश होते हैं
वातावरण ठंडी हो जाती है और गर्मी कम हो जाती है।
ആകാശത്തിൽ മേഘങ്ങൾ ഉരുണ്ടു കൂടുന്നു
ഇടിമിന്നൽ മിന്നിത്തിളങ്ങുന്നു
കുളങ്ങളിലും, നദികളിലും, കിണറുകളിലും വെള്ളം നിറയുന്നു
ചെടികളും, വൃക്ഷങ്ങളും പച്ചപ്പ് കൊണ്ട് നിറയുന്നു
വയലുകളിൽ ധാന്യങ്ങൾ വിളയുന്നു
എല്ലാ ജീവജാലങ്ങളും സന്തോഷിക്കുന്നു
അന്തരീക്ഷം തണുപ്പുള്ളതാകുന്നു, ചൂട് കുറയുന്നു.

प्रश्न 3.
आनेवाले दिनों के लिए (വരും ദിവസങ്ങളിൽ)?
उत्तर :
“बूँद बूँद नहीं बरतेंगे,
तो बूँद बूँद को तरसेंगे ।

ഓരോ തുള്ളിയും സൂക്ഷിച്ച് ഉപയോഗിച്ചില്ലെങ്കിൽ
ഓരോ തുള്ളിക്കും വേണ്ടി ദയനീയമായി കെയേണ്ടി വരും.

प्रश्न 4.
कोलैश तैयार करें।
‘बरसात की झाँकियाँ’ (മഴക്കാലത്തേക്ക് ഒരു എത്തിനോട്ടം)
उत्तर :
छात्र अपने-अपने मन के अनुसार बरसात के विविध चित्र एकत्रित करके कोलैश तैयार करें.
കുട്ടികൾ അവരവരുടെ ഇഷ്ടമനുസരിച്ച് മഴക്കാലത്തെ ചിത്രങ്ങൾ കണ്ടെത്തി, അത് ഏകോപിപ്പിച്ച് കൊളാഷ് തയ്യാറാക്കുക.

प्रश्न 5.
चित्र पहचानें बक्सा भरें (ചിത്രം മനസ്സിലാക്കി ബോക്സിൽ എഴുതുക)
वर्ग पहेली (പദപ്രശ്നം)

बारिश कैसे मिली परीक्षाकेंद्रित प्रश्न और उत्तर

प्रश्न 1.
किसका बहुत प्यास लगी है?
ആർക്കാണ് ഒരുപാട് ദാഹിച്ചത്?
उत्तर :
बगुल को बहुत प्यास लगी है।
കൊക്കിനാണ് ഒരുപാട് ദാഹിച്ചത്.

प्रश्न 2.
बगुल और बंदर किस से मिले?
കൊക്കും കുരങ്ങനും ആരെയാണ് കണ്ടുമുട്ടിയത്?
उत्तर :
बगुल और बंदर बिल्ली से मिले ।
കൊക്കും കുരങ്ങനും പൂച്ചയെയാണ് കണ്ടുമുട്ടിയത്.

प्रश्न 3.
बिल्ली ने क्या कहा?
പൂച്ച എന്താണ് പറഞ്ഞത്?
उत्तर :
बिल्ली ने कहा कि बारिश बादल के पास है।
മഴ മേഘങ്ങൾക്കടുത്തുണ്ട്. എന്നാണ് പൂച്ച പറഞ്ഞത്.

प्रश्न 4.
किसने कहा कि बादल बूढे बबूल में फँसा है?
ആരാണ് പറഞ്ഞത് മേഘം വയസ്സായ ബബൂൽ മരത്തിൽ അകപ്പെട്ടിരിക്കുകയാണെന്ന്?
उत्तर :
बकरी ने कहा कि बादल बूढ़े बबूल में फँसा है।
ആടാണ് പറഞ്ഞത് മേഘം വയസ്സായ ബബൂൽ മരത്തിൽ അകപ്പെട്ടിരിക്കുകയാണെന്ന്

प्रश्न 5.
बूढा बबूल कहाँ रहता है?
വയസ്സായ ബബൂൽ മരം എവിടെയാണുള്ളത് ?
उत्तर :
बूढा बबूल बाडी के पास रहता है।
വയസ്സായ ബബൂൽ മരം ഫലവൃക്ഷത്തോട്ടത്തിന്റെ അടുത്താണുള്ളത്.

प्रश्न 6.
अंत में सब जानवर कहाँ पहूँचे?
അവസാനം എല്ലാ മൃഗങ്ങളും എവിടെയെത്തി?
उत्तर :
अंत में सब जानवर बूढे बबूल के पास पहूँचे।
അവസാനം എല്ലാ മൃഗങ്ങളും വയസ്സായ ബബുൽ മരത്തിന്റെ അടുത്തെത്തി.

प्रश्न 7.
चित्र देखें नाम लिखें (Page No: 19)
19-ാം പേജിലെ ചിത്രം കണ്ടതിനു ശേഷം അതിലുള്ള മൃഗങ്ങളുടെ പേരെഴുതുക.
उत्तर :
बंदर, चिडिया, बैल, मेंढक, बकरी और बिल्ली।
കുരങ്ങൻ, പക്ഷി, കാള, തവള, ആട്, പൂച്ച

प्रश्न 8.
चित्र देखें (Page No: 16) और नमूना के अनुसार वाक्य लिखिए ।
16-ാം പേജിലെ ചിത്രം കണ്ടിട്ട് മാതൃക അനുസരിച്ച് വാക്യങ്ങൾ എഴുതുക.
चित्र में एक बंदर है |
(ചിത്രത്തിൽ ഒരു കുരങ്ങനുണ്ട്
………………………………………..
………………………………………..
………………………………………..
………………………………………..
………………………………………..
………………………………………..
उत्तर :
चित्र में एक बगुला है। (ചിത്രത്തിൽ ഒരു കൊക്കുണ്ട്)
चित्र में एक पेड है। (ചിത്രത്തിൽ ഒരു മരമുണ്ട്.
चित्र में एक नदी है । (ചിത്രത്തിൽ ഒരു നദിയുണ്ട്.
चित्र में एक बिल्ली है । (ചിത്രത്തിൽ ഒരു പൂച്ചയുണ്ട്)
चित्र में एक पहाड है। (ചിത്രത്തിൽ ഒരു പർവ്വതമുണ്ട്.)

प्रश्न 9.
चित्र घटना के अनुसार घटनाओं को क्रमबद्ध करके लिखिए |
ചിത്രത്തിലെ സംഭവത്തിനനുസരിച്ച് സംഭവങ്ങളെ ക്രമത്തിലാക്കി എഴുതുക.

पतंग पेड पर फँस गयी है।
പട്ടം മരത്തിൽ അകപ്പെട്ടിരിക്കുന്നു.
सीना पतंग को बचाने की कोशिश करती है।
സീന പട്ടത്തെ രക്ഷിക്കാൻ ശ്രമിക്കുന്നു.
सीना पतंग उड़ाती है।
സീന പട്ടം പറപ്പിക്കുന്നു.
सीना की पतंग वापस मिली।
സീനയുടെ പട്ടം അവൾക്ക് തിരികെ ലഭിച്ചു.
उत्तर :
सीना पतंग उड़ाती है।
पतंग पेड पर फँस गयी है।
सीना पतंग को बचाने की कोशिश करती है।
सीना की पतंग वापस मिली ।

प्रश्न 10.
निम्नलिखित कविता पढकर उससे तुकांत शब्त चुनकर लिखिए ।
താഴെ തന്നിരിക്കുന്ന കവിത വായിച്ചിട്ട് അതിൽ നിന്നും
തുകാത് (Rhyming Words) തിരഞ്ഞെടുത്തെഴുതുക.
कोयल ने मीठे गीत सुनाए,
पेडों पर प्यारे पत्ते आए ।
चहचहाती चिडिया उडती,
कल-कल करती नदियाँ बहती ।
पानी में तो आयी मछली,
चारों और फैली हरियाली |
उत्तर :
सुनाए – आए
उडती – बहती
मछली-हरियाली

प्रश्न 11.
कौन बोला ? (ആര് പറഞ്ഞ ബ്രായ്ക്കറ്റിൽ നിന്ന് തിരഞ്ഞെടുത്തെഴുതുക
क)भैया, बहुत प्यास लगी है। (बगूला, बंदर)
സഹോദരാ, വളരെയധികം ദാഹം തോന്നുന്നു

(ख) चलो, बारिश को ढूँढें । (बगुला, बंदर)
വരൂ, മഴയെ അന്വേഷിക്കാം

ग) पानी कहीं नहीं। (बगुला, बंदर)
വെള്ളം എവിടെയുമില്ല

घ) बारिश मिलेगी बादल के पास ( बगुला, बिल्ली )
മഴ മേഘങ്ങളുടെ അടുത്തു നിന്ന് ലഭിക്കും

ङ) बादल बूढे बबूल में फँसा है। (बिल्ली, बकरी)
മേഘം വയസ്സായ ബബുൽ മരത്തിൽ അകപ്പെട്ടിരിക്കുന്നു

च) बबूल उधर बाडी के पास हे । (बकरी, बैल)
ബബുൽ അവിടെ ഫലവൃക്ഷത്തോട്ടത്തിന്റെ അടുത്താണ്.

छ) बाप रे! बेचारा काँटों में फँसा हुआ है। (बकरी, बंदर)
അയ്യോ! പാവം മുള്ളുകൾക്കിടയിൽ കുടുങ്ങി കിടക്കുകയാണ്.
വരൂ രക്ഷിക്കാം
चलो, बचाएँ। (बंदर / बाकी जानवर )
उत्तर :
क) बगुला
ख) बंदर
ग) बंदर
घ) बिल्ली
ङ) बकरी
च) बैल
छ) बंदर
ज) बाकी जानवर

प्रश्न 12.
कविता को क्रमबद्ध करके लिखिए |
കവിത ക്രമത്തിലാക്കി എഴുതുക.
बादल दानी, बादल दानी,
चम चम चमकी बिजली रानी
खूब झमाझम बरसो पानी
उठी गगन में घटा सुहानी |
उत्तर :
चम – चम चमकी बिजली रानी
उठी गगन में घटा सुहानी
बादल दानी, बादल दानी
खूब झमाझम बरसो पानी

प्रश्न 13.
पाठभाग के अनुसार कविता की पंक्तियाँ पूरा करें।
കവിതയുടെ വരികൾ പാഠഭാഗത്തിനനുസരിച്ച് പൂരിപ്പിക്കുക)
प्यासी चिडिया, प्यासी गौया,
खाली है सब ताल तलैया
……………………………….
……………………………….
……………………………….
……………………………….
उत्तर :
सूखे घाट पडी हे नैया
हुई मछलियों को हैरानी
बादल दानी, बादल दानी

प्रश्न 14.

क) बारिश का मौसम सबको पसंद है। चित्र देखकर किन्हीं दो घटनाओ का वर्णन कीजिए
മഴക്കാലം എല്ലാവർക്കും ഇഷ്ടമാണ്. ചിത്രം കണ്ടിട്ട് അതിലെ ഏതെങ്കിലും രണ്ട് സംഭവങ്ങൾ എഴുതുക.

ख) चित्र से दो जानवरों का नाम चुनकर लिखिए ।
ചിത്രത്തിൽ നിന്ന് രണ്ട് മൃഗങ്ങളുടെ പേരെഴുതുക.
उत्तर :
(क) लडका बारिश में खेलता है।
ആൺകുട്ടി മഴയിൽ കളിക്കുന്നു

ख) लडकी छतरी पकडकर नाव जल में डालकर खेलती है।
പെൺകുട്ടി കുട പിടിച്ച് വെള്ളത്തിൽ വള്ളം ഒഴുക്കി കളിക്കുന്നു.

कुत्ता और मेंढक – പട്ടിയും തവളയും

प्रश्न 15.
नीचे दिए गए गद्द खण्ड पढें और प्रश्नों को उत्तर दीजिए ।
താഴെ തന്നിരിക്കുന്ന ഗദ്യ ഭാഗം വായിച്ച് അതിനുത്തരം നൽകുക.

इस साल भी बारिश कम है । नदी-नालों में पानी कम हो रहा है। मनुष्य पेड काटते हैं, पहाड़ों को तोडते हैं । इनसे प्रकृति का नाश होते हैं। बारिश के कम हो जाने का यही कारण है।
ഈ വർഷവും മഴ കുറവാണ്. നദികളിലും തോടുകളിലും വെള്ളം കുറഞ്ഞു കൊണ്ടിരിക്കുന്നു. മനുഷ്യൻ മരങ്ങൾ വെട്ടുന്നു, പർവ്വതങ്ങൾ പൊട്ടിക്കുന്നു. ഇതിലൂടെ, പ്രകൃതിക്ക് നാശം സംഭവിക്കുന്നു. മഴ കുറയാനുള്ള കാരണം തന്നെ ഇതാണ്.

क) प्रकृति का नाशे कैसे होते हैं ?
പ്രകൃതിയുടെ നാശം എങ്ങനെ സംഭവിക്കുന്നു?
उत्तर :
मनुष्य पेड काटते हैं और पहोडों को तोडते हैं। इनसे प्रकृति का नाश होते हैं।

ख) सही प्रस्ताव चुनकर लिखिए ।
ശരിയായ പ്രസ്താവന തിരഞ്ഞെടുത്തെഴുതുക.

1) पेड लगाने से बारिश कम होती है।
മരങ്ങൾ വച്ച് പിടിപ്പിക്കുന്നതിലൂടെ മഴ കുറയുന്നു
2) प्रकृति के नाश से बारिश कम होती है।
പ്രകൃതിയിലുണ്ടാകുന്ന നാശം മൂലമാണ് മഴ കുറയുന്നത്.
उत्तर :
प्रकृति के नाश से बारिश कम होति है

प्रश्न 16.
‘प्रकृति संरक्षण’ पर एक पोस्टर तैयार कीजिए ।
പ്രകൃതി സംരക്ഷണത്തെക്കുറിച്ച് ഒരു പോസ്റ്റർ തയ്യാറാക്കുക.
उत्तर :

प्रश्न 17.
कवितांश पढ़ें और उत्तर लिखें ।
കവിത വായിച്ചിട്ട് ഉത്തരം എഴുതുക.

बरसा पानी रिमझिम रिमझिम
खेलें आओ मिलकर हम ।
मेघ गरजता घड घड घड
मोर नाचता छम छम छम ।
मेंढक बोला टर टर टर
लगता नहीं मुझे अब डर ।
भर गये सारे ताल – तलैया,
नाच रहे मिलता – ता – थैया ।

1) पानी कैसे बरसता है ? മഴ എങ്ങനെ വെയ്യുന്നു?
(घड – घड / रिमझिम रिमझिम)
उत्तर :
रिमझिम – रिमझिम

2) मोर कैसे नाचता है? മയിൽ എങ്ങനെയാണ് നൃത്തം ചെയ്യുന്നത്?
(छम छम / टर – टर)
उत्तर :
छम – छम

3) कविता से दो तुकांत शब्द चुनकर लिखिए ।
കവിതയിൽ നിന്ന് രണ്ട് തുകാന്ത് വാക്കുകൾ തിരഞ്ഞെടുത്തെഴുതുക.
उत्तर :
टर – डर, तलैया – थैया

प्रश्न 18.
खंभे से उचित अक्षर चुनकर लिखिए ।
ബ്രായ്ക്കറ്റിൽ നിന്ന് ഉചിതമായ അക്ഷരം തിരഞ്ഞെടുത്ത് പൂരിപ്പിക്കുക.
(चा, श, बे, द, ल)
बार – – – – –
बा – – – – ल
बबू – – –
ब – – – या
– – – – चारा
उत्तर :
बारिश
बादल
बबूल
बचारा
बेचारा

प्रश्न 19.
चित्र के आधार पर विवरण तैयार करें ।
ചിത്രത്തെ അടിസ്ഥാനമാക്കി വിവരിക്കുക.

उत्तर :
लडके चित्र में पतंग उडाते हैं।
ചിത്രത്തിൽ ആൺകുട്ടികൾ പട്ടം പറപ്പിക്കുന്നു.

एक लडकी भी पतंग उड़ाती है।
ഒരു പെൺകുട്ടിയും പട്ടം പറപ്പിക്കുന്നു

एक कुत्ता वहाँ बैठा है।
ഒരു നായ അവിടെ ഇരിക്കുന്നു.

एक लडकी बेंच पर बैठी है।
ഒരു പെൺകുട്ടി ബെഞ്ചിൽ ഇരിക്കുന്നു.

चित्र में एक पेड है।
ചിത്രത്തിൽ ഒരു മരമുണ്ട്

प्रश्न 20.
चित्र के स्थान पर उचित शब्द चुनकर पूरा करें।
ചിത്രത്തിന്റെ സ്ഥാനത്ത് ഉചിതമായ വാക്ക് തിരഞ്ഞെടുത്ത് പൂരിപ്പിക്കുക.
(बंदर, हाथी, बिल्ली)

उत्तर :
हाथी, बंदर, बिल्ली

प्रश्न 21.
उलटा पुलटा शब्द को क्रमबद्ध करके लिखिए:
തിരിഞ്ഞു മറിഞ്ഞുമുള്ള വാക്കുകൾ ക്രമത്തിലാക്കുക
ली बि ज
या चि डि
ह या रि ली
न कि सा
उत्तर :
बिजली
चिडिया
हरियाली
किसान

प्रश्न 22.
सही मिलान करो (ശരിയായി പൊരുത്തപ്പെടുത്തുക)

म्याऊँजी	बबूल
बूढ़ा	बकरी
बहनजी	बिल्ली

उत्तर :

म्याऊँजी	बिल्ली
बूढ़ा	बबूल
बहनजी	बकरी

प्रश्न 23.
बगुला और बंदर के बीच का वार्तालाप तैयार करें।
കൊക്കും കുരങ്ങനും തമ്മിലുള്ള സംഭാഷണം തയ്യാറാക്കുക)
बगुला : भैया, बहुत प्यास लगी है
बंदर – : ………….
बगुला : तो क्या करें ?
बंदर – : …………
उत्तर :
बगुला : भैया, बहुत प्यास लगी है।
बंदर : पानी कहीं नहीं है ।
बगुला : तो क्या करें ?
बंदर : चलो, बारिश को ढूँढें।

बारिश कैसे मिली Summary in Malayalam & Hindi

Barish Kaise Mili Summary in Malayalam & Hindi

भैया बहुत प्यास लगी है। क्या करें ?
पानी कहीं नहीं है। चलो बारिश को ढूँढें ।
സഹോദരാ, വളരെ ദാഹം തോന്നുന്നു. എന്തു ചെയ്യും?
വെള്ളം എവിടെയുമില്ല. വരൂ, മഴയെ അന്വേഷിക്കാം.
उत्तर :
वे बिल्ली से मिले ।
म्याऊँ जी, बारिश कहाँ मिलेगी ?
बारिश मिलेगी बादल के पास ।
അവർ പൂച്ചയെ കണ്ടുമുട്ടുന്നു
മ്യാവൂ ജി, നിങ്ങൾക്ക് എവിടെ മഴ ലഭിക്കും?
മേഘങ്ങൾക്ക് സമീപം മഴ കാണപ്പെടും.

चलते-चलते वे बकरी के पास पहूँचे।
बहनजी, कहीं बादल को देखा है ?
हाँ, वह तो बूढे बबूल में फँसा है।
നടന്നു നടന്ന് അവർ ആടിന്റെ അടുത്ത് എത്തി
സഹോദരീ, എവിടെയെങ്കിലും മേഘത്തെ കണ്ടോ?
ഉവ്വ്, ആ മേഘം വയസ്സായ ബബൂൽ മരത്തിൽ കുടുങ്ങി കിടക്കുകയാണ്.

आगे उन्होंने बैल को देखा ।
दादा जी बूढा बबूल कहाँ रहता है ?
उधर ….. बाडी के पास ।
മുന്നോട്ട് ചെന്നപ്പോൾ അവർ കാളയെ കണ്ടുമുട്ടി.
മുത്തശ്ശാ, ആ വയസ്സായ ബബൂൽ മരം എവിടെയാണ് ള്ള്ളത്?
അവിടെ ………… ഫലവൃക്ഷത്തോട്ടത്തിന്റെ അടുത്ത്

अंत में वे बूढे बबूल के पास पहूँचे ।
बाप रे! बेचारा काँटों में फँसा हुआ है।
चलो, बचाएँ ।
सबने मिलकर बादल को बचाया ।
അവസാനം അവർ വയസ്സായ ബബുൽ മരത്തിന്റെ അടുത്തെത്തി.
അയ്യോ! ആ പാവം മുള്ളുകൾക്കിടയിൽ കുടുങ്ങി കിടക്കുകയാണ്.
വരൂ, നമുക്ക് രക്ഷിക്കാം.
എല്ലാവരും ചേർന്ന് മേഘത്തെ രക്ഷിച്ചു.

पानी बरसने लगा…..
सब नाचने लगा…
മഴ പെയ്യാൻ തുടങ്ങി
എല്ലാവരും നൃത്തം ചെയ്യാൻ തുടങ്ങി.

बारिश कैसे मिली शब्दार्थ

• बगुला – കൊക്ക്, Heron
• बंदर – കുരങ്ങൻ, monkey
• ढूँढना – അന്വേഷിക്കുക, to search
• बूढा – വയസ്സായ old
• बबूल – ഒരു ഔഷധ വൃക്ഷം, a medicinal plant
• बेचारा – പാവം, poor
• कॉटा – മുള്ള്, thorn
• नाचना – നൃത്തം ചെയ്യുക to dance
• बकरी – ആട്, goat
• बैल – കാള, Ox
• भैया – സഹോദരാ
• बहुत – ഒരുപാട്, ധാരാളം
• प्यास – ദാഹം
• के पास – അടുത്ത്
• चलते – चलते – നടന്നു . നടന്നു
• फँसना – കുടുങ്ങുക
• उधर – അവിടെ
• बचाना – രക്ഷിക്കുക

Practicing with SCERT Kerala Syllabus 6th Standard Hindi Textbook Solutions Unit 1 Chapter 1 बादल दानी Hindi Poem Question Answer Notes Summary in Malayalam & Hindi improves language skills.

Badal Dani Hindi Poem Class 6 Question Answer Notes Summary

SCERT Class 6 Hindi Unit 1 Chapter 1 Question Answer Kerala Syllabus बादल दानी

Badal Dani Hindi Poem Question Answer

प्रश्न 1.
‘जीवन बरसता है’ ऐसा क्यों कहते हैं?
ജീവനാണ് ചെയ്യുന്നത്. ഇങ്ങനെ എന്തു കൊണ്ടാണ്
उत्तर:
क्योंकि जीवन का आधार पानी है।

കാരണം ജീവന്റെ അടിസ്ഥാനം വെള്ളം ആണ്

प्रश्न 2.
बारिश से कौन कौन खुरा होते हैं ?
മഴ കൊണ്ട് ആരൊക്കെയാണ് സന്തോഷിക്കുന്നത്?
उत्तर:
लड़का, लडकी, मेंढक, हिरण, बतख, फूल, मछली, बंदर, प्याध पतंग सब बारिश से खुश होते हैं ।
മത്സ്യം, കുരങ്ങൻ, തുമ്പി തുടങ്ങി എല്ലാവരും മഴ കാരണം സന്തോഷിക്കുന്നു.

