Class 7

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 11 Perimeter and Area MCQ Questions

Multiple Choice Questions :

Question 1.
Out of the following figures which has the greatest perimeter ?
(The length and breadth of each rectangle is equal)
(i) (ii) (iii)

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv).
Answer:
(c) (iii)

Question 2.
Perimeter of a rectangle = 2 (…………)
(a) Length + Breadth
(b) Length × Breadth
(c) Length ÷ Breadth
(d) Length – Breadth.
Answer:
(a) Length + Breadth

Question 3.
Area of rectangle = …………….
(a) Length + Breadth
(b) Length × Breadth
(c) Length – Breadth
(d) Length ÷ Breadth.
Answer:
(b) Length × Breadth

Question 4.
Circumference of a circle = …………….
(a) πr
(b) 2πr
(c) πr²
(d) 2πr².
Answer:
(b) 2πr

Question 5.
Area of a circle =
(a) 2πr
(b) πr
(c) πr²
(d) 2πr²
Answer:
(c) πr²

Question 6.
The area of a square park whose perimeter is 320 m will be :
(a) 3200 m²
(b) 640 m²
(c) 6400 m²
(d) 6400 m
Answer:
(c) 6400 m2

Question 7.
The area of a rectangular plot is 440 m² and its length is 22 m. Its breadth will be :
(a) 20 m
(b) 40 m
(c) 44 m
(d) 21 m.
Answer:
(a) 20 m

Question 8.
The radius ofa circle is 14 cm. Its area will be:
(a) 88 cm²
(b) 616 cm²
(c) 196 cm²
(d) 56 cm²
Answer:
(b) 616 cm²

Fill in the blanks :

Question 1.
Perimeter of a rectangle = 2 (………….)
Answer:
Length + Breadth

Question 2.
Area of rectangle = ………….
Answer:
Length × Breadth

Question 3.
Perimeter of a square = ………….
Answer:
4 × side

Question 4.
Area of a square = ………….
Answer:
(side)2

Question 5.
Area of a triangle = ………….
Answer:
\(\frac {1}{2}\) × Base × height

Write True/False :

Question 1.
Area of a square with side 10 cm is 20 cm² (True/False)
Answer:
False

Question 2.
One are has 100 m² (True/False)
Answer:
True

Question 3.
One hectare has 1000 m² (True/False)
Answer:
False

Question 4.
Area of circle is πr² square unit (True/False)
Answer:
True

Question 5.
Brahmgupta gave the formule for the area of a cyclic quadrilateral (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Hindi Book Solutions Chapter 2 धूल का फूल Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Hindi Chapter 2 धूल का फूल

Hindi Guide for Class 7 PSEB धूल का फूल Textbook Questions and Answers

(क) भाषा-बोध

1. शब्दार्थ

सरपट – तेज़ चाल से
खलिहान = कटी हुई फसल रखने का स्थान
विपन्नता = गरीबी
मजूरी = मज़दूरी
व्यर्थ = बेकार
मुटियार = नवयुवती
उत्थान = उन्नति, बढ़ती
जन्नत = स्वर्ग
निहारते = देखते
प्रांगण = आंगन
पुष्प-गुच्छ = फूलों का गुलदस्ता
गुंजायमान = गूंजता हुआ

2. इन मुहावरों के अर्थ लिखते हुए वाक्य प्रयोग करें:

खुशी के आँसू छलकना ________________ ___________________________
मन बल्लियों उछलना _____________ _________________________
नाम रोशन करना __________________ ______________________
गले लगाना ________________ _______________________
फूले न समाना _______________ __________________________
उत्तर:
खुशी के आँसू छलकना (बहुत अधिक प्रसन्न होना) – परीक्षा में प्रथम आने का समाचार मिलते ही नीलम की आँखों में खुशी के आँसू छलकने लगे।
मन बल्लियों उछलना(बहुत प्रसन्न होना) – बहुत दिनों बाद गाँव जाते हुए प्रशांत का मन बल्लियों उछलने लगा।
नाम रोशन करना (प्रसिद्धि मिलना) – कलक्टर बन कर प्रशांत ने अपने माता-पिता का नाम रोशन कर दिया।
गले लगना (प्रेम से भेंटना)  -माता ने पुत्र को आशीर्वाद देते हुए बाँहों में भरकर गले लगा लिया।
फूले न समना (बहुत प्रसन्न होना) – एशिया-कप हॉकी में विजय प्राप्त कर भारतीय खिलाड़ी फूले न समा रहे थे।

3. विपरीत शब्द लिखें: 

मौन = ……………..
ऊबड़-खाबड़ = ……………
ज़रूरी = ……………..
शहर = …………….
सुनसान = ……………….
मजदूर = ……………..
आमदन = ………………
उत्थान = ……………
उत्तर:
शब्द विपरीत शब्द
मौन = मुखर
ऊबड़-खाबड़ = सीधी-सपाट
ज़रूरी = गैर-ज़रूरी
शहर = गाँव
सुनसान = आबाद
मजदूर = मालिक
आमदन = खर्च
उत्थान = पतन

4. दो-दो पर्यायवाची शब्द लिखें :

तरक्की = ……………
शिक्षा = ……………
अध्यापक = ……………
मेहनत = …………….
शिष्य = …………….
उत्तर:
शब्द पर्यायवाची शब्द
तरक्की = वृद्धि, बढ़ती
शिक्षा = परामर्श, सलाह, तालीम, सबक
अध्यापक = शिक्षक, गुरु
मेहनत = परिश्रम, श्रम
शिष्य = चेला, विद्यार्थी

प्रयोगात्मक व्याकरण

गुरु की शरण = गुरुशरण
माँ और बाप = माँ-बाप

उपर्युक्त पदों में गुरु की शरण को गुरुशरण तथा माँ और बाप को माँ-बाप रूप में संक्षेप में लिख सकते हैं। इस प्रकार शब्दों के मेल से नए शब्द बन जाते हैं।
अतः परस्पर सम्बन्ध रखने वाले दो या दो से अधिक शब्दों के मेल से जब कोई नया सार्थक शब्द बनता है तो उस मेल को समास कहते हैं।

समास करने के बाद जो शब्द बनता है उसे समस्तपद कहते हैं। समस्तपद को इसके शब्द खण्डों में अलग-अलग करने की विधि को विग्रह कहते हैं। जैसे :

गुरुशरण (समस्त पद) = गुरु की शरण (विग्रह)

विशेष:
समस्त पद के दो पद होते हैं- पूर्व पद और उत्तर पद। पहले पद को पूर्व पद तथा बाद को उत्तर पद कहते हैं। जैसे-गुरु (पूर्व पद), शरण (उत्तर पद)

पदों की प्रधानता के आधार पर समास के चार भेद होते हैं :

1. अव्ययीभाव समास
2. तत्पुरुष समास
3. द्वंद्व समास
4. बहुब्रीहि समास

(क)	समस्त पद	विग्रह	जिस अर्थ में अव्यय यहाँ प्रयुक्त हुआ
(1)	बेरोज़गार	रोज़गार के बिना	‘बे’ का प्रयोग के बिना’ अर्थ में हुआ है।
(2)	आजीवन	जीवन तक	‘आ’ का प्रयोग तक के अर्थ में हुआ है।
(3)	यथानियम	नियम के अनुसार	‘यथा’ का प्रयोग ‘अनुसार’ के अर्थ में हुआ है।

यहाँ समस्त पद में ‘बे’, ‘आ’ तथा ‘यथा’ अव्यय हैं तथा इसके मेल से पूर्ण पद ही अव्यय बन गया है।

