CBSE Class 7

These NCERT Solutions for Class 7 Hindi Vasant Chapter 1 हम पंछी उन्मुक्त गगन के Questions and Answers are prepared by our highly skilled subject experts.

हम पंछी उन्मुक्त गगन के NCERT Solutions for Class 7 Hindi Vasant Chapter 1

Class 7 Hindi Chapter 1 हम पंछी उन्मुक्त गगन के Textbook Questions and Answers

कविता से

प्रश्न 1.
हर तरह की सुख सुविधाएँ पाकर भी पक्षी पिंजरे में बंद क्यों नहीं रहना चाहते?
उत्तर:
हर तरह की सुख सुविधाएँ पाकर भी पक्षी पिंजरे में बंद इसलिए नहीं रहना चाहते क्योंकि उन्हें स्वतंत्रता प्यारी है। वह कष्ट उठाकर भी स्वच्छंद आकाश में उड़ना पसंद करते हैं। पिंजरे में बंद रहकर मिलने वाली कोई भी सुविधा उसे सुख नहीं दे सकती। उन्हें अपनी उड़ान में कोई बाधा पसंद नहीं है। वे तो उन्मुक्त गगन के पंछी हैं।

प्रश्न 2.
पक्षी उन्मुक्त रहकर अपनी कौन-कौन सी इच्छाएँ पूरी करना चाहते हैं ?
उत्तर:

1. पक्षी उन्मुक्त रहकर खुले आकाश में गाना चाहते हैं।
2. वे नदी एवं झरनों का बहता जल पीना चाहते हैं।
3. वे अपनी स्वाभाविक गति से उड़ना चाहते हैं।
4. वे नीले आकाश में उड़कर अनार के दानों रूपी तारों को चुगना चाहते हैं।

प्रश्न 3.
भाव स्पष्ट कीजिएया तो क्षितिज मिलन बन जाता/या तनती साँसों की डोरी।
उत्तर:
भाव-पक्षी हर हाल में आकाश में उड़ना चाहते हैं। वे उड़ते हुए अपने लक्ष्य तक पहुँचना चाहते हैं अथवा अपने इस प्रयास में अपने आपको पूरी तरह थका डालना चाहते हैं।

कविता से आगे

प्रश्न 1.
बहुत से लोग पक्षी पालते हैं-
(क) पक्षियों को पालना उचित है अथवा नहीं ? अपने विचार लिखिए।
(ख) क्या आपने या आपकी जानकारी में किसी ने कभी कोई पक्षी पाला है? उसकी देख-रेख किस प्रकार की जाती होगी, लिखिए।
उत्तर:
(क) सभी को स्वतंत्र रहने का अधिकार है। यदि हम पक्षियों को पालेंगे तो यह उनकी स्वतंत्रता का हनन होगा। हमें उनको स्वच्छंद आकाश में उड़ने देना चाहिए। आकाश में उड़ते और चहचहाते हुए पक्षी भले लगते हैं। यदि कोई हमें कैद करके रखे तो हमें कैसा लगेगा ? हमें यह अवश्य सोचना चाहिए।

(ख) हमारे पड़ोस के घर में तोता पाला हुआ है। वे उस तोते को पिंजरे में बंद रखते हैं। तोता पिंजरे में ही उनके द्वारा दिया गया अन्न व जल ग्रहण करता है। तोते के इस बंधन को देखकर तरस आता है।

प्रश्न 2.
पक्षियों को पिंजरे में बंद करने से केवल उनकी आजादी का हनन ही नहीं होता, अपितु पर्यावरण भी प्रभावित होता है। इस विषय पर दस पंक्तियों में अपने विचार लिखिए।
उत्तर:
पक्षियों को पिंजरे में बंद रखने से उनकी आज़ादी का हनन होता है। हमें यह अधिकार बिल्कुल भी नहीं है कि हम किसी को कैद करें। ईश्वर ने उनको भी पैदा किया है। उनको भी स्वच्छंद रहने का अधिकार दिया है। मनुष्य इस सृष्टि का सबसे दुष्ट प्राणी है। यह अपनी थोड़ी-सी खुशी के लिए दूसरों को कष्ट देता है। पक्षियों के बंधन में रहने से पर्यावरण को हानि होती है। वन के अंदर बाघों का होना और उपवनों में पक्षियों का चहचहाना किसे अच्छा नहीं लगता होगा। पक्षी पर्यावरण को संतुलित बनाकर रखते हैं। पक्षियों को देखकर हम मौसम के बारे में कई जानकारियाँ प्राप्त कर सकते हैं-जैसे-गौरेया जब रेत में खिलवाड़ करती है तो इसे वर्षा होने का संकेत माना जाता है। पक्षी जब किसी वृक्ष के फल को खाते हैं तो वे अपनी बीट द्वारा उस फल के बीजों को दूसरे स्थान पर ले जाते हैं। बरगद-पीपल जैसे वृक्षों की उत्पत्ति में पक्षी बहुत सहायक हैं। ये वृक्ष छायादार होने के साथ-साथ पर्यावरण को भी शुद्ध बनाते हैं।

अनुमान और कल्पना

प्रश्न 1.
क्या आपको लगता है कि मानव की वर्तमान जीवन-शैली और शहरीकरण से जुड़ी योजनाएँ पक्षियों के लिए घातक हैं ? पक्षियों से रहित वातावरण में अनेक समस्याएँ उत्पन्न हो सकती हैं। इन समस्याओं से बचने के लिए हमें क्या करना चाहिए ? उक्त विषय पर वाद-विवाद प्रतियोगिता का आयोजन कीजिए।
उत्तर:
अध्यापक छात्रों के बीच इस विषय पर वाद-विवाद प्रतियोगिता का आयोजन करें।

प्रश्न 2.
यदि आपके घर के किसी स्थान पर किसी पक्षी ने अपना आवास बनाया है और किसी कारणवश आपको अपना घर बदलना पड़ रहा है तो आप उस पक्षी के लिए किस तरह के प्रबंध करना आवश्यक समझेंगे? लिखिए।
उत्तर:
यदि उस पक्षी ने उस स्थान पर अंडे दे रखे हैं तो हम कुछ दिन इंतजार करेंगे। जब अंडों से निकले बच्चे उड़ान भरने लायक हो जाएँगे तब उसके घोंसले को किसी दूसरे स्थान पर रख देंगे तथा हम उनकी देखभाल भी करते रहेंगे।

भाषा की बात

प्रश्न 1.
स्वर्ण-शृंखला और लाल किरण-सी में रेखांकित शब्द गुणवाचक विशेषण हैं। कविता से ढूँढ़कर इस प्रकार के तीन और उदाहरण लिखिए।
उत्तर:
उन्मुक्त गगन, पुलकित पंख, बहता जल, कटुक-निबौरी, कनक-कटोरी, नीले नभ, सीमाहीन-क्षितिज, आकुल-उड़ान।

प्रश्न 2.
“भूखे-प्यासे’ में द्वंद्व समास है। इन दोनों शब्दों के बीच लगे चिह्न को सामासिक चिह्न (-) कहते हैं। इस चिह्न से ‘और’ का संकेत मिलता है, जैसे-भूखे-प्यासे = भूखे और प्यासे।
इस प्रकार के दस अन्य उदाहरण खोजकर लिखिए।
उत्तर:
माता-पिता = माता और पिता
भाई-बहिन = भाई और बहिन
अच्छा-बुरा = अच्छा और बुरा
सुख-दुःख = सुख और दुःख
आकाश-पाताल = आकाश और पाताल
भीम-अर्जुन = भीम और अर्जुन
सीता-गीता = सीता और गीता
पाप-पुण्य = पाप और पुण्य
दिन-रात = दिन और रात।
दूध-दही = दूध और दही।