प्रश्न 3.
ताल – लय के साथ गाएँ

1. यह गीत कौन गाता होगा ?
ഈ പാട്ട് ആരായിരിക്കും പാടിയിട്ടുണ്ടായരിക്കുക?
उत्तर:
लड़का और लड़की
ആൺകുട്ടിയും പെൺകുട്ടിയും

2. इसका प्रसंग क्या हो सकता है ?
ഇതിന്റെ സന്ദർഭം എന്തായിരിക്കും?
उत्तर:
सूखे के बाद बारिश के आगमन ।
ഉണക്കിന് ശേഷം മഴയുടെ വരവ്

3. वहाँ कौन कौन होंगे ?
അവിടെ ആരൊക്കെ കാണുമായിരിക്കും?
उत्तर:
लड़का, लडकी, मेंढक, हिरण, बतख, फूल, मछली, बंदर |
ലഡക, ലഡക്കി, മെന്തക്, ഹിരണ്, ബതഖ്, ഫൂൾ, മഛലി, ബന്ദർ |

4. वे क्या कया करते होंगे ?
അവർ എന്തൊക്കെ ആയിരിക്കും ചെയ്യുന്നത്?
उत्तर:
वे मेघ से कहते हैं कि सागर से जल लेकर गाँव – गाँव में पानी बरसने कों ।
അവർ മേഘത്തോട് സമുദ്രത്തിൽ നിന്നും ജലം എടുത്തുകൊണ്ട് വന്ന് ഗ്രാമങ്ങൾ തോറും ചെയ്യാൻ പറയുന്നു

बादल दानी Hindi Poem व्याकरण भाग

नीचे की पंक्तियाँ पढें ।
चम – चम चमकी बिजली रानी
उठा गगन में घटा सुहानी

1. रेखांकित शब्दों की समानता क्या है ?
അടിവരയിട്ട വാക്കുകളിലെ സാമ്യം എന്താണ്?
उत्तर:
दोनों शब्दों के अंत में एक ही वर्ण आया है। इसे तुकांत कहते हैं।
രണ്ട് വാക്കുകളുടെയും അവസാനം ഒരേ അക്ഷരം വന്നിരിക്കുന്നു. ഇതിനെ തുകാന്ത് എന്ന് വിളിക്കുന്നു. तुकांत की परिभाषा (Rhyming Words’ Definition), തുകാന്ത് എന്നാൽ എന്താണ്?

काव्य पंक्तियों के अंतिम भाग में पायी जाने वाली वर्णों की समानता को तुकांत कहते हैं।
കവിതയിലെ വരികളുടെ ആവസാന ഭാഗത്ത് ഒരേ അകചരത്തിൻ്റെ സാമ്യത്തെ തുക്കാത് അഥവാ rhyming എന്ന് വിളിക്കുന്നു

तुकांत शब्द कविता को सुंदर बनाते हैं।
തുകാന്ത് വാക്കുകൾ കവിതയെ സുന്ദരമാക്കുന്നു.

ज़रा ये पंक्तियाँ पढ़ें ।
നമുക്ക് ഈ വരികൾ ഒന്ന് വായിക്കാം.

बारिश का मौसम है आया,
हम बच्चों के मन को भाया ।
इन पंक्तियों के तुकांत शब्द पहचानें।
उनके नीचे रेखा खींचें ।

ഈ വരികളിലെ തുകാന്ത് വാക്കുകൾ മനസ്സിലാക്കുക.
അതിന്റെ അടിയിൽ വരയിടുക.

बारिश का मौसम है: आया,
हम बच्चों के मन को भाया ।
हम भी ऐसी पंक्तियाँ जोड़ें।
നമുക്ക് ഈ കവിതയിൽ കുറച്ച് വരികൾ കൂടി എഴുതി ചേർക്കാം

बारिश का मौसम

बारिश का मौसम है आया,
हम बच्चों के मन को भाया ।
आसमान में बादल आया,
नदी नालों में पानी लाया ।
कोयल ने मीठे गीत सुनाए,
पेडों पर प्यारे पत्ते – आए ।
चह – चहाती चिडिया उडती,
कल-कल करती नदियाँ बहती,
पानी में तो आयी मछली
चारों ओर फैली हरियाली ।

ऐसे जोडे चुनकर लिखें। (Page No: 13)
പാഠഭാഗത്ത് നിന്ന് തുകാന്ത് വാക്കുകൾ അവസാനിക്കുന്ന വരികൾ എഴുതാം.

बादल दानी, बादल दानी,
खूब झमाझम बरसो पानी ।
प्यासी चिडिया, प्यासी गैया,
खाली है सब ताल तलैया ।
हुई मछलियों को हैरानी,
बादल दानी, बादल दानी ।

बादल दानी Hindi Poem परीक्षाकेंन्द्रित प्रश्न और उत्तर

प्रश्न 1.
पहले चित्र में (Page No. 9) कौन कौन हैं ?
ആദ്യത്തെ ചിത്രത്തിൽ നമുക്ക് ആരെയൊക്കെ കാണാൻ സാധിക്കുന്നു.
उत्तर :
लडका लडकी, मेघ, हिरण, पेड, चिडिया, मोर, साँप, मेंढक, और खरगोश ।
ആൺകുട്ടി, പെൺകുട്ടി, മേഘം, മാൻ, മരം, പക്ഷി, മയിൽ, പാമ്പ്, തവള, മുയൽ.

प्रश्न 2.
सब जीव क्या कर रहे हैं?
എല്ലാ ജീവജാലങ്ങളും എന്തു ചെയ്യുകയാണ്?
उत्तर :
सब जीव मेघ से पानी माँग रहे हैं।
എല്ലാ ജീവജാലങ്ങളും മേഘങ്ങളിൽ നിന്ന് ജലം ചോദിച്ചു

प्रश्न 3.
दूसरे चित्र में (Page No. 10) हम क्या-क्या देखते हैं ?
രണ്ടാമത്തെ ചിത്രത്തിൽ നമ്മൾ എന്തൊക്കെ കാണുന്നു?
उत्तर :
लडका लडकी, मेंढक, फूल वाली पौधा, बतख, हिरण, मछली, बंदर और व्याध पतंग।
ആൺകുട്ടി, പെൺകുട്ടി, തവള, പൂച്ചെടി, താറാവ്, മാൻ, മത്സ്യം, കുരങ്ങ്, ഡ്രാഗൺഫ്ലൈ.

प्रश्न 4.
‘बादल दानी’ कविता का संदेश क्या है ?
ബാദൽ ദാനി കവിതയുടെ സന്ദേശം എന്താണ്?
उत्तर :
जल ही जीवन है। जल के बिना जीवन संभव नहीं पानी का संरक्षण करना हमारा कर्तव्य है ।
ജലം തന്നെയാണ് ജീവിതം. ജലമില്ലെങ്കിൽ ജീവിതമില്ല. ജലം സംരക്ഷിക്കുക എന്നത് നമ്മുടെ കർത്തവ്യമാണ്.

प्रश्न 5.
सागर से कौन जल भर लाते हैं?
സമുദ്രത്തിൽ നിന്ന് ജലം ആരാണ് കൊണ്ടുവരുന്നത്?
उत्तर :
सागर से बादल जल भर लाते हैं।
സമുദ്രത്തിൽ നിന്ന് മേഘമാണ് ജലം കൊണ്ടു വരുന്നത്.

प्रश्न 6.
बादल कहाँ से जल भर लाते हैं।
മേഘം എവിടെ നിന്നാണ് ജലം കൊണ്ടുവരുന്നത്?
उत्तर :
बादल समुद्र से जल भर लाते हैं ?
മേഘം സമുദ്രത്തിൽ നിന്നാണ് വെള്ളം കൊണ്ടുവരുന്നത്

प्रश्न 7.
बादल कहाँ रस बरसाते हैं?
മേഘം എവിടെയാണ് ജലം പൊഴിക്കുന്നത്?
उत्तर :
बादल गाँव-गाँव में रस बरसाते हैं।
മേഘം ഗ്രാമങ്ങൾ തോറും ജലം പൊഴിക്കുന്നു.

प्रश्न 8.
बादल किसकी प्यास बुझाते हैं?
മേഘം ആരുടെ ദാഹമാണ് ശമിപ്പിക്കുന്നത്?
उत्तर :
बादल सब जीवो की प्यास बुझाते हैं ।
മേഘം എല്ലാ ജീവജാലങ്ങളുടെയും ദാഹം ശമിപ്പിക്കുന്നു.

प्रश्न 9.
कौन धानी हो जाती है ?
ആരാണ് പച്ചപ്പായി മാറുന്നത്?
उत्तर :
धरती धानी हो जाती है।
ഭൂമിയാണ് പച്ചപ്പായി മാറുന്നത്.

प्रश्न 10.
खेतों में क्या महके?
വയലുകളിൽ എന്താണ് സുഗന്ധം പരത്തുന്നത്?
उत्तर :
खेतों में धान की फसल महके ।
വയലുകളിൽ ധാന്യം വിളയുന്നതിന്റെ സുഗന്ധമാണ് പരക്കുന്നത്

प्रश्न 11.
किसान का मन कब खुश हो जाता है ?
കർഷകന്റെ മനസ്സ് എപ്പോഴാണ് സന്തോഷിക്കുന്നത്?
उत्तर :
किसान का मन मेढों तक पानी देखकर खुश हो जाता है
കർഷകന്റെ മനസ്സ് വയൽവരമ്പു വരെ വെള്ളം നിറഞ്ഞതു കണ്ട് സന്തോഷിക്കുന്നു.

प्रश्न 12.
बादल को ‘बादल दानी’ क्यों कहा है ?
മേഘത്തെ ദാനിയായ മേഘം എന്ന് വിളിക്കുന്നത് എന്തുകൊണ്ടാണ്?
उत्तर :
बादल पानी का दान करता है। इसलिए बादल को ‘बादल दानी’ कहा हैं।

മേഘം ജലം ദാനം ചെയ്യുന്നു. അതുകൊണ്ടാണ് മേഘത്തെ ദാനിയായ മേഘം എന്ന് വിളിക്കുന്നത്.

प्रश्न 13.
बादल दानी’ से कैसे बरसने को कहा है?
ദാനിയായ മേഘത്തോട് എങ്ങനെയാണ് ചെയ്യാൻ പറയുന്നത്?
उत्तर :
बादल दानी’ से खूब झमाझम बरसने को कहा है।
ദാനിയായ മേഘത്തോട് നല്ല ശക്തിയായി പെയ്യാനാണ് പറയുന്നത്.

प्रश्न 14.
बारिश के आने पर प्रकृति में क्या-क्या परिवर्तन आता है?
മഴ വരുമ്പോൾ പ്രകൃതിയിലുണ്ടാകുന്ന മാറ്റങ്ങൾ എന്തൊക്കെയാണ്?
उत्तर :
सब जीवों की प्यास बुझ जाती है।
धरती हरी-भरी हो जाती है।
खेतों में फसलों की सुगंध फैल जाती है।
घाट की राह पर हरियाली लहराती है।
किसानों का मन खुश हो जाता है।

प्रश्न 15.
सहि मिलान करें
क) बिजली – खाली है
ख) घटा – बरसो पानी
ग) ताल-तलैया – चम चम चमकी
घ) चिडिया – उठी गगन में
ङ) बादल दानी – प्यासी
उत्तर :
क) बिजली – चम चम चमकी
ख) घटा – उठी गगन में
ग) ताल-तलैया – खाली है
घ) चिडिया – प्यासी
ङ) बादल दानी – बरसो पानी

प्रश्न 16.
पाठभाग से तुकांत शब्द ढूँढकर लिखिए ।

• रानी – सुहानी
• दानी –
• गैया –
• हैरानी –
• लाते –
• धानी –
• महके –
• पानी –

उत्तर :

• रानी – सुहानी
• दानी – पानी
• गैया – नलैया
• हैरानी – दानी
• लाते – बरसाते
• धानी – दानी
• मह – लहके, चहके
• पानी – दानी

प्रश्न 17.
समान अर्थवाले शब्द
(नैया, चिडिया, सागर, गैया, घटा, धरती)
क) समुद्र – सागर
ख) भूमि –
ग) मेघ –
घ) नाव –
ङ) पक्षी –
च) गाय –
उत्तर :
क) समुद्र – सागर
ख) भूमि – धरती
ग) मेघ – घटा
घ) नाव – नैया
ङ) पक्षी – चिड़िया
च) गाय – गैया

चित्रविवरण कीजिए: (ചിത്രം വിവരിക്കുക)

प्रश्न 1.
चित्र में एक लडका और लडकी बादलों के ऊपर बैठकर नीचे की ओर खुशी से देखता है।
चिडिया एक सूखा पेड पर बैठकर ऊपर की ओर देखता है।
मेंढक, मोर, हिरण, साँप और खरगोश ऊपर की ओर देखता है।
वे सब बारिश की प्रतीक्षा में है।

ചിത്രത്തിൽ ഒരു ആൺകുട്ടിയും പെൺകുട്ടിയും മേഘങ്ങൾക്കു മുകളിലിരുന്ന് സന്തോഷത്തോടെ താഴേക്ക് നോക്കുകയാണ്.

പക്ഷി ഒരു ഉണങ്ങിയ മരത്തിലിരുന്ന് മുകളിലേക്ക് നോക്കുകയാണ് തവളയും മയിലും മാനും പാമ്പും മുയലുമെല്ലാം മുകളിലേക്ക് നോക്കുന്നു.

ഇവരെല്ലാം മഴയ്ക്കായുള്ള പ്രതീക്ഷയിലാണ്.

प्रश्न 2.
मोर, चिडिया और मेंढक का एक वार्तालाप तैयार कीजिए ।
മയിലും, പക്ഷിയും, തവളയും തമ്മിലുള്ള സംഭാഷണം തയ്യാറാക്കുക.
उत्तर :
मोर : देखो, देखो ऊपर की तरफ देखो ।
മയിൽ : നോക്കൂ നോക്കൂ മുകളിലേയ്ക്ക് നോക്കൂ.
मेंढक : क्या है?
തവള : എന്താണ്?
चिडिया : एक लडका और लडकी बादल के ऊपर बैठे हैं।
പക്ഷി : ഒരു ആൺകുട്ടിയും പെൺകുട്ടിയും കൂടി മേഘങ്ങൾക്കു മകളിൽ ഇരിക്കുന്നു.
मोर : वे क्या कर रहे हैं ?
മയിൽ : അവൻ എന്തു ചെയ്യുകയാണ്?
मेंढक : वे बारिश को लेकर आते हैं।
തവള : അവർ മഴയുമായി വരികയാണ്.
चिडिया : अच्छा, चलो, बारिश में नहाने के लिए तैयार करो।
പക്ഷി : രി വരൂ, മഴയിൽ കുളിക്കാൻ തയ്യാറായിക്കൊള്ളൂ.
मेंढक और मोर : हम सब तैयार हैं।
തവളയും മയിലും : ഞങ്ങൾ എല്ലാവരും തയ്യാറാണ്.

प्रश्न 3.
निम्नलिखित पद्दांश को पढ़कर कविता की पंक्तियों को वाक्य में लिखिए |
താഴെ തന്നിരിക്കുന്ന പദ്യഭാഗം വായിച്ചിട്ട് കവിതയിലെ വരികൾ വാക്യത്തിലെഴുതാം

सागर से भर भर जल लाते,
गाँव – गाँव में रस बरसाते,
सब जीवों की प्यास बुझाते,
धरती हो जाती है धानी,
बादल दानी, बादल दानी
उत्तर :
बादल सागर से जल भरकर लाते हैं और गाँव – गाँव में रस बरसाते हैं। बादल सब जीवों की प्यास बुझाते हैं। इससे यह धरती धानी हो जाती है ।
മേഘം സമുദ്രത്തിൽ നിന്നും വെള്ളം നിറച്ചു കൊണ്ടു വന്ന് ഗ്രാമങ്ങൾ തോറും മഴ പെയ്യിക്കുന്നു. മേഘം എല്ലാ ജീവജാലങ്ങളുടെയും ദാഹം ശമിപ്പിക്കുന്നു. ഇതിനാൽ ഈ ഭൂമി പച്ചപ്പ് നിറഞ്ഞതാകുന്നു.

प्रश्न 4.
वर्ग पहेली बुझाएँ
उत्तर :

प्रश्न 5.
कवितांश पढें ओर अनुबद्ध प्रश्नों के उत्तर लिखें
കവിതാഭാഗം വായിച്ച് അതുമായി ബന്ധപ്പെട്ടുള്ള ചോദ്യങ്ങൾക്ക് ഉത്തരം എഴുതുക.
वर्षा रानी वर्षा रानी
रिमझिम रिमझिम बरसो पानी
ताल – तलैया भर लो रानी
…………………………………….
…………………………………….
…………………………………….

(क) कवितांश से तुकांत शब्द चुनकर लिखें।
കവിതാഭാഗത്ത് നിന്ന് തുകാന്ത് വാക്കുകൾ കണ്ടെത്തി എഴുതാം.
उत्तर :
रानी – पानी

(ख) कवितांश के साथ पंक्तियाँ जोडें ।
കവിതാഭാഗത്ത് വരികൾ എഴുതിച്ചേർക്കാം.
उत्तर :
गाँव- गाँव में तुम बरसा पानी,
वर्षा रानी, वर्षा रानी ।

बादल दानी Hindi Poem Summary in Hindi

Badal Dani Hindi Poem Summary in Hindi

बादल दानी
कोई फर्क नहीं पड़ता कि मौसम क्या हो सकता है, बारिश का दिन भी बहुत राहत और आराम देता है और हमारी आत्मा को शांत करता है। बरसात के दिनों का आनंद लेने के लिए कोई आयु सीमा नहीं है। लगभग हर उम्र के लोग समान रूप से इसका आनंद लेते हैं। इस प्रकार, कई कारणों से बारिश के दिन बहुत महत्वपूर्ण होते हैं।

बारिश के मौसम से ज़मीन, तालाबों या नदियों में पानी भर जाता है, जिससे सिंचाई करने में मदद मिलती है। पानी से बिजली उत्पादन भी बढ़ता है । इससे अर्थव्यवस्था पर अच्छा प्रभाव पड़ता है । पृथ्वी पर जल पाया जाता है इसी कारण से इसे ब्रह्मांड का एक अनोखा ग्रह कहा जाता है । जल के कारण ही पृथ्वी पर मनुष्य की जाति विकसित हो रही है। जीने के लिए मनुष्य पशु पेड़ पौधों सभी को जल की ज़रूरत होती है। अगर किसी कारण जल समाप्त हो जाता है तो कोई भी जीव जंतु जीवित नहीं रह पाएगा क्योंकि सभी जीने के लिए जल का उपयोग करते हैं।

जब गर्म नम हवा, ठंडे और उच्च दबाव वाले वातावरण के संपर्क में आती है तब वारिश होती है, गर्म हवा अपने अंदर ठंडी हवा से ज़्यादा पानी जमा कर सकती है, और जब ये हवा अपने जमा पानी के साथ ऊँचाई की और जाती है तो ठंडे जलवायु से जाकर मिल जाती है और अपने अन्दर जमा पानी के भारी हो जाने से उसे नीचे गिरा देती है।

Badal Dani Hindi Poem Summary in Malayalam

बादल दानी Hindi Poem Summary in Malayalam

കാലാവസ്ഥ എന്തുതന്നെയായാലും, മഴയുള്ള ഒരു ദിവസം നമ്മുടെ ആത്മാവിന് ആശ്വാസം നൽകുന്നു. മഴക്കാലം ആസ്വദിക്കാൻ പ്രായപരിധിയില്ല. മിക്കവാറും എല്ലാ പ്രായത്തിലുമുള്ള ആളുകൾ ഇത് ഒരു പോലെ ആസ്വദിക്കുന്നു. അതിനാൽ മഴയുള്ള ദിവസങ്ങൾ പല കാരണങ്ങളാൽ വളരെ പ്രധാനമാണ്.

മഴക്കാലത്ത് നിലത്തോ കുളങ്ങളിലോ നദികളിലോ വെള്ളം നിറയുന്നു. ഇത് ജലസേചനത്തിന് സഹായിക്കുന്നു. ജലത്തിൽ നിന്നുള്ള വൈദ്യുതി ഉൽപാദനവും വർധിക്കുന്നു. ഇത് സമ്പദ് വ്യവസ്ഥയിൽ നല്ല സ്വാധീനം ചെലുത്തുന്നു.

ഭൂമിയിൽ ജലം കാണപ്പെടുന്നു. ഇക്കാരണത്താൽ അതിനെ പ്രപഞ്ചത്തിന്റെ അതുല്യ ഗ്രഹം എന്ന് വിളിക്കുന്നു. ഭൂമിയിൽ മനുഷ്യരാശി വികസിക്കുന്നത് ജലം കൊണ്ടാണ്. മനുഷ്യർക്കും മൃഗങ്ങൾക്കും മരങ്ങൾക്കും ചെടികൾക്കും ജീവിക്കാൻ വെള്ളം ആവശ്യമാണ്. ചില കാരണങ്ങളാൽ വെള്ളം തീർന്നുപോയാൽ, ഒരു ജീവജാലത്തിനും അതിജീവിക്കാൻ കഴിയില്ല. കാരണം എല്ലാവരും ജീവിക്കാൻ വെള്ളം ഉപയോഗിക്കുന്നു.

ചൂടുള്ള ഈർപ്പമുള്ള വായു തണുത്തതും ഉയർന്ന മർദവുമായ അന്തരീക്ഷവുമായി സമ്പർക്കം പുലർത്തുമ്പോൾ 29 ചെയ്യുന്നു. തണുത്ത വായുവിനേക്കാൾ കൂടുതൽ ജലം ഉള്ളിൽ സംഭരിക്കാൻ ചൂടുള്ള വായുവിന് കഴിയും. ഈ വായു അതിന്റെ അടിഞ്ഞുകൂടിയ വെള്ളവുമായി ഉയരത്തിലേക്ക് പോകുമ്പോൾ, അത് തണുത്ത കാലാവസ്ഥയുമായി കലരുകയും അതിനുള്ളിൽ അടിഞ്ഞു കൂടിയ വെള്ളത്തിന്റെ ഭാരം കാരണം അത് താഴേക്ക് വീഴുകയും ചെയ്യുന്നു.