अतएव जिस समस्त पद में पूर्वपद प्रधान हो और अव्यय हो और समास होने पर पूर्ण पद ही अव्यय बन जाए, वह अव्ययी भाव समास कहलाता है।
अन्य उदाहरण-आमरण-मरने तक, निडर-डर के बिना, भरपेट-पेट भर कर, आजन्म-जन्म भर, प्रति पल-हर पल, बेखबर-बिना खबर के, अनजान-जाने बिना आदि।
तत्पुरुष समास को समझने के लिए कारक का ज्ञान अपेक्षित है। अत: पहले कारकों को समझते हैं।

हमें साहब ने रहने के लिए घर दिया।

यदि इस वाक्य को इस ढंग से लिखें-‘हमें साहब रहने घर दिया’ तो वाक्य में आए शब्दों का एक-दूसरे से सम्बन्ध नहीं प्रकट होता और न ही अर्थ स्पष्ट होता है।
इसलिए वाक्य में आए ने, के लिए चिह्न वाक्य के अन्य शब्दों का परस्पर सम्बन्ध जोड़ते हैं।

अतएव संज्ञा या सर्वनाम के जिस रूप से उनका सम्बन्ध क्रिया तथा वाक्य के दूसरे शब्दों में जाना जाए, उसे कारक कहते हैं।

विशेष :- वाक्य में प्रयुक्त ‘के, ने, से, के लिए’ कारक चिह्नों को परसर्ग भी कहते हैं।

(1) शरण ने क्षमा माँगी।
इस वाक्य में क्षमा माँगने का काम शरण ने किया अर्थात् कर्ता शरण है। अतः शरण ने में कर्ता कारक है।

(2) प्रशांत उच्च पद को प्राप्त हुआ।
इस वाक्य में प्राप्त हुआ क्रिया है, प्रशांत कर्ता है तथा क्रिया का फल पद पर पड़ रहा है। अतः पद को में कर्म कारक है।

अतएव वाक्य में जिस संज्ञा या सर्वनाम पर क्रिया का फल पड़ता है, उसे कर्म कारक कहते हैं।

(3) प्रशांत गाड़ी से गाँव आया।
इस वाक्य में गया क्रिया का साधन गाड़ी है। अतः गाड़ी से में करण कारक है।

अतएव कर्ता जिस साधन की मदद से क्रिया सम्पन्न करता है, उसे करण कारक कहते हैं।

(4) हम पढ़ने के लिए विद्यालय जाते थे।
इस वाक्य में जाना क्रिया का कार्य पढ़ने के लिए है, अतः यहाँ सम्प्रदान कारक है।

अतएव जिस संज्ञा या सर्वनाम के लिए कुछ किया जाए उसे सम्प्रदान कारक कहते हैं।

(5) हमारा पूरा परिवार गाँव से शहर आ गया।
इस वाक्य में गाँव से पद से अलग होने का अर्थ स्पष्ट हो रहा है, इसलिए यहाँ अपादान कारक है।

अतः जिस संज्ञा से पृथक्ता अर्थात् अलग होने का भाव प्रकट हो, उसे अपादान कारक कहते हैं।

इसके अतिरिक्त किसी से सीखने, लगाने, डरने, बचाने, तुलना करने, माँगने, निकलने तथा दूरी का भाव दर्शाने में भी अपादान कारक होता है।

(6) नसीब का लड़का कलक्टर बन गया।
इस वाक्य में नसीब का लड़का से पिता-पुत्र का सम्बन्ध प्रकट हो रहा है अतः यहाँ सम्बन्ध कारक है।

अतएव जहाँ दो संज्ञाओं या सर्वनामों का आपस में सम्बन्ध प्रकट हो, वहाँ सम्बन्ध कारक होता है।

(7) प्रशांत पहले गाँव में रहता था।
इस वाक्य में गाँव में पद में रहना क्रिया के आधार का पता चलता है, यहाँ अधिकरण कारक है।