काव्यांश की सप्रसंग व्याख्या एवं अर्थग्रहण-संबंधी प्रश्नोत्तर

1. हम पंछी ……………………………… मैदा से।
शब्दार्थः पंछी-पक्षी, उन्मुक्त गगन-खुला आकाश, पिंजरबद्ध-पिंजरे में बँधकर (कैद होकर), कनक-सोना, पुलकित-प्रसन्नचित्त, कटुक-कड़वी, निबौरी-नीम का फल, कनक कटोरी-सोने की कटोरी।

प्रसंग- प्रस्तुत पंक्तियाँ हमारी पाठ्य पुस्तक ‘बसंत भाग-2 में संकलित कविता’ ‘हम पंछी उन्मुक्त गगन के से ली गई हैं। इसके लेखक ‘श्री शिव मंगल सिंह ‘सुमन’ जी हैं। कवि ने इस पंक्तियों में स्वतंत्रता के महत्त्व को दर्शाया है। उनका कहना है कि एक पक्षी भी स्वतंत्रता के महत्त्व को भली-भाँति जानता है वह किसी भी प्रकार का बंधन स्वीकार नहीं करता।

व्याख्या- पक्षी कहते हैं कि हम स्वच्छंद आकाश में विचरण करने वाले हैं। यदि हमको पिंजरे में कैद करके रखा जाएगा तो हमारा गायन जो हमारी चहचहाट के रूप में प्रकट होता है, वह समाप्त हो जाएगा। हमको स्वतंत्रता का जीवन पसंद है। पिंजरे में बंद करके हमको चाहे कितनी भी सुविधाएँ क्यों न दी जाएँ हमारे लिए वे व्यर्थ हैं। यदि हमको सोने के पिंजरे में रखा जाए तो भी हम पंख फड़फड़ाकर स्वतंत्र होने की हर संभव कोशिश करेंगे भले ही हमारे कोमल पंख पिंजरे की तीलियों से टकराकर टूट जाएँ।

पक्षी आगे कहते हैं कि हम तो बहता हुआ जल पीने वाले हैं, जो स्वतंत्र रहकर ही मिल सकता है। यदि हमको पिंजरे में बंद किया तो हम भूखे-प्यासे मर जाएँगे परंतु पिंजरे में मिलने वाली सुख-सुविधाओं को स्वीकार नहीं करेंगे। हमारे लिए तो सोने की कटोरी में मिलने वाले मैदे के पकवान से कहीं बेहतर नीम की कड़वी निबौरी है जिसको हम स्वतंत्रता पूर्वक ग्रहण करते हैं।

अर्थग्रहण संबंधी प्रश्नोत्तर

प्रश्न 1.
पक्षी पिंजरे में कैद होकर क्यों नहीं गा पाएँगे ?
उत्तर:
पक्षी खुले आकाश में विचरण करते हैं। पिंजरे में कैद होना उनको पसंद नहीं। स्वतंत्र रहकर ही उनका स्वभाविक गायन हो सकता है। पिंजरे में कैद रहकर तो वे अपनी चहचहाट भूल जाते हैं।

प्रश्न 2.
पक्षियों के पंख क्यों टूट जायेंगे ?
उत्तर:
पक्षी आजाद होने के लिए अपने पंख फड़फड़ाएँगे उनके पंख पिंजरे की तीलियों से टकराकर टूट जाएँगे। वे फिर भी पिंजरे से बाहर निकलने का प्रयास करते रहेंगे।

प्रश्न 3.
पक्षी कनक कटोरी की मैदा से नीम की कड़वी निबौरी को क्यों अच्छा मानते हैं ?
उत्तर:
सोने की कटोरी के मैदे को खाने के लिए उन्हें अपनी स्वतंत्रता गंवानी पड़ेगी जबकि नीम की कड़वी निबौरी को वे स्वतंत्र रहकर, स्वच्छंद आकाश में उड़कर अपनी इच्छानुसार ग्रहण कर सकते हैं। आजादी से जो मिल जाता है उससे बढ़कर कोई वस्तु नहीं होती

2. स्वर्ण-शृंखला ……………………………. केदाने।
शब्दार्थः स्वर्ण शृंखला-सोने की जंजीर, गति-चाल, फुनगी-वृक्ष की शाखाओं का ऊपरी सिरा, अरमान-दिल की इच्छा, तारक-तारे।

प्रसंग- प्रस्तुत पंक्तियाँ ‘शिव मंगल सिंह सुमन’ रचित कविता ‘हम पंछी उन्मुक्त गगन के’ से ली गई हैं। कवि ने यहाँ पिंजरे में कैद पक्षी की मनोव्यथा का चित्रण किया है कि वे बंधन में पड़कर कैसे अपनी स्वभाविकता खो बैठे हैं। उनके मन के अरमान मन में ही रह गए।

व्याख्या- पक्षी कहते हैं कि सोने की जंजीरों में बँधकर हम अपनी स्वाभाविकता खो बैठे हैं। हम अपनी गति और आकाश में उड़ना बिल्कुल भूल गए। स्वच्छंद होकर उड़ने का जो सुख था अब वह केवल स्वप्न की ही बात रह गई। हम कैसे वृक्ष की शाखाओं की चोटियों पर बैठकर झूला झूलते थे, अब स्वप्न में ही स्वतंत्रता के इस सुख को अनुभव करते हैं।

पक्षी कहते हैं कि हमारे भी अरमान थे कि हम आकाश में स्वच्छंद होकर विचरण करें। हम नीले आकाश में सीमाओं तक जाकर उसको छूना चाहते थे। हम भी सूर्य की किरण के समान अपनी लाल चोंच को खोलकर आकाश में अनार के दानों रूपी तारों को चुगें। परंतु बंधन में पड़ जाने के कारण हमारी यह इच्छा हमारे मन में ही रह गई।

अर्थग्रहण संबंधी प्रश्नोत्तर

प्रश्न 1.
पक्षी अपनी गति और उड़ान क्यों भूल गए ?
उत्तर:
पक्षी लालच के कारण बंधन में पड़ गए, उनको पिंजरे में कैद कर लिया गया। इस कारण से उनकी स्वाभाविकता ही समाप्त हो गई और वे अपनी चाल व उड़ान सब भूल गए।

प्रश्न 2.
पक्षी किस प्रकार के स्वप्न देखते हैं ?
उत्तर:
पक्षी वृक्षों की शाखाओं की फुनगियों पर झूला झूलने का स्वप्न देखते हैं क्योंकि पिंजरे में बंद रहकर तो वे कुछ भी नहीं कर पाते।

प्रश्न 3.
पक्षियों के क्या-क्या अरमान थे ?
उत्तर:
पक्षियों के अरमान थे कि वे उड़ते हुए नीले आकाश में अंतिम छोर तक जाकर उसे छुएँ और सूर्य की किरण के समान अपनी लाल-चोंच से अनार के दानों जैसे तारों को चुनें।

3. होती सीमाहीन …………………………….. न डालो।
शब्दार्थः सीमाहीन-जिसकी कोई सीमा (हद) न हो, क्षितिज-जहाँ धरती आकाश मिलते हुए प्रतीत हों, होड़ा-होड़ी-प्रतिस्पर्धा, नीड़-घोंसला, आश्रय-सहारा, आकुल-बेचैन, विघ्न-बाधा।

प्रसंग- प्रस्तुत पंक्तियाँ हमारी पाठ्य पुस्तक ‘बसंत भाग-2 में संकलित कविता ‘हम पंछी उन्मुक्त गगन के ‘ से ली गई हैं। इसके लेखक ‘श्री शिव मंगल सिंह सुमन’ जी हैं। पक्षी उन्मुक्त गगन में उड़ना चाहते हैं। वे मनुष्य से अपेक्षा करते हैं कि वे उनको आश्रय भले ही न दें परंतु उनको स्वच्छंद होकर खुले आकाश में उड़ने दें।

