S.chand publication New Learning Composite mathematics solution of class 7 Chapter 2 Fractions and Decimals Exercise 2I

 Exercise 2I

Question 1

Multiply :

(a) $\begin{array}{r}0.007 \\\times \quad 5 \\\hline \end{array}$

Sol :

$\begin{array}{r}0.007 \\\times 5 \\\hline 0035\end{array}$

(b) $\begin{array}{r}0.009 \\\times \quad 0.7 \\\hline\end{array}$

Sol :

$\begin{array}{r}0.009 \\\times \quad 0.7 \\\hline 0.0063\end{array}$

(c) $\begin{array}{r}4.05 \\\times \quad 8 \\ \hline\end{array}$

Sol :

$\begin{array}{r}4.05 \\\times 8 \\\hline 32.40 \\\hline\end{array}$

(d) $\begin{array}{r}2.06 \\\times 0.03 \\\hline\end{array}$

Sol :

$\begin{array}{r}2.06 \\\times 0.03 \\\hline 0.0618\end{array}$

Question 2

(a) 39.6×0.8

Sol : 31.68

(b) 7.003×1.6

Sol : 11.2048

(c) 897×0.008

Sol : 7.176

Question 3

(a) 1.025×0.0048

Sol :  0.00492

(b) 0.0035×0.726

Sol : 0.0025410

(c) 7.5089×0.47

Sol : 3.529183

Sanskrit translation of class 8 chapter 12 कः रक्षति कः रक्षितः in hindi

 कः रक्षति कः रक्षितः (कौन रक्षा करता है किसकी रक्षा की जाये।)

पाठ का परिचय

[प्रस्तुत पाठ स्वच्छता तथा पर्यावरण सुधार को ध्यान में रखकर सरल संस्कृत में लिखा गया एक संवादात्मक पाठ है। हम अपने आस-पास के वातावरण को किस प्रकार स्वच्छ रखें कि नदियों को प्रदूषित न करें, वृक्षों को न काटें, अपितु अधिकाधिक वृक्षारोपण करें और धरा को शस्यश्यामला बनाएँ। प्लास्टिक का प्रयोग कम करके पर्यावरण संरक्षण में योगदान करें। इन सभी बिन्दुओं पर इस पाठ में चर्चा की गई है। पाठ का प्रारंभ कुछ मित्रों की बातचीत से होता है, जो सांयकाल में दिनभर की गर्मी से व्याकुल होकर घर से बाहर निकले हैं -]

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(क) (ग्रीष्मौ सायंकाले विद्युदभावे प्रचण्डोष्मणा पीडितः वैभवः गृहात् निष्क्रामति)

वैभवः – अरे परमिन्दर्! अपि त्वमपि विद्युदभावेन पीडितः बहिरागतः?

परमिन्दर् – आम् मित्र! एकतः प्रचण्डातपकालः अन्यतश्च विद्युदभावः परं बहिरागत्यापि पश्यामि यत् वायुवेगः तु सर्वथाऽवरुद्धः।

सत्यमेवोक्तम् प्राणिति पवनेन जगत् सकलं, सृष्टिर्निखिला चैतन्यमयी।

क्षणमपि न जीव्यतेऽनेन विना, सर्वातिशायिमूल्यः पवनः॥

विनयः – अरे मित्र! शरीरात् न केवलं स्वेदबिन्दवः अपितु स्वेदधाराः इव प्रस्रवन्ति स्मृतिपथमायाति शुक्लमहोदयैः रचितः श्लोकः। तप्तर्वाताघातैरवितुं लोकान् नभसि मेघाः,

आरक्षिविभागजना इव समये नैव दृश्यन्ते॥

सरलार्थ-

गर्मी के मौसम में शाम के समय बिजली के चले जाने पर बहुत तेज गर्मी से परेशान वैभव घर से बाहर निकलता है।)

वैभव – अरे परमिंदर ! क्या तुम भी बिजली के चले जाने से परेशान होकर बाहर आ गए हो। 

परमिंदर – हाँ मित्र ! एक तो बहुत तेज गर्मी का समय है दूसरा बिजली चली गई है, परन्तु बाहर आने के बाद देखता हूँ कि  हवा की गति भी पूरी तरह से रुक गई है। 

सत्य ही कहा है –

वायु से ही  प्राणवान है अर्थात जीवित है, पूरी सृष्टि वायु के कारण ही सजीव है।

इसके बिना अर्थात वायु के बिना क्षण भर के लिए भी जीवित नहीं रहा जा सकता है। सबसे अधिक मूल्यवान हवा ही है। 

विनय – अरे मित्र ! शरीर से न केवल पसीने की बूंदें बल्कि पसीने की नदियाँ है। शुक्ल महोदय के द्वारा रचित श्लोक याद आ रहा है। 

तपती हुई हवा के आघात से लोगों को बचाने के लिए आकाश में बादल भी सुरक्षा विभाग के लोगों की तरह दिखाई नहीं दे रहे है। अर्थात जिस प्रकार जरुरत समय सुरक्षा  लोग दिखाई नहीं देते वैसे ही गर्मी  समय आकाश  नहीं दिखाई देते हैं।  

शब्दार्थ-

प्रचण्ड-भयंकर।

बहिः-बाहर।

आगतः-आ गया।

प्रचण्ड-तीव्र।

अन्यतः-और भी।

आगत्य-आकर।

अवरुद्धः-रुक गया।

प्राणिति-जीवित है (Survives)।

सकलम्-सारा।

निखिला-सम्पूर्णं (Whole)।

जीव्यते-जीवित है।

सर्वातिशायि-सबसे बढकर।

स्वेदबिन्दवः-पसीने की बूंदें।

प्रस्रवन्ति-बह रही हैं।

तप्तैः-गर्म।

वाताघातैः-लू के द्वारा।

अवितुम्-रक्षा करने के लिए।

नभसि-आकाश में। आरक्षिः-पुलिस।

दृश्यन्ते-दिखाई पड़ते हैं।

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(ख) परमिन्दर् – आम् अद्य तु वस्तुतः एव