अतएव जहाँ संज्ञा या सर्वनाम शब्द के आधार का पता चले उसे अधिकरण कारक कहते हैं।

(8) अरे शरण गाड़ी जल्दी चलाओ।
इस वाक्य में अरे शरण ! को सम्बोधन किया गया है, इसलिए यहाँ सम्बोधन कारक है।

अतएव संज्ञा या सर्वनाम के जिस रूप से किसी को पुकारने, बुलाने, सुनाने या सावधान करने का भाव प्रकट हो, वहाँ सम्बोधन कारक होता है। आइए, अब तत्पुरुष समास को समझते हैं।

(ख) समस्त पद विग्रह
पदप्राप्त पद को प्राप्त

उपर्युक्त समास में समस्त पद बनाते समय पूर्वपद (पद) के साथ आए परसर्ग (को) का लोप हो गया है। इसके उत्तरपद (प्राप्त) प्रधान है।
अतएव जिस समास में उत्तर पद प्रधान हो उसे तत्पुरुष समास कहते हैं। पद बनाते समय पूर्वपद के साथ आने वाले परसर्ग का लोप हो जाता है।

अन्य उदाहरण

समस्त पद = विग्रह
यशप्राप्त = यश को प्राप्त
समस्त पद = विग्रह
भावविह्वल = भाव से विह्वल
पाठशाला = पढ़ने के लिए शाला
धनहीन = धन से हीन
विद्याभ्यास = विद्या का अभ्यास
सिरदर्द = सिर में दर्द

(ख) विचार-बोध

1. प्रश्नों के उत्तर एक या दो वाक्यों में लिखें:

प्रश्न 1.
प्रशान्त का मन बल्लियों क्यों उछल रहा था ?
उत्तर:
प्रशान्त का मन बल्लियों इसलिए उछल रहा था क्योंकि वह कई वर्षों के बाद अपने गाँव जा रहा था।

प्रश्न 2.
गाँव की ओर जाते हुए उसे किन-किन लोगों की याद आने लगी ?
उत्तर:
गाँव की ओर जाते हुए पंच जी के खेत, दीनू ग्वाले की गाय-भैंसें, सब्बू कुम्हार के चाक की याद आने लगी।

प्रश्न 3.
उसके अध्यापक का क्या नाम था ?
उत्तर:
उस के अध्यापक का नाम मास्टर आदित्य प्रकाश था।

प्रश्न 4.
हर माँ-बाप का क्या सपना होता है ?
उत्तर:
हर माँ-बाप का यह सपना होता है कि उसकी संतान पढ़-लिखकर कुछ बन जाए।

प्रश्न 5.
प्रशान्त का क्या सपना था? यह सपना उसने कैसे पूरा किया ?
उत्तर:
प्रशान्त का सपना था कि एक दिन ज़रूर वह कुछ बनेगा। यह सपना उस ने खूब पढ़ कर पूरा किया।

प्रश्न 6.
लड़कियों की शिक्षा के सम्बन्ध में उसके क्या विचार थे ?
उत्तर:
लड़कियों की शिक्षा के सम्बन्ध में उसके विचार थे कि लड़कियाँ घर का श्रृंगार होती हैं। उन्हें खूब पढ़ाना चाहिए, क्योंकि वे परिवार का आधार होती हैं।

प्रश्न 7.
प्रशान्त गाँव में क्यों आया था ?
उत्तर:
प्रशान्त गाँव में गाँव की पाठशाला का दर्जा बढ़ाने का आदेश लेकर आया था।

2. इन प्रश्नों के उत्तर चार-पाँच वाक्यों में लिखें:

प्रश्न 1.
गरीबी में रहते हुए भी प्रशान्त ने अपने लक्ष्य को कैसे प्राप्त किया?
उत्तर:
प्रशान्त गरीबी में रहते हुए भी पढ़ाई की तरफ बहुत ध्यान देता था। वह मास्टर आदित्य प्रकाश जी की बातें सुनकर धन्य हो जाता था। पढ़ाई का खर्च चलाने के लिए वह छुट्टियों में छोटा-मोटा आमदनी वाला काम कर लेता था। वह खूब पढ़ कर अफसर बनना चाहता था। उसे विश्वास था कि बड़े बनने की कुंजी विद्या है। इसलिए वह मेहनत से पढ़ता था और कर्म को पूजा मानता था। इस प्रकार परिश्रमपूर्वक पढ़-लिख कर उसने अपना लक्ष्य प्राप्त किया और कलक्टर बन गया।

प्रश्न 2.
आपका क्या लक्ष्य है ? अपने लक्ष्य को प्राप्त करने के लिए आप क्या करेंगे?
उत्तर:
मेरे जीवन का लक्ष्य आदर्श अध्यापक बनना है, जो अपने विद्यार्थियों को विद्या के प्रति सच्ची लगन पैदा करके उन्हें भावी भारत का सच्चा एवं अनुशासित नागरिक बना सके। इसके लिए मैं खूब मेहनत से पढंगा। बी०ए० करने के बाद अध्यापक के प्रशिक्षण के लिए बी०एड्० की परीक्षा उत्तीर्ण करके किसी अच्छे विद्यालय में शिक्षक का पद ग्रहण कर विद्यार्थियों को सर्वगुण सम्पन्न बनाने का प्रयास करूँगा।

3. इस कहानी में कई बिन्दुओं को छुआ गया है जैसे :

…………………… गाँवों से शहर की ओर पलायन
…………………… ग्रामीण लोगों की दशा/गरीबी/यथास्थिति
…………………… लक्ष्य प्राप्त करना
…………………… लड़कियों की शिक्षा के प्रति सोच
…………………… गाँव के प्रति प्यार
…………………… सम्बन्धों की आत्मीयता
…………………… अध्यापकों का सम्मान
…………………… इन बिंदुओं पर विचार-विमर्श करें।
उत्तर:
विद्यार्थी आपस में विचार-विमर्श करें।

PSEB 7th Class Hindi Guide धूल का फूल Important Questions and Answers

निम्नलिखित प्रश्नों के उत्तर उचित विकल्प चुनकर लिखिए

प्रश्न 1.
‘धूल का फूल’ किसकी कहानी है ?
(क) खेत मज़दूर के पुत्र की
(ख) खेत की
(ग) खेत मजदूर की पुत्री की
(घ) नाले की
उत्तर:
(क) खेत मज़दूर के पुत्र की

प्रश्न 2.
खेत मजदूर का लड़का पढ़-लिखकर क्या बनता है ?
(क) चपड़ासी
(ख) कलक्टर
(ग) अध्यापक
(घ) पुलिस कप्तान
उत्तर:
(ख) कलक्टर

प्रश्न 3.
गाड़ी चलाते हुए कौन सोच रहा था ?
(क) करण
(ख) गुरचरण
(ग) गुरशरण
(घ) गौरव
उत्तर:
(ग) गुरशरण

प्रश्न 4.
कुम्हार का क्या नाम था ?
(क) सब्बू
(ख) चौधरी
(ग) कब्बू
(घ) चेतन
उत्तर:
(क) सब्बू

प्रश्न 5.
बड़ा बनने की कुंजी क्या है ?
(क) शरीर की ताकत
(ख) ज़मीन
(ग) धन
(घ) विद्या
उत्तर:
(घ) विद्या

प्रश्न 6.
कलक्टर प्रशांत क्या आदेश लेकर आया था?
(क) गाँव खाली कराने का
(ख) गाँव के विद्यालय का दर्जा बढ़ाने का
(ग) ज़मीन लेने का
(घ) ज़मीन देने का
उत्तर:
(ख) गाँव के विद्यालय का दर्जा बढ़ाने का