व्याख्या- पक्षी खुले आकाश में उड़ने की कामना करते हुए कहते हैं कि यदि हम खुले आकाश में उड़ते तो हमारा मुकाबला सीमाहीन क्षितिज से होता। हमारे दोनों पंख आगे बढ़ने के लिए एक-दूसरे से अधिक बल लगाते। हम उस स्थल पर पहुँच जाते जहाँ यह धरती और आकाश मिलता हुआ दिखाई देता है। ऐसा करने में हम थककर चूर हो जाते और हमारी साँस फूलने लगती।

पक्षी मनुष्य से कहते हैं कि हे मनुष्य! आप हमें पेड़ की टहनी पर भले ही घोंसला न बनाने दो और हमसे पेड़ की टहनी का आश्रय भी छीन लो। हमें इस बात का इतना दुःख नहीं होगा। बस हम तो यह चाहते हैं कि जब ईश्वर ने हमें उड़ने के लिए पंख दिए हैं तो हमें खुले आकाश में उड़ने दिया जाए पिंजरे में बंद करके हमारी इस उड़ान में बाधा मत बनो।

अर्थग्रहण संबंधी प्रश्नोत्तर

प्रश्न 1.
पक्षी क्या चाहते हैं ?
उत्तर:
पक्षी चाहते हैं कि उनको सीमाहीन आकाश में उड़ने दिया जाए। वे उड़ते हुए क्षितिज तक जाना चाहते हैं।

प्रश्न 2.
पक्षी मनुष्य से क्या अपेक्षा करता है ?
उत्तर:
पक्षी मनुष्य से अपेक्षा करता है कि वे चाहे उनके आश्रय को नष्ट कर दे परंतु भगवान ने उनको उड़ने के लिए पंख दिए हैं अतः उनको स्वच्छंद आकाश में उड़ने दें। वे उनकी उड़ान में किसी भी प्रकार की बाधा न डालें।

These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions

NCERT In-text Question Page No. 158

Question 1.
Find the percentage of children of different heights for the following data.
Answer:

Question 2.
A shop has the following number of shoe pairs of different sizes.
Size 2 : 20
Size 3 : 30
Size 4 : 28
Size 5 : 14
Size 6 : 8
Write this information in tabular form as done earlier and find the percentage of each shoe size available in the shop.
Answer:

NCERT In-text Question Page No. 159 & 160

Question 1.
A collection of 10 chips with different colours is given.
Answer:

Question 2.
Mala has a collection of bangles. She has 20 gold bangles and 10 silver bangles. What is the percentage of bangles of each type? Can you put it in the tabular form as done in the above example?
Answer:
We have

Note: We can use percentages for comparing various quantities.

NCERT In-text Question Page No. 160

Question 1.
Look at the examples below and in each of them, discuss which is better for comparison.
In the atmosphere, 1 g of air contains:
(i) .78g Nitrogen – 78 % Nitrogen
.21 g Oxygen or 21 % Oxygen
.01g Nitrogen – 1 % Other Gas
(ii) A shirt has:
\(\frac{2}{5}\) Cotton 60% Cotton
or
\(\frac{2}{5}\) Polyester 40% Polyester
Answer:
The second one (i.e. percentage) is better for comparison in both the cases.

NCERT In-text Question Page No. 161

Question 1.
Convert the following to per cents:
(a) \(\frac{12}{16}\)
(b) 3.5
(c) \(\frac{49}{50}\)
(d) \(\frac{2}{2}\)
(e) 0.05
Answer:
(a) \(\frac{12}{16}\) = (\(\frac{12}{16}\) x 100) % = ( \(\frac{3}{4}\) x 100) %
= (3 x 25)% = 75%

(b) 3.5 = (3.5 x 100)% = 350%

(c) \(\frac{49}{50}\) ( \(\frac{49}{50}\) x 100)%
= (\(\frac{49}{50}\) x 100)%
= (49 x 2)% = 98%

(d) \(\frac{2}{2}\) = ( \(\frac{2}{2}\) x 100)%
= (1 x 100)%
= 100%
(e) 0.05 = (0.05 x 100) % = 5 %

Question 2.
(i) Out of 32 students, 8 are absent.
What per cent of the students are absent?
(ii) There are 25 radios, 16 of them are out of order. What per cent of radios are out of order?
(iii) A shop has 500 parts, out of which 5 are defective. What per cent are defective?
(iv) There are 120 voters, 90 of them voted yes. What per cent voted yes?
Answer:

= (3 x 25)% = 75%
75% voters voted ‘yes’

NCERT In-text Question Page No. 162

Question 1.
Fill in the blanks
(i) 35% + ……………. % = 100%.
(ii) 64% + 20% + ……………. % = 100%
(iii) 45% = 100% – ……………. %.
(iv) 70% = ……………. % – 30%
Answer:
(i) 100% – 35% = 65%
35% + 65% = 100%
(ii) 64%+ 20% =84%
100% – 84% = 16%
64% + 20% + 16% = 100%
(iii) 100%-45% =55%
45% = 100% – 55%
(iv) 70% + 30% = 100%
70% = 100% – 30%

Question 16.
If 65% of students in a class have a bicycle, what per cent of the student do not have bicycles?

Answer:
Part of students having bicycles = 65%
Remaining part of students
= 100% – 65% = 35%
Thus, 35% of students do not have bicycles.

Question 17.
We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what per cent are mangoes?
Answer:
Quantity of apples = 50%
Quantity of oranges = 30%
Quantity of mangoes in the basket
= 100% – (50% + 30%)
= 100% -80% = 20%
Thus, the percent of mangoes in the basket = 20%

NCERT In-text Question Page No. 163

Question 1.
What per cent of these figures are shaded?

Answer:
(i) Fraction which is shaded
\(=\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)=\left(\frac{1+1+1}{4}\right)=\frac{3}{4}\)
Percentage of shaded part
= \(\frac { 3 }{ 4 }\) x 100% = 75%
(ii) Fraction of diagram which is shaded
\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{2+1+1}{8}=\frac{4}{8}=\frac{1}{2}\)
Percentage of diagram which is shaded = \(\frac { 1 }{ 2 }\) x 100% = 50%

NCERT In-text Question Page No. 164

Question 1.
Find
(a) 50% of 164
(b) 75% of 12]
(c) 12 \(\frac { 1 }{ 2 }\) % of 64
Answer:
(a) 50% of 164 = \(\frac { 50 }{ 100 }\) x 164
= \(\frac { 1 }{ 2 }\) x 164 = 82

(b) 75% of 12 = \(\frac { 75 }{ 100 }\) x 12 = \(\frac { 3 }{ 4 }\) x 12
= 3 x 3 = 9

(c) 12\(\frac { 1 }{ 2 }\) % of 64 = \(\frac { 25 }{ 2 }\)% or 64
= \(\frac{2}{100} \times 64=\frac{25}{100} \times \frac{64}{2}\)
= \(\frac { 1 }{ 8 }\) x 64 = 8

Question 2.
8% children of a class of 25 like getting wet in the rain. How many children like getting wet in the rain?
Answer:
Number of children who like getting wet in rain
= 8% of 25 = \(\frac { 8 }{ 100 }\) x 25 = 2

Question 3.
9 is 25% of what number?
Answer:
Let the required be P
25% of P = 9
\(\frac { 25 }{ 100 }\) x P = 9
25 P = 9 x 100
P = \(\frac { 9 x 100 }{ 25 }\) = 9 x 4 = 36
The required number is 36.