निदाघतापतप्तस्य, याति तालु हि शुष्कताम्।

पुंसो भयादितस्येव, स्वेदवज्जायते वपुः॥

जोसेफः – मित्राणि! यत्र-तत्र बहुभूमिकभवनानां, भूमिगतमार्गाणाम्, विशेषतः

मैट्रोमार्गाणां, उपरिगमिसेतूनाम् मार्गेत्यादीनां निर्माणाय वृक्षाः कर्त्यन्ते

तर्हि अन्यत् किमपेक्ष्यते अस्माभिः? वयं तु विस्मृतवन्तः एव

एकेन शुष्कवृक्षण दह्यमानेन वह्निना।

दह्यते तद्वनं सर्वं कुपुत्रेण कुलं यथा॥

सरलार्थ –

परमिंदर – आज तो वास्तव में (अर्थात वास्तव में ही आकाश में बादल दिखाई नहीं दे रहे)

तेज गर्मी के ताप से मनुष्य तालु सुख जाता है। भयभीत मनुष्य का शरीर पसीने से तरबतर हो जाता है।  

जोसेफ – मित्र ! यहाँ – वहाँ पृथ्वी पर भवनों का, भूमिगत मार्गों का, विशेष रूप से ऊपर से मैट्रो के मार्गों के पुलों इत्यादि के निर्माण के अत्यधिक वृक्ष काटे जाते हैं। अवश्य ही हमसे क्या अपेक्षा की जाती है?  हम तो भूल ही गए –

एक सूखे हुए वृक्ष के द्वारा पूरा वन जला दिया जाता है उसी प्रकार कुपुत्र के द्वारा पूरा कुल का ही नाश हो जाता है। 

शब्दार्थ-

वस्तुतः-वास्तव में।

निदाघ-गर्मी।

याति-प्राप्त होता है।

शुष्कताम्-सूखापन।

पुंसः-मनुष्य का।

भयादितस्य-भयभीत।

वपुः-शरीर।

स्वेदवत्-पसीने से तर।

उपरिगामि-ऊपर से जाने वाले।

कर्त्यन्ते-काटे जाते हैं।

शुष्क-सूखा।

विस्मृतवन्तः- भूल गए हैं।

वह्निना-अग्नि के द्वारा।

दह्यमानेन-जलाए जाते हुए।

शक्ष्येम-सकेंगे।

आगच्छन्तु-आओ।

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(ग) परमिन्दर् – आम् एतदपि सर्वथा सत्यम्! आगच्छन्तु नदीतीरं गच्छामः। तत्र चेत्

काञ्चित् शान्तिं प्राप्तुं शक्ष्येम।

(नदीतीरं गन्तुकामाः बालाः यत्र-तत्र अवकरभाण्डारं दृष्ट्वा वार्तालापं कुर्वन्ति)

जोसेफः – पश्यन्तु मित्राणि यत्र-तत्र प्लास्टिकस्यूतानि अन्यत् चावकरं प्रक्षिप्तमस्ति।

कथ्यते यत् स्वच्छता स्वास्थ्यकरी परं वयं तु शिक्षिताः अपि अशिक्षिता

इवाचरामः अनेन प्रकारेण…. वैभवः – गृहाणि तु अस्माभिः नित्यं स्वच्छानि क्रियन्ते परं किमर्थं स्वपर्यावरणस्य

स्वच्छतां प्रति ध्यानं न दीयते। विनयः पश्य-पश्य उपरितः इदानीमपि अवकरः मार्गे क्षिप्यते।

(आहूय) महोदये! कृपां कुरू मार्गे भ्रमद्भ्यः । एतत् तु सर्वथा अशोभनं कृत्यम्।

अस्मत्सदृशेभ्यः बालेभ्यः भवतीसदृशैः एवं संस्कारा देयाः ।

रोजलिन् – आम् पुत्र! सर्वथा सत्यं वदसि! क्षम्यताम्। इदानीमेवागच्छामि। (रोजलिन् आगत्य बालैः साकं स्वक्षिप्तमवकर मार्गे विकीर्णमन्यदवकर चापि सङ्गृह्य अवकरकण्डोले पातयति)

सरलार्थ –

परमिंदर – हाँ ये बिलकुल सही है। चलो नदी के किनारे चलते हैं। वहाँ कुछ शांति प्राप्त  सकेंगे। 

(नदी किनारे जाने के इच्छुक बालक यहाँ – वहाँ गंदगी के ढेर देखकर बातचीत करते हैं।)

जोसेफ – मित्र देखो यहाँ – वहाँ  प्लास्टिक का थैला/थैलियाँ और अन्य दूसरा कचरा भी फेंका हुआ है। कहा जाता है कि स्वच्छता स्वास्थ्य के लिए लाभकारी होती है। परन्तु हम शिक्षित होते हुए भी अनपढ़ों की तरह आचरण करते हैं इस प्रकार …….