बादल दानी Hindi Poem पद्यांश का अवतरण

चम चम चमकी बिजली रानी
उठी गंगन में घटा सुहानी
बादल दानी, बादल दानी
खूब झमझम बरसो पानी

മിന്നിമിന്നി തിളങ്ങുന്ന ഇടിമിന്നൽ
ആകാശത്ത് ഉയർന്നു നിൽക്കുന്ന ഭംഗിയുള്ള മേഘം
ദാനിയായ മേഘമേ, ദാനിയായ മേഘമേ
നീ ധാരാളമായി പെയ്തിറങ്ങിയാലും

प्यासी चिडिया, प्यासी गैया,
खाली है सब ताल – तलैया
सूखे घाट पडी है नैया
हुई मछलियों को हैरानी, बादल दानी बादल दानी

ദാഹിച്ചിരിക്കുന്ന പക്ഷികൾ, ദാഹിച്ചിരിക്കുന്ന പശുക്കൾ
എല്ലാ കുളങ്ങളും വറ്റി വരണ്ടിരിക്കുന്നു
വറ്റി വരണ്ട് കടവുകളിൽ എങ്ങും പോകാനാകാതെ
കിടക്കുന്ന തോണി
മത്സ്യങ്ങൾ വളരെ വിഷമത്തിലാണ്
ദാനിയായ മേഘമേ, ദാനിയായ മേഘമേ

सागर से भर – भर जल लाता
गाँव- गाँव में रस बरसाते
सब जीवों की प्यास बुझाते
धरती हो जाती है धानी
बादल दानी बादल दानी

സമുദ്രത്തിൽ നിന്നും നിറയെ വെള്ളം കൊണ്ടുവന്ന്
ഗ്രാമങ്ങൾ തോറും സന്തോഷം പൊഴിക്കുന്നു
എല്ലാ ജീവജാലങ്ങളുടെയും ദാഹം ശമിപ്പിക്കുന്നു
ഭൂമി മുഴുവൻ പച്ചപ്പ് നിറഞ്ഞതാകുന്നു
ദാനിയായ മേഘമേ, ദാനിയായ മേഘമേ

फसल धान की खेतों : महके
राह घाट हरियाली लहके
मन किसान का गाए चहके
देख – देख मेढों तक पानी
बादल दानी बादल दानी
खूब झमझम बरसो पानी ।

വയലുകളിൽ ധാന്യങ്ങളുടെയും വിളകളുടെയും സുഗന്ധം പരക്കുന്നു
കടവുകളിലേക്കുള്ള വഴികളിൽ പച്ചപ്പ് നിറയുന്നു
കർഷകരുടെ മനസ്സ് സന്തോഷം കൊണ്ട് നിറഞ്ഞ് അവർ പാട്ട് പാടുന്നു
വയൽ വരമ്പ് വരെ വെള്ളം നിറയുന്നത് കണ്ട്
ദാനിയായ മേഘമേ, ദാനിയായ മേഘമേ
ധാരാളമായി നീ പെയ്തിറങ്ങിയാലും

बादल दानी Hindi Poem कवि परिचय बाबुलाल शर्मा प्रेम

हिन्दी साहित्य के सुप्रसिद्ध कवि श्री बाबुलाल शर्मा प्रेम | हिन्दी के बालकविता के सशक्त कवि थे। उनका जन्म १ मई १९३५ को उत्तर प्रदेश के हरदोई जिले के नेवदा में हुआ। वे ‘प्रेम’ उपनाम से जाने जाते हैं। बचपन से ही वे साहित्य रचना की ओर आकृष्ट हुए हैं। छोटी-छोटी कविताएँ लिखना उनका मनपसंद कार्य हैं। बच्चों के लिए साहित्य लेखन में वह मात्र २० साल की उम्र से ही सक्रिय हो गए थे। उन्होंने करीब २० हज़ार रचनाएँ लिखीं, जिसको प्रमुख पत्र-पत्रिकाओं ने प्रकाशित किया। उनकी रचनाएँ विभिन्न शैक्षणिक बोर्ड में पढाई जा रही है । वरिष्ट बाल साहित्यकार बाबुलाल शर्मा का गुरुवार को आलमबाग के एक निजी अस्पताल में निधन हो गया। वह बीते एक अगस्त को अपने निवास स्थान की छत से गिर गए थे, तभी से वह अस्पताल में भर्ती था । वह ८५ वर्ष के थे। उन्हें हाल ही में घोषित हुए उत्तर प्रदेश हिन्दी संस्थान के प्रतिष्ठित बाल भारती साहित्य सम्मान के लिए नामित किया गया था, जिसका वितरण हिन्दी दिवस पर होना था । ‘अंचल का फूल’ उनका प्रथम कविता संग्रह है। याद के बादल, फिर हँसेगे फूल, टके टके की वात (कहानी संग्रह) आदि उनकी रचनाएँ हैं ।

ഹിന്ദി സാഹിത്യത്തിലെ സുപ്രസിദ്ധനായ കവിയായി രുന്നു. ശ്രീ ബാബുലാൽ ശർമാ പ്രേം. അദ്ദേഹം ഹിന്ദിയിലെ കുട്ടികൾക്കായുള്ള ഉത്തർപ്രദേശിലെ കവിതകൾ എഴുതുന്നതിൽ പ്രസിദ്ധനായിള്ള അദ്ദേഹത്തിൻ്റെ ജനനം, 1935m ഹർദോയി ജില്ലയിലെ നവ ദ യിലായിരുന്നു. കുട്ടിക്കാലം മുതൽ തന്നെ സാഹിത്യരചനയിലേക്ക് അദ്ദേഹം ആകൃഷ്ടനായി തീർന്നു.

ചെറിയ ചെറിയ കവിതകൾ എഴുതുക. അദ്ദേഹത്തിന്റെ മനസ്സിന് ഇഷ്ടമുള്ള കാര്യമായിരുന്നു. കുട്ടികൾക്കായുള്ള സാഹിത്യ രചനയിൽ തന്നെ അദ്ദേഹം 20-ാം വയസ്സു മുതൽ സജീവമായി. ഏകദേശം ഇരുപതിനായിരം രചനകൾ അദ്ദേഹം നിർവ്വഹിച്ചിട്ടുണ്ട്. ഇതിൽ പ്രധാനപ്പെട്ട രചനകൾ എല്ലാം തന്നെ മാസികയിൽ പ്രസിദ്ധീകരിച്ചിട്ടുണ്ട്. അദ്ദേഹത്തിന്റെ രചനകൾ എല്ലാം തന്നെ വിഭിന്ന വിദ്യാഭ്യാസ മേഖലകളിൽ പഠന വിഷയമാക്കിയിട്ടുണ്ട്. മുതിർന്ന ബാലസാഹിത്യകാരനായ ശ്രീ ബാബുലാൽ ശർമ, ആലംബാഗിലെ ഒരു പ്രൈവറ്റ് ഹോസ്പിറ്റലിൽ ഒരു വെള്ളിയാഴ്ചയാണ് മരിച്ചത്. അദ്ദേഹം കഴിഞ്ഞ ഒരു ഓഗസ്റ്റ് മാസം തന്റെ താമസസ്ഥലത്തിന്റെ മുകളിൽ നിന്നും. താഴെ വീഴുകയുണ്ടായി. അപ്പോൾ തന്നെ അദ്ദേഹത്തെ ആശുപത്രിയിൽ പ്രവേശിപ്പിച്ചിരുന്നു. അദ്ദേഹത്തിന് അപ്പോൾ 85 വയസ്സായിരുന്നു. അദ്ദേഹത്തിനെ ഉത്തർ പ്രദേശിലെ ഹിന്ദി ഇൻസ്റ്റിറ്റ്യൂട്ടിലെ പ്രശസ്തമായ ബാലഭാരതി സാഹിത്യ പുരസ്കാരത്തിനായി നാമനിർദ്ദേശം ചെയ്യപ്പെട്ടിട്ടുണ്ടായിരുന്നു. അതിന്റെ വിതരണം ഹിന്ദി ദിവസത്തിൽ നൽകാനിരുന്നതാണ് ‘ആഞ്ചൽ കാ ഫൂൽ അദ്ദേഹത്തിന്റെ ആദ്യത്തെ കവിതാ സമാഹാരമാണ്. യാദ് കേ ഫൂൽ, ഫിർ ഹസേംഗേ ഫൂൽ, ടക്കേ – ടക്കേ കി ബാത്ത് (കഥാ സമാഹാരം) തുടങ്ങിയ പ്രധാനപ്പെട്ട രചനകളാണ്.

बादल दानी शब्दार्थ

• बादल – മേഘം clouds
• चमकना – തിളങ്ങുന്ന shining
• बिजली – മിന്നൽ lightning
• गगन – ആകാശം sky
• घटा – മേഘം cloud
• सुहानी – സുഖം തരുന്ന pleasing
• बरसना – പെയ്യുക to rain
• पानी – ജലം water
• प्यासी – ദാഹമുള്ള thirsty
• चिड़िया – പക്ഷി bird
• गैया – പശു cow
• खाली – ശൂന്യമായ empty
• ताल-तलैया – കുളം pond
• घाट – കടവ് ferry
• नैया – തോണി boat
• मछली – മത്സ്യം fish
• हैरानी – വിഷമം worry
• सागर – സമുദ്രം sea
• गाँव – ഗ്രാമം village
• धरती – ഭൂമി earth
• धानी – പച്ചപ്പ് greenary
• फसल – വിളവ് crop
• धान – ധാന്യം grains
• खेत – വയൽ field
• महके – സുഗന്ധം പരത്തുന്ന emitting fragrance
• राह – വഴി way
• हरियाली – പച്ചയായ greenary
• किसान – കർഷകൻ farmer
• मेढों – വയൽവരമ്പ് field ridge
• खूब – നല്ലതുപോലെ
• सूखा – ഉണങ്ങിയ
• भरना – നിറയ്ക്കുക
• जल – വെള്ളം
• लाना – കൊണ്ടുവരിക
• जीव – ജീവജാലങ്ങൾ

Teachers recommend solving Kerala Syllabus 9th Standard Physics Question Paper Set 4 to improve time management during exams.

Kerala Syllabus Std 9 Physics Model Question Paper Set 4 English Medium

Time: 1½ Hours
Max Score: 40 Marks

Instructions

• First 15 minutes is given as cool off time. This time is to be spent for reading and understanding the questions.
• Answer the questions according to the directions.
• Score and time to be considered while answering.

I. Answer any three questions from 1 to 4. (1 score each) (3 × 1 = 3)

Question 1.
Impossible position-time graph is

Answer:
B

Question 2.
Which of these figures representing the path of light through a glass slab is correct?

Answer:
B

Question 3.
Give an example of inertia of motion
Answer:

• When a bus stops suddenly, the passengers tend to fall forward.
• A participant in long jump competition, runs some distance and then jumps.
• A ball set rolling on a horizontal floor keeps moving forward.

Question 4.
1 kgwt = ______ N
Answer:
1 kg wt = 9.8 N

II. Answer any Seven questions from 5 to 13. (2 score each) (7 × 2 = 14)

Question 5.
Describe a scenario where displacement is zero, but distance is not.
Answer:
If a person walks 5 m north and then 5 m south through the same path, their displacement is zero because they end up at their starting point. However, the distance travelled is 10 m, as it is the total path length covered. This illustrates how displacement considers direction, while distance does not.

Question 6.
Write any two examples of circular motion.
Answer:

• Motion of Artificial satellites
• Whirling a stone tied to a string
• Rotation of fan blades

Question 7.
Describe the benefits of using small prisms in reflectors of motorcycles.
Answer:
Many small prisms are used in reflectors used in motor cycles. Light falling on the prisms undergoes total internal reflection and reflects back without loosing the intensity of light. The image can be seen with greater visual clarity.

Question 8.
Observe the figure

When the cad Is suddenly struck. what happens to the coin? Why?
Answer:
a) The coin falls in to the glass. It is due to the inertia of rest of the coin.

Question 9.
Raju placed a pencil in a trough of water as shown in figure.

a) Can you see a change in the position of the portion of the pencil under water?
b) Why does the ray of light coming after reflection from the pencil undergo a deviation?
Answer:
a) Yes
b) The ray of light from the immersed part of the pencil entering air from water is deviated at the surface of separation. It undergoes refraction.

Question 10.
Classify the following instances into inertia of motion and inertia of rest
a) A fan continue to rotate for a while even after it is switched off.
b) It is possible to strike out a single carrom coin without disturbing others, from a set of coins stacked one over the other.
Answer:
a-Inertia of motion
b-Inertia of rest

Question 11.
Write some traffic rules for pedestrians to follow.
Answer:

• Pedestrians should walk along the right side of the road.
• Cross the road only at the zebra crossing, obeying the traffic signal.
• Use sidewalks.
• Avoid crossing between parked cars.

Question 12.
A car starting from rest travels 100 m in 5 s with uniform acceleration. Find the acceleration of the car?
Answer:
u = 0 m/s
t = 5s
s = 100 m
a = ?
s = ut + \(\frac { 1 }{ 2 }\)at²
100 = 0 × 5 + \(\frac { 1 }{ 2 }\) × a × 5²
a = \(\frac{100 \times 2}{25}\) = 8 m/s²

Question 13.
State true or false.
a) The value of g will be the same everywhere on the earth.
b) The factors that influence the value of g are mass of the earth and radius of the earth.
Answer:
a) False
b) True

III. Answer any five questions from 14 to 19. (3 score each) (5 × 3 = 15)

Question 14.
The velocity-time graph of an object in straight line motion is given.

Identify the regions of the graph where the object is moving with:
a) acceleration.
b) uniform velocity.
c) deceleration.
Answer:
a) Region OA
b) Region AB
c) Region BC

Question 15.
The figure shows the motion of an object along a circular path.

a) What do you mean by uniform circular motion?
b) Write any one example for uniform circular motion.
c) If this object loses centripetal force at the point A while in circular motion, draw the path of its movement.
Answer:
a) If the object swept equal distance at equal intervals of time in a circular path it is uniform circular motion.
b) The earth revolves around the sun.
c)

Question 16.
A car moving with uniform acceleration and changes its velocity from 10 m/s to 50 m/s in 4 seconds, then
a) Find out the acceleration of car?
b) Find out the displacement in this time?
Answer:
a) a = \(\frac { v-u }{ t }\)
t = 4s
a = \(\frac{50-10}{4}\)
a = 10 m/s²

b) s = ut + \(\frac { 1 }{ 2 }\) at²
= 10 × 4 + \(\frac { 1 }{ 2 }\)(10 × 4²)
= 40 + 80
= 120 m

Question 17.
Observe the figure. A ray of light enters from medium 1 to medium 2

a) Find out which medium has greater optical density.
b) Which medium has the greater refractive index?
c) What is the relation between optical density and refractive index of a medium?
Answer:
a) Medium-1
b) Medium-1
c) As the refractive index of a medium increases, optical density also increases.

Question 18.
State true or false.
a) All internal forces are unbalanced forces.
b) All internal forces are balanced forces.
c) All balanced forces are internal forces.
Answer:
a) False
b) True
c) False

Question 19.
Differentiate between mass and weight.
Answer:
Table

IV. Answer any two questions from 20 to 22. (4 score each) (2 × 4 = 8)

Question 20.
The path of light through different media is shown in the following figure. Analyse them and answer the following questions.

a) Find out critical angle of glass.
b) Which figure represents the total internal reflection?
c) Explain how total internal reflection takes place in glass by referring the above figures.
Answer:
a) 42°
b) R
c) When a ray of light passes from a medium of higher optical density (glass) to a medium of lower optical density (air) at an angle of incidence greater than the critical angle, the ray is reflected back to the same medium without undergoing refraction. This is total internal reflection.

Question 21.
Observe the figure. A, B, C are three positions on the surface of the earth and D is the centre of the earth.

a) Is the value of mass or weight of an object is same at all positions on the surface of the earth.
b) What is the weight of an object at D?
c) On the surface of the earth at which positions among A, B and C the value of acceleration due to gravity be maximum. Justify your answer.
Answer:
a) Mass of an object is same at all positions on the surface of the earth but weight of the object varies.

b) Zero

c) Value of acceleration due to gravity is – maximum at A. Acceleration due to gravity, ‘g’ depends on the radius of the earth. As the radius decreases ‘g’ increases. The distance to A (Pole) is less than that of other positions.

Question 22.
“A man jumps from a boat to the shore”
a) Identify the action and reaction in the above situation.
b) State the law related to the given situation.
c) Write another example related to this law.
Answer:
a) Action – The boat moves backward. Reaction – The man lands on the shore.

b) Newton’s third law of motion.
The law states that for every action there is an equal and opposite reaction.

c) Recoil of a gun
Rowing a boat

You can Download Decimal Operations Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Decimal Operations Text Book Questions and Answers

Triangle Problem Textbook Page No. 109

Anup made a triangle with three sticks of length 4 centimetres each. What is the perimeter of this triangle?
How did you do it?
Suma made a triangle using 4.3 centimetre sticks instead.
What is the perimeter?
4.3 + 4.3 + 4.3 = 12.9 cm.

Instead of adding again and again, we only compute 3 times 4.3
How do we find it?
4.3 centimetres mean 43 millimetres. And 3 times 43 millimetres is 43 × 3 = 129 millimetres.
This is 12.9 millimeters.
There’s another way of doing this:
4.3 = 4\(\frac{3}{10}\) = \(\frac{43}{10}\)
So, 3 times \(\frac{43}{10}\) is
\(\frac{43}{10}\) × 3 = \(\frac{129}{10}\) = 12.9 cm.
That is, 4.3 × 3 = 12.9

Cloth problem

To make a shirt for a boy in the class, 1.45 metres of cloth is needed, on average.
How much cloth is needed to make shirts for the 34 boys in the class?
We must calculate 34 times 1.45.
1.45 metres mean 145 centimetres; And 34 times 145 is
145 × 34 = 4930
How much metres is 4930 centimetres?
\(\frac{4930}{100}\) metre = 49.30 metres

How about writing all measurements as tractions?
1.45 = 1\(\frac{45}{100}\) = \(\frac{145}{100}\)
1.45 × 34 = 1\(\frac{45}{100}\) × 34 = \(\frac{145}{100}\) × 34 = \(\frac{4930}{100}\)
We can write it as a decimal.
\(\frac{4930}{100}\) = 49.30 = 49.3
Thus 1.45 × 34 = 49.3

The area of a square of side 1 centimetre 1 square centimetre and the area of a square of side 1 millimetre is 1 square millimetre. 1 centimetre is 10 millimetres. So in the bigger square we can stack 10 smaller squares each along the length and breadth 10 × 10 = 100 small squares in all. So the smaller square is \(\frac{1}{100}\) of the bigger square. That means 1 sq.mm. = \(\frac{1}{100}\) sq.cm

Area

We know how to calculate the area of a rectangle of length 8 centimetres and height 6 centimetres. What about a rectangle of length 8.5 centimetres and breadth 6.5 centimetres? The lengths in millimetres are 85 and 65. So area is 85 × 65 = 5525 square millimetres. How do we change it into square centimetres?
1 square millimetre = \(\frac{1}{100}\) square centimetre.
5525 square millimetres = \(\frac{5525}{100}\) = 55.25 square centimetres.

How about writing all measurements as fractions?
8.5 centimetres = 8\(\frac{5}{10}\) centimetres = \(\frac{85}{10}\) centimetres
6.5 centimetres = 6\(\frac{5}{10}\) centimetres = \(\frac{65}{10}\) centimetres
Area is \(\frac{85}{10}\) × \(\frac{65}{10}\) square centimetres.
\(\frac{85}{10}\) × \(\frac{65}{10}\) = \(\frac{5525}{100}\) = 55.25
Thus area is 55.25 square centimetres.
Let’s write the computation using numbers only.
8.5 × 6.5 = 55.25

Textbook Page No. 111

Question 1.
The sides of a square are of length 6.4 centimeters. What is its perimeter?
Answer:
25.6 centimeters.
Explanation:
6.4 centimeter is length of one side,
The perimeter of a square is P = 6.4 + 6.4 + 6.4 + 6.4 = 25.6 centimeters.
There’s another way of doing this is,
Perimeter of a square is P = 4 x side
P = 4 x 6.4
P = 25.6 centimeters.

Question 2.
3 rods of length 6.45 meters each are laid end to end. What is the total length?
Answer:
19.62 meters.
Explanation:
6.45 centimeter is length of one side.
The perimeter of a triangle is P = 6.45 + 6.45 + 6.45 = 19.62 meters.
There’s another way of doing this is,
Perimeter of a square is P = 3 x side
P = 3 x 6.45
P = 19.62 centimeters

Question 3.
A bag can be filled with 4.575 kilograms of sugar. How much sugar can be filled in 8 such bags?
Answer:
36.6 kg
36.6 kilograms
Explanation:
A bag can be filled with 4.575 kilograms of sugar.
8 bags of sugar = 8 x 4.575 = 36.6 kg.

Question 4.
The price of one kilogram of rice is 34.50 rupees. How much money do we need to buy 16 kilograms?
Answer:
552 rupees.
Explanation:
The price of one kilogram of rice is 34.50 rupees.
Cost of 16 kilograms = 16 x 34.50 = 552 rupees.

Question 5.
6 bottles are filled with the coconut oil in a can. Each bottle contains 0.478 liters. How much oil was in the can, in liters?
Answer:
2.868 liters.
Explanation:
6 bottles are filled with the coconut oil in a can.
Each bottle contains 0.478 liters.
Total oil in the can, in liters was = 6 x 0.478 = 2.868 liters

Question 6.
The length and breadth of a rectangular room are 8.35 meters and 3.2 meters. What is the area of that room?
Answer:
26.72 meter square.
Explanation:
Area of rectangle = length x breadth
length of a rectangle room = 8.35 meters
breadth of a rectangle room = 3.2 meters
Area = 8.35 x 3.2 = 26.72 meter square.

Multiplication

What is the meaning of 4.23 × 2.4?
4.23 × 2.4 = \(\frac{423}{100}\) × \(\frac{24}{10}\) = \(\frac{423 \times 24}{1000}\)
To compute this, we have to multiply 423 by 24 and then divide by 1000.
423 × 24 = 10152
\(\frac{423 \times 24}{1000}\) × \(\frac{10152}{1000}\) = 10.152

In the answer, how many digits are there after the decimal point? Why three?
Look at the fraction form of the answer. The denominator is 1000, right?
How did we get this 1000?

Look at the denominator of the fractions we multiplied.
So how do we complete 4.23 × 0.24?
First find 423 × 24 = 10152.

Now how many digits are there after the decimal point in the product?
If we write 4.23 × 0.24 as a fraction, what would be the denominator of the product?
4.23 as a fraction has denominator 100.
0.24 as a fraction has denominator 100.What about the denominator of the product?
So, 4.23 × 0.24 = \(\frac{10152}{10000}\) = 1.0152

Like this, how do we do 2.45 × 3.72?
First calculate 245 × 372.
Now we must find out the number of digits after the decimal point.
What is the denominator of 2.45 as a fraction.
And of 3.72?
What is the denominator of the product?
So,
2.45 × 3.72 = 9.1140 = 9.114

Answer:
0.1 × 0.1 = 0.01
0.01 × 0.01 = 0.0001
0.001 × 0.001 = 0.000001
0.0001 × 0.0001 = 0.00000001
Explanation:
To multiply a decimal number by a decimal number,
we first multiply the two numbers ignoring the decimal points.
Then place the decimal point in the product, in such a way that decimal places in the product is equal to the sum of the decimal places in the given numbers as shown above.