निम्नलिखित रिक्त स्थानों की पूर्ति उचित विकल्पों से कीजिए

प्रश्न 1.
कलक्टर के चाचा का नाम ………… था ।
(क) चरण सिंह
(ख) विक्रम सिंह
(ग) करण सिंह
(घ) चेतन सिंह
उत्तर:
(क) चरण सिंह

प्रश्न 2.
कलक्टर का नाम …………… था।
(क) अविरल
(ख) प्रशांत
(ग) चहल
(घ) वेदांत
उत्तर:
(ख) प्रशांत

प्रश्न 3.
प्रशांत के पिता का नाम …………. था ।
(क) अब्दुल्ला
(ख) कासिम
(ग) सुजान सिंह
(घ) नसीब
उत्तर:
(घ) नसीब

प्रश्न 4.
प्रशांत के मास्टर का नाम ………… था ।
(क) आदित्य
(ख) विश्वास
(ग) विवेक शर्मा
(घ) कर्म सिंह
उत्तर:
(क) आदित्य

दिए गए शब्दों का सही अर्थ मिलान कीजिए

प्रश्न 1.
सरपट:
साँप का पेट
सिर का पेट
तेज चाल से
उत्तर:
तेज़ चाल से

प्रश्न 2.
प्रांगण:
आंगन
पराग का कण
परात
उत्तर:
आंगन

प्रश्न 3.
मजूरी:
मंजूरी
मज़दूरी
मंजर
उत्तर:
मजदूरी

प्रश्न 4.
उत्थान:
उनका
उन्नति
उतना
उत्तरः
उन्नति

धूल का फूल Summary

धूल का फूल पाठ का सार

‘धूल का फूल’ एक ऐसे खेत-मज़दूर गरीब पिता के पुत्र की कहानी है, जो अपने परिश्रम तथा इच्छा शक्ति के बल पर पढ़-लिखकर कलक्टर बन जाता है। ऊबड़-खाबड़ सड़क पर गाड़ी चलाते हुए गुरशरण सोच रहा था कि न मालूम क्यों साहब इधर दौरे पर आए हैं ? जब उसने साहब से पूछा कि अभी और कितनी दूर जाना है तो साहब ने उसे चलते रहने को कहा।

साहब का मन बहुत प्रसन्न था। वे बरसों बाद अपने गाँव जा रहे थे। वे बीस-पच्चीस साल पहले के पंच जी के खेत, दीनू ग्वाले की गाय-भैंसें, सब्बू कुम्हार के चाक आदि की बातें सोच कर भाव-विभोर हो रहे थे। उन्हें गन्ने का रस पीना, गर्म गुड़ खाना, ऊधम मचाना, ककड़ियाँ-खरबूजे खाना, पाठशाला में आदित्य मास्टर जी से पढ़ना आदि याद आ रहा था। उसके पिता दूसरों के खेतों में मजदूरी करते थे तथा माँ के साथ वह खेतों पर खाना ले जाता था। उसके पिता को शहर में चपरासी की नौकरी मिली तो सारा परिवार गाँव से शहर आ गया। जहाँ आकर गाड़ी में बैठे अफसर को देखकर उसका मन भी उन जैसा बनने की इच्छा करता और वह सोचता कि जब लाल बहादुर शास्त्री, लिंकन, एडीसन जैसे बड़े बन सकते हैं, तो वह क्यों नहीं?

सरकारी स्कूल दूर थे फिर भी वह पढ़ता गया क्योंकि उसे विश्वास था कि बड़े बनने की कुंजी विद्या ही है और विद्याधन मेहनत के बिना नहीं मिलता। वह कर्म को पूजा मानने लगा। वह शरण को गाड़ी धीरे चलाने के लिए कहता है और शरण से यह जान कर कि उसका लड़का तो पढ़ने जाता है परन्तु वह अपनी लड़की को स्कूल नहीं भेजता तो उसे समझाता है कि लड़की को भी पढ़ाओ क्योंकि लड़कियाँ घर का श्रृंगार होती हैं, वे ही परिवार का आधार हैं। उसने गाड़ी की खिड़की से मुटियारों को सिर पर बोझ लेकर जाते, खेतों में कम्बाइन चलाते तथा स्त्रियों-बच्चों को झोला लिए अनाज की बालियाँ चुनते देखा तो सोचने लगा कि आजादी के इतने सालों बाद भी गरीब की वही दशा है जो उस के ज़माने में थी। तभी उसे एक बुजुर्ग दिखाई दिए। उस ने गाड़ी रुकवाई और उनके पैर स्पर्श किए। वे चाचा चरण सिंह थे।