वैभव – हम घरों को तो प्रतिदिन साफ़ करते हैं किन्तु किसलिए अपने पर्यावरण की स्वच्छता की और ध्यान नहीं दिया जाता है। 

विनय – देखो – देखो ऊपर से अभी भी मार्ग में कूड़ा-करकट डाला जा रहा है। 

(बुलाकर के) महोदय, कृपा करें मार्ग में ऐसे कूड़े को मत फैलाओ, ये तो हमेशा ही अशोभनीय कार्य है। अर्थात  रास्ते में कूड़ा – करकट फैंकना सही बात नहीं है। हमारे जैसे बालकों को आप जैसी महिलाएँ द्वारा इस प्रकार संस्कार दिए जायेंगे। अर्थात बड़ी महिलाओं  द्वारा छोटे बच्चों  प्रकार के संस्कार नहीं जाने चाहिए। 

रोजलिन – हाँ पुत्र ! बिल्कुल सही कह रहे हो। माफ़ कर दीजिये। अभी आती हूँ। 

(रोजलिन आ कर बालकों के साथ अपने द्वारा फेंका गया कचरे को एक साथ इकठ्ठा करके कूड़ेदान में डालती है। 

शब्दार्थ- अवकर-कूड़ा।

प्रक्षिप्तम्-फेंक दिया।

आचरामः-आचरण करते हैं।

दीयते-दिया जाता है।

उपरितः-ऊपर से।

भ्रमद्भ्यः -भ्रमण करते हुए। क

कत्यम्-कार्य।

क्षम्यताम्-क्षमा करिए।

अवगच्छामि-जानती हूँ।

कण्डोले-टोकरी में।

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(घ) बालाः – एवमेव जागरूकतया एव प्रधानमन्त्रिमहोदयानां स्वच्छताऽभियानमपि गतिं प्राप्स्यति।

विनयः – पश्य पश्य तत्र धेनुः शाकफलानामावरणैः सह प्लास्टिकस्यूतमपि

खादति। यथाकथञ्चित् निवारणीया एषा। (मार्गे कदलीफलविक्रेतारं दृष्ट्वा बालाः कदलीफलानि क्रीत्वा धेनुमाह्वयन्ति भोजयन्ति च, मार्गात् प्लास्टिकस्यूतानि चापसार्य पिहिते अवकरकण्डोले क्षिपन्ति)

परमिन्दर् – प्लास्टिकस्य मृत्तिकायां लयाभवात् अस्माकं पर्यावरणस्य कृते महती क्षतिः भवति। पूर्वं तु कार्पासेन, चर्मणा, लौहेन, लाक्षया, मृत्तिकया, काष्ठेन वा निर्मितानि वस्तूनि एव प्राप्यन्ते स्म। अधुना तत्स्थाने प्लास्टिकनिर्मितानि वस्तूनि एव प्राप्यन्ते।

सरलार्थ-

बालक – इस प्रकार जागरूकता से ही प्रधानमंत्री महोदय का स्वच्छता अभियान भी गति प्राप्त करेगा। 

हम सभी के सहयोग से स्वच्छता अभियान सफल हो पायेगा। 

विनय – देखो देखो वहाँ जो गाय है सब्जियों और फलों के छिलकों के साथ प्लास्टिक की थैलियाँ भी खा रही हैं। इसको किसी भी तरह रोकना चाहिए। 

(मार्ग में केले बेचने वाले को देखकर बच्चे केले खरीद कर गाय को बुलाते है और खिलते है। रास्ते से प्लास्टिक की थैलियों को हटाकर ढके हुए कूड़ेदान में डालते है।)

परमिंदर – प्लास्टिक जो मिटटी में नष्ट नहीं होने के कारण हमारे पर्यावरण की बहुत अधिक हानि होती है। पहले तो कपास से, चमड़े से, लोहे से, लाख से, मिटटी से तथा काठ से बानी हुई ही वस्तुएँ प्राप्त होती थी।  अब उसके स्थान पर प्लास्टिक निर्मित वस्तुएँ ही प्राप्त होती हैं। 

शब्दार्थ-

प्राप्स्यति-प्राप्त करेगा।

आवरणैः-छिलकों।

यथाकथञ्चित्-जैसे-तैसे।

निवारणीया-हटाना चाहिए।

कदली-केला।

अपसार्य-हटाकर।

पिहित-ढके हुए।

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(ङ) वैभवः – आम् घटिपट्टिका, अन्यानि बहुविधानि पात्राणि, कलमेत्यादीनि सर्वाणि नु प्लास्टिकनिर्मितानि भवन्ति।

जोसैफः – आम् अस्माभिः पित्रोः शिक्षकाणां सहयोगेन प्लास्टिकस्य विविधपक्षाः विचारणीयाः। पर्यावरणेन सह पशवः अपि रक्षणीयाः। (एवमेवालपन्तः सर्वे नदीतीरं प्राप्ताः, नदीजले निमज्जिताः भवन्ति गायन्ति च

सुपर्यावरणेनास्ति जगतः सुस्थितिः सखे।

जगति जायमानानां सम्भवः सम्भवो भुवि॥5॥

सर्वे – अतीवानन्दप्रदोऽयं जलविहारः।

सरलार्थ-

हाँ घड़ी का पट्टा और अन्य प्रकार के बर्तन, पेन/कलम आदि सब कुछ ही तो प्लास्टिक से बना हुआ होता है। 

जोसेफ – हाँ हमारे माता – पिता एवं गुरुजनों के सहयोग से प्लास्टिक के अनेक पहलुओं पर विचार करना चाहिए। पर्यावरण के साथ पशुओं की भी रक्षा करनी चाहिए। 

(इस प्रकार बातचीत करते हुए सभी नदी के किनारे पहुँच गए और नदी के जल में स्नान किया तथा गाते हैं – 

स्वच्छ पर्यावरण के द्वारा ही जगत की सुंदर  स्थिति है। संसार में उत्पन्न होने वालों की उत्पति पृथ्वी पर ही है। सभी अत्यधिक आनंद के साथ जल विहार करते हैं। 

शब्दार्थ-

मृत्तिकायां-मिट्टी में।

क्षतिः-हानि। कार्पासेन-कपास से।

चर्मणा-चमड़े से। लाक्षया-लाख से।

काष्ठेन-काठ से। आलपन्तः-बात करते हुए।

निमज्जिताः-स्नान किया।

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S.chand publication New Learning Composite mathematics solution of class 7 Chapter 9 Pairs and Angles Exercise 9B

 Exercise 9B

Question 1

Name the pair of angles marked as A for alternate, B for corresponding and C for co-interior.