Textbook Page No. 113

Question 1.
Calculate the products below:

i) 46.2 × 0.23
Answer:
10.626
Explanation:
If we write 46.2 × 0.23 as a fraction,
46.2 as a fraction has denominator 10.
0.23 as a fraction has denominator 100.
= \(\frac{462}{10}\) × \(\frac{23}{100}\)
= \(\frac{462 \times 23}{1000}\)
To compute this, we have to multiply 462 by 23 and then divide by 1000.
462 × 23 = 10,626
= \(\frac{10626}{1000}\) = 10.626
So, 46.2 × 0.23 = 10.626

ii) 57.52 × 31.2
Answer:
1794.624
Explanation:
If we write 57.52 × 31.2 as a fraction,
57.52 as a fraction has denominator 100.
31.2 as a fraction has denominator 10.
= \(\frac{5752}{100}\) × \(\frac{312}{10}\)
= \(\frac{5752 \times 312}{1000}\)
To compute this, we have to multiply 5752 by 312 and then divide by 1000.
5752 × 312 = 17,694,624
= \(\frac{17,94,624}{1000}\) = 1794.624
So, 57.52 × 31.2 = 1794.624

iii) 0.01 × 0.01
Answer:
Explanation:
If we write 0.01 × 0.01 as a fraction,
0.01 as a fraction has denominator 100.
0.01 as a fraction has denominator 100.
= \(\frac{1}{100}\) × \(\frac{1}{100}\)
= \(\frac{1 \times 1}{10000}\)
To compute this, we have to multiply 1 by 1 and then divide by 10000.
1 × 1 = 1
= \(\frac{1}{10000}\) = 0.0001
So, 0.01× 0.01 = 0.0001

iv) 2.04 × 2.4
Answer:
4.896
Explanation:
If we write 2.04 × 2.4 as a fraction,
2.04 as a fraction has denominator 100.
2.4 as a fraction has denominator 10.
= \(\frac{204}{100}\) × \(\frac{24}{10}\)
= \(\frac{204 \times 24}{1000}\)
To compute this, we have to multiply 204 by 24 and then divide by 1000.
204 × 24 = 4896
= \(\frac{4896}{1000}\) = 4.896
So, 2.04 x 2.4 = 4.896

v) 2.5 × 3.72
Answer:
9.3
Explanation:
If we write 2.5 × 3.72 as a fraction,
2.5 as a fraction has denominator 10.
3.72 as a fraction has denominator 100.
= \(\frac{25}{10}\) × \(\frac{372}{100}\)
= \(\frac{25 \times 372}{1000}\)
To compute this, we have to multiply 25 by 372 and then divide by 1000.
25 × 372 = 9300
= \(\frac{9300}{1000}\) = 9.3
So, 2.5× 3.72 = 9.3

vi) 0.2 × 0.002
Answer:
0.0004
Explanation:
If we write 0.2 × 0.002 as a fraction,
0.2 as a fraction has denominator 10.
0.002 as a fraction has denominator 1000.
= \(\frac{2}{10}\) × \(\frac{2}{1000}\)
= \(\frac{2 \times 2}{10000}\)
To compute this, we have to multiply 2 by 2 and then divide by 10000.
2 × 2 = 4
= \(\frac{4}{10000}\) = 0.0004
So, 0.2 × 0.002 = 0.0004

Question 2.
Given that 3212 × 23 = 73876, find the products below, without actually multiplying?

i) 321.2 × 23 = _____
Answer:
7387.6,
Explanation:
Given 3212 × 23 = 73876,
In 321.2 × 23 first find out the number of digits after the decimal point.
(1 + 0) = 1
So, 321.2 × 23 = 7387.6

ii) 0.3212 × 23 = _____
Answer:
7.3876
Explanation:
Given 3212 × 23 = 73876,
In 0.3212 × 23 first find out the number of digits after the decimal point.
(4 + 0) = 4
So, 0.3212 × 23 = 7.3876

iii) 32.12 × 23 = ____
Answer:
738.76
Explanation:
Given 3212 × 23 = 73876,
In 32.12 × 23 first find out the number of digits after the decimal point.
(2 + 0) = 2
So, 32.12 × 23 = 738.76

iv) 32.12 × 0.23 = ____
Answer:
7.3867
Explanation:
Given 3212 × 23 = 73876,
In 32.12 × 0.23 first find out the number of digits after the decimal point.
(2 + 2) = 4
So, 32.12 × 0.23 = 7.3876

v) 3.212 × 23 = ____
Answer:
73.876
Explanation:
Given 3212 × 23 = 73876,
In 3.212 × 23 first find out the number of digits after the decimal point.
(3 + 0) = 3
So, 3.212 × 23 = 73.876

vi) 321.2 × 0.23 = _____
Answer:
73.876
Explanation:
Given 3212 × 23 = 73876,
In 321.2 × 0.23 first find out the number of digits after the decimal point.
(1 + 2) = 3
So, 321.2 × 0.23 = 73.876

Question 3.
Which of the products below is equal to 1.47 × 3.7?
i) 14.7 × 3.7
ii) 147 × 0.37
iii) 1.47 × 0.37
iv) 0.147 × 37
v) 14.7 × 0.37
vi) 0.0147 × 370
vii) 1.47 × 3.70
Answer:
Option iv, v and vii has the equal products.
Explanation:
first find out the number of digits after the decimal point, then add to the product.
i) 14.7 × 3.7 = 54.39 (1 + 1 = 2)
ii) 147 × 0.37 = 54.39 (0 + 2 = 2)
iii) 1.47 × 0.37 = 0.5439 (2 + 2 = 4)
iv) 0.147 × 37 = 5.439 (3 + 0 = 3)
v) 14.7 × 0.37 = 5.439 (1 + 2 = 3)
vi) 0.0147 × 370 = 0.5439 (4 + 0 = 4)
vii) 1.47 × 3.70 = 0.5439 (2 + 2 = 4)
So, iv, v and vii has the equal products.

Question 4.
A rectangular plot is of length 45.8 meters and breadth 39.5 meters .What is its area?
Answer:
1809.10 meter square.
Explanation:
Area of rectangle = length x breadth
length of a rectangle plot= 45.8 meters
breadth of a rectangle plot = 39.5 meters
Area = 45.8 x 39.5 = 1809.10 meter square.

Question 5.
The price of petrol is 68.50 rupees per liter. What is the price of 8.5 liters?
Answer:
582.25 liters.
Explanation:
The price of petrol is 68.50 rupees per liter.
The price of 8.5 liters = 68.50 x 8.5 = 582.25 L

Question 6.
Which is the largest product among those below.
i) 0.01 × .001
ii) 0.101 × 0.01
iii) 0.101 × 0.001
iv) 0.10 × 0.001
Answer:
Option (ii)
Explanation:
The largest product among those below are,
i) 0.01 × .001 = 0.00001 (2 + 3 = 5)
ii) 0.101 × 0.01 = 0.00101 (3 + 2 = 5)
iii) 0.101 × 0.001 = 0.000101 (3 + 3 = 6)
iv) 0.10 × 0.001 = 0.00010 (2 + 3 = 5)
So, the largest product is 0.101 x 0.01 = 0.00101

It is easy to calculate these products;
384 × 10
230 × 100

Now calculate these products:

• 125 × 10
Answer:
1250
Explanation:
First multiply the number by ignoring zeros.
125 x 1 = 125
Then add zero to the product.
125 + 0 = 1250
So, 125 x 10 = 1250

• 4.2 × 10
Answer:
42
Explanation:
To multiply a decimal by 10,
move the decimal point in the multiplication by one place to the right.
4.2 x 10 = 42

• 13.752 × 10
Answer:
137.52
Explanation:
To multiply a decimal by 10,
move the decimal point in the multiplication by one place to the right.
13.752 x 10 = 137.52

• 4.765 × 100
Answer:
476.5
Explanation:
To multiply a decimal by 100,
move the decimal point in the multiplication by two places to the right.
So, 4.765 x 100 = 476.5

• 3.45 × 100
Answer:
345
Explanation:
To multiply a decimal by 100,
move the decimal point in the multiplication by two places to the right.
3.45 x 100 = 345

• 14.572 × 100
Answer:
1457.2
Explanation:
To multiply a decimal by 100,
move the decimal point in the multiplication by two places to the right.
14.572 x 100 = 1457.2

• 1.345 × 1000
Answer:
1345
Explanation:
To multiply a decimal by 1000,
move the decimal point in the multiplication by three places to the right.
1345 x 1000 = 1345

• 2.36 × 1000
Answer:
0.236
Explanation:
To multiply a decimal by 1000,
move the decimal point in the multiplication by three places to the right.
2.36 x 1000 = 0.236

• 1.523 × 1000
Answer:
1523
Explanation:
To multiply a decimal by 1000,
move the decimal point in the multiplication by three places to the right.
1.523 x 1000 = 1523

Have you found out an easy way to multiply decimals by numbers 10,100,1000 and so on?
Answer:
Yes, just by moving the places towards the right.
Explanation:
When a decimal number is multiplied by 10, 100 or 1000,
the digits in the product are the same as in the decimal number,
but the decimal point in the product is shifted to the right as many places as there are zeros.

Let’s divide! Textbook Page No. 114

4 girls divided a 12 meter long ribbon among them. What length did each get?
It is not difficult to calculate this.
How about a 13 meter long ribbon?
12 meter divided into 4 equal parts give 3 meter long pieces; the remaining 1 meter divided into 4 gives \(\frac{1}{4}\) meter. Altogether 3\(\frac{1}{4}\) meters.

So, each gets 3\(\frac{1}{4}\) meters
We can write this as 13 ÷ 4 = 3\(\frac{1}{4}\)
We can also write it as a decimal.
\(\frac{1}{4}\) meter means 25 centimeter; that is, 0.25 meters.
So, instead of 3\(\frac{1}{4}\) metrer, we can write 3.25 meters.

Look at this problem;

A square is made with a 24.8 centimeter long rope. What is the length of its side?
To find the length of a side, 24.8 must be divided into four equal parts.
24.8 centimeters means 24 centimeters and 8 millimeters.
24 centimeters divided into four equal parts give 6 centimeters each.
The remaining 8 millimeters divided into four equal parts give 2 millimeters each.
Thus the length of a side is 6 centimeters and 4 millimeters, that is 6.2 centimeters.
This problem also we can write using numbers only.
24.8 ÷ 4
The way we found the answer can also be written using just numbers.

24.8 mean 24 and 8 tenths. Dividing each by 4 gives 6 and 2 tenths; that is 6.2
These operations can be written in short hand as shown on the right.

A line of length 13.2 centimeters is divided into 3 equal parts .What is the length of each part?
We first divide 12 centimeters of 13.2 centimeters into 3 equal parts, getting 4 centimeter long parts; 1 centimeter and 2 millimeters remaining.
That is, 12 millimeters are left.

Dividing this into 3 equal parts gives 4 millimeters each. So, 13.2 centimeters divided into 3 equal parts give 4 centimeters and 4 millimeters as the length of a part.
That is 4.4 centimeters.
How about writing this as a division of numbers?
13.2 ÷ 3 = 4.4

How did we do this?
13.2 mean 13 and 2 tenths. in this, dividing 13 by 3 gives quotient 4 and remainder 1.

Changing this 1 to tenths and adding them to the 2 tenths already there, we get 12 tenths. 12 divided by 3 gives 4.
Thus we get 4 and 4 tenths; that is 4.4.
These operations also we can write in shorthand.

Let’s look at another problem:

4 people shared 16.28 kilograms of rice. How much does each get?
If 16 kilograms is divided in to 4 equal parts, how much is each part?
0.28 kilograms means 280 grams.
What if we divide 280 grams into 4?
So, how much does each get?
How about writing this using only numbers?
16.28 ÷ 4 = 4.07
16.28 means 16 and 2 tenths and 8 hundredths.
16 divided by 4 gives 4.

Changing 2 tenths to 20 hundredths and adding to the original 8 hundredths give 28 hundredths.
28 divided by 4 gives 7
So the total quotient is 4 and 7 hundredths.
That is 4.07.
The operation can be written like this:

25.5 kilograms of sugar is packed into 6 bags of the same size. How much is in each bag?
24 kilograms divided into 6 equal parts give 4 kilograms each. The remaining 1.5 kilograms, changed to grams are 1500 grams.
Dividing this into 6 equal parts gives 1500 ÷ 6 = 250 grams.

So one bag contains 4 kilograms and 250 grams; that is 4.250 kilograms.
We usually write this as 4.25 kilograms.
As numbers, we find
25.5 ÷ 6 = 4.25
The method of finding the answer can also be written using only numbers.
25.5 means 25 and 5 tenths.
25 divided by 6 gives 4 and remainder 1.
The remaining 1, changed to tenths and added to the original 5 tenths give 15 tenths; divided this by 6 gives 2 tenths and remainder 3 tenths.

These 3 tenths can be changed into 30 hundredths ; and this divided by 6 gives 5 hundredths.
What then is the total quotient?
4 and 2 tenths and 5 hundredths
That is ,4.25
Let’s write these operations in shorthand.

Textbook Page No. 118

Question 1.
The total amount of milk given to the children in a school for the 5 days of last week is 132.575 liters. How much was given on average each day?
Answer:
26.515 liters
Explanation:
Total milk given in last 5 days = 132.575 liters
Number of days = 5
Average milk given on each day = Total milk given by number of days.
= 132.575 ÷ 5
= 26.515 liters

Question 2.
8 people shared 33.6 kilograms of rice. Sujitha divided her share into three equal parts and gave one part to Razia. How much did Razia get?
Answer:
1.4 kg
Explanation:
Number of people = 8
Total rice shared = 33.6 kilograms.
No. of kgs each person got = 33.6 ÷ 8 = 4.2 kg
Sujitha divided her share into three equal parts = 4.2 ÷ 3 = 1.4 kg
Razia gets 1.4 kg share of Sujitha.

Question 3.
A ribbon of length 0.8 meters is divided into 16 equal parts. What is the length of each part’?
Answer: 5 cm
Explanation:
A ribbon of length 0.8 meters is divided into 16 equal parts.
1 m = 100 cm
0.8 m = 100 x 0.8 = 80 cm
The length of each part = 80 ÷ 16 = 5 cm

Question 4.
Do the problems below:
i) 54.5 ÷ 5
Answer:
10.9
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern.

ii) 14.24 ÷ 8
Answer:
1.78
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

iii) 56.87 ÷ 11
Answer:
5.17
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

iv) 3.1 ÷ 2
Answer:
1.55
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

v) 35.523 ÷ 3
Answer:
11.841
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

vi) 36.48 ÷ 12
Answer:
3.4
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a patter

vii) 16.56 ÷ 9
Answer:
1.84
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

viii) 32.454 ÷ 4
Answer:
8.1135
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

ix) 425.75 ÷ 25
Answer:
17.03
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

Question 5.
Given 105.728 ÷ 7 = 15.104, find the answer to the problems below, with out actual division.
i) 1057.28 ÷ 7
Answer:
151.04
Explanation:
1057.28 mean 1057 and 2 tenths, 8 hundredths.
in this, dividing 1057 by 7 gives quotient 151.
Changing this 2 tenths to 20 hundredths and adding to the original 8 hundredths gives 28 hundredths.
Then, we get 28 ÷ 7 = 4
Thus we get 151 and 2 tenths and hundredths; that is 151.04.

ii) 1.05728 ÷ 7
Answer:
0.15104
Explanation:
1.05728 mean 1 and 0 tenths, 5 hundredths, 7 thousandths, 2 ten thousandths and 8 lakhs.
count the number of decimals and move the decimals from right in the quotient.
we get 1.057281 ÷ 7 0.15104

Question 6.
A number multiplied by 9 gives 145.71.  What is the number?
Answer:
16.19
Explanation:
Let the number be x.
9x = 145.71
x = \(\frac{145.71}{9}\)
= \(\frac{145.71 × 100}{9 × 100}\)
= \(\frac{14571}{900}\)
= 16.19
So, 16.19 x 9 = 145.71

16.34 ÷ 10 = 163.4
25.765 ÷ 100 = _____.
347.5 ÷ 100 = ______.
238.4 ÷ 1000 = _____.
What have you found out about dividing a number in decimal form by 10, 100, 1000 and so on?
Answer:
When we divide a decimal by 10, 100 and 1000,
the place value of the digits decreases.
The digits move to the right since the number gets smaller,
but the decimal point does not move.
Explanation:
When we observe the below division, there is no change in decimal places.
16.34 ÷ 10 = 163.4
25.765 ÷ 100 = 0.25765
347.5 ÷ 100 = 3.475
238.4 ÷ 1000 = 0.2384

Other Divisions

A rope of length 8.4 meters is cut into 0.4 meter long pieces. How many pieces can we make?
8.4 meters is 840 centimeters and 0.4 meter is 40 centimeters. So the number of pieces is 840 ÷ 40 = 21
We can write this as
8.4 ÷ 0.4 = 21
What does this mean?
8.4 is 21times 0.4
How about doing this with fractions?
84 = \(\frac{84}{10}\), 0.4 = \(\frac{4}{10}\)
\(\frac{84}{10}\) ÷ \(\frac{4}{10}\) means, finding out the number, \(\frac{4}{10}\) of which is \(\frac{84}{10}\).

And we know that it is \(\frac{10}{4}\) times \(\frac{84}{10}\).
That is \(\frac{84}{10}\) ÷ \(\frac{4}{10}\) = \(\frac{84}{10}\) × \(\frac{10}{4}\) = 21
That is, \(\frac{84}{10}\) ÷ \(\frac{4}{10}\) = \(\frac{84}{10}\) ÷ \(\frac{10}{4}\) = 21
Can we compute 36.75 ÷ 0.5 like this?
36.75 = \(\frac{3675}{100}\), 0.5 = \(\frac{5}{10}\)
\(\frac{3675}{100}\) ÷ \(\frac{5}{10}\) = \(\frac{3675}{100}\) × \(\frac{10}{5}\) = \(\frac{735}{10}\)
That is, 36.75 ÷ 0.5 = 73.5
We can also write \(\frac{36.75}{0.5}\) = 73.5
So how do we find \(\frac{48.72}{0.12}\)?
\(\frac{48.72}{0.12}\) = 48.72 ÷ 0.12 = \(\frac{4872}{100}\) ÷ \(\frac{12}{100}\)
= \(\frac{4872}{100}\) × \(\frac{100}{12}\)
= \(\frac{4872}{12}\)
= 406

Textbook Page No. 119

Question 1.
The area of a rectangle is 3.25 square meters and its length is 2.5 centimeters. What is its breadth’?
Answer:
1.3 meters
Explanation:
Area of rectangle = length x breadth
The area of a rectangle is 3.25 square meters,
length is 2.5 centimeters.
breadth = \(\frac{3.25}{2.5}\)
b = 1.3 meters

Question 2.
A can contains 4.05 liters of coconut oil. It must be filled in to 0.45 liter bottles. How many bottles are needed?
Answer:
9 bottles.
Explanation:
A can contains 4.05 liters of coconut oil.
Capacity of one bottle = 0.45 liters
Number of bottles required = \(\frac{4.05}{0.45}\)
Divided the numerator and denominator by 100
= \(\frac{405}{45}\) = \(\frac{81}{9}\)
= 9 bottles.

Question 3.
Calculate the quotients below:

i) \(\frac{35.37}{0.03}\)
Answer:
1,179
Explanation:
\(\frac{35.37}{0.03}\) = 35.37 ÷ 0.03
= \(\frac{3537}{100}\) ÷ \(\frac{3}{100}\)
= \(\frac{3537}{100}\) × \(\frac{100}{3}\)
= \(\frac{3537}{3}\)
= 1179

ii) \(\frac{10.92}{2.1}\)
Answer:
52
Explanation:
\(\frac{10.92}{2.1}\) = 10.92 ÷ 2.1
= \(\frac{1092}{100}\) ÷ \(\frac{21}{10}\)
= \(\frac{1092}{100}\) × \(\frac{10}{21}\)
= \(\frac{10920}{210}\)
= 52

iii) \(\frac{40.48}{1.1}\)
Answer:
3,680
Explanation:
\(\frac{40.48}{1.1}\) = 4048 ÷ 11
= \(\frac{4048}{100}\) ÷ \(\frac{11}{10}\)
= \(\frac{4048}{100}\) × \(\frac{10}{11}\)
= \(\frac{40480}{11}\)
= 3680

iv) \(\frac{0.045}{0.05}\)
Answer:
0.9
Explanation:
\(\frac{0.045}{0.05}\) = 0.045 ÷ 0.05
= \(\frac{45}{1000}\) ÷ \(\frac{5}{100}\)
= \(\frac{45}{1000}\) × \(\frac{100}{5}\)
= \(\frac{45}{50}\)
= 0.9

v) 0.001 ÷ 0.1
Answer:
0.01
Explanation:
0.001 ÷ 0.1
= \(\frac{1}{1000}\) ÷ \(\frac{1}{10}\)
= \(\frac{1}{1000}\) × \(\frac{10}{1}\)
= \(\frac{1}{100}\)
= 0.01

vi) 5.356 ÷ 0.13
Answer:
41.2
Explanation:
5.356 ÷ 0.13
= \(\frac{5356}{1000}\) ÷ \(\frac{13}{100}\)
= \(\frac{5356}{1000}\) × \(\frac{100}{13}\)
= \(\frac{5356}{130}\)
= 41.2

vii) \(\frac{0.2 \times 0.4}{0.02}\)
Answer:
4
Explanation:
\(\frac{0.2 \times 0.4}{0.02}\)
\(\frac{0.08}{0.02}\) = 4

viii) \(\frac{0.01 \times 0.01}{0.001 \times 0.1}\)
Answer:
1
Explanation:
\(\frac{0.01 \times 0.01}{0.001 \times 0.1}\)
\(\frac{0.0001}{0.0001}\) = 1

Question 4.
12125 divided by which number gives 1.2125?
Answer:
10,000
Explanation:
Let number be divided by x.
\(\frac{12125}{x}\) = 1.2125
x = \(\frac{12125}{1.2125}\)
x = 10,000
12125 divided by 10,000 gives 1.212

Question 5.
0.01 multiplied by which number gives 0.00001?
Answer:
0.001
Explanation:
Let the number be multiplied by x.
0.01x = 0.00001
x = \(\frac{0.00001}{0.01}\)
multiply both numerator and denominator with 100
x = \(\frac{0.00001}{0.01}\) x \(\frac{100}{100}\)
x = 0.001
0.01 x 0.001 = 0.00001

Fractions and decimals

Fractions written as decimals are of denominators 10,100, 1000 and so on.
For some fractions, we can first change the denominator into one of these and then write in decimal form. For example,
\(\frac{1}{2}\) = \(\frac{5}{10}\) = 0.5
\(\frac{1}{4}\) = \(\frac{25}{100}\) = 0.25
\(\frac{3}{4}\) = \(\frac{75}{100}\) = 0.75

How do we write \(\frac{1}{8}\) in decimal form?
8 = 2 × 2 × 2
So, multiplying 8 by three 5’s we can make it a product of 10’s.
8 × (5 × 5 × 5) = (2 × 2 × 2) × (5 × 5 × 5)
= (2 × 5) × (2 × 5) × (2 × 5)
= 10 × 10 × 10 = 1000
5 × 5 × 5 = 125, right? So
\(\frac{1}{8}\) = \(\frac{125}{8 \times 125}\) = \(\frac{125}{1000}\) = 0.125
In much the same way,
\(\frac{5}{8}\) = \(\frac{5 \times 125}{8 \times 125}\) = \(\frac{625}{1000}\) = 0.625

How about \(\frac{1}{40}\) ?
40 = (2 × 2 × 2) × 5
To get a product of 10’s we have to multiply 40 by two 5’s; that is
40 × 25 = (2 × 2 × 2 × 5) × (5 × 5)
= (2 × 5) × (2 × 5) × (2 × 5)
= 10 × 10 × 10
= 1000
So,
\(\frac{1}{40}\) = \(\frac{25}{40 \times 25}\) = \(\frac{25}{1000}\) = 0.025
And \(\frac{21}{40}\)?
\(\frac{21}{40}\) = \(\frac{21 \times 25}{40 \times 25}\) = \(\frac{525}{1000}\) = 0.525

Similarly, since 125 × 8 = 1000, we can write
\(\frac{121}{125}\) = \(\frac{121 \times 8}{125 \times 8}\) = \(\frac{968}{1000}\) = 0.968
Thus we can find the decimal form of any fraction whose denominator is a multiple of 2’s and 5’s.