उन्होंने उसे नहीं पहचाना तो उसने स्वयं ही अपना परिचय दिया कि वह उनका प्रशांत है। चाचा को भी याद आया नसीब का पुत्र प्रशांत। वह आज यहाँ गाँव के विद्यालय का दर्जा बढ़ाने का आदेश लेकर आया था। पाठशाला रंगोली, रंग-बिरंगी झंडियों से सजी हुई थी। पुष्प-गुच्छों से उसका स्वागत हुआ। अपने प्रिय अध्यापक आदित्य प्रकाश के पैरों को स्पर्श करने के लिए जैसे ही प्रशांत झुका कि उन्होंने उसे बाँहों में भरकर गले लगा लिया और वे फूले नहीं समा रहे थे कि उनका गरीब प्रशांत कलक्टर प्रशांत बन गया है। वही सरकार की ओर से पाठशाला का दर्जा बढ़ाने का आदेश लाया है।

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

1. Find the circumference of circle whose
(i) Radius (r) = 21 cm
(ii) Radius (r) = 3.5 cm
(iii) Diameter = 84 cm
Solution:
(i) Given radius (r) = 21 cm
circumference of circle = 2πr
= 2 × \(\frac {22}{7}\) ×21
= 132 cm

(ii) Given radius (r) = 3.5 cm
Circumference = 2πl
= 2 × \(\frac {22}{7}\) × 3.5
= 22 cm

(iii) Given Diameter (d) = 84 cm
radius (r) = \(\frac{d}{2}=\frac{84}{2}\)
= 42 cm
Circumference = 2πr
= 2 × \(\frac {22}{7}\) × 42
= 264 cm

2. If the circumference of a circular sheet is 176 m, find its radius.
Solution:
Given circumference of circular sheet = 176 m
Let radius = r
So 2πr = 176
r = \(\frac{176}{2 \pi}\)
\(\frac{176}{2 \times \frac{22}{7}}\)
= 28 m

3. A circular disc of diameter 8.4 cm is divided into two parts what is the perimeter of each semicircular part ?
Solution:
Given diameter of a circular disc = 8.4 cm
radius (r) = \(\frac{8.4}{2}\) = 4.2 cm
Perimeter of semicircular part = πr + 2r
= \(\frac {22}{7}\) × 4.2 + 2 × 4.2
= 22 × 0.6 + 8.4
= 21.6 cm

4. Find the area of the circle having
(i) Radius r = 49 cm
(ii) Radius r = 2.8 cm
(iii) Diameter = 4.2 cm
Solution:
(i) Given radius (r) = 49 cm
Area of circle = πr²
= \(\frac {22}{7}\) × 49 × 49
= 7546 cm²

(ii) Given radius (r) = 2.8 cm
Area of circle = πr²
= \(\frac {22}{7}\) × 2.8 × 2.8
= 24.64 cm²

(iii) Given diameter (d) = 4.2 cm
radius (r) = \(\frac{d}{2}=\frac{4.2}{2}\)
Area of circle = πr²
= \(\frac {22}{7}\) × 2.1 × 2.1
= 13.86 cm²

5. A gardener wants to fence a circular garden of radius 15 m. Find the length of wire, if he makes three rounds offense. Also, find the cost of wire if it costs ₹ 5 per meter (