Sol :

(a) A

(b) B

(c) C

(d) A

(e) C

(f) B

(g) B

(h) A

Question 2

Give examples of each angle pair.

(a) corresponding angles

Ans. ∠1 and ∠3, ∠2 and ∠4, ∠5 and ∠7, ∠6 and ∠8

(b) alternate angles

Ans. ∠2 and ∠7, ∠6 and ∠3

(c) co-interior angles

Ans. ∠2 and ∠3, ∠6 and ∠7

(d) vertically opposite angles.

Ans. ∠1 and ∠6, ∠2 and ∠5, ∠3 and ∠8, ∠4 and ∠7

Question 3

Name the type of angle pair shown in each letter.

Sol :

(a) Corresponding ∠S

(b) Alternate ∠S

(c) Co interior ∠S

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 9 Pairs and Angles Exercise 9A

 Exercise 9A

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Q1 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 1

What is measure of complement of 380

Sol : 90°-38°=52°

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Q2 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 2

What is measure of supplement of 170°

Sol : 180°-170°=10°

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Q3 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 3

Tell whether the angles shown below are complementary, supplementary or neither.

Sol : 

(a) Complementary

(b) Supplementary

(c) Neither 

(d) Neither

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Q4 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 4

For each angle in group A, find its complement in group B. (Example 20° and 70°)

A: 20°, 10°, 35°, 45°, 60°, 85°, 65°

B: 25°, 45°, 30°, 5°, 70°, 80°, 55°

Sol :

20° and 70°

10° and 80°

35° and 55°

45° and 45°

60° and 30°

85° and 5°

65° and 25°

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Q5 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 5

For each angle in group A, find its supplement in group B.

A: 140°, 70°, 179°, 165°, 55°, 45°, 80°

B: 100°, 15°, 125°, 110°, 1°, 40°, 135°

Sol :

140° and 40°

70° and 110°

179° and 1°

165° and 15°

55° and 125°

45° and 135°

80° and 100°

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Q6 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 6

Two complementary angles are in the ratio 7:8, find the angles.

Sol :

Let, the angles be 7x and 8x

7x+8x=90

15x=90

$x=\frac{90}{15}$

∴x=6

Complementary angles are 42 and 48

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Q7 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 7

Two supplementary angles are in the ratio 3:7, Find the angles.

Sol :

Let , the angles are 3x and 7x

3x+7x=180

10x=180

$x=\frac{180}{10}$

∴x=18

∴The angles are 54 and 126

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Q8 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 8

Tell whether the angles are only adjacent, adjacent and form a linear pair or not adjacent.

(a) ∠1 and ∠4

Sol : Adjacent and form linear pair

(b) ∠2 and ∠3

Sol : Adjacent and form a linear pair

(c) ∠3 and ∠4

Sol : Only adjacent

(d) ∠3 and ∠1

Sol : Not adjacent 

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Q9 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 9

Find the measures of the lettered angles.

(a) 90°

(b) 180°-108°=72°

∴P=72°

(c) 180°-39°=141°

∴x=141°

(d) 7b+2b=180°

9b=180°

$b=\frac{180^{\circ}}{9}$

b=20°

∴7b=(7×20)=140°

∴2b=(2×20)=40°

(e) 180°-(62°+53°)

=180°-115°

y=65°

(f) b+b+b=180°

3b=180°

$b=\frac{180^{\circ}}{3}$

∴b=60°

(g) 139°+d=180°

∴d=180°-139°=41°

c+49°=180°

∴c=180°-49°=131°

(h) f+32°=90°

∴f=90°-32°=58°

65°+g=90°

g=90°-65°=25°

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Q10 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 10

Find  

(a) b, if a=110°

b=360°-110°=250°

(b) b, if $a=1\frac{1}{4}$ rt ∠s

$a=1\frac{1}{4} \times 90 ^{\circ}$

$=\frac{5}{4} \times 90^{\circ}$

=112.5°

∵a+b=360° (sum of angles around a point)

∴b=360°-112.5°=247.5°

$=\frac{11}{4}$ right angle

$=2\frac{3}{4}$ rt  ∠s

(c) a, if b=258°

a=360°-258°=102°

(d) b, if b-3a=40°

b-3a=40°

or b=40°+3a

b=360°-a

40°+3a=360°-a

3a+a=360°-40°

4a=320°

$a=\frac{320^{\circ}}{4}$

∴a=80°

∴b=360°-80°=280°

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Q11 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 11

Find the values of a,b,c and d

(a) 

a=360°-(70°+60°)

=360°-130°=230°

(b) 

b=360°-(125°+59°+110°)

=360°-294°=66°

(c) 

4c+5c=180°

9c=180°

$c=\frac{180^{\circ}}{9}=20^{\circ}$

∴c=20°

4c=80°

5c=100°

(d)

3d+62°+62°+105°+47°=360°

3d=360°-(62°+62°+105°+47°)

3d=360°-276°

3d=84°

∴d=28°

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Q12 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 12

Name the pairs of vertical angles

Sol :

∠3 and ∠6

∠2 and ∠5

∠1 and ∠4

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Q13 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 13

How many pairs of vertical angles are in the diagram ?

Sol : 4

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Q14 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 14

Find the values of each variable.

(a) a=60°

(b) ∠a=∠c$=\frac{360^{\circ}-(25^{\circ}+25^{\circ})}{2}$

$=\frac{360^{\circ}-50^{\circ}}{2}=\frac{310^{\circ}}{2}$

=155°

(c) y=x

2y+y=180°

3y=180°

$y=\frac{180^{\circ}}{3}$

∴y=60°

(d)

p+76°=180° (Linear Pair)

p=180°-76°=104°

∠p=∠r=104° (Vertically Opposite Angles )

4q=76° (Vertically Opposite Angles )

$q=\frac{76}{4}$

q=19°

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Q15 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 15

In the diagram on the right, ∠1 = 50° and ∠2 = 84°, find the measures of ∠3, ∠4, ∠5 and ∠6.