Now look at this problem:
24 kilograms of sugar are packed into 25 packets of the same size. How much does each packet contain?
24 kilograms means 24000 grams. So each packet contains \(\frac{24000}{25}\) grams.
\(\frac{24000}{25}\) = 960

Thus each packet contains 960 grams or 0.96 kilograms.
We can do this in a different way. Each packet contains \(\frac{24}{25}\) kilograms.
\(\frac{24}{25}\) = \(\frac{24 \times 4}{25 \times 4}\) = \(\frac{96}{100}\) = 0.96
So, one packet contains 0.96 kilograms.

Textbook Page No. 121

Question 1.
Find the decimal forms of the fractions below:

i) \(\frac{3}{5}\)
Answer:
0.6
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.
\(\frac{3}{5}\) = 0.6

ii) \(\frac{7}{8}\)
Answer:
0.875
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

iii) \(\frac{5}{16}\)
Answer:
0.3125
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

iv) \(\frac{3}{40}\)
Answer:
0.075
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

v) \(\frac{3}{32}\)
Answer:
0.09375
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

vi) \(\frac{61}{125}\)
Answer:
0.488
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

Question 2.
Write the answer to the questions below in decimal form.
(i) 3 liters of milk is used to fill 8 identical bottles. How much does each bottle contain?
Answer:
0.375 liters.
Explanation:
3 liters of milk is used to fill 8 identical bottles.
Each bottle contains = \(\frac{3}{8}\)
= 0.375 liters.

(ii) A 17 meter long string is cut into 25 equal parts. What is the length of each part?
Answer:
0.68 meters
Explanation:
A 17 meter long string is cut into 25 equal parts.
The length of each part = \(\frac{17}{25}\)
= 0.68 meters

(iii) 19 kilograms of rice is divided among 20 people. How much does each get?
Answer:
0.95 kilograms.
Explanation:
19 kilograms of rice is divided among 20 people.
Total kilograms of rice each get = \(\frac{19}{20}\)
= 0.95 kg

Question 3.
What is the decimal form of \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\)?
Answer:
0.9375
Explanation:
\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\)
numerators are same denominators are different, so find the LCM of denominators
= \(\frac{(1 ×16) + (1 × 8) + (1 × 4) + (1×  2)}{32}\)
= \(\frac{16 + 8 + 4 + 2}{32}\)
= \(\frac{30}{32}\)
= 0.9375

Question 4.
A two digit number divided by another two digit number gives 4.375.What are the numbers’?
Answer:
The two digit numbers are 70 and 16.
Explanation:
Given number is 4.375
convert the decimal number to whole number,
\(\frac{4375}{1000}\)
find the factors of above fraction,
factors of 4375 = 5 x 5 x 5 x 5 x 7
factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5
Divide the common factors in both the numbers,
\(\frac{35}{8}\)
multiply both numerator and denominator with 2,
\(\frac{35 × 2}{8 × 2}\) = \(\frac{70}{16}\) = 4.375

Question 1.
What is the volume of a rectangular block of length 25.5 centimeters, breadth 20.4 centimeters and height 10.8 centimeters?
Answer:
5618.16 cm³
Explanation:
Given, length 25.5 centimeters, breadth 20.4 centimeters and height 10.8 centimeters.
Volume of a cuboid = l x b x h
V = 25.5 x 20.4 x 10.8 = 5618.16 cm³

Question 2.
The heights of three boys sitting on a bench are 130.5 centimeters 128.7 centimeters and 134.6 centimeters .What is the average height’?
Answer:
131.26 centimeters.
Explanation:
The heights of three boys sitting on a bench are,
130.5 centimeters 128.7 centimeters and 134.6 centimeters.
The average height of 3 boys = \(\frac{130.5 + 128.7 + 134.6}{3}\)
= 393.8 ÷ 3 = 131.26

Question 3.
Calculate \(\frac{4 \times 3.06}{3}\).
Answer:
4.08
Explanation:
\(\frac{4 \times 3.06}{3}\)
\(\frac{12.24}{3}\) = 4.08

Question 4.
The price of 22 pencils is 79.20 rupees. What is the price of 10 pencils’?
Answer:
36 rupees
Explanation:
The price of 22 pencils is 79.20 rupees.
Price of each pencil = 79.20 ÷ 22 = 3.60 rupees
The price of 10 pencils = 3.60 x 10 = 36 rupees.

Question 5.
Calculate the following:

i) \(\frac{2.3 \times 3.2}{0.4}\)
Answer:
18.4
Explanation:
\(\frac{2.3 \times 3.2}{0.4}\)
\(\frac{7.36}{0.4}\) = 18.4

ii) \(\frac{0.01 \times 0.001}{0.1 \times 0.01}\)
Answer:
0.01
Explanation:
\(\frac{0.01 \times 0.001}{0.1 \times 0.01}\)
\(\frac{0.00001}{0.001}\) = 0.01

Question 6.
Dividing 0.1 by which number gives 0.001?
Answer:
100
Explanation:
Let the number to be divided is x
0.1 ÷  x = 0.001
0.1 = 0.001x
x = \(\frac{0.1}{0.001}\)
multiply both numerator and denominator with 1000.
x = \(\frac{100}{1}\)
x = 100

You can Download Joining Angles Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles

Joining Angles Text Book Questions and Answers

Joining Angles Textbook Page No. 123

Your geometry box has two set squares. Each of them has three angles.

Look at an angle drawn with a corner of a set square.

How much is ∠CAB?
What if we draw another angle on top of this, using a corner of the other set square?
Answer:
∠CAB = 30
Explanation:
By observing the figure triangle ABC, we came to conclude that,
as ∠CAB is 30 degrees, ∠ABC is 90 degrees and angle ∠BCA is 60 degrees
These set squares come in two usual forms,
both right triangles one with 90-45-45 degree angles,
the other with 30-60-90 degree angles.
Combining the two forms by placing the hypotenuses together will get 15° and 75° angles.

How much is ∠DAC?
And ∠DAB?
Answer:
75°

Explanation:
By observing set square, the figure angle ∠DAB,
we conclude that, as ∠DAB is 75 degrees, ∠CAB is 30 degrees and angle ∠DAC is 45 degrees.
∠DAC = 45° And ∠DAB = 75°

 

Answer:
∠DAB = 75°

Explanation:
By observing set square, the figure angle ∠DAB,
we conclude that, as ∠DAB is 75 degrees, ∠CAB is 30 degrees and angle ∠DAC is 45 degrees.
∠DAC = 45° And ∠DAB = 75°

Now suppose we draw angles as shown below:

How much ∠DAC?
Like this, angles of what different measures can we draw using the two set squares?
Answer:
15°
Explanation:
These set squares come in two usual forms, both right triangles
one with 90-45-45 degree angles, the other with 30-60-90 degree angles.
Combining the two forms by placing the hypotenuses together will get 15° and 75° angles.

In each picture below, two angles are given. Calculate the third angle as a sum or difference;

∠DAC = ____ – ____ = _____
Answer:
40°
Explanation:
∠DAC = 40°
∠CAB = 60°
∠DAB = 40° + 60° = 100°

∠DAB = ____ – ____ = _____
Answer:
75°
Explanation:
∠DAB = 30° + 60° = 90°

∠DAC = ____ – ____ = _____
Answer:
50°
Explanation:
Given;
∠DAC = 80°; ∠CAB = 30°
∠DAC = 80° – 30° = 50°

∠DAC = ____ – ____ = _____
Answer:
75°
Explanation:
Given;
∠DAB = 120°; ∠CAB = 45°
∠DAC = 120° – 45° = 75°

Two sides Textbook Page No. 126

Draw a line and then a perpendicular at one end.

We have noted that such an angle measures 90°.
Now extend the horizontal line a bit to the left;

Now there is another angle on the left of the vertical line also. What is its measure?
Perpendicular means straight up, not leaning to the left or right.
So the angle on the left is also 90°.

Now let’s draw a slanted line through the foot of the perpendicular.

How much is the angle on the left of this slanted line’?
A bit more than 90°, right’?
How much more’?

By how much is the angle on the right less than 90°?

Now can’t we calculate the angle on the left also?

see this picture

How much is the angle on the left of the slanted line? Imagine a perpendicular through the point where the lines meet.
By how much is the angle on the right less than 90°?
So, by how much is the angle on the left more than 90°?
Thus the angle on the left is 90° + 40° = 130°.

Joining setsquares Textbook Page No. 128

The picture shows two identical setsquares placed together;

What are the measures of the angles of this triangle?
Answer:
60°
Explanation:
The sum of the equilateral triangle is 120°
60° + 60° + 60° = 120°
60° in three corners as shown in the below figure

In the pictures below, two angles are marked and one of them is given. Find the other:

Answer:
80°
Explanation:

As we know a straight line is having 180° angle.
∠AOB = 180°
∠AOC = ∠AOB – ∠COB
∠AOC = 180° – 100° = 80°

Answer:
140°
Explanation:

As we know a straight line is having 180° angle.
∠AOB = 180°
∠AOC = ∠AOB – ∠COB
∠AOC = 180° – 40° = 140°

 

Answer:
130°
Explanation:

as we know a straight line is having 180° angle
∠AOB = 180°
∠AOC = ∠AOB – ∠COB
∠AOC = 180° – 50° = 130°

 

Answer:
100°
Explanation:

As we know a straight line is having 180° angle
∠AOB = 180°
∠AOC = ∠AOB – ∠COB
∠AOC = 180° – 80° = 100°

Meeting lines

See these pictures:

All of them show two lines meeting ; and each has two angles, one on the left and one on the right.
In the first picture, both angles are 90°. In the second ,the angle on the right is less than 90° and the angle on the left is greater than 90°; in the third picture , it is the other way round.

In the second and third picture, the angle on one side is that much more than 90° as the angle on the other side is less than 90°.
So, the sum of the angles on either side is 90° + 90° = 180°, right?

Thus we can write this as a general principle:
When two lines meet, the sum of the angles on either side is 180°.

Two such angles, made by two lines meeting is called a liner pair. So we can state this principle like this also:
The sum of the angles in a linear pair is 180°.

Textbook Page No. 130

Question 1.
How much is ∠ACE in the picture below?

Answer:
105°
Explanation:
as we know a straight line is having ∠ACB 180° angle
∠ACB = 180°
∠ACE = ∠ACB – ∠ECB
∠ACE = 180° – 75° = 105°

Question 2.
What is the angle between the lines in this picture?

Answer:
75°
Explanation:
Sum of the angles = 180°
with the help of a protractor we can measure the angles as 45° and 60°
180° – (45° + 60°)
180° – 105° = 75°

Question 3.
In the picture below, ∠ACE = ∠BCD. Find the measure of each.

Answer:
150°
Explanation:
Sum of the angles = 180°
∠ACE + ∠BCD = 180° – ∠DCE
180° – 30° = 150°
∠ACE + ∠BCD = 150°
∠ACE = 75°
∠ACE = ∠BCD = 75°

Question 4.
One angle of a linear pair is twice the other. How much is each?
Answer:
60° and 120°
Explanation:
Sum of the angles = 180°
If one angle of linear is 60°
twice the other is 60° + 60° = 120°

Question 5.
The angles in a linear pair are consecutive odd numbers. How much is each’?
Answer:
89° and 91°
Explanation
A linear pair are consecutive odd number.
x + x + 2 = 180°
2x + 1 = 180°
2x = 178°
x = 89°
The two consecutive odd numbers are 89° and and 91°

So, required two angles that are consecutive odd numbers and are in a linear pair are 89° and 91°.

Crossing lines

See the picture:

How much is the angle on the left?
What if we extend the top line downwards crossing the horizontal line?

Now we have two angles below also. What are their measures?
The angles above and below, on the right of the slanted line, form a linear pair, right’?

Thus don’t we get one angle below?
Like this, the angle above and below on the left also form a linear pair.

Thus we get the angle below on the left also. Let’s look at all the angles together:

Linear pair Textbook Page No. 131

Draw the line AB and a point C on it. Draw a circle centered at C. Mark a point D on the circle.

Join CD. Now we can hide the circle. By choosing Angle and clicking on B, C, D in order, we get the measure of ∠BCD. In the same way, click on A, C, D to get ∠ACD.

Using Move change the position of D. How do the angles change? Look at the sum of BCD and ACD.
Answer:
Yes,
Explanation:
There is no change in the angles after the move position of D.
The sum of the angle ∠BCD + ∠ACD = 180° degrees

Textbook Page No. 132

Some pictures showing two lines crossing each other are given below. One of the four angles so formed is given. Calculate the other three and write them in the pictures.

Answer:
The angles three are 135°, 45°, 135° as shown in the below figure,

Explanation:
∠AOB is straight angle is equal to 180°
∠BOD and ∠AOD  are adjacent angles,
the sum of ∠BOD and ∠AOD is 180°
∠BOD = 45°
∠AOD = 180° – 45°
∠AOD = 135°
As opposite angle of a bisecting line at one center point is equal.
∠COB and ∠AOD are equal to 135°
∠BOD and ∠AOC are equal to 45°

Answer:
The three angles are 60°, 120°, 60° as shown in the below figure,

Explanation:
∠AOB is straight angle is equal to 180°
∠BOD and ∠AOD  are adjacent angles,
the sum of ∠BOD and ∠AOD is 180°
∠AOD = 120°
∠BOD = 180° – 120°
∠BOD = 60°
as opposite angle of a bisecting line at one center point is equal.
∠COB and ∠AOD are equal to 120°
∠BOD and ∠AOC are equal to 60°

Answer:
90 degrees as shown below,

Explanation:
∠AOB is straight angle is equal to 180°
∠BOD and ∠AOD  are adjacent angles and the sum of ∠BOD and ∠AOD is 180° degrees
∠BOD = 90°
∠AOD = 180° – 90°
∠AOD = 90°
as opposite angle of a bisecting line at one center point is equal
∠COB and ∠AOD are equal to 90°
∠BOD and ∠AOC are equal to 90°

Near and opposite

The picture shows the four angles made by the line CD crossing the line AB:

We can pair these four angles in various ways. Of these, four are linear pairs. which are they?
Answer:

• ∠APC
• ∠BPC
•∠APD
•∠DPB
These are nearby angles in the picture. What about the other two pairs?
Answer:

•∠APC, ∠BPD are opposite angles
•∠APC, ∠BPC are opposite angles

• ∠APC, ∠BPD
• ∠APD, ∠BPC

They are not nearby angles; they are opposite angles. What is the relation between them?

Look at ∠APC and ∠BPD. If we add ∠BPC to any of these, we get 180°. In other words, each of these is ∠BPC subtracted from 180°.
So, ∠APC = ∠BPD.
Similarly, can’t you see that the other pair of opposite angles are also equal?

This we write as a general principle;
The opposite angles formed by two lines crossing each other are equal.

We can combine the general result on nearby and opposite angles.
Of the four angles formed by two lines crossing each other, the sum of the nearby angles is 1800, the opposite angles are equal.

Draw a circle center at a point A. Mark four points B, C, D, E on the circle.

Draw the lines BD and CE. Now hide the circle.
Use Angle to mark the four angles in the picture.
Use Move to change of B, C, D, E.
Observe what happens to the opposite angles.

In the picture above the sum of the green and red angles is 180° and the sum of the green and blue angles is also 180°. So the red and blue angles are equal. Can you see that the green and yellow angles are equal?
Answer:
Yes, the red and blue angles are equal.
Explanation:
As opposite angles of a bisecting two lines at one point, the opposite angles are equal.
So, the green and yellow angles are equal.
Angles that are opposite to each other when two lines cross, are also known as vertical angles, because the two angles share the same corner.
Opposite angles are also congruent angles, when they are equal or have the same measurement.

Question 1.
Two picture of lines passing through a point are given below. Some of the angles are given. Calculate the other angles marked and write in the figure:

Answer:

Explanation:
As A and A’ and X and X’ are straight lines and the straight angle is 180 degrees.
With reference to the center point and bisecting lines of BB’ and CC’ angles are noted.
Same as in the figure XX’ bisects the lines YY’ and ZZ’ at common center the angles are noted. Opposite angles are opposite to each other when two lines cross, are also called vertical angles, because the two angles share the same corner.
Opposite angles are also congruent angles, because they are equal or have the same measurement.

Question 2.
Of the four angles made by two lines crossing each other, one angle is half of another angle. Calculate all four angles.
Answer:
The four angles are 60°, 120°, 60°, 120°.
Explanation:
Linear pair,
∠a + ∠b = 180°.
∠c + ∠d = 180°.
∠a = ∠c
∠b = ∠d
Let ∠a = 2x
∠b = half of ∠a
∠b = x
∠a + ∠b = 180°.
2x + x = 180°.
3x = 180°.
x = 180/3
x = 60°.
∠b = 60°, ∠c = 60°.
∠a = 2x
∠a = 2 x 60
∠a = 120°.
So, the four angles are 60°, 120°, 60°, 120°.

Question 3.
Of the four angles formed by two lines crossing each other, the sum of two angles is 100°. Calculate all four angles.
Answer:
Four angles are 50°, 130°, 50°, 130°
Explanation:
Linear pairs lie on the same line.
Sum of the four angles = 180°.
∠A + ∠C = 100°.
x + x = 100°.
2x = 100°.
x = 100/2 = 50°.
Linear pair,
∠A + ∠B = 180°.
50° + ∠B = 180°.
∠B = 180°- 50°.
∠B = 130°
Linear pair,
∠C + ∠D = 180°.
50° + ∠D = 180°.
∠D = 180°- 50°.
∠D = 130°
So, the four angles are 50°, 130°, 50°, 130°

You can Download Letter Math Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Letter Math Text Book Questions and Answers

Addition and Subtraction Textbook Page No. 149

Mary is now 4 years old; and her brother Johny is 8.
What would be Mary’s age after 2 years?
And Johny’s age?
What were their ages 3 years ago?
Can you fill up the blanks in the table.


In this, how do we compute Johny’s age from Mary’s age?
Adding 4 to Mary’s age gives Johny’s age, right?
We can shorten it like this:
Johny’s age = Mary’s age +4
There’s a trick to shorten it further. Let’s write m for Mary’s age and
j for Johny’s age. Then we can write
j = m + 4
Answer:

Explanation:
Difference between Johny’s age and Mary’s age = Johny’s age – Mary’s age
= 8 – 4
= 4.
Johny’s age = Difference between Johny’s age and Mary’s age + Mary’s age
=> 1 + 4
=> 5 years.
Johny’s age – Difference between Johny’s age and Mary’s age = Mary’s age
=> 6 – 4
=> 2 years.
Johny’s age – Difference between Johny’s age and Mary’s age = Mary’s age
=> 7 – 4
=> 3 years.
Johny’s age = Difference between Johny’s age and Mary’s age + Mary’s age
=> 5 + 4
=> 9 years.

Here, the letter m which stands for Mary’s age can be any of the numbers 1, 2, 3 and so on. Accordingly we get
the numbers 5, 6, 7 and so on as j.
Another problem:
See this figure;

However we draw the slanted line, what is the relation connecting the angles on the left and right?
We can write it like this:

The sum of the angles on the left and right is 180°.
What if we write the measure of the angle on the left as l° and the measure of the angle on the right as r°, then we can shorten this as:
l + r = 180

One fact different ways Textbook Page No. 150

We can say the same fact in different ways.

1. Johny is 4 years older than Mary.
2. Mary is 4 years younger than Johny.
3. The difference in ages between Johny and his younger sister Mary is 4.

When we write such relations using letters also, we can put them in different ways. If we write j for Johny’s age and m for Mary’s age, the statements above become

1. j = m + 4
2. m = j – 4
3. j – m = 4

In what all ways can we state the relationship between the angles a line makes on the two sides of
another line?
Can you write each of these using letters?7*

Now look at this figure:

A figure of four sides. Drawing a line from one corner to the opposite comer, we can split it into two triangles;

What about a five sided figure?

We can draw lines from one comer to two other corners to get three triangles.

And a six sided figure?

Answer:

Explanation:
A six sided figure can be drawn by joining the edges and making two straight lines in between the lines.

Changing angles 

We can use a slider to make an angle which can be changed as we like
Choose slider and click on the Graphic view. In the window which opens up, choose Integer and give Min = 0, Max = 180. Click Apply. We get a slider named n.

Mark two points A, B.

Choose Angle with Given size and click on A and B in order. In the window which opens up, give n° as the size of the angle and click OK. We get a new point A’.
Draw the lines BA and BA’. As we move the slider, the size of angle B changes.

Draw seven sided and eight sided figures like this. From one particular comer, draw lines to other corners to split them into triangles. Make a table like this:

Answer:

Explanation:
Number of lines + 1 = Number of Triangles.
=> 3 + 1 = 4.
=> 4 + 1 = 5.
=> 5 + 1 = 6.

In a 12 sided figure, how many such lines can be drawn from one particular comer?
Answer:
9 lines can be drawn from one particular comer.

Explanation:
Number of Sides – 3 = Number of lines
=> 12 – 3
=> 9.

• What is the relation connecting the number of sides and number of lines in general?
Answer:
The relation connecting the number of sides and number of lines in general is they are directly connected to eachother.

Explanation:
The number of lines and number of sides are directly related to eachother in forming the number of shapes.

• What is the relation connecting the number of sides and the number of triangles?
Answer:
The number of lines is the relation connecting the number of sides and the number of triangles.

Explanation:
The relation connecting the number of sides and the number of triangles is the number of lines used.

• What is the relation connecting the number of lines and the number of triangles?
Answer:
Number of sides used to be formed is the relation connecting the number of lines and the number of triangles.

Explanation:
Number of sides connects the number of lines and the number of triangles.

Writing the number of sides as s, the number of lines as l and the number of triangles as t, how do we write these relations?
This is how Sneha wrote them:

• s – 3 = l
• t + 2 = s
• t – 1 = l

In what other ways can we write them?
Try!
Answer:
Other ways can we write them are:
t = s – 2.
t = l + 1.

Explanation:
t + 2 = s
=> t = s – 2.
t – 1 = l
=> t = l + 1.

Regular polygons Textbook Page No. 152

Polygons with equal sides and equal angles are called regular polygons.

We can easily draw such figures using Geogebra. We use Regular polygon to do this.

Choose this and click on two positions. In the window, which opens up give the number of sides and click OK.

Now look at this problem:

A shopkeeper decides to sell something for a price 1oo rupees more than what he bought it for. If the had bought it for 500 rupees, at what price would he sell it? What if the price at which he bought is 600 rupees? Here what is the relation between the prices at which he bought and the price at which he sells?

This hundred rupee increase in price is called the profit in the sale. If he wants a profit of 150 rupees, what would be the relation between how much the shopkeeper bought it for and how much he sells it for? What if he wants a profit of 200 rupees? Write down these relations using letters.