Question 4.
75% of what number is 15?
Answer:
Let the required number be p
75% of P = 15
= \(\frac { 75 }{ 100 }\) x P = 15
\(\frac { 3 }{ 4 }\) x P = 15
\(\frac { 3P }{ 4 }\) = 15
3P = 15 x 4
P = \(\frac{15 \times 4}{3}\) = 5 x 4 = 20
The required number is 20

NCERT In-text Question Page No. 166

Question 1.
Divide 15 sweets between Manu and Sonu so that they get 20% and 80% of them respectively.
Answer:
Share of Manu = 20% of 15 sweets
= \(\frac { 20 }{ 100 }\) x 15 sweets
= 3 sweets
Share of Sonu = 80% of 15 sweets
= \(\frac { 80 }{ 100 }\) x 15
= \(\frac { 4 }{ 5 }\) x 15
= 4 x 3 = 12 sweets
Manu share = 3 sweets and Sonu share =12 sweets.

Question 1.
If the angles of a triangle are in the ratio 2:3:4, find the value of each angle.
Answer:
We know that sum of three angles of a triangle =180°
Let the three angles be 2x, 3x, and 4x
2x +3x + 4x = 180°
9x = 180°
x = \(\frac { 180° }{ 9 }\) = 20°
The three angles are 2(20° ); 3(20° ) and 4(20°)
i.e. 40°, 60° and 80°

NCERT In-text Question Page No. 167

Question 1.
Find percentage of increase or decrease
(a) Price of shirt decreased from ₹ 80 to ₹ 60
(b) Marks in a test increased from 20 to 30
Answer:
(a) Initial price of the shirt = ₹ 80
Decrease price of the shirt = ₹ 60
Decrease in the price
= ₹80 – ₹60 = ₹20
Per cent of decrease in price 20
= \(\frac { 20 }{ 80 }\) x 100%
= ₹ 25%

(b) Initial marks = 20
Increased marks = 30
Increase in marks = 30 – 20 = 10
Percent of increase in marks
= \(\frac { 10 }{ 20 }\) x 100% = 50%

Question 2.
My mother says in her childhood petrol was ₹ 1 a litre. It is ₹ 52 per litre today. By what per centage has the price gone up?
Answer:
Initial Price = ₹ 1
Increase price = ₹ 52
Increase in price = ₹52 – ₹1 = ₹51
Percentage of increase in petrol
= \(\frac { 51 }{ 1 }\) x 100%
= 5100%

NCERT In-text Question Page No. 169

Question 1.
A shopkeeper bought a chair for ₹ 375 and sold it for ₹ 400. Find the gain percentage.
Answer:
C.P of a chair = ₹ 375
S.P of a chair = ₹ 400
Profit = S.P – C.P
= ₹ 400 – ₹ 375 = ₹ 25
Profit percentage = \(\frac { Profit }{ C.P }\) x 100%
= \(\frac { 25 }{ 375 }\) x 100%
= \(\frac { 1 }{ 15 }\) x 100% = \(\frac { 20 }{ 3 }\)%
= 6\(\frac { 2 }{ 3 }\)%

Question 2.
Cost of an item is₹50 It was sold with a profit of 12%. Find the selling price.
Answer:
We have
CP of the item = ₹50
Profit % = 12%
Profit = 12% of ₹50
= ₹\(\frac { 12 }{ 100 }\) x 50 = ₹6
SP = CP + Profit
= ₹50 + ₹6 = ₹56

Question 3.
An article was sold for ₹250 with a profit of 5%. What was its cost price?
Answer:
Let the cost price be ₹ x.
Profit % = 5%
∴ Profit = 5% of x = ₹ \(\frac { 5 }{ 100 }\) x x =₹ \(\frac { x }{ 20 }\)
∴ SP = CP + Profit
∴ ₹ 250 = ₹ x + \(\frac { x }{ 20 }\) [∵ ₹ = ₹ 250]
or ₹ 250 = ₹ \(\frac { 21 }{ 20 }\)x
or x = \(\frac{250 \times 20}{21}=\frac{5000}{21}=238 \frac{2}{21}\)
Thus cost price of the article = ₹ 238 \(\frac { 2 }{ 21 }\)

Question 4.
An item was sold for ₹ 540 at a loss of 5% what was its cost price?
Answer:
Let the cost rice be ₹ x
Loss% = 5%
Loss = 5% of ₹ x
= \(\frac { 5 }{ 100 }\) x x = \(\frac { x }{ 20 }\)
Selling price = Cost price – Loss
540 = x – \(\frac { x }{ 20 }\)
540 = \(\frac{20 x-x}{20}=\frac{19 x}{20}\)
19x = 540 x 20
x = \(\frac{540 \times 20}{19}\)
= \(\frac{10800}{19}=568 \frac{8}{19}\)
or ₹ 568.40 (approx.)
Cost price of the article
= ₹ 568\(\frac{8}{19}\) or ₹ 568.42

NCERT In-text Question Page No. 170

Question 1.
₹ 10,000 is invested at 5% interest rate p.a. Find the interest at the end of one year.
Answer:
Principal (P) = ₹ 10,000
Rate (R) = 5% p.a.
Time (T) = 1 year

Question 2.
₹ 3,500 is given at 7% p.a. rate of interest. Find the interest which will be received at the end of two years.
Answer:
Principal (P) = ₹ 3,500; Rate (R) = 7% p.a. Time (T) = 2 years
∴ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\frac{3,500 \times 7 \times 2}{100}\)
= ₹ 35 x 7 x 2 = ₹ 490.

Question 3.
₹ 6,050 is borrowed at 6.5% rate of interest p.a. Find the interest and the amount to be paid at the end of 3 years.
Answer:
Principal (P) = ₹ 60,50
Rate (R) = 6.5% p.a.
Time (T) = 3% years
∴ Simple interest

Since, Amount = Principal + Interest
= ₹ 6,050 + ₹ 1,17.75 = ₹ 7,229.75
Thus, the amount to be paid at the end of 3 years = 7,229.75.

Question 4.
₹ 7000 is borrowed at 3.5% rate of interest p.a. borrowed for 2 years. Find the amount to be paid at the end of the second year.
Answer:
Principal (P) = ₹ 7,000
Rate = 3.5% p.a.
Time (T) = 2 years
∴ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\begin{aligned}
&7,000 \times 3.5 \times 2\\
&0
\end{aligned}\) = ₹ 70 x \(\frac{35}{10}\) x 2
= ₹ 7 x 35 x 2 = ₹ 490
Since, Amount = Principal + Interest
Amount to be paid at th end of 2nd year = ₹ 7,000 + ₹ 490 = ₹ 7,490.

NCERT In-text Question Page No. 171

Question 1.
You have ₹ 2,400 in you account and the interest rate is 5%. After how many years would you earn ₹ 240 as interest.
Answer:
Here, principal (P) = 2,400
Rate (R) = 5% p.a.
Time (T) = ?
Simple interest = ₹ 2,40
∵ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
∴ 240 = \(\frac{2400 \times 5 \times \mathrm{T}}{100}\)
∴ T = \(\frac{240 \times 100}{2400 \times 5}\) = 2 years
Thus, an interest of ? 240 will be obtained after 2 years.

Question 2.
On a certain sum, the interest paid after 3 years is ₹ 450 at 5% rate of interest per annum. Find the sum.
Answer:
Let the sum be ‘P’
450 = \(\frac{3 \mathrm{P}}{20}\)
Interest (I) = ₹ 450
3P = 450 x 20
Rate (R) = 5% p.a
P = \(\)
Time (T) = 3 years = 150 x 20
I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ 3000
450 = \(\frac{\mathrm{P} \times 5 \times 3}{100}\)
The required sum is ₹ 3000.

These NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions

NCERT In-text Question Page No. 140 and 141
Question 1.
Lengths of the sides of the triangles are indicated. By applying the SSS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form:


Answer:
(i) InΔABC and ΔPQR such that
AB = 1.5 cm; PQ =1.5 cm
∴ AB = PQ
BC = 2.5 cm; QR = 2.5 cm
∴ BC = QR
AC = 2.2 cm and PR = 2.2 cm
∴ AC = PR
∴ The three sides of ΔABC are equal to the three sides of ΔPQR.
∴ ΔABC ≅ ΔPQR
( By SSS congruence criterion)

(ii) In ΔDEF and ΔLMN
DE = 3.2 cm, MN = 3.2
∴ DE = NM
EF = 3 cm; LM = 3 cm
∴ EF = LM
DF = 3.5 cm, LN = 3.5 cm
∴ DF = LN
∴ The three sides of ΔDEF are equal to three sides of ALMN.
ΔDEF ≅ ΔNML
(By SSS congruence criterion)

(iii) In ΔABC and ΔPQR,
∴ AC = 5 cm, PR = 5 cm
∴ AC = PR
BC = 4 cm, PQ = 4 cm
∴ BC = PQ
AB = 2 cm, QR = 2.5 cm
AB ≠ QR
AB ≠ QR
ΔABC and ΔPQR are not congruent,

(iv) We have ΔABD and ΔADC., such that
AB = 3.5 cm, AC = 3.5 cm
∴ AB = AC
BD = 2.5 cm, CD = 2.5 cm
AD is common ( AD = AD)
since three sides of ΔABD are equal to the three sides of ΔACD
∴ The two triangles are congruent.
ΔABD ≅ ΔACD
(SSS Congruence criterion)

Question 2.
In the given figure, AB = AC and D is the mid- point of \(\overline{\mathrm{BC}}\)
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC? Give reasons.
(iii) Is ∠B = ∠C ? Why?

Answer:
(i) Here, AB = AC and D is the midpoint of \(\overline{\mathrm{BC}}\)
∴ BD = DC
In ΔADB and ΔADC,
AB = AC, BD = DC
AD = AD (common)

(ii) The three sides of ΔADB are equal to the three sides of ΔADC
By SSS rule of congruence criterion
ΔADB ≅ ΔADC

(iii) Since ΔADB ≅ ΔADC,
∠B = ∠C (CPCT)

Question 3.
In the given figure, AC = BD and AD = BC. Which of the following statement is meaningfully written?
(i) ΔABC ≅ ΔABD
(ii) ΔABC ≅ ΔBAD

Answer:
Have AC = BD and AD = BC
In the given triangles ABC and ABD
AB = AB (common)
BD = AC (Given)
AD = BC (Given)
Three sides of ΔABD are equal to the three sides of ΔABC .So the two triangles are congruent,
∴ A ↔ B; B ↔ A; D ↔ C
(i) ΔABC ≅ ΔABD is not true or meaningless.
(ii) ΔABC ≅ ΔBAD is true or meaningful.

NCERT In-text Question Page No. 143 & 144
Question 1.
Which angle is included between the sides \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{EF}}\) is ΔDEF.
Answer:
In ADEF, the angle between \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{EF}}\), is ∠DEF.

Question 2.
By applying SAS congruence rule, you want to establish that ΔPQR ≅ ΔFED. It is given that PQ = FE and RP = DF. What additional information is needed to establish the congruence?
Answer:
We have ΔPQR = ΔFED
[Using SAS congruence rule]
∴ P ↔ F, Q ↔ E, R ↔ D
and PQ = FE and RP = DF.

Question 3.
The given measures of some parts of the triangles are indicated. By applying SAS congruence rule, state the pairs of congruence triangles, if any in each case. In case of congruent triangles, write them in symbolic form.


Answer:
(i) ΔABC and ΔDEF are congruent
AB = 2.5 cm; DE = 2.5 cm
∴ AB = DE
AC = 2.8 cm, DF = 2.8 cm
∴ AC = DF
∠A = 80°; ∠D= 70°
∴ ∠A ≠∠D
∴ ΔABC and ΔDEF are not congruent.

(ii) In ΔDEF and ΔPQR
EF = 3 cm; QR = 3 cm
∴ EF = QR
FD = 3.5 cm; PQ = 3.5 cm
∴ FD = PQ
∠F = 40°, ∠Q = 40°
∴ ∠F = ∠Q
∴ Two sides of ΔADE and their included angle are equal to the corresponding sides and their included angle of ΔPQR
∴ The two triangles are congruent.
ΔDEF ≅ ΔPQR
(SAS Congruence rule)

(iii) In ΔABC and ΔPQR.
AC = 2.5 cm, PR = 2.5 cm
∴ AC = PR
BC = 3 cm; PQ = 3 cm
∴ BC = PQ
∠C = 35° = ∠P = 35°
∴ ∠C = ∠P
∴ Two sides of ΔABC and their included angle are equal to the two corresponding sides and their included angle of ΔPQR.
∴ ΔABC ≅ ΔRQP

(iv) In the ΔPRS and ΔQPR
PQ = 3.5 cm, SR = 3.5 cm
∴ PQ = SR
PR = PR (PR is common)
∠QPR = 30°∠PRS = 30°
∴ Two sides of ΔQPR and their included angle are equal to the corresponding sides and their included angle of ΔPRS.
∴ ΔPQR ≅ ΔRPS
(SAS congruence rule)

Question 4.
In the given Figure, \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) bisect each other at O.
(i) State the three pairs of equal parts in two triangles AOC and BOD.
(ii) Which of the following statements are true?
(a) ΔAOC ≅ ΔDOB
(b) ΔAOC ≅ ΔBOD

Answer:
Since \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) bisect each other at O.
AO = BO and CO = DO
∠AOC = ∠BOD
(vertically opposite angles are equal)

(i) In ΔAOC and ΔBOD
we have, AO = BO and CO = DO
∠AOC = ∠BOD

(ii) From the above given relation
ΔAOC ≅ ΔBOD
(By SAS congruence rule)
(a) The statement ΔAOC ≅ ΔDOB is false.
(b) The statement ΔAOC ≅ ΔBOD is true.

NCERT In-text Question Page No. 145 & 146
Question 1.
What is the side included between the angles M and N of ΔMNP?
Answer:
In AMNP, the side included between M and N is \(\overline{\mathrm{MN}}\).

Question 2.
You want to establish ΔDEF ≅ ΔMNP, using the ASA congruence rule. You are given that ∠D = ∠M and ∠F = ∠P. What information is needed to establish the congruence? (Draw a rough figure and then try!)
Answer:
To establish ΔDEF ≅ ΔMNP, using ASA congruence rule, we need the side containing ∠D and ∠F to be equal to the side containing ∠M and ∠P.

i. e. We need \(\overline{\mathrm{DF}}=\overline{\mathrm{MP}}\).

Question 3.
In the given figure, PL⊥OB and PM⊥OA such that PL = PM. Prove that ΔPLO ≅ ΔPMO

Answer:
In ΔPLO and ΔPMO, we have
∠PLO = ∠PMO = 90°
(Given)
OP = OP (Hypotenuse – common side)
PL = PM (Given)
By using RHS Congruency, we get
ΔPLO ≅ ΔPMO

Question 4.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence rule. In case of congruence, write it in symbolic form.
ΔDEF
(i) ∠D = 60°, ∠F = 80°, ∠DF = 5 cm
(ii) ∠D = 60°, ∠F = 80°, DF = 6cm
(iii) ∠E = 80°, ∠F = 30°, EF = 5 cm

ΔPQR
(i) ∠Q = 60°, ∠R = 80°, QR = 5cm
(ii) ∠Q = 60°, ∠R = 80°, QP = 6cm
(iii) ∠P = 80°, PQ = 5 cm, ∠R = 30°
Answer:
(i) Since ∠D = ∠Q (each = 60°)
∠F = ∠R (each = 80°)
Included side DF = Included side QR [each = 5 cm]
∴ Using ASA congruence rule, we can say that two triangles are congruent.
Also, D ↔ Q, F ↔ R and E ↔ P
∴ ΔDEF ≅ ΔQPR

(ii) Here, QP is not the included side.
∴ The given triangles are not congruent.

(iii) Here, PQ is not the included side.
∴ The given triangles are not congruent.