Sol :

If ∠1 = 50° then, ∠4=50°

If ∠2 = 84° then ,∠5=84°

∴∠3+∠6=360°-(50°+50°+84°+84°)

=360°-268°=92°

∴∠3$=\frac{92^{\circ}}{2}$=46°

and ∠6=46°

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Q16 | Ex-9A |Class 7 |S.Chand | New Learning Composite maths |Pairs and Angles |myhelper

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Question 16

Find the angles in each of the following.

(a) The angles are supplementary and the larger is 20° less than 3 times the smaller.

Sol :

Let, the smaller angle be x

ATQ,

x+(3x-20°)=180°

4x-20°=180°

4x=180°+20°

$x=\frac{200^{\circ}}{4}=50^{\circ}$

∴The smaller angle=50°

∴The larger angle=(3×50°)-20

=150°-20°=130°

(b) The angles are complementary and the larger is 15° more than twice the smaller.

Sol :

Let , the smaller angle be x

ATQ,

x+(2x+15°)=90°

3x=90°-15°

3x=75°

$x=\frac{75}{3}=25$

∴The smaller angle=25°

The larger angle=(2×25°)+15°

=50°+15°=65°

(c) The angles are adjacent and form an angle of 120°. The larger is 20° less than 3 times the smaller.

Sol :

Let, the smaller angle be x

The larger angle be (3x-20°)

ATQ,

x+(3x-20°)=120°

4x=120°+20°

$x=\frac{140^{\circ}}{4}$

x=35°

∴The smaller angle=35°

The larger angle=(3×35°)-20°

=105°-20°=85°

(d) The angles are vertical and complementary.

Sol :

Let, the angle be x and y

ATQ,

x=y

∴x+y=90°

y+y=90°

2y=90°

y=45°

∴The vertical angles are 45° and 45°

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 8 Percentage and Its Applications Exercise 8H

 Exercise 8H

Question 

Find the missing value.

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Q1 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(1) I if P = Rs. 800, R = 5% p.a., T = 2 years.

Sol :

$I=\frac{\text{PRT}}{100}$

$=\frac{800 \times 5 \times 2}{100}$

=80

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Q2 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(2) I if P = Rs. 5000, R = 6.5% p.a. T = 3 years.

Sol :

$I=\frac{\text{PRT}}{100}$

$=\frac{5000 \times 65\times 3}{100}$

=975

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Q3 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(3) R is I = Rs. 364 P = Rs. 1300, T = 7 years.

Sol :

$I=\frac{\text{PRT}}{100}$ or $=R\frac{100I}{PT}$

$=\frac{100 \times 364}{1300 \times 7}$

=4

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Q4 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(4) P if I = Rs. 440, R = 5%, T = 4 years

Sol :

$I=\frac{\text{PRT}}{100}$ or  $P=\frac{100I}{RT}$

$=\frac{100 \times 440}{5 \times 4}$

=2200

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Q5 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(5) T if I = Rs. 455, R = 7%, P = Rs. 1300

Sol :

$I=\frac{\text{PRT}}{100}$ or  $T=\frac{100I}{PR}$

$=\frac{100 \times 455}{1300 \times 7}$

=5

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Q6 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(6) After 2 years, the credit balance in a saving account earning simple interest was Rs. 585.75. The original amount was Rs. 550. What was the interest rate?

Sol :

Simple Interest(585.75-550)=35.75

Interest Rate (R)$=\frac{100I}{PT}$

$=\frac{100 \times 35}{550 \times 2}$

=3.25

Ans : Interest Rate was 3.25%

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Q7 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(7) In what time will Rs. 1860 amount to Rs. 2641.20 at simple interest of 12% per annum.

Sol :

SI=2641.20-1860=781-20

$I=\frac{PRT}{100}$

or $T=\frac{100I}{PR}$

$=\frac{100\times 781.20}{1860\times 12 \times 100}$

=3.5

Ans : 3.5 years

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Q8 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(8) Find the simple interest on Rs. 7300 from 10 May 2017 to 10 September 2017 at 5% per annum.

Sol :

$I=\frac{PRT}{100}$

$=\frac{7300 \times 5\times \frac{123}{365}}{100}$

=123

Ans : 123

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Q9 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(9) A sum of Rs. 400 amounts to Rs. 480 in 4 years. What will it amount ti, if the rate of interest is increased by 2%?

Sol :

I=480-400=80

$I=\frac{PRT}{100}$
or $R=\frac{100I}{PT}$
$=\frac{100 \times 80}{400 \times 4}$
=5

If rate if interest increased 2% per annum,

New (R1)=(5+2)=7%

Then, $I_{1}=\frac{P_{1}R_{1}T_{1}}{100}$
$=\frac{400 \times 4 \times 7}{100}=112$

Therefore, New amount=(P+1)

=400+112=512

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Q10 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(10) On a certain sum, the simple interest at the end of 6-1/4 years becomes 3/8 of the sum. Find the rate of interest.

Sol :

$I=\frac{PRT}{100}$

$\frac{3}{8}p=\frac{PR\left(\frac{25}{4}\right)}{100}$

$\frac{3}{8}=\frac{R}{16}$

$R=\frac{16\times 3}{8}=6$

Ans : Rate of interest is 6%

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Q11 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(11) A sum doubles itself in 8 years at simple interest. Find the rate of interest per annum.

Sol :

Let , the sum of money be x

After 8 years amount becomes =2x

S.I=2x-x=x

∴Rate per annum$=\frac{S.I \times 100}{P\times T}$
$=\frac{x\times 100}{x\times 8}=\frac{100}{8}$
=25.5

Ans : Rate of interest is 12.5%

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Q12 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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(12) In how many years will the simple interest on a sum of money be equal to the principal at the rate of 16-2/3% per annum?