How do we write in general, the relation between the price at which a shopkeeper buys something, the price at which he sells it and the profit? Adding the profit to the price at which it is bought gives the price at which it is sold.
Writing the price at which it is bought as b, the profit as p and the price at which it is sold as s, we get
s = b + p

In what other ways can you write this relation?
Answer:
In a transaction, the selling price is greater than the cost price, it means we earn a profit.

Explanation:
The selling price of it as S, the profit bought of it as profit P, the cost price for what it is purchased as C.
=> Profit = Selling price -Cost price.
=> P = SP – CP.

Question 1.
For getting books through post, 25 rupees has to be added to the price of the book, as postage. Fill in the blanks of the table below:

In what all ways can we say the relation between the price of a book and the total cost? Write these using letters. What if the postage is 30 rupees? What if it is 35 rupees?

Now if the postage also changes according to the price of the book, in what all ways can we say this relation?
Write them using letters.
Answer:
Profit = Selling price – Cost price.

Explanation:
If the selling price(SP) is lesser than the cost price (CP), whatever difference you get between the two is the loss suffered. Similarly, Loss is C.P. – S.P. Always remember that you calculate profit or loss on the cost price.
Total cost (TC) = Price of the postage(CP) + Postage (P)
=> Total cost (TC) – Price of the postage(CP) = Postage (P)

Question 2.
Make a table showing the number of girls, the number of boys and the total number of children
in each class of your school. What are the different ways of stating the relation between these numbers? Write these using letters.
Answer:
C = G + B = 63.

Explanation:
Number of girls in a class (G)= 31.
Number of boys in a class (B) = 32.
Total number of students in class (C) = Number of girls in a class (G)+ Number of boys in a class (B)
= 31 + 32
= 63.

Question 3.
The sides of a triangle are 4 centimetres, 6 centimetres and 8 centimetres. What is its perimeter?
Denoting the length of the sides as a, b, c and the perimeter as p, how do we write the relation between them?
Answer:
Perimeter of the triangle (p) = 18 cm.

Explanation:
Side of the triangle (a) = 4 cm.
Side of the triangle (b) = 6 cm.
Side of the triangle (c) = 8 cm.
Perimeter of the triangle (p) = Side of the triangle (a) + Side of the triangle (b) + Side of the triangle (c)
= 4 + 6 + 8
= 10 + 8
= 18 cm.

Textbook Page No. 154

We can draw regular polygon with as many sides as we like, using a slider.
Choose slider and click. In the window which opens up, choose Integer and type 3 for min. We get a slider named n.

Choose Regular polygon and click on two positions. In the window which opens up, give n as the number of sides instead of a specific number

Click and drag the dot on the slider to change the number n The number of sides of the polygon changes accordingly.

Letter multiplication

Rani is making triangles with matchsticks:

How many triangles are there in the picture?

How many matchsticks are used to make them?
How did you calculate it?
Did you add repeatedly as 3 + 3 + 3 + 3 = 12?
Or multiply as 3 × 4 = 12?

How many matchsticks do we need to make 10 triangles like this?
In general, the number of matchsticks is three times the number of triangles.
How about writing this in shorthand, using letters?
If we take the number of triangles as t and the number of matchsticks as m, what is the relation between the numbers t and m?
m = 3 × t

When we write numbers as letters, we don’t usually write the multiplication sign; that is, we omit the multiplication in 3 × t and write it simply as 3t. Thus if we take the number of matchsticks Rani needs to make t triangles as m, then we usually write the relation between the numbers m and t as
m = 3t
Now let’s see how many triangles can be made with 45 matchsticks. The number of matchsticks is three times the number of triangles. So the number of triangles is a third of the number of matchsticks.
So with 45 matchsticks, we can make \(\frac{45}{3}\) = 15 triangles.
In general, the number of triangles is the number of matchsticks divided by 3.

We can write this also using letters.
t = n ÷ 3.
We usually wnte this as t = \(\frac{m}{3}\).

Inside a circle Textbook Page No. 155

We have seen in the lesson Angles how regular polygons can be drawn by dividing a circle into equal parts.
Let’s see how we can do this in GeoGebra.
Make an Integer slider named n. Draw a circle centred at a point A and mark a point B on it. Choose Angle with
Given Size and click on B, A in order.
In the window which opens up. give the angles as (360/n)°. We get a new point B on the circle. Choose Regular
Polygon and click on B, B’. Give the number of sides as n.

Question 1.
How many squares are there in the picture? How many matchsticks are used to make it? How many matchsticks are needed to make five squares like this? In what different ways can we state relation between the number of squares and number of matchsticks?
Write them using letters.

Answer:
Total number of matchsticks used = Number of squares in the figure × Number of matchsticks used in a square.
Number of matchsticks used to make 5 square = 20.

Explanation:
Number of squares in the figure = 3.
Number of matchsticks used in a square = 4.
=> Total number of matchsticks used = Number of squares in the figure × Number of matchsticks used in a square
= 3 × 4
= 12.
Number of squares needed to make = 5.
Number of matchsticks used to make 5 square = Number of squares needed to make × Number of matchsticks used in a square
= 5 × 4
= 20.

Question 2.
All children in the school bought pens from the co-operative store, at 5 rupees each. Write in the table, how much the children of various classes paid.

In what all ways can we say the relation between the number of children and the amount they paid? Write them using letters.
Answer:
Total amount 6A childen paid = 170 rupees.
Total amount 6B childen paid = 160 rupees.
Total amount 6C childen paid = 180 rupees.

Explanation:
Number of children 6A = 34.
Number of children 6B = 32.
Number of children 6C = 36.
Cost of each pen children bought = 5 rupees.
Total amount 6A childen paid = Number of children 6A × Cost of each pen children bought
= 34 × 5
= 170 rupees.
Total amount 6B childen paid = Number of children 6A × Cost of each pen children bought
= 32 × 5
= 160 rupees.
Total amount 6C childen paid = Number of children 6A × Cost of each pen children bought
= 36 × 5
= 180 rupees.

Question 3.
What is the perimeter of a square of side 5 centimetres. What about a square of perimeter 6 centimetres? In what all ways can we say the relation between the length of a side and perimetre of a square? Write all these using letters.
Answer:
Perimeter of a square (p)= 4 × Side of the square (s)

Explanation:
Side of the square (s)= 5 cm.
Perimeter of a square (p)= 4 × Side of the square (s)
= 4 × 5
= 20 cm.
Perimeter of a square (p) = 6 cm.
Perimeter of a square (p)= 4 × Side of the square (s)
=> 6 = 4 × Side of the square (s)
=> 6 ÷ 4 = Side of the square (s)
=> 1.5 cm = Side of the square (s)

Question 4.
How much money does 5 ten rupee notes make? What about 7 ten rupee notes? In what all ways can we say the number of ten rupee notes and the total amount? Using t to denote the number of ten rupee notes and a to denote the total amount, in what all ways can we write this relation?
Answer:
Total amount 5 ten rupee notes make (a) = 50 rupees.
Total amount 7 ten rupee notes make (a) = 70 rupees.

Explanation:
One rupee note = 1 rupees.
Ten rupee note (t) = 10 rupees.
Total amount 5 ten rupee notes make (a) = 5 × Ten rupee note (t)
= 5 × 10
= 50 rupees.
Total amount 7 ten rupee notes make (a) = 7 × Ten rupee note (t)
= 7 × 10
= 70 rupees.

Multiplication again

What is the area of a rectangle of length 5 centimetres and breadth 3 centimetres?
How about a rectangle of length 5\(\frac{1}{2}\) centimetres and breadth 3\(\frac{1}{4}\) metres?
Whatever be the length and breadth, area is their product, right?
How do we write this using letters?

Taking the length as l centimetres, breadth as b centimetres and area as a square centimetres,
a = l × b = lb
See how we have omitted the multiplication sign here also.

Like this, the volume of a rectangular block is the product of its length, breadth and height.
This also we can write using letters. Taking length as l centimetres, breadth as b centimetres, height as h centimetres and volume as v cubic centimetres, we can write
v = lbh

Textbook Page No. 157

Using the angles drawn in changing angle, we can draw a nice picture.
First change the values of slider n from 0 to 360. Use Circle with Centre through Point to draw a circle centred at A’ through A. Right click on the circle and choose trace on.

Now right click on the slider and choose Animation on. Don’t you get a picture like this?

Question 1.
What is the total price of 5 pens, each of price 8 rupees? What is the price of 10 notebooks, each of price 12 rupees?

i. In what all ways can we say the relation between the price of something, the number bought and the total price?
Answer:
Total price (t) = Cost of each pen × Number of pens = 40 rupees.
Total price (t) = Cost of each notebook × Number of notebooks = 120 rupees.

Explanation:
Cost of each pen (p) = 8 rupees.
Number of pens (n) = 5.
Total price (t) = Cost of each pen × Number of pens
= 8 × 5
= 40 rupees.
Cost of each notebook (p) = 12 rupees.
Number of notebooks (n) = 10.
Total price (t) = Cost of each notebook × Number of notebooks
= 12 × 10
= 120 rupees.

ii. Taking the price of an article as p, their number as n and the total price as t, in what all ways can we write the relations between p, n and t ?
Answer:
Total price of the articles (t) = Cost of each article (p) × Number of articles (n)

Explanation:
Total price of the articles (t) = Cost of each article (p) × Number of articles (n)
=> Total price of the articles (t) – Cost of each article (p)= Number of articles (n)
=> Total price of the articles (t) ÷ Number of articles (n) = Cost of each article (p)

Question 2.
One litre of kerosene weighs 800 grams.

i. What is the weight of 2 litres of keroscene?
Answer:
Total weigh of kerosene = 1,600 liters.

Explanation:
Weigh of one litre of kerosene = 800 grams.
Number of liters of kerosene = 2.
Total weigh of kerosene = Weigh of one litre of kerosene × Number of liters of kerosene
= 800 × 2
= 1,600 liters.

ii. What is the weight of \(\frac{1}{2}\) litre of keroscene?
Answer:
Total weigh of kerosene = 400 liters.

Explanation:
Weigh of one litre of kerosene = 800 grams.
Number of liters of kerosene = \(\frac{1}{2}\)
Total weigh of kerosene = Weigh of one litre of kerosene × Number of liters of kerosene
= 800 × \(\frac{1}{2}\)
= 400 liters.

iii. What is the weight of 1 millilitre of keroscene?
Answer:
Weight of 1 millilitre of keroscene = 0.8.

Explanation:
Weigh of one litre of kerosene = 800 grams.
1 milliliters of kerosene = ??
Conversation:
1 liter = 1000 milliliters.
=> ?? = 1 milliliter
=> 1 × 1 = ?? × 1000
=> 1 ÷ 1000 = ??
=> 800 ÷ 1000 = ??
=> 0.8 milliliters= ??

iv. Taking the weight of v millilitres of kersone as w grams, write a relation between v and w.
Answer:
Weight of millilitres of kersone (v) = Weight of kersone ÷ 1000.

Explanation:
Weight of millilitres of kersone = (v)
Weight of kersone = w grams.
Conversation:
1 liter = 1000 milliliters.
=> Weight of millilitres of kersone (v) = Weight of kersone ÷ 1000.

Question 3.
One cubic centimetre of iron weighs 7.8 grams.

i. Taking the volume of an iron object as v cubic centimetres and weight w grams, write a relation between v and w.
Answer:
Density = w/v.

Explanation:
Weigh of one cubic centimetre of iron = 7.8 grams.
volume of an iron object = v cubic centimetres.
Weight of iron = w grams.
Weight is the amount of matter an object contains, while volume is how much space it takes up.

ii. Taking the length, breadth and height of a rectangular block of iron as l, b, h and its weight as w, write a relation between l, b, h and w.
Answer:
A relation between l, b, h and w is that we get to know the area of the rectangular block of iron.

Explanation:
Length of the rectangular block of iron = l.
Breadth of the rectangular block of iron = b.
Height of the rectangular block of iron = h.
Weight of the rectangular block of iron = w.
=> Area of four walls = 2 × (lh + bh)
$=>2(l+b)h$

Sequence rule

Look at these numbers:
1, 1, 2, 3, 5, 8, ………….
Can you say what the next number is?
For any three consecutive numbers a, b, c of this sequence, we must have a + b = c
Now try writing some more numbers of this. It is called Fibonacci sequence.

Multiplication and addition

Ravi has 3 ten rupee notes and a one rupee coin; Lissy has 5 ten rupee notes and a one rupee coin.
How much money does Ravi have?
And Lissy?
How did you calculate?
Similarly how much does 25 ten rupee notes and a one rupee coin make?
(10 × 25) + 1 = 251
In general, how much money does some ten rupee notes and a one rupee coin make?
We have to multiply the number of notes by 10 and add 1, right?

Let’s write it using letters.
Take the number of ten rupee notes as t.
So how much money is t ten rupee notes and a one rupee coin?
What about 8 ten rupee notes and 7 one rupee coins?
What is the general method to calculate the amount of money, some ten rupee notes and some one rupee coins make?
Multiply the number of notes by 10 and add the number of coins.
How do we write this using letters?
t ten rupee notes and c coins make 10t + c rupees.

Question 1.
How much money does 8 ten rupee notes and 2 five rupee notes make? What about 7 ten rupee notes and 4 five rupee notes?

i. How do we say the relation between the number of ten rupee notes, the number of five rupee notes and the total amount?
Answer:
The relation between the number of ten rupee notes, the number of five rupee notes and the total amount is ten rupee notes are 4 times greater than five rupee notes.

Explanation:
Number of ten rupee notes = 8.
Number of five rupee notes = 2.
Total amount = (Number of ten rupee notes × 10) + (Number of five rupee notes × 2)
= (8 × 10) + (5 × 2)
= 80 + 10
= 90 rupees.
Number of seven rupee notes = 7.
Number of five rupee notes = 4.
Total amount = (Number of ten rupee notes × 10) + (Number of five rupee notes × 2)
= (7 × 7) + (5 × 4)
= 49 + 20
= 69 rupees.

ii. Taking the number of ten rupee notes as t, the number of five rupee notes as f and the total amount as a how do we write the relation between t, f and a?
Answer:
The relation between t, f and a is a direct relationship.

Explanation:
Number of ten rupee notes = t.
Number of five rupee notes = f.
Total amount = a.
Total amount = (Number of ten rupee notes × 10) + (Number of five rupee notes × 5)
=> a = (t × 10) + (f × 5)

Question 2.
The price of a pen is 7 rupees and the price of a notebook is 12 rupees.

i. What is the total price of 5 pens and 6 notebooks?
Answer:
Total price of pens = 35 rupees.
Total price of notebooks = 72 rupees.

Explanation:
Price of a pen = 7 rupees.
Number of pens = 5.
Total price of pens = Price of a pen × Number of pens
= 7 × 5
= 35 rupees.
Price of a notebook = 12 rupees.
Number of notebooks = 6.
Total price of notebooks = Price of a notebook × Number of notebooks
= 12 × 6
= 72 rupees.

ii. What about 12 pens and 7 notebooks?
Answer:
Total price of pens = 84 rupees.
Total price of notebooks = 84 rupees.

Explanation:
Price of a pen = 7 rupees.
Number of pens = 12.
Total price of pens = Price of a pen × Number of pens
= 7 × 12
= 84 rupees.
Price of a notebook = 12 rupees.
Number of notebooks = 7.
Total price of notebooks = Price of a notebook × Number of notebooks
= 12 × 7
= 84 rupees.

iii. What is the relation between the number of pens, the number of books and the total price?
Answer:
The number of pens, the number of books and the total price is a direct relationship.

Explanation:
The number of pens, the number of books and the total price is a direct relationship as in number of pens increases the total price increases and same with the number of books.

iv. Taking the numbers of pens as p, the number of notebooks as n and total price as t, how do we write the relation between them?
Answer:
The relation between them is that if any one increases among them automaticaaly the other one increases.

Explanation:
Numbers of pens = p.
Number of notebooks = n.
Total price = t.
If number of pens increases the total price of pens will increase and same with the number of notebooks.

Question 3.
What length of wire is neeed to make a triangle of sides 10 centimetres each? To make a square of side 10 centimetres?

i. What is the total length needed to make 5 such triangles and 6 such squares?
Answer:
Total length of wire is needed to make triangles = 50 cm.
Total length of wire is needed to make squares = 60 cm.

Explanation:
Length of wire is needed to make a triangle = 10 centimetres each.
Number of triangles = 5.
Total length of wire is needed = Length of wire is needed to make a triangle × Number of triangles
= 10 × 5
= 50 cm.
Length of wire is needed to make a square = 10 centimetres each.
Number of squares= 6.
Total length of wire is needed = Length of wire is needed to make a square × Number of squares
= 10 × 6
= 60 cm.

ii. What about 4 triangles and 3 squares?
Answer:
Total length of wire is needed to make triangles = 40 cm.
Total length of wire is needed to make squares = 30 cm.

Explanation:
Length of wire is needed to make a triangle = 10 centimetres each.
Number of triangles = 4.
Total length of wire is needed = Length of wire is needed to make a triangle × Number of triangles
= 10 × 4
= 40 cm.
Length of wire is needed to make a square = 10 centimetres each.
Number of squares= 3.
Total length of wire is needed = Length of wire is needed to make a square × Number of squares
= 10 × 3
= 30 cm.

iii. What is the relation between the number of triangles, the number of squares and the total length of wire?
Answer:
The number of triangles, the number of squares and the total length of wire is direct relationship.

Explanation:
The number of triangles, the number of squares and the total length of wire is that if number of triangles or squares increases the total length of them will increases.

iv. Taking the number of triangles as t, the number of squares as s and the total length as l , how do we write the relation between t, s and l?
Answer:
The relation between t, s and l is that they are directly propotionate to eachother.

Explanation:
Number of triangles = t.
Number of squares = s.
Total length = l.
If number of triangles or squares increases the total length also increases.

Addition and multiplication

Four friends went to buy pens and note books. The price of a pen is 8 rupees and the price of a notebook is 12 rupees. The shopkeeper calculated like this:
Price of 4 pens is 8 × 4 = 32 rupees
Price of 4 notebooks is 12 × 4 = 48 rupees
Total 80 rupees
The friends calculated like this:
The amount one has to spend is 8 + 12 = 20 rupees
Total expense 20 × 4 = 80 rupees
Another problem: we want to make a rectangle with eerkkil bits; of length 5\(\frac{1}{2}\) centimetres and breadth 3\(\frac{1}{2}\) centimetres. What is the total length of eerkkil needed?

We can calculate the total length as
5\(\frac{1}{2}\) + 3\(\frac{1}{2}\) + 5\(\frac{1}{2}\) + 3\(\frac{1}{2}\) = 18
Or, we can calculate it as two eerkkil bits of length 5\(\frac{1}{2}\) centimetres and two of length 3\(\frac{1}{2}\) centimetres;
(2 × 5\(\frac{1}{2}\)) + (2 × 3\(\frac{1}{2}\)) = 11 + 7 = 18

There is a third way: take it as two eerkkil bits of length 5\(\frac{1}{2}\) + 3\(\frac{1}{2}\) centimetres:
(2 × 5\(\frac{1}{2}\) + 3\(\frac{1}{2}\)) = 2 × 9 = 18

Which is the easiest?

So if we take the length of a rectangle as l, breadth as b and perimeter as p, the relation between l, b andp can be written in various ways as
p = l + b + l + b
p = 2l + 2b
p = 2 (l + b)
Usually, the last one is the easiest to use.

For example, we can quickly calculate the perimeter of a rectangle of length 27 centimetres and breadth 43 centimetres as 2 × (27 + 43) = 140 centimetres.

Question 1.
There are 25 children in one room and 35 in another. 5 biscuits are to be given to each. How many biscuits are needed?
i. What if the number of children are 20 and 40?
Answer:
Total number of biscuits needed = 300.

Explanation:
Number of children on one room = 20.
Number of children on another room = 40.
Number of biscuits given to each = 5.
Total number of biscuits needed = (Number of children on one room + Number of children on another room) × Number of biscuits given to each
= (20 + 40) × 5
= 60 × 5
= 300.

ii. If we take the number of children in the first room as f in the second room as s and the total number of biscuits as t, in what all ways can we write the relation between f s and t? What if each is given 6 biscuits instead of 5?
Answer:
If there is a increase in the price of each biscuit the total number of biscuits needed also increases.

Explanation:
Number of children in the first room = f.
Number of children in the second room = s.
Total number of biscuits = t.
Total number of biscuits will be effected if there is a increase in the price of each biscuit increases.

iii. If the number of biscuits given to each is taken as b, in what all ways can we write the relation between f s, t and b?
Answer:
t = (f + s) × b.

Explanation:
Number of biscuits given to each = b.
Number of children in the first room = f.
Number of children in the second room = s.
Total number of biscuits = t.
Total number of biscuits needed = (Number of children on one room + Number of children on another room) × Number of biscuits given to each

Question 2.
In the picture, M is the point right at the middle of AC.

What is the length of AM?
Answer:
length of AM = 4 cm.

Explanation:
Length of AC = AB + BC
=> 6 + 2
= 8 cm.
M is the point right at the middle of AC.
=> Length of AM = Length of AC ÷ 2
=> 8 ÷ 2
= 4 cm.

i. If a 5 centimetre long line is extended by 4 centimetres, at what distance from an end point of the longer line is the point exactly at its middle?
Answer:
Length of the line from an end point of the longer line is the point exactly at its middle =4.5 cm.

Explanation:
Length of long line = 5 cm.
Length of the line extended = 4 cm.
Total length of the line = Length of long line + Length of the line extended
= 5 + 4
= 9 cm
Length of the line from an end point of the longer line is the point exactly at its middle = Total length of the line ÷ 2
= 9 ÷ 2
= 4.5 cm.

ii. What if a 7\(\frac{1}{2}\) centimetre long line is extended by 2\(\frac{1}{2}\) centimetres?
Answer:
Length of the line from an end point of the longer line is the point exactly at its middle = 5 cm.

Explanation:
Length of long line = 7\(\frac{1}{2}\) cm.
Length of the line extended = 2\(\frac{1}{2}\) cm.
Total length of the line = Length of long line + Length of the line extended
= 7\(\frac{1}{2}\) + 2\(\frac{1}{2}\)
= {[(7 × 2) + 1] ÷ 2} + {[(2 × 2) + 1] ÷ 2}
= [(14 + 1) ÷ 2] + [(4 + 1) ÷ 2]
= \(\frac{15}{2}\) + \(\frac{5}{2}\)
= (15 + 5) ÷ 2
= \(\frac{20}{2}\)
= 10 cm.
Length of the line from an end point of the longer line is the point exactly at its middle = Total length of the line ÷ 2
= 10 ÷ 2
= 5 cm.

iii. A line of length l centimetres is extended by e centimetres. The mid point of the long line, is m centimetres away from one of its end points. What is the relation between l, e and m?
Answer:
There is a direct relationship between l, e and m.

Explanation:
Line of length = l centimetres.
Line of length is extended = e centimetres.
Mid point of the long line = m centimetres.
The relation between if line length increases the mid point distance to the end point also increases.