Question 5.
In the given figure ray AZ bisects ∠DAB as well as ∠DCB.
(i) State the three pairs of equal parts in triangles BAC and DAC.
(ii) Is ΔBAC ≅ ΔDAC? Give reasons.
(iii) Is AB = AD? justify your answer.
(iv) Is CD = CB? Give reasons.

Answer:
(i) AC is the bisector of ∠DAB as well as of ∠DCB
∴ ∠DAC = ∠BAC and ∠DCA = ∠BCA
In ΔBCA and ΔDAC equal parts are
AC = AC (Common)
∠DAC = ∠BAC (AC is a bisector)
∠DCA = ∠BCA (AC is a bisector)

(ii) From the above relation, the two triangles are congruent (using ASA congruence rule)
A ↔ A C ↔ C and D ↔ B
∴ ΔADC ≅ ΔABC
or
ΔBAC ≅ ΔDAC

(iii) Since ΔBAC ≅ ΔDAC The corresponding parts are equal.
∴ AB = AD

(iv) Since ΔBAC ≅ ΔDAC
∴ CD = CB (Corresponding parts are equal)

NCERT In-text Question Page No. 148
Question 1.
In the given figure measures of some parts of triangles are given. By applying R.H.S congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form.


Answer:
(i) In the right ΔPQR and right ΔDEF Hypotenuse PR = Hypotenuse DF (each 6 cm)
PQ = 3 cm, DE = 2.5 cm
PQ ≠ DE
∴ ΔPQR is not congruent to ΔDEF

(ii) Right ΔABC and right ΔABD we have two right triangles ABC and ABD such that
Hypotenuse AB = Hypotenuse BA (common)
AC = 2 cm, BD = 2 cm
∴ AC = BD ∠C = ∠D = 90°
Buy using RHS congruence rule, we can say that the two right triangles are congruent
∴ ΔABC ≅ ΔBAD

(iii) Right ΔABC and Right ΔACD, We have right ΔABC and right ΔACD such that Hypotenuse AC = Hypotenuse AC (common) side \(\overline{\mathrm{AB}}\) = side \(\overline{\mathrm{AD}}\) (each 3.6 cm)
∴ The two right triangles are congruent.
∴ ΔABC ≅ ΔADC

(iv) Right ΔPQS and Right ΔPRS
Hypotenuse PQ = Hypotenuse PR (each 3 cm)
side PS = side PS (common)
∴ using R.H.S congruence rule that two right triangles are congruent.
∴ i.e., ΔPQS ≅ ΔPRS

Question 2.
It is to be established by RHS congruence rule that ΔABC ≅ ΔRPQ. What additional information is needed, if it is given that ∠B = ∠P = 90° and AB = RP?
Answer:
We have ΔABC ≅ ΔRPQ
Since, ∠B = 90° ⇒ Side AC is hypotenuse and ∠P = 90° ≅ Side RQ is hypotenuse.
The required information needed is Hypotenuse
AC = Hypotenuse RQ.

Question 3.
In the given figure, BD and CE are altitudes of ΔABC such that BD = CE.
(i) State the three pairs of equal parts in ΔCBD and ΔBCE.
(ii) Is ΔCBD ≅ ΔBCE? why or why not?
(iii) Is ∠DCB = ∠EBC? why or why not?

Answer:
(i) Hypotenuse BC = Hypotenuse BC
(common)
side BD = side CE (Given)
∠BEC = ∠BDC= 90°
The above are the three equal parts in ΔCBD and ΔBCE

(ii) ∠D = ∠E and ∠B = ∠C
∴ ΔCBD ≅ ΔBCE (By RHS)

(iii) Since ΔCBD ≅ ΔBCE
∠DCB =∠EBC
( ∵ their corresponding parts are equal)

Question 4.
ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC? why or why not?
(iii) Is ∠B = ∠C? why or why not?
(iv) Is BD = CD? why or why not?

Answer:
(i) The three pairs of equal parts in ΔADB and ΔADC are
AD = AD (common)
Hypotenuse AB = Hypotenuse AC
(Given)
∠ADB = ∠ADC (each 90°)

(ii) A ↔ A, B ↔ C, D ↔ D
∴ ΔADB ≅ ΔADC (RHS)

(iii) ΔADB ≅ ΔADC
As, corresponding parts are equal.
∴ ∠B = ∠C

(iv) ΔADB ≅ ΔADC
As corresponding parts are equal.
BD = CD

These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.3

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent ¡n each case,
(a) Gardening shears bought for 250 and sold for ₹ 325 .
(b) A refrigerator bought for ₹ 12,000 and sold at 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Answer:
(a) Cost Price (CP) = ₹ 250
Selling Price = ₹ 325
Profit = S.P – C.P
= ₹ 325 – 250
= ₹ 75
Profit% = \(\frac{75}{250}\) x 100%
= 3 x 100%
= 30%
Profit = ₹ 75; Profit % = 30%.

(b) Cost Price (CP) = ₹ 12,000
Selling Price (SP) = ₹ 13,500
Profit = SP – CP
= ₹ 13500 – ₹ 12000
= ₹ 1500
Profit% = \(\frac{\text { Profit }}{\text { C.P. }}\) x 100
= \(\frac{1500}{12000} \times\) x 100%
= \(\frac{150}{12} \%\)
= 12.5%
Profit = ₹ 1500 and Profit% = 12.5%.

(c) Cost Price (CP) = ₹ 2500
Selling Price (SP) = ₹ 3000
Profit = SP – CP
= ₹ 3000 – 2500
= ₹ 500
Profit% = \(\frac{\text { Profit }}{\text { C.P. }}\) x 100
= \(\frac{500}{2500}\)
= \(\frac{5}{25}\) x 100
= 5 x 4 = 20%
Profit = ₹ 500
and Profit% = 20%.

(d) Cost Price (CP) = ₹ 250
Selling Price (SP) = ₹ 150
Loss = CP – S.P
= ₹ 250 – 150
= ₹ 1oo
Loss% = \(\begin{aligned}
&\text { Loss } \\
&\hline \text { C.P. }
\end{aligned}\) x 100
= ₹ \(\frac{100}{250}\) x 100%
= \(\frac{10}{25}\) x 100%
= 10 x 4 = 40%
Loss = ₹ 100; loss% = 40%.

Question 2.
Convert each part of the ratio to percentage:
(a) 3 : 1
(b) 2 : 3 :5
(c) 1 : 4
(d) 1 : 2 : 5
Answer:
(a) 3:1
Total parts of the ratio = 3 + 1 = 4
Percentage of the first part
= \(\frac { 3 }{ 4 }\) x 100% = 3 x 25% = 75%
Percentage of the second part
= \(\frac { 1 }{ 4 }\) x 100 = 25%

(b) 2:3:5
Total parts of the ratio
= 2 + 3 + 5 = 10
Percentage of the first part
= \(\frac { 2 }{ 10 }\) x 100%
= 2 x 10% = 20%
Percentage of the second part
= \(\frac { 3 }{ 10 }\) x 100% = 3 x 10% = 30%
Percentage of the third part
= \(\frac { 5 }{ 10 }\) x 100% = 5 x 10% = 50%
1st part = 20%, 2nd part = 30%; 3rd part = 50%

(c) 1:4
Total parts of the ratio = 1+4 = 5
Percentage of the first part
= \(\frac { 1 }{ 5 }\) x 100% = 20%
Percentage of the second part
= \(\frac { 4 }{ 5 }\) x 100% = 80%
Percentage of first part = 20%
Percentage of second part = 80%