Sol :
Let, Principle is P and Interest is also P then $P=\frac{PRT}{100}$
$1=\frac{50 \times T}{3\times 100}$
$T=\frac{100\times 3}{50}$
=6

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 8 Percentage and Its Applications Exercise 8G

 Exercise 8G

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Q1 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 1

Find the profit or loss per cent.

(a) C.P. = Rs. 500, S.P. = Rs. 650

Sol :

Profit=(650-500)=150

∴Profit %$=\left(\frac{150}{500}\times 100\right)$=30%

(b) C.P. = Rs. 900, S.P = Rs. 828

Sol :

Loss=900-828=72

∴Loss %$=\left(\frac{72}{900}\times 100\right)$=8%

(c) C.P. = Rs. 325, overheads = Rs. 25, S.P = Rs. 490

Sol :

Total CP=325+25=350

∴Profit %$=\left(\frac{490-350}{350}\times 100\right)$%

$=\left(\frac{140}{350}\times 100\right)$=40%

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Q2 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 2

 Find the S.P. of an article if its

(a) C.P. = Rs. 80, profit = 15%

Sol :

Profit$=80 \times \frac{15}{100}$=12

∴SP=80+12=92

(b) C.P = Rs. 950, loss = 8%

Sol :

Loss$=950\times \frac{8}{100}$=76%

SP=950-76=874

(c) C.P = Rs. 190.50, octroi = Rs. 20.50, transportation charges = Rs. 59, loss = 20%

Sol :

Total CP=190.50+20.50+59=270

∴SP$=\frac{100-20}{100}\times 270$

$=\frac{80}{100}\times 270$

=216

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Q3 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 3

Find the C.P. if

(a) S.P. = Rs. 2380, profit = 19%

Sol :

CP$=\left(\frac{100}{100+\text{gain %}}\times \text{SP}\right)$

$=\frac{100}{100+19}\times 2380$

$=\frac{100}{119}\times 2380$

=2000

(b) S.P. = Rs. 510, loss = 15%

Sol :

CP$=\left(\frac{100}{100+\text{gain %}}\times \text{SP}\right)$

$=\left(\frac{100}{100-15}\times 510\right)$

$=\frac{100}{85} \times 510$

=600

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Q4 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 4

 A person buys a book for Rs. 200 and sells it for Rs. 225. What will be his gain per cent?

Sol :

Gain%$=\left(\frac{225-200}{200}\times 100\right)$%

$=\frac{25}{200}\times 100$%

=12.5%

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Q5 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 5

A calculator is bought for Rs. 350 and sold at a gain of 15%. What is the selling price of calculator?

Sol :

∴S.P$=\frac{100+15}{100}\times 350$

$=\frac{115}{10}\times 35$

=402.50

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Q6 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 6

By selling a cellphone for Rs. 2400 a shopkeeper makes a profit of 25%. What would be his profit percentage if he sells it for Rs. 2040?

Sol :

C.P$=\frac{100}{100+25}\times 2400$

$=\frac{100}{125}\times 2400$

=1920

Profit %$=\left(\frac{2040-1920}{1920}\times 100\right)$%

$=\left(\frac{120}{1920}\times 100\right)$%

=6.25%

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Q7 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 7

The owner of a cell phone charges his customer 23% more than the cost price. Find the cost price, if a customer paid Rs. 7011 for his cellphone.

Sol :

C.P$=\frac{100}{100+23}\times 7011$

$=\frac{100}{123}\times 7011$

=5700

Cost price of the phone 5700

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Q8 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 8

By selling an article for Rs. 100, a man gains Rs. 15. Find his gain %.

Sol :

C.P=100-15=85

Gain %$=\left(\frac{15}{85}\times 100\right)$%

=17.65%

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Q9 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 9

A trader sells an article and loses 12 and 1/2%. Find the ratio of cost price of the selling price.

Sol :

Let, the CP of an article be 100

∴SP of an article $=\left(100-12\frac{1}{2}\right)$

=100-12.5=87.5

Ratio of CP and SP= 100 : 87.5

=8 : 7

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Q10 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 10

A person purchased 10 dozen pens at the rate of Rs. 40 per dozen. On checking, he found that 20 pens were not working. In order to earn 25% profit, at what price should he sell each pen?

Sol :

10 dozen pens =10×12=120

Rate of per pen $=\frac{40}{12}$ 

Out of 120 pen 20 pens not working 100 pens working

Price of 100 pens$=\frac{20}{6}\times 100$

=333.3

≃334

25% profit on 334$=\frac{25}{100}\times 334$

=83.5

≃84

25% Profit on 100 pens =(334+84)=418

25% profit on each pen$=\frac{418}{100}$

=4.18

≃5

Ans : He should sell each pen at 5 Rs

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Q11 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 11

When a plot is sold for Rs. 18,700, owner loses 15%. At what price must the plot be sold in order to gain 15%?

Sol :

CP of plot$=\frac{100}{100-15}\times 18700$

$=\frac{100}{85} \times 18700$

=22000

To gain 15% , SP of plot$=\frac{100+15}{100}\times 22000$

=115×220=25300

SP of the plot is 25300

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Q12 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 12

By selling a small plot for Rs. 80,000, I gain 25 per cent. What per cent will i lose if i sell it for Rs. 56,000?

Sol :

CP$=\frac{100}{100+25}\times 80000$

$=\frac{100}{125} \times 80000$

=64000

∴Lose percentage$=\left(\frac{64000-56000}{64000}\times 100\right)$%

$=\left(\frac{8000}{64000}\times 100\right)$%

=12.5 %

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Q13 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 13

A man purchased a car for Rs. 1,35,000 and spent Rs. 25,000 on repairs. At what price was the car sold, if he suffered 10% loss on it?