Question 3.
The length of a rectangle is 4 centimetres and its breadth is 3 centimetres. The length is increased by 2 centimetres to make a large rectangle.

i. What is the area of the large rectangle? If the length is increased by 3 centimetres, what is the area of the large rectangle?
Answer:
Area of large rectangle = 42 square cm.

Explanation:
Length of the rectangle = 4 cm.
Breadth of the rectangle = 3 cm.
Area of rectangle = Length of the rectangle  × Breadth of the rectangle
= 4 × 3
= 12 square cm.
length is increased by 3 centimetres
=> Length of the rectangle = 4 cm + 3 cm = 7 cm.
Breadth of the rectangle = 3 cm + 3 cm = 6 cm.
Area of large rectangle = Length of the rectangle  × Breadth of the rectangle
= 7 × 6
= 42 square cm.

ii. Taking the length and breadth of the original rectangle as l centimetres and b centimetres, the increase in length as i centimetres and the area of the large rectangle as a square centimetres, in what all ways can we write the relation between l, b, i and a?
Answer:
The relation between l, b, i and a is that if there is any change in length and breadth of rectangle will automatically increases the area of the rectangle.

Explanation:
Length of the original rectangle = l centimetres.
Breadth of the original rectangle = b centimetres.
Increase in length
=> length increased = i centimetres.
Area of the large rectangle = a square centimetres.
If there is any decrease or increase in the length and breadth, the area of the rectangle also changes.

You can Download How Much of Hundred? Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 9 How Much of Hundred?

How Much of Hundred? Text Book Questions and Answers

Sale Textbook Page No. 135

See the ad?
The table shows the original price of some of the things in this shop. We want to calculate the new prices: How?
It’s 10 rupees less for every 100 rupees. So to compute the reduction in price, we must first find out how many 100’s are there in the original price and then multiply it by 10.

For example , the original price of a fan is 1200 rupees.
That is 12 hundreds: so the reduction in price is
12 × 10 = 120 rupees
We can do both operations together:
\(\frac{1200}{100}\) × 10 = 120
So the price of a fan now is 1200 – 120 = 1080 rupees.
Similarly, can’t you compute the prices of others now?

Answer:
Clock = 450 rupees
Iron = 720 rupees
CFL Bulb = 225 rupees
Mixer = 2250 rupees
Cooker = 1260 rupees

Explanation:
Clock :
the original price of a clock is 500 rupees.
It’s 10 rupees less for every 100 rupees.
That is 5 hundreds: so the reduction in price is
5 × 10 = 50 rupees
We can do both operations together:
\(\frac{500}{100}\) × 10 = 50
So the price of a clock now is 500 – 50 = 450 rupees.

Iron :
the original price of a Iron is 800 rupees.
It’s 10 rupees less for every 100 rupees.
That is 8 hundreds: so the reduction in price is
8 × 10 = 80 rupees
We can do both operations together:
\(\frac{800}{100}\) × 10 = 80
So the price of a iron now is 800 – 80 = 720 rupees.

CFL Bulb :
the original price of a CF Bulb is 250 rupees.
It’s 10 rupees less for every 100 rupees.
That is 2 hundred and 50 fifty rupees
so the reduction in price is
2.5 × 10 = 25 rupees
We can do both operations together:
\(\frac{250}{100}\) × 10 = 25
So the price of a CF Bulb now is 250 – 25 = 225 rupees.

Cooker :
the original price of a cooker is 1400 rupees.
It’s 10 rupees less for every 100 rupees.
That is 14 hundreds: so the reduction in price is
14 × 10 = 140 rupees
We can do both operations together:
\(\frac{1400}{100}\) × 10 = 140
So the price of a cooker now is 1400 – 140 = 1260 rupees.

Mixer :
the original price of a mixer is 2500 rupees.
It’s 10 rupees less for every 100 rupees.
That is 25 hundreds: so the reduction in price is
25 × 10 = 250 rupees
We can do both operations together:
\(\frac{2500}{100}\) × 10 = 250
So the price of a mixer now is 2500 – 250 = 2250 rupees.

Loans

A co-operative bank offers agricultural loans. It must be paid back in one year; and for every hundred rupees, 12 rupees more should be given.
See the amounts of loan some people have taken out:

Calculate how much each should pay back.
To know how much more each should give back, we must find out how many hundreds are there in the loan and multiply it by 12.
As before, we need only divide by 100 and multiply by 12.
For example, Raji’s loan is 1550 rupees.
To calculate how much more she has to pay back, we divide 1550 by 100 and multiply by 12:
\(\frac{1550}{100}\) × 12 = 186
So, Raji has to pay back 1550 + 186 = 1736 rupees.
Like this, calculate how much each has to pay back.

To know how much more each should give back,
we must find out how many hundreds are there in the loan and multiply it by 12.
As before, we need only divide by 100 and multiply by 12.

Sabu’s loan is 4000 rupees.
To calculate how much more she has to pay back, we divide 4000 by 100 and multiply by 12:
\(\frac{4000}{100}\) × 12 = 480
So, Sabu has to pay back 4000 + 480 = 4480 rupees.

Suma’s loan is 5500 rupees.
To calculate how much more she has to pay back, we divide 5500 by 100 and multiply by 12:
\(\frac{5500}{100}\) × 12 = 660
So, Suma has to pay back 5500 + 660 = 6160 rupees.

Gokul’s loan is 3750 rupees.
To calculate how much more she has to pay back, we divide 3750 by 100 and multiply by 12:
\(\frac{3750}{100}\) × 12 = 450
So, Gokul has to pay back 3750 + 450 = 4200 rupees.

Nabeel’s loan is 3800 rupees.
To calculate how much more she has to pay back, we divide 3750 by 100 and multiply by 12:
\(\frac{3800}{100}\) × 12 = 456
So, Nabeel has to pay back 3800 + 456 = 4256 rupees.

Percent

In the first problem. the reduction in price is 10 rupees for each hundred rupees.
We say that there is 10 percent reduction in price.
10 percent is written as 10%.

In the loan problem, 12 rupees more for every hundred rupees should be paid back.
That is, 12%( 12 percent) more should be paid back.

The word per means for each. The cent part comes from the Latin word centum meaning hundred.

Donations Textbook Page No. 137

Each month Joseph donates 8% of his earnings to the Medical Aid Fund. He earned 12000 rupees in January. How much should he donate that month?
8 percent means ,8 for each hundred. So we must find out the number of hundreds in 12000 and multiply it by 8:
\(\frac{12000}{100}\) × 8 = 120 × 8 = 960
So, Joseph would donate 960 rupees that month.
This can also be calculated as 12000 × \(\frac{8}{100}\).
That is, \(\frac{8}{100}\) of 12000.

Joseph’s friend Ah donates 12% of his monthly earnings. In January he earned 15000 rupees. How much would he give.
We can think of 12% as 12 for each hundred and calculate
\(\frac{15000}{100}\) × 12
Or we can think of 12% as \(\frac{12}{100}\) and calculate.
15000 × \(\frac{12}{100}\)
Do it any way you like.

Answer:
18,000
Explanation:
Joseph’s friend Ah donates 12% of his monthly earnings.
In January he earned 15000 rupees.
Total amount he gave = \(\frac{15000}{100}\) × 12
= 150 x 12 = 18,000
Or we can think of 12% as \(\frac{12}{100}\)
15000 × \(\frac{12}{100}\)
= \(\frac{1,800,000}{100}\)
= 18,000

Textbook Page No. 138

Question 1.
See the ad.

Sheela bought cloths worth 1800 rupees. How much should she pay?

Answer:
1260 rupees,
Explanation:
30 percent means ,30 for each hundred.
So we must find out the number of hundreds in 1800 and multiply it by 30:
Sheela bought cloths worth 1800 rupees
\(\frac{1800}{100}\) × 30 = 18 x 30 = 540
So, Sheela amount to be paid after discount is
1800 – 540 = 1260 rupees

Question 2.
Johny save 15% of his earnings each month. In January he got 32000 rupees. How much would he save?

Answer:
4800 rupees,
Explanation:
Johny save 15% of his earnings each month, in January he got 32000 rupees.
15 percent means ,15 for each hundred.
So we must find out the number of hundreds in 32000 and multiply it by 15:
\(\frac{32000}{100}\) × 15 = 320 x 15 = 4800 rupees

Question 3.
A TV manufacturer decides to raise prices by 5% next month. The price of a model is 26000 rupees now. What would be its price next month?

Answer:
27300 rupees,
Explanation:
A TV manufacturer decides to raise prices by 5% next month.
The price of a model is 26000 rupees now.
5 percent means ,5 for each hundred.
So we must find out the number of hundreds in 26000 and multiply it by 5:
\(\frac{26000}{100}\) × 5 = 260 x 5 = 1300 rupees
A TV manufacturer decides to raise prices by 5% next month = 1300 rupees
The TV price in the next month is,
26000 + 1300 = 27300 rupees

Question 4.
A car manufacturer decides to lower prices by 2% from next month. What would be the price next month for a car, now priced at 250000 rupees?

Answer:
245000 rupees
Explanation:
A car manufacturer decides to lower prices by 2% next month.
The price of a model is 250000 rupees now.
2 percent means ,2 for each hundred.
So we must find out the number of hundreds in 250000 and multiply it by 2:
\(\frac{250000}{100}\) × 2 = 2500 x 2 = 5000 rupees
A car manufacturer decides to lower prices by 2% next month = 5000 rupees
The car price in the next month is,
250000 – 5000 = 245000 rupees

Question 5.
A company pays 8% of a month’s salary as festival allowance. How much festival allowance would a person whose salary is 12875 rupees get?

Answer:
13905 rupees
Explanation:
A company pays 8% of a month’s salary as festival allowance.
The price of a model is 12875 rupees now.
8 percent means ,8 for each hundred.
So we must find out the number of hundreds in 12875 and multiply it by 8:
\(\frac{12875}{100}\) × 8 = 128.75 x 8 = 1030 rupees
festival allowance would a person whose salary is 12875 rupees get is,
12875 + 1030 = 13905 rupees

Another percent

In a school, 240 children took an exam and 40% of them got A grade in all subjects.
What does it mean?
There is no sense in saying that 40 children out of every 100 got A grade.
Here it means, \(\frac{40}{100}\) of the total number of children got A grades in all subjects.
That is, the number of children who got A grade is
240 × \(\frac{40}{100}\) = 96

What is 20% of 60%? What about 60% of 20. Are 40% of 30 and 30% of 40 equal?

Answer:
Yes, 40% of 30 and 30% of 40 equal.
Explanation:
20% of 60%
(20 x 60)% = \(\frac{20 x 60}{100}\) = 12
60% of 20
20 × \(\frac{60}{100}\) = 12
40% of 30
30 × \(\frac{40}{100}\) = 12

Let’s look at another problem:

There are 40 children in a class and 50% of them are boys. How many boys are there in this class?
50% of the class are boys means, \(\frac{50}{100}\) of the total number of children in the class are boys.
That is, \(\frac{1}{2}\) of the total number, or half the total.
So there are 20 boys in the class.

Election

There are 1200 voters in a panchayat ward and 80% of them voted in an election. How many people voted’?

The number of persons who voted is \(\frac{80}{100}\) of the total number of voters.
So the number of persons voted is \(\frac{80}{100}\) of 1200.
That is, 1200 × \(\frac{80}{100}\) = 960

Textbook Page No. 140

Question 1.
In a company, 46% of the workers are women. It has 300 workers in all. How many of them are women?

Answer:
138 are women.
Explanation:
The number of workers in a company are 300,
46% of the workers are women.
300 x \(\frac{46}{100}\) = 138

Question 2.
In a class, 20% of the children are members of the Math Club. There are 35 children in the class. How many are members of the Math Club?

Answer:
7 member are in the Math Club.
Explanation:
In a class, 20% of the children are members of the Math Club.
20 percent means ,20 for each hundred.
So we must find out the number of hundreds in 1450 and multiply it by 54
35 x \(\frac{20}{100}\) = 7

Question 3.
In an election, the candidate who won got 54% of the votes. 1450 votes were polled. How many votes did the winner get?

Answer:
783 votes.
Explanation:
54 percent means ,54 for each hundred.
So we must find out the number of hundreds in 1450 and multiply it by 54
In an election, 54% the candidate who won got 54% of the votes
1450 x \(\frac{54}{100}\) = 783

Question 4.
The price of a car is 530000 rupees now. The manufacturer decides to reduce the price by 2% next month. What would be the reduction in price? What would be the new price?

Answer:
519400 rupees.
Explanation:
2 percent means ,2 for each hundred.
So we must find out the number of hundreds in 530000 and multiply it by 2
The price of a car is 530000 rupees now.
The manufacturer decides to reduce the price by 2% next month.
530000 x \(\frac{2}{100}\) = 10600
530000 – 10600 = 519400

Question 5.
1300 children took the Nu MATS test and 65%of them scored more than 25. How many are they?

Answer:
845 children scored more then 25.
Explanation:
65 percent means ,65 for each hundred.
So we must find out the number of hundreds in 1300 and multiply it by 13
1300 x \(\frac{65}{100}\) = 845
845 children scored more then 25

The other percent

60% of the workers in a company are women.
What all things do we know from this statement?
\(\frac{60}{100}\) of the total number of workers are women.
So what fraction of the workers are men? \(\frac{40}{100}\), right?
That is 40% of the workers are men.

In other words, \(\frac{3}{5}\) of the workers are women and \(\frac{2}{5}\) of the workers are men (How come?)

There were 320 children in the sub-district camp for scout and guides. 55% of them were guides and the rest, scout. How many scouts were there?
The percent of scouts is 1oo – 55 = 45.
So the number of scouts is 320 × \(\frac{45}{100}\)
This is easy to calculate, isn’t it?

Answer:
Yes,
Explanation:
The percent of scouts is 1oo – 55 = 45.
So the number of scouts is
320 × \(\frac{45}{100}\) = 144

Textbook Page No. 141

Question 1.
Of the 420 children in a school, 5% did not come one day. How many came to school that day?

Answer:
399 children came to school.
Explanation:
5 percent means ,5 for each hundred.
So we must find out the number of hundreds in 4.20 and multiply it by 5.
420 x \(\frac{5}{100}\) = 21
420 – 21 = 399

Question 2.
There are 280 plants in Sabu’s garden 70% of them are flowering. How many are non flowering?

Answer:
84 plants are non flowering.
Explanation:
70 percent means ,70 for each hundred.
So we must find out the number of hundreds in 2.80 and multiply it by 70.
280 x \(\frac{70}{100}\) = 196
280 – 196 = 84

Question 3.
There are 480 vehicles in a parking lot. Of these, 45% are motorbikes and 4% are cars. The rest are mini buses. How many mini buses are there?

Answer:
245 mini buses.
Explanation:
45% + 4% = 49%
100% – 49% = 51%
51% vehicles are mini buses.
480 x \(\frac{51}{100}\) = 244.8
So, there are 245 mini buses.

How many in all?

In a compound, there are 32 coconut trees and they form 50% of the total number of trees.
How many trees are there in all?
50% of the trees means \(\frac{50}{100}\) = \(\frac{1}{2}\) of the trees.
Thus, half the trees are coconut palms and so the total number of trees is double the number of coconut palms.
That is the total number of trees = 2 × 32 = 64.

In the sub-district Math fair, 60% of the children were girls. The actual number of girls was 108. How many children were there in all?

The number of girls is \(\frac{60}{100}\) = \(\frac{3}{5}\) of the total.
This means \(\frac{3}{5}\) of the total number is 108.
So the total number is \(\frac{5}{3}\) times 108.
That is, 108 × \(\frac{5}{3}\) = 180.
Thus we see that 180 children were there at the math fair.

Textbook Page No. 142

Question 1.
26 children of a class got A grade in an exam. It is 65% of the total number in the class. How many are there in the class?

Answer:
40 children.
Explanation:
26 children of a class got A grade in an exam.
It is 65% of the total number in the class means \(\frac{65}{100}\)x = 26
Total number of children in the class
x = \(\frac{100}{65}\) x 26
x = 26 \(\frac{20}{13}\)
x = 40

Question 2.
Jayan spent 8400 rupees in a month for food and it is 35% of his earnings. How much did he earn that month?

Answer:
24,000 rupees.
Explanation:
Jayan spent 8400 rupees in a month for food and it is 35% of his earnings.
Monthly amount earned by Jayan be x.
35% of x = 8400
\(\frac{35}{100}\)x = 8400
x = 8400\(\frac{20}{7}\)
x = 24000.

Question 3.
32 teachers of a school are male and they form 40% of the total number of teachers. How many teachers are there in the school?

Answer:
80 teachers.
Explanation:
Let the total number of teachers = x
Number of male teachers = 32
40% of the total no. of teachers = male teachers.
40% of x = 32
\(\frac{40}{100}\)x = 32
\(\frac{2}{5}\)x = 32
x = 32\(\frac{5}{2}\)
x = \(\frac{160}{2}\)
x = 80

Percent of percent

A man spends 20% of his earnings on education and 25% of this amount on books. What percent of his total earnings does he spend on books’?
It is \(\frac{25}{100}\) of \(\frac{20}{100}\) of the total earnings.
\(\frac{25}{100}\) of \(\frac{25}{100}\) means
= \(\frac{20}{100}\) × \(\frac{25}{100}\)
= \(\frac{1}{5}\) \(\frac{25}{100}\) = \(\frac{5}{100}\)
Thus the amount spend on books is 5% of the total earnings.
Now what percent of a number is 40% of its 30% percent’?

Answer:
12% of a number is 40% of its 30% percent.
Explanation:
It is \(\frac{40}{100}\) of \(\frac{30}{100}\) of the total earnings.
\(\frac{40}{100}\) of \(\frac{30}{100}\) means
= \(\frac{30}{100}\) × \(\frac{40}{100}\)
= \(\frac{3}{10}\) \(\frac{2}{5}\)
= \(\frac{3}{25}\) x 100
= \(\frac{300}{25}\)
= 12
Thus the amount spend on books is 12% of the total earnings.

Changing percent

A shop offers 20% reduction in prices. Ravi bought a shirt worth 400 rupees from this shop. How much should he pay?
He need only pay \(\frac{20}{100}\) of 400 less.
400 × \(\frac{20}{100}\) = 80
So, he must pay
400 – 80 = 320 rupees.
There is another way to do this.
The reduction in price is 20% of 400.
So he needs to pay only 80% of 400
80% of 400 = 400 × \(\frac{80}{100}\) = 320 rupees.

Now look at another problem:

In a school, there were 800 children last year and this year, the number is 12% more. How many children are there now?

The increase is 800 × \(\frac{12}{100}\) = 96
So then we can calculate the number of children now.
This we can do in a different way.
800 + (800 × \(\frac{12}{100}\)) = 800 × (1 + \(\frac{12}{100}\))
= 800 × \(\frac{112}{100}\) = 896
We can say \(\frac{112}{100}\) times is 112 percent(112%).

Area

If the length and breadth of a rectangle are increased by 10%, by how much percent would the area increase? What if length is increased by 10% and the breadth decreased by 10%?

Answer:
If the length and breadth of a rectangle are increased by 10%,
Area = 21%
If length is increased by 10% and the breadth decreased by 10%,
Area = 1%
Explanation:
Let length = 100; Breadth = 100
Area = 100 x 100 = 10000
After increasing length and breadth 10%
length = 110; Breadth = 110
Area = 110 x 110 = 12100
Increase in Area = 12100 – 10000 = 2100
% increase in area = \(\frac{2100}{10000}\) x 100 = 21%
if length is increased by 10% and the breadth decreased by 10%
length = 110; Breadth = 90
Area = 110 x 90 = 9900
% increase in area = \(\frac{9900}{100}\) = 99%
% of change in new area = 100 – 99 = 1%

Textbook Page No. 143

Question 1.
The price of a bicycle was 3400 rupees last month. Now it is reduced by 15%. What is the price now?

Answer:
2,890 rupees.
Explanation:
Cost of bicycle = 3400 rupees last month.
,Now it is reduced by 15%.
Total price = \(\frac{15}{100}\) x 3400
= 15 x 34
= 510 rupees.
Cost of bicycle now = Cost of bicycle last month – Amount reduced.
= 3400 – 510 = 2890

Question 2.
A watch priced at 3680 rupees is now sold for 20% less. What is the price now?

Answer:
2,944 rupees.
Explanation:
Initial cost of watch = 3680 rupees.
sold for 20% less.
Amount reduced = 3680 x \(\frac{20}{100}\)
= 2 x 368 = 736
Price of watch now = Initial price – Amount reduced.
= 3680 – 736
= 2944 rupees.

Question 3.
The amount of rainfall this year is calculated to be 20% more than last year. Last year’s rainfall was 230 centimeters. What is the amount of rainfall this year?

Answer:
276 cm.
Explanation:
The amount of rainfall this year is calculated to be 20% more than last year.
Last year’s rainfall was 230 centimeters.
Amount increases in rainfall = 20% of 230 cm
= \(\frac{20}{100}\) x 230
= 46 cm
Total amount of rainfall this year = last year rainfall + increase of rainfall this year.
= 230 + 46
= 276 cm

Question 4.
A person earned 12000 rupees last month and it is 6% more this month. How much is this month’s earning?

Answer:
12,720 rupees.
Explanation:
Amount earned last month = 12000 rupees.
Increase 6% more in this month.
Increased amount = 6% of 12000 rupees.
= \(\frac{6}{100}\) x 12000
= 720 rupees.
Amount earned this month = Amount earned last month + Incresed amount
= 12000 + 720
= 12720 rupees.

Fractional percent

We have said that 25% of something means \(\frac{25}{100}\) of that number;
which means \(\frac{1}{4}\) of that.
What about 125% of something’?
\(\frac{125}{100}\) times it; or 1\(\frac{1}{4}\) of it.
Thus percent of something means a fraction of it or certain times it.

We can put it in a different manner:
10% means 10 times \(\frac{1}{100}\)
20% means 20 times \(\frac{1}{100}\)
25% means 25 times \(\frac{1}{100}\)
60% means 60 times \(\frac{1}{100}\)

In this manner, 12\(\frac{1}{2}\) times of \(\frac{1}{100}\) can be said to be 12\(\frac{1}{2}\) %
What fraction is this?
\(\frac{1}{100}\) × 12\(\frac{1}{2}\) = \(\frac{1}{100}\) × \(\frac{25}{2}\) = \(\frac{1}{8}\)
Thus, 12\(\frac{1}{2}\) % of something means \(\frac{1}{8}\) of that.
12\(\frac{1}{2}\) % can also be written 12.5%
So, what does 33\(\frac{1}{3}\) % mean?
33\(\frac{1}{3}\) of \(\frac{1}{100}\).
\(\frac{1}{100}\) × 33\(\frac{1}{3}\) = \(\frac{1}{100}\) × \(\frac{100}{3}\) = \(\frac{1}{3}\)
So, 33\(\frac{1}{3}\)% of something means \(\frac{1}{3}\) of that.