(d) 1:2:5
Total parts of the ratio = 1 + 2 + 5 = 8
Percentage of the first part
= \(\frac { 1 }{ 8 }\) x 100% = \(\frac { 25 }{ 2 }\)% = 12\(\frac { 1 }{ 2 }\) % or 12.5%
Percentage of the second part
= \(\frac { 2 }{ 8 }\) x 100% = 25%
Percentage of the third part
= \(\frac { 5 }{ 8 }\) x 100% = \(\frac { 125 }{ 2 }\) %
= 62\(\frac { 1 }{ 2 }\)% or 62.5%
Percentage of the first part = 12\(\frac { 1 }{ 2 }\) %
Percentage of the second part = 25%
Percentage of the third part = 62\(\frac { 1 }{ 2 }\) %
= \(\frac { 2 }{ 35 }\) x 100%
= \(\frac{2 \times 20}{7} \%=\frac{40}{7} \%=5 \frac{5}{7} \%\)
Increase percentage = 5\(\frac{5}{7}\)

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Answer:
Initial population = 25000.
Decreased population = 24500
Decrease in population = 25000 – 24500 = 500
Percentage of decrease = \(\frac{\text { Decrease in population }}{\text { Initial population }}\) x 100
= \(\frac{500}{25000}\) x 100% = \(\frac{50 \times 1}{25}\) = 2%

Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase?
Answer:
Initial cost = ₹ 3,50,000
Increased cost = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – 3,50,000
= ₹ 20,000
Increase% = \(\frac{\text { Increase in amount }}{\text { Initial cost }}\) x 100
= \(\frac{20,000}{3,50,000}\) x 100%
= \(\frac{2}{35}\) x 100%
= \(\frac{2 \times 20}{7} \%=\frac{40}{7} \%=5 \frac{5}{7} \%\)
Increase percentage = 5\(\frac{5}{7} \%\)

Question 5.
I buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Answer:
Cost price of the T.V = ₹ 10,000
Profit = 20% of C.P = \(\frac{20}{100}\) x 10,000
= \(\frac { 1 }{ 5 }\) x 10000 = ₹ 2000 5
Selling price of the T.V = ₹ 10,000 + ₹ 2000 = ₹ 12,000
Hence, I get ₹ 12,000

Question 6.
Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Answer:
Let the cost price of the washing machine be Y
Selling price of the washing machine = ₹ 13,500
Loss = 20% of C.P 20
= \(\frac{20}{100}\) x x
= \(\frac{1}{5}\) x x = \(\frac{x}{5}\)
Selling price = Cost price – Loss
13500 = x – \(\frac{x}{5}\)
= \(\frac{5 x-x}{5}=\frac{4 x}{5}\)
\(\frac{4 \mathrm{x}}{5}\) = 13500
x = \(\frac{13500 \times 5}{4}\)
= ₹ 3,375 x 5 = ₹ 16,875
She bought it for ₹ 16,875.

Question 7.
(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Answer:
(i) The ratio of calcium : carbon :
oxygen = 10 : 3 : 12
Total ratios = 10 + 3 + 12 = 25
Percentage of carbon in chalk
= \(\frac { 3 }{ 25 }\) x 100% = 3 x 4% = 12%

(ii) Let the weight of the stick of chalk be ‘x’ g
12% of the chalk mixture is carbon 12% of x = 3
\(\frac{12}{100}\) x x = 3
x = \(\frac{3 \times 100}{12}=\frac{100}{4}\) = 25g
Weight of the chalk stick = 25 g

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Answer:
Cost price of the book = ₹ 275
Loss = 15% of C.P
S.P of the book = ₹ 275 – ₹ 41.25
= \(\frac{15}{100}\) x 275 = ₹ 233.75
= ₹ \(\frac{15}{4}\) x 11
S.P of the book = ₹ 233.75
= ₹ 41.25

Question 9.
Find the amount to be paid at the end of 3 years in each case:
(a) Principal = ₹ 1200 at 12% p.a.
(b) Principal = ₹ 7500 at 5% p.a.
Answer:
(a) Principal (P) = ₹ 1200
Rate (R) = 12%
Time (T) = 3 years
T ,T. P x R x T
Interest (I) = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\frac{1200 \times 12 \times 3}{100}\)
= ₹ 12 x 12 x 3
= ₹ 432
Amount = Principal + Interest
= ₹ 1200 + 432
= ₹ 1632
Amount to be paid at the end of 3 years = ₹ 1632

(b) Principal (P) = ₹ 7500
Rate (R) = 5% p.a.
Time (T) = 3 years
Interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\frac{7500 \times 5 \times 3}{100}\)
= ₹ 75 x 5 x 3
= ₹ 1125
Amount= Principal + Interest
= ₹ 7500+ ₹ 1125 = ₹ 8625
Amount to be paid = ₹ 8625

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 is 2 years?
Answer:
Principal (P) = ₹ 56000
280 = 560 x R x 2
Rate (R) = R
R = \(\frac{280}{560 \times 2}\) %
Time (T) = 2 years
Interest (I) = ₹ 280
R = \(\frac{1}{2 \times 2} \%=\frac{1}{4} \%\)
or 0.25%
Interest (I) = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
Thus, rate of interest = \(\frac{1}{4}\) % or 0.25%
280 = \(\frac{56000 \times \mathrm{R} \times 2}{100}\)

Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Answer:
Let Principal be ‘P’
Rate (R) = 9% p.a.
45 = \(\frac{\mathrm{P} \times 9 \times 1}{100}\)
Interest (I) = ₹ 45
P x 9 = 45 x 100
Time (T) = 1 year
P = \(\frac{45 \times 100}{9}\)
Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ 500
The sum borrowed is ₹ 500

These NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.2

Question 1.
Which congruence criterion do you use in the following?
(a) Given : AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF

(b) Given : ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ

(c) Given:
∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ΔLMN ≅ ΔGFH

(d) Given : EB = DB
∠A = ∠C = 90°
AE = BC
So, ΔABE ≅ ΔCDB

Answer:
(a) Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
By SSS Congruence Criterion

(b) Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
By SAS Congruence Criterion

(c) Given:
∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
ΔLMN ≅ ΔGFH
By ASA Congruence Criterion

(d) Given: EB = DB
AE = BC
∠A = ∠C = 90°
So, ΔABE ≅ ΔCDB
By RHS Congruence Criterion.

Question 2.
You want to show that ΔART ≅ ΔPEN,
(a) If you have to use SSS criterion, then you need to show

(i) AR =
(ii) RT =
(iii) AT =

(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
(i) RT = and
(ii) PN =

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ∠RAT =
(ii) ∠ATR =
Answer:
ΔART ≅ ΔPEN
(a) (i) AR = PE
(ii) RT = EN
(iii) AT = PN

(b) Given: ∠T = ∠N
(i) RT = EN
(ii) PN = AT

(c) (i) ∠RAT = ∠EPN
(ii) ∠ATR = ∠PNE

Question 3.
You have to show that ΔAMP = AMQ. In the following proof, provide the missing reasons.

Answer:

Question 4.
In ΔABC, ∠A =30°, ∠B = 40° and ∠C= 110°
In APQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A Student says that ΔABC = ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
Answer:
No, he is not justified because AAA is not a congruence criterion.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ΔRAT ≅ ?

Answer:
We have N ↔T
O ↔ A
W ↔ R
∴ ΔRAT ≅ ΔWON

Question 6.
Complete the congruence statement:
ΔBCA ≅ ?

ΔQRS ≅ ?

(i) We have
A ↔ A
B ↔ B
T ↔ C
∴ ΔBCA ≅ ΔBTA

(ii) We have
R ↔ P
Q ↔ T
S ↔ Q
∴ ΔQRS ≅ ΔTPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that
(i) the triangles are congruent.
(ii) the triangles are not congruent.
What can you say about their perimeters?
Answer:

(i) Area of ΔABC = \(\frac { 1 }{ 2 }\) × 4 × 3 sq cm = 6 sq cm
Area of ΔCDE = = 6 sq cm
Perimeter of ΔABC = (3 + 4 + 5) cm = 12 cm
Perimeter of ΔCDE = (3 + 4 + 5) cm = 12 cm
The two triangles are congruent.
(∵ Perimeter of ΔABC = Perimeter of ΔCDE)
(ii)

Perimeter of ΔPQR = (3 + 4 + 5) cm = 12cm
Perimeter of ΔPRS = (4 + 3.5 + 4) cm = 11.5 cm
∴ The two triangles are not congruent.
(∵ Perimeter of ΔPRS ≠ Perimeter of ΔPQR)

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Answer:
In ΔABC and ΔDEF

AB = 2 cm DF = 2cm
∴ AB = DF
BC = 4 cm, ED = 4 cm
∴ BC = ED
AC = 3 cm, EF = 3 cm
∴ AC = EF
∠BAC = ∠EDF
∠ABC = ∠DEF
But ΔABC is not congruent to ΔDEF.