Sol :

Total CP=135000+25000

=160000

Loss=10%

∴Then profit$=160000 \times \frac{10}{100}$

=16000

∴Total SP=160000-16000=144000

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Q14 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 14

Sudhir earned a profit of 20% by selling 60 apples at the rateof Rs. 42.50 for 5 apples. Find the total cost at which the apples were bought.

Sol :

Rate of 60 apples$=4250 \times \frac{60}{5}$

=2510

∴CP of apples$=\left(\frac{100}{100+20}\times 510\right)$

$=\left(\frac{100}{120}\times 510\right)$

=425

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Q15 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 15

The cost price of 16 pens is equal to the selling price of 12 pens. Find the gain or loss percent?

Sol :

$\frac{SP}{CP}=\frac{16}{12}=\frac{4}{3}$

Loss %$=\frac{4-3}{3}\times 100=\frac{1}{3}\times 100=33\frac{1}{3}$

Ans : Loss$=33\frac{1}{3}$%

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Q16 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 16

If the selling price of 4 articles is equal to the cost price of 5 articles. Find the profit percent?

Sol :

5CP=4SP

$\frac{CP}{SP}=\frac{4}{5}$

∴Profit%$=\frac{5-4}{4}\times 100$

$=\frac{1}{4}\times 100$

=25

Ans : Profit =25%

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Q17 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 17

A fruit seller buys 700 oranges at the rate of  Rs. 500 for 100 oranges and another variety of 500 oranges at the rate of Rs. 700 for 100 oranges and sells them at Rs. 84 per dozen. Find the profit per cent.

Sol :

CP of 700 oranges$=\left(\frac{500\times 700}{100}\right)$

=3500

CP of 500 oranges of another variety$=\left(\frac{700}{100}\times 500 \right)$

=3500

∴Total CP=3500+3500=7000

Total number of oranges purchased=700+500=1200

∴Total SP$=\left(\frac{84}{12}\times 1200\right)$

=8400

∴Profit%$=\left(\frac{8400-7000}{7000}\times 100\right)$%

$=\left(\frac{14000}{7000}\times 100\right)$%

=20%

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Q18 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 18

Some articles were bought at 6 for Rs. 5 and sold at 5 for Rs. 6. Find the gain per cent.

Sol :

Let, Number of articles bought =30

∴CP of 30 articles $=\left(\frac{5}{6}\times 30\right)$

=25

∴SP of 30 articles$=\left(\frac{6}{5}\times 30\right)$

=36

∴Gain%$=\left(\frac{36-25}{25}\times 100\right)$%

$=\left(\frac{11}{25} \times 100\right)$%=44%

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Q19 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 19

An article is sold at a gain of 15%. Had it been sold for Rs. 27 more, the profit would have been 20%. Find the cost price of the article.

Sol :

100 units CP

15% gain SP₁=115

20% gain SP₂=120

So, 5 unit =27

1 unit $=\frac{27}{5}$

∴100$=\frac{27}{5}\times 100$=540

CP of the article 540

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Q20 | Ex-8G |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

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Question 20

A dealer sold two types of goods for Rs. 10,000 each. On one of them, he lost 20% and on the other, he gained 20%. Find his gain or loss per cent on the entire transaction.

Sol :

Total SP=10000+10000=20000

Since, we know that

CP on loss$=SP \times \frac{100}{100-20}$

CP on profit$=SP\times \frac{100}{100+20}$

∴Total CP$=\left[10000\times \frac{100}{100-20}\right]+\left[10000\times \frac{100}{100+20}\right]$

$=\left(10000\times \frac{100}{80}\right)+\left(10000 \times \frac{100}{120}\right)$

$=50000\times \left(\frac{1}{4}+\frac{1}{6}\right)$

$=50000\times \frac{3+2}{12}=\frac{62500}{3}$

∴Loss$=\frac{62500}{3}-20000$

$=\frac{62500-60000}{3}=\frac{2500}{3}$

∴Loss %$=\left(\dfrac{\frac{250000}{3}}{\frac{62500}{3}} \times 100\right)$%

$=\left(\frac{2500}{62500}\times 100\right)$=4%

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 8 Percentage and Its Applications Exercise 8F

 Exercise 8F

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Q1 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 1

55 per cent of the students in a school are girls. What percentage of the students are boys? Find the number of students, if there are 216 boys.

Sol :

∴Percentage of boys=100-55=45%

45% of student=216

1% of student$=\frac{216}{45}$

45% of student$=\frac{216 \times 100}{45}$=480

∴Total number of student=480

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Q2 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 2

If 10% is deducted from a bill, Rs. 7200 remains to be paid. How much is the bill?

Sol :

∴Remaining bill=(100-10)=90%

90% bill=7200

1% bill $=\frac{7200}{90}$

100% bill $=\frac{7200 \times 100}{90}$=8000

∴Total bill=8000

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Q3 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 3

Calculation gives the length of a line segment as 6.4 cm, while from drawing and measurement it is obtained as 6.6 cm. Find the error per cent.

Sol :

∴Error length=(6.6-6.4)=0.2 cm

In 6.4 cm the error is 0.2 cm

In 1 cm the error is $=\frac{0.2}{6.4}$ cm

In 100 cm the error is $=\frac{0.2\times 100}{6.4}=\frac{25}{8}=3\frac{1}{8}$ cm

∴Error Percentage is $3\frac{1}{8}$%

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Q4 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 4

Atul deposited 30% of his income in a bank. After spending 50% of the remaining, he had Rs. 10,500 left with him. What is his income?

Sol :

Remaining income after depositing in bank=100-30=70%

50% of 70%=35%

35% of income is 10500

1% of income is $\frac{10500}{35}$

100% of income is $=\frac{10500 \times 100}{35}$=30000

∴Atul's income is 30000

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Q5 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 5

Manish obtained 498 marks out of 600 marks and Aravind scored 640 marks out of 800 marks. Whose performance is better?