Textbook Page No. 144

Question 1.
Explain each percent below as a fraction of something.
i) 6\(\frac{1}{4}\)%

Answer:
\(\frac{1}{16}\) of that or 6.25%.
Explanation:
6\(\frac{1}{4}\) of \(\frac{1}{100}\).
\(\frac{25}{4}\) × \(\frac{1}{100}\)
= \(\frac{25}{400}\)
= \(\frac{1}{16}\)
=6.25%
So, 6\(\frac{1}{4}\)% of something means \(\frac{1}{16}\) of that or 6.25%.

ii) 6\(\frac{2}{3}\)%

Answer:
\(\frac{1}{15}\) of that or 6.67%.
Explanation:
6\(\frac{2}{3}\) of \(\frac{1}{100}\).
\(\frac{20}{3}\) × \(\frac{1}{100}\)
= \(\frac{20}{300}\)
= \(\frac{2}{30}\)
= \(\frac{1}{15}\)
= 6.67%
So, 6\(\frac{2}{3}\)% of something means \(\frac{1}{15}\) of that or 6.67%.

iii) 8\(\frac{1}{3}\)%

Answer:
\(\frac{1}{1200}\) of that or 0.0012%.
Explanation:
8\(\frac{1}{3}\) of \(\frac{1}{100}\).
\(\frac{25}{3}\) × \(\frac{1}{100}\)
= \(\frac{25}{300}\)
= \(\frac{1}{12}\)
= 8.33%
So, 8\(\frac{1}{3}\)% of something means \(\frac{1}{12}\) of that or 8.33%.

iv) 16\(\frac{2}{3}\)%

Answer:
\(\frac{5}{30}\) of that or 16.6%.
Explanation:
16\(\frac{2}{3}\) of \(\frac{1}{100}\).
\(\frac{50}{3}\) × \(\frac{1}{100}\)
= \(\frac{50}{300}\)
= \(\frac{5}{30}\)
= 16.6%
So, 16\(\frac{2}{3}\)% of something means \(\frac{5}{30}\) of that or 16.6%.

v) 62\(\frac{1}{2}\)%

Answer:
\(\frac{31}{50}\) of that or 62%.
Explanation:
62\(\frac{1}{2}\) of \(\frac{1}{100}\).
\(\frac{124}{2}\) × \(\frac{1}{100}\)
= \(\frac{3131}{50}\)
= 62%
So, 62\(\frac{1}{2}\)% of something means \(\frac{124}{200}\) of that or 62%.

vi) 66\(\frac{2}{3}\)%

Answer:
\(\frac{2}{3}\) of that or 66.6%.
Explanation:
66\(\frac{2}{3}\) of \(\frac{1}{100}\).
\(\frac{200}{3}\) × \(\frac{1}{100}\)
= \(\frac{200}{300}\)
= \(\frac{2}{3}\)
= 66.6%
So, 66\(\frac{2}{3}\)% of something means \(\frac{2}{3}\) of that or 66.6%.

vii) 83\(\frac{1}{3}\)%

Answer:
\(\frac{5}{6}\) of that or 83.3%.
Explanation:
83\(\frac{1}{3}\) of \(\frac{1}{100}\).
\(\frac{250}{3}\) × \(\frac{1}{100}\)
= \(\frac{250}{300}\)
= \(\frac{25}{30}\)
= \(\frac{5}{6}\)
= 83.3%
So, 83\(\frac{1}{3}\)% of something means \(\frac{5}{6}\) of that or 83.3%.

Fraction and percent

We have seen that any percent can be explained as a fraction. On the other hand, can we express every fraction of something as a percent?
For that, we look at percents in a different way.
For example,
10% means 10 times \(\frac{1}{100}\).
We can put it differently;
10% means of 10 times \(\frac{1}{100}\)
Similarly, we can say
20% means \(\frac{1}{100}\) of 20 times.
25% means \(\frac{1}{100}\) of 25 times.
12\(\frac{1}{2}\) % means of 12\(\frac{1}{2}\) times.
That is the fraction expressing a percent as a part or times of something is \(\frac{1}{100}\) of the number given as a percent.

So the percent number is 100 times this fraction.
For example, let’s compute what percent of something, \(\frac{2}{5}\) of it gives.
\(\frac{2}{5}\) is \(\frac{1}{100}\) of the percent number.
So, the percent number is 100 times \(\frac{2}{5}\)
\(\frac{2}{5}\) × 100 = 40
Thus, \(\frac{2}{5}\) of something is 40% of it.

Now look at this problem:

In a school, 120 children appeared for SSLC examination. 110 children qualified for higher studies. What part of the children appeared for examination qualified?
\(\frac{110}{120}\) = \(\frac{11}{25}\)
That is, this fraction is the \(\frac{1}{100}\) parts of the percent of children qualified. Then percent of qualified children is 100 times of this. That is,
\(\frac{11}{25}\) × 100 = 91\(\frac{2}{3}\)
Therefore, 91\(\frac{2}{3}\)% of children is qualified for higher studies.

Question 1.
There are 750 children in a school and 450 of them are girls. What is the percent of girls?
Answer:
60%
Explanation:
There are 750 children in a school.
450 of them are girls.
The percent of girls = \(\frac{450}{750}\) × 100
= \(\frac{9}{15}\) × 100
= \(\frac{900}{15}\)
=  60%

Question 2.
Rafi earns 20000 rupees a month and he spends 6400 rupees from this on food. What percent of his earnings is this?
Answer:
percent of his earnings is 32 or 32%
Explanation:
Rafi monthly earnings = 20000 rupees.
Rafi spent on food = 6400 rupees.
% of his earnings = (Amount spent on food ÷ Rafi monthly earnings) x 100
= \(\frac{6400}{20000}\) x 100
= \(\frac{64}{200}\) x 100
= 32%

Question 3.
Jameel’s salary was 20000 rupees last month and 21000 rupees this month. By what percent has the salary increased?
Answer:
Salary increase 5%
Explanation:
Jameel’s last month salary = 20000 rupees.
Jameel’s this month salary = 21000 rupees.
Increase in salary = 21000 – 20000 = 1000 rupees
percent of increased salary = \(\frac{1000}{20000}\) x 100
= \(\frac{10}{2}\)
= 5%

Question 4.
Of 600 grams of sugar, 500 grams is used up. What percent is left?
Answer:
16.6% or 17%
Explanation:
Total grams of sugar = 600gm
Used sugar = 500 gm
Sugar left = 600 – 500 = 100 gm
% of sugar left = \(\frac{100}{600}\) x 100
= \(\frac{100}{6}\)
= \(\frac{50}{3}\)
= 16.6% or 17%

Question 5.
The sides of a square are increased by 10% to make a larger square. By what percent is the area increased?
Answer:
21% of area is increased.
Explanation:
Let the side of the smaller square is x cm.
Area of smaller square is (x cm)².
Side of the larger square is increased by 10%
Side is increased by 10% of x.
= \(\frac{10}{100}\)x
= \(\frac{x}{10}\)cm
Side of larger square = x + \(\frac{x}{10}\)
= \(\frac{11x}{10}\)
Area of larger square = (\(\frac{11x}{10}\))²
= (\(\frac{121x}{100}\))²
Increased area = Area of larger square – area of smaller square
= \(\frac{21}{100}\)x² cm²
Percentage of increased area = (Increased area ÷ Area of the smaller square) x 100
= (\(\frac{21}{100}\)x² cm²÷ x² cm²) x 100
= \(\frac{21}{100}\) x 100
= 21%

Question 6.
Ajayan’s salary is 25% more than Vijayan’s salary. By what percent of Ajayan’s salary is Vijayan’s salary less’?
Answer:
20% less than Ajayan’s salary.
Explanation:
Let Vijayan’s salary be 100
Ajayan’s salary = \(\frac{125}{100}\) x 100
= 125
Vijayan’s salary is \(\frac{100 X 100}{125}\) % of Ajayan’s salary.
= 80% of Ajayan’s salary.
Vijayan’s salary = 100 – 80 = 20% less than Ajayan’s

Textbook Page No. 147

Question 1.
Express the percent each fraction below indicates.

i) \(\frac{3}{8}\)
Answer:
37.5%
Explanation:
\(\frac{3}{8}\) x 100
= \(\frac{3x 100}{8}\)
= 37.5 %

ii) \(\frac{7}{20}\)
Answer:
15%
Explanation:
\(\frac{3}{20}\) x 100
= \(\frac{3x 100}{20}\)
= 15 %

iii) \(\frac{2}{3}\)
Answer:
66.67%
Explanation:
\(\frac{2}{3}\) x 100
= \(\frac{2 x 100}{3}\)
= 66.67%

iv) \(\frac{28}{25}\)
Answer:
112%
Explanation:
\(\frac{28}{25\) x 100
= \(\frac{28 x 100}{25}\)
= 112%

v) 2\(\frac{1}{3}\)
Answer:
33.34%
Explanation:
\(\frac{1}{3}\) x 100
= \(\frac{1 x 100}{3}\)
= 33.34%

Question 2.
What is the difference between 40% of 60 and 60% of 40?
Answer:
Both are same 24%
Explanation:
40% of 60
\(\frac{40}{100}\) x 60
= \(\frac{40 x 60}{100}\)
= 24
60% of 40
\(\frac{60}{100}\) x 40
= \(\frac{60 x 40}{100}\)
= 24
the difference between 40% of 60 and 60% of 40 are same.

Question 3.
In a school there are 1240 children and 30% of them are girls. How many boys are there in the school?
Answer:
372 boys.
Explanation:
In a school there are 1240 children,
30% of them are girls.
Total boys in the school = \(\frac{30}{100}\) x 1240
= \(\frac{1240 x 30}{100}\)
= 124 x 3
= 372 boys.

Question 4.
If 40% of 20 is added to 30% of 50, we get 50% of a number. What is this number?
Answer:
46
Explanation:
If 40% of 20 is added to 30% of 50,
we get 50% of a number
\(\frac{40}{100}\) x 20
= \(\frac{40 x 20}{100}\)
= 8
\(\frac{30}{100}\) x 50
= \(\frac{30 x 50}{100}\)
= 15
= 8+15 = 23
50% of a number is  23
100 % is 2 x 23 = 46

Question 5.
23 percent of a number is 69. What is the number?
Answer:
300
Explanation:
23 percent of a number is 69
x . \(\frac{23}{100}\)  = 69
23 x = 69 x 100
x = \(\frac{69 x 100}{23}\)
x = 300

Question 6.
10 percent of a number is 1.5. What is the number?
Answer:
300
Explanation:
10 percent of a number is 1.5.
x . \(\frac{10}{100}\)  = 1.5
23 x = 69 x 100
x = \(\frac{69 x 100}{23}\)
x = 300

Question 7.
The price of an article was 1800 rupees last month. The price is reduced by 10% this month .The shopkeeper says this price would be increased by 10% next month. What would be the price next month?
Answer:
1782 rupees.
Explanation:
The price of an article was 1800 rupees last month.
The price is reduced by 10% this month.
Amount reduced = \(\frac{10}{100}\) x 1800 = 180
Cost of article this month = 1800 – 180 = 1620
The shopkeeper says this price would be increased by 10% next month.
The price for next month = \(\frac{10}{100}\) x 1620 = 162
Cost of article for next month = cost of article this month + amount increased
= 1620 + 162 = 1782

Question 8.
Kannan has 600 rupees. He gave 50% of this to Thomas. Thomas gave 33\(\frac{1}{3}\) % of what he got to Hamza. How much did Hamza get?
Answer:
Hamza get 100 rupees.
Explanation:
Money Kannan has = 600 rupees
He gave 50% of this to Thomas.
Amount of money Kannan gives to Thomas = \(\frac{50}{100}\) x 600 = 300
Now Thomas has 300 rupees.
Thomas gave 3\(\frac{1}{3}\) % of what he got to Hamza.
% of money Thomas given to Hamza = 33\(\frac{1}{3}\)%
Amount of money Thomas given to Hamza = 33\(\frac{1}{3}\)% x 300
= \(\frac{100}{3}\) x \(\frac{1}{100}\) x 300
= 100

Question 9.
All children in class 7 passed the math exam. Details of grades are given below.

Fill in the blanks.
Answer:

Explanation:
As we know total percentage is 100%
40 + 30 + 25 + x = 100
95 + x = 100
x = 100 – 95
x = 5%
Number of children in Grade D
5% = 9
25% = x
by cross multiplying the above two
Number of children in Grade C
x 5% = 9 x 25%
x = \(\frac{9 × 25}{5}\)
x = 45
Number of children in Grade B
x 25% = 45 x 30%
x = \(\frac{45 × 30}{25}\)
x = 54
Number of children in Grade A
x 30% = 54 x 40%
x = \(\frac{54 × 40}{30}\)
x = 72

You can Download Statistics Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 11 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 11 Statistics

Statistics Text Book Questions and Answers

Bar graphs Textbook Page No. 165

Remember how we drew bar graphs to show numerical information? See the bar graph below.
It shows the amount of money each class in a school donated to the Snehasparsham medical aid fund.

• What is the total amount donated?
• Which class gave the most?
• And the least?

What other things should we get from this table?
Answer:
The other things we get from the data given are What is the average amount denoted? and
How many classes gave the funds?

Explanation:
From the data given above, the other things we get:
What is the average amount denoted?
How many classes gave the funds?

The table below shows the number of children in each class in a school.

Let’s draw a bar graph of this.
The height of each bar should be according to the number of children. If we take it as 1 centimetre per child, what would be the heights?
So, what length per child would be a convenient scale?
What other things should we decide to draw the graph?

• The width of a bar
• Distance between bars
  

Now draw a graph in your notebook.
We get the number of children in each class from this graph and we get the amount each class donated from the first graph.

For example, in class VA, there are 25 children and they donated 600 rupees. So how much did each child of this class give on average?
Calculate the average amount for other classes also.

• Which class has the highest average?
Answer:
VIIA class has the highest average.

Explanation:
Number of students of VA = 25.
Total amount donated by VA = 600 rupees.
Amount denoted by each student = Total amount donated by VA ÷ Number of students of VA
= 600 ÷ 25
= 24.
Number of students of VB = 30.
Total amount donated by VB = Amount denoted by each student × Number of students of VB
= 30 × 24
= 720.
Number of students of VIA = 30.
Total amount donated by VIA = Amount denoted by each student × Number of students of VIA
= 30 × 24
= 720.
Number of students of VIB = 20.
Total amount donated by VIB = Amount denoted by each student × Number of students of VIB
= 20 × 24
= 480.
Number of students of VIIA = 40.
Total amount donated by VIIA = Amount denoted by each student × Number of students of VIIA
= 40 × 24
= 960.
Number of students of VIIB = 35.
Total amount donated by VIIB = Amount denoted by each student × Number of students of VIIB
= 35 × 24
= 840.
Total amount donated = 600 + 720 + 480 + 960 + 840
= 3600.
Total number of students = 25 + 30 + 30 + 20 + 40 + 35
= 180.
Average amount donated = Total amount donated ÷ Total number of students
= 3600 ÷ 180
= 20.
Average of VA = 600 ÷ 20 = 30.
Average of VB = 720 ÷ 20 = 36.
Average of VIA = 720 ÷ 20 = 36.
Average of VIB = 480 ÷ 20 = 24.
Average of VIIA = 960 ÷ 20 = 48.
Average of VIIB = 840 ÷ 20 = 42.

• And the lowest?
Answer:
VIB Class the lowest.

Explanation:
Number of students of VA = 25.
Total amount donated by VA = 600 rupees.
Amount denoted by each student = Total amount donated by VA ÷ Number of students of VA
= 600 ÷ 25
= 24.
Number of students of VB = 30.
Total amount donated by VB = Amount denoted by each student × Number of students of VB
= 30 × 24
= 720.
Number of students of VIA = 30.
Total amount donated by VIA = Amount denoted by each student × Number of students of VIA
= 30 × 24
= 720.
Number of students of VIB = 20.
Total amount donated by VIB = Amount denoted by each student × Number of students of VIB
= 20 × 24
= 480.
Number of students of VIIA = 40.
Total amount donated by VIIA = Amount denoted by each student × Number of students of VIIA
= 40 × 24
= 960.
Number of students of VIIB = 35.
Total amount donated by VIIB = Amount denoted by each student × Number of students of VIIB
= 35 × 24
= 840.
Total amount donated = 600 + 720 + 480 + 960 + 840
= 3600.
Total number of students = 25 + 30 + 30 + 20 + 40 + 35
= 180.
Average amount donated = Total amount donated ÷ Total number of students
= 3600 ÷ 180
= 20.
Average of VA = 600 ÷ 20 = 30.
Average of VB = 720 ÷ 20 = 36.
Average of VIA = 720 ÷ 20 = 36.
Average of VIB = 480 ÷ 20 = 24.
Average of VIIA = 960 ÷ 20 = 48.
Average of VIIB = 840 ÷ 20 = 42.

In class VI, 20 children got A grade, 50 got B grade, 20 got C grade, 15 got D grade and 5 got E grade. Draw a bar graph showing this.
Answer:

Explanation:
Number of students of VI – Y-axis representation.
Types of Grades – X- axis representation.
20 children got A grade,
50 children got B grade,
20 children got C grade,
15 children got D grade,
5 children got E grade.

Double bar

The bar graph below shows the number of boys and girls present in class 5 of a school, from 1^st to 5^th of June.

Complete the table, based on this.

Answer:

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.

• On which day was the least number of children present?
Answer:
On 3/6/15 was the least number of children present.

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.

• On which day was the most number of boys present? And the least?
Answer:
On 5/6/15 was the most number of boys present and 1/6/15 the least.

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.
Most boys – 5/6/15.
Least boys – 1/6/15

• What about girls?
Answer:
On 4/6/15 was the most number of girls present and 3/6/15 the least.

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.
Most girls – 4/6/15.
Least girls – 3/6/15

• On which day was the difference in number between boys and girls the most?
Answer:
1/6/15, 3/6/15 and 4/6/15 these days the difference in number between boys and girls the most.

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.
Total boys and girls = 49 + 52 + 47 + 51 + 54 = 253.
Difference on 1/6/15:
26 – 23 = 3.
Difference on 2/6/15:
26 – 26 = 0.
Difference on 3/6/15:
25 – 22 = 3.
Difference on 4/6/15:
27 – 24 = 3.
Difference on 5/6/15:
28 – 26 = 2.
There is no specific date for the difference in number between boys and girls the most. Major difference of 3 is on three days of 3.

100 grams of rice is taken per child for lunch. How much rice was used each day during this week?
Answer:
Total quantity of rice children nedded for a week = 1,77,100 grams.

Explanation:
Number of grams of rice is taken per child for lunch = 100.
Number of days in a week = 7.
Total number of boys and girls = 49 + 52 + 47 + 51 + 54 = 253.
Number of grams each children eats for lunch = Number of grams of rice is taken per child for lunch × Total number of boys and girls
= 100 × 253
= 25,300 grams.
Total quantity of rice children nedded for a week = Number of grams each children eats for lunch × Number of days in a week
= 25,300 × 7
= 1,77,100 grams.

Textbook Page No. 168

Question 1.
The table shows the number of notebooks sold at the school store during six months.

Draw a bar graph of this.
Answer:

Explanation:
June – 140 note books.
July – 130 note books.
August – 150 note books.
September – 160 note books.
October – 120 note books.
November – 150 note books.

Question 2.
The table shows the various expenses of George’s family for the last month.

Draw a bar graphs of this. Write down some facts we get from the graph.
Answer:

Explanation:
Less expense the family invests on Travelling.
Most of their expenses is consumption of food.

Question 3.
The table below shows the amount of electricity used at Soumya’s home during last year.

i. How many units were used in all last year?
Answer:
Total number of units used in the last year = 2100 KW.

Explanation:
Number of units used in:
Month = January, February = 340 KW.
Month = March, April = 440 KW.
Month = May, June = 410 KW.
Month = July, August = 290 KW.
Month = September, October = 300 KW.
Month = November, December = 320 KW.
Total number of units used in the last year = 340 KW + 440 KW + 410 KW + 290 KW + 300 KW + 320 KW = 2100 KW.

ii. What is the average use every two months?
Answer:
Average use of January, February = 340 KW ÷ 2 = 170 KW.
Average use of  March, April = 440 KW ÷ 2 = 220 KW.
Average use of May, June = 410 KW ÷ 2 = 205 KW.
Average use of July, August = 290 KW ÷ 2 = 145 KW.
Average use of September, October = 300 KW ÷ 2 = 150 KW.
Average use of November, December = 320 KW ÷ 2 = 160 KW.

Explanation:
Number of units used in:
Month = January, February = 340 KW.
Average use of January, February = 340 KW ÷ 2 = 170 KW.
Month = March, April = 440 KW.
Average use of  March, April = 440 KW ÷ 2 = 220 KW.
Month = May, June = 410 KW.
Average use of May, June = 410 KW ÷ 2 = 205 KW.
Month = July, August = 290 KW.
Average use of July, August = 290 KW ÷ 2 = 145 KW.
Month = September, October = 300 KW.
Average use of September, October = 300 KW ÷ 2 = 150 KW.
Month = November, December = 320 KW.
Average use of November, December = 320 KW ÷ 2 = 160 KW.

iii. During which two months was the use closest to the average?
Answer:
Average use of January, February of 170 KW is the use closest to the average.

Explanation:
Total number of units used in the last year = 2100 KW.
Average use in last year = 2100 KW ÷ 12 = 175.
Average use of January, February = 340 KW ÷ 2 = 170 KW.
Average use of  March, April = 440 KW ÷ 2 = 220 KW.
Average use of May, June = 410 KW ÷ 2 = 205 KW.
Average use of July, August = 290 KW ÷ 2 = 145 KW.
Average use of September, October = 300 KW ÷ 2 = 150 KW.
Average use of November, December = 320 KW ÷ 2 = 160 KW.

Question 4.
The bar graph shows the percent of those who voted in some wards in a panchayath election.

The table shows the total number of voters.

Calculate the actual number of men and women who voted in each ward.
Answer:
Ward 1: Actual number of total = 558 + 456 = 1,014.
Ward 2: Actual number of total = 632 + 756 = 1,388.
Ward 3: Actual number of total = 640 + 532 = 1,172.
Ward 4: Actual number of total = 680 + 810 = 1,490.
Ward 5: Actual number of total =576 + 666 = 1,242.

Explanation:
Ward 1:
Number of men = 620 × 90%
= 620 × 90 ÷ 100
= 62 × 9
= 558.
Number of Women = 570 × 80%
= 570 × 80 ÷ 100
= 57 × 8
= 456.
Actual number of total = 558 + 456 = 1,014.
Ward 2:
Number of men = 790 × 80%
= 790 × 80 ÷ 100
= 79 × 8
= 632.
Number of Women = 840 × 90%
= 840 × 90 ÷ 100
= 84 × 9
= 756.
Actual number of total = 632 + 756 = 1,388.
Ward 3:
Number of men = 800 × 80%
= 800 × 80 ÷ 100
= 80 × 8
= 640.
Number of Women = 760 × 70%
= 760 × 70 ÷ 100
= 76 × 7
= 532.
Actual number of total = 640 + 532 = 1,172.
Ward 4:
Number of men = 850 × 80%
= 850 × 80 ÷ 100
= 85 × 8
= 680.
Number of Women = 900 × 90%
= 900 × 90 ÷ 100
= 90 × 9
= 810.
Actual number of total = 680 + 810 = 1,490.
Ward 5:
Number of men = 720 × 80%
= 720 × 80 ÷ 100
= 72 × 8
= 576.
Number of Women = 740 × 90%
= 740 × 90 ÷ 100
= 74 × 9
= 666.
Actual number of total =576 + 666 = 1,242.