Question 9.
If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Answer:
Given ΔABC = ΔPQR
∴ A ↔ P; B ↔ Q and C ↔ R
Two angles ∠B and ∠C of ΔABC are respectively equal to two angles ∠Q and
∠R of ΔPQR
If BC = QR then ΔABC ≅ ΔPQR (using ASA congruence criterion)
We use ASA congruence criterion.

Question 10.
Explain why AABC = AFED

Answer:
ZB = ZE (each 90°)
ZA = ZF (Given)
ZC = ZD
(3^rd angle are equal) BC = ED (Given)
Two angles ZB and ZC and included side BC of AABC are respectively equal to the angle ZE and ZD and the included side ED of ADEF.
∴ ΔABC ≅ ΔFED (ASA)

These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.2

Question 1.
Convert the given fractional numbers to per cents
(i) \(\frac{1}{9}\)
(b) \(\frac{5}{4}\)
(c) \(\frac{3}{40}\)
(d) \(\frac{2}{7}\)
Answer:

Question 2.
Convert the given decimal fractions to per cents.
(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35
Answer:
(a) 0.65 : \(\frac{64}{100}=\frac{65}{100}\) × 100% = 65%
(b) 2.1 = \(\frac{21}{10}=\frac{21}{10}\) × 100% = 210%
(c) 0.02 = \(\frac{2}{100}=\frac{2}{100}\) × 100% =2 %
(d) 12.35 = \(\frac{1235}{100}=\frac{1235}{100}\) × 100% = 1235

Question 3.
Estimate what part of the figures is shaded and hence find the per cent which is shaded.

Answer:
(i) \(\frac{1}{4}\) part is shaded.
\(\frac{1}{4}=\frac{1}{4}\) × 100% = % = 25 %
This coloured part is 25%.

(ii) 3 parts out of 5 parts are shaded so,
\(\frac{3}{5}\) part is shaded
\(\frac{3}{5}=\frac{3}{5}\) × 100% = 3 × 20% = 60%
This coloured part is 60%.

(iii) Here 3 parts out of 8 parts are shaded 3
so, \(\frac{3}{8}\) part is shaded.
\(\frac{3}{8}=\frac{3}{8}\) × 100% = \(\frac{3}{2}\) x 25%
= 37\(\frac { 1 }{ 2 }\)% or 37.5%
This 37\(\frac { 1 }{ 2 }\) part is shaded.

Question 4.
Find:
(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹ 2500
(d) 75% of 1 kg
Answer:
(a) 15% of 250 = \(\frac{15}{100}\) of 250
= \(\frac{15}{100} \times 250=\frac{15 \times 250}{100}=\frac{75}{2}\)
= 37\(\frac{1}{2}\) or 37.5
∴ 15% of 250 = 37 \(\frac{1}{2}\) or 37.5 2

(b) 1% of 1 hour = \(\frac{1}{100}\) × 60 minutes
= \(\frac{1 \times 60}{100}=\frac{3}{5}\) minutes
= \(\frac { 3 }{ 5 }\) × 60 seconds = 36 seconds
∴ 1% of 1 hour = 36 seconds

(c) 20% of ₹ 2500 = \(\frac { 20 }{ 100 }\) of
2500 = \(\frac { 20 }{ 100 }\) x 2500
= ₹ \(\frac{20 \times 2500}{100}\) = ₹ 500
∴ 20% of ₹ 2500 = ₹ 500

(d) 75% of 1 kg = 75% of 1000 g
= \(\frac{75}{100} \times 100 \mathrm{~g}=\frac{75 \times 1000}{100}\) = 750 g
∴ 75% of 1 kg = 750g

Question 5.
Find the whole quantity if
(a) 5% of it is 600.
(b) 12% of it is ₹ 1080.
(c) 40% of it is 500 km.
(d) 70% of it is 14 minutes.
(e) 8% of it is 40 litres.
Answer:
(a) 5% of a quantity is 600.
Let the quantity be x
5% of x = 600
\(\frac { 5 }{ 100 }\) × x = 600
x = \(\frac{600 \times 100}{5}\)
= 120 × 100
= 12000
Thus the required quantity is 12000.

(b) 12% of it is ₹ 1080
Let the required amount be x
12% of x = 1080
\(\frac { 12 }{ 100 }\) × x = 1080
x = \(\frac{1080 \times 100}{12}\)
= \(\frac{1080 \times 25}{3}\)
The required amount = ₹ 9000
= 360 × 25 = ₹ 9000

(c) 40% of it is 500 km
Let the total quantity be x
40% of x = 500
\(\frac { 4 }{ 100 }\) × x = 500
x = \(\frac{500 \times 100}{40}\) km
= 50 × 25 km
= 1250 km
The required quantity is 1250 km.

(d) 70% of it is 14 minutes
Let the required time be x
70% of x = 14 minutes
\(\frac{70}{100}\) × x = 14
x = \(\frac{14 \times 100}{70}\)
= 20 minutes
Thus, the required quantity is 20 minutes.

(e) 8% of a quantity is 40 litres
Let the quantity be x
8% of x =40 litres
\(\frac { 8 }{ 100 }\) × x = 40
x = \(\frac{40 \times 100}{8}\) = 500litres
The required quantity is 500 litres.

Question 6.
Convert given per cents to decimal fractions and also to fractions in simplest forms:
(a) 25%
(b) 150%
(c) 20%
(d) 5%
Answer:
(a) 25% = \(\frac{25}{100}=\frac{1}{4}\)
Thus, 20% = \(\frac{1}{4}\) = 0.25

(b) 150% = \(\frac{150}{100}=\frac{3}{2}\)
Thus 150% = \(\frac{3}{2}\) (or) 1\(\frac{1}{2}\) = 1.5

(c) 20% = \(\frac{20}{100}=\frac{1}{5}\) = 0.2
Thus 20% = \(\frac{1}{5}\) = 0.2

(d) 5% = \(\frac{5}{100}=\frac{1}{20}\)
Thus 5% = \(\frac{1}{20}\) = 0.05

Question 7.
In a city, 30% are females, 40% are males and remaining are children. What per cent are children?
Answer:
Females are 30% and males are 40%
Number of children = 100% – (30% + 40%)
= 100% – 70% = 30%
Thus, children are 30% of the population.

Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Answer:
Total number of voters = 15,000
Part of voters who voted = 60%
Part of voters who did not vote = 100% – 60% = 40%
40% of 15000 = \(\frac{40}{100}\) × 15000
= 40 × 150 = 6000
Thus, 6000 voters did not vote.

Question 9.
Meeta saves ₹ 4000 from her salary. If this is 10% of her salary. What is her salary?
Answer:
Meeta saving = ₹ 4000
Let the salary be ₹ x
Saving = 10% of x
10% of x =4000
\(\frac{10}{100}\) × x = 4000
x = \(\frac{4000 \times 100}{10}\)
= ₹ 40,000
Meeta salary = ₹ 40,000/-

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Answer:
Total number of matches played = 20
Part of matches won = 25%
25% of 20 = \(\frac{25}{100}\) × 20 = \(\frac{25 \times 20}{100}\) = 5
The team won 5 matches.