Sol :

∴Percentage of Manish score$=\left(\frac{498}{600} \times 100\right)$=83 %

∴Percentage of Arvind's score$=\left(\frac{640}{800}\times 100\right)$=80%

Ans : Manish's performance is better than Arvind performance

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Q6 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 6

In a graduate class of 200, 40% are women and 1/5 becomes lecturers. If the number of men who became lecturers is twice that of women, calculate approximate percentage of men who became lecturers.

Sol :

Number of women$=\frac{40}{100}\times 200$=80

Number of women who become lecturer$=80\times \frac{1}{5}$=16

Number of men who becomes lecturer=16×2=32

Number of men=200-80=120

∴Percentage of men who becomes lecturer$=\left(\frac{32}{120}\times 100\right)$

=26.6%

≃27%

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Q7 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 7

The price of an article is reduced by 25%. In order to retain the original price, by what percentage should the present price be increased?

Sol :

Let the price 100

25% reduced new price 75

So the price raised should be $=\left(\frac{25}{75}\times 100\right)$=33.33%

Ans : The present price increased 33.33%

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Q8 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 8

In a mixture of milk and water, the proportion of water by weight is 75%. If in the 60 g mixture, 15 g of water is added, what would be the percentage of water?

Sol :

In 60 g mixture , weight of water$=\left(\frac{75}{100}\times 60\right)$=45 gm

15 g water is added , now weight of water=45+15=60gm

Now , Percentage of water $=\left(\frac{60}{75}\times 100\right)$=80%

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Q9 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 9

One litre of water is evaporated from a 6 litre solution containing 4% sugar. Find the percentage of sugar in the remaining solution.

Sol :

Quality of sugar in 6 litre solution $=\left(\frac{4}{100}\times 6\right)$=0.24 litre

1 litre of water evaporated from a 6 litre solution

Now, quantity of solution=(6-1)=5 litres

Percentage of sugar in remaining solution$=\left(\frac{0.25}{5}\times 100\right)$=4.8%

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Q10 | Ex-8F |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 10

State Electricity Board gives 15% discount on electricity bills, if it is paid before due date. A person gets Rs. 54 as discount. What was the actual amount of the bill?

Sol :

State Electricity Board gives 15 Rs discount on 100

State Electricity Board gives 1 Rs discount on $=\frac{100}{15}$

State Electricity Board gives 54 Rs discount on $=\frac{100\times 54}{15}$=360

Ans : Actual deposit of bill is 360

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 8 Percentage and Its Applications Exercise 8E

Exercise 8E

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Q1 | Ex-8E |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 1

Ravi’s class has 18 boys and 12 girls. What is its percentage composition?

Sol :
∴Percentage of boys$=\frac{18}{30}\times 100$=60%

∴Percentage of girls$=\frac{12}{30}\times 100$=40%

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Q2 | Ex-8E |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 2

Chalk contains 10% calcium, 3% carbon and 12% oxygen. Find the amount of carbon and calcium (in grams) in 2 and $\frac{1}{2}$ kg of chalk.

Sol :

Amount of carbon$=\left(2\frac{1}{2}\times \frac{3}{100}\right)$ kg

$=\frac{6}{2}\times \frac{3}{100}=\frac{3}{40}$

∴Amount of calcium$=\left(2\frac{1}{2}\times \frac{10}{100}\right)$ kg

$=\left(\frac{5}{2} \times \frac{18}{100}\right)=\frac{1}{4}$

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Question 3

An alloy contains 4% tin, 1% zinc and the remaining copper.

(a) What is the percentage composition of the alloy?

Sol :

Percentage of tin=4%

Percentage of zinc=1%

Percentage of copper=(100-5)=95%

(b) Find the mass (in kg) of each metal in 5 tonnes of the alloy.

Sol :

5 tonnes=5000 kg

Mass of tin $=\left(5000 \times \frac{4}{100}\right)$=200 kg

Mass of zinc $=\left(5000 \times \frac{1}{100}\right)$=50 kg

Mass of copper $=\left(5000 \times \frac{95}{100}\right)$=4750 kg

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Question 4

A packet of cake mix has the composition of its contents shown on the label. Calculate the percentage composition of the cake mix.

Nutritional  value (per 50 g serve)
Protein	3.5 g
Fat	5.5 g
Carbohydrates	27 g
Other ingredients	14 g

Sol :

∴Percentage of protein$=\frac{3.5}{50} \times 100$=7% 

∴Percentage of fat$=\frac{5.5}{50} \times 100$=11% 

∴Percentage of carbohydrates$=\frac{27}{50} \times 100$=54% 

∴Percentage of other ingredients$=\frac{14}{50} \times 100$=28% 

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Q5 | Ex-8E |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 5

The average composition (by weight) of milk is shown in the table.

Ingredients	Percentage
Carbohydrate	5
Protein	3
Fat	4
Water	87
Other	1

(a) How many grams are there of each component in 1 kg of milk?

Sol :

[1 kg=1000 gm]

∴Quantity of carbohydrates$=\left(1000 \times \frac{5}{100}\right)$=50 gm

∴Quantity of protein$=\left(1000\times \frac{3}{100}\right)$=30gm

∴Quantity of fat$=\left(1000 \times \frac{4}{100}\right)$=40 gm

∴Quantity of water$=\left(1000 \times \frac{87}{100}\right)$=870 gm

∴Quantity of other ingredient$=\left(1000 \times \frac{1}{100}\right)$=10 gm

(b) How many grams of proteins are there in 7.5 kg of milk?

Sol :

7.5 kg=7500 gm

∴Quantity of protein$=\left(7500\times \frac{3}{100}\right)$=225 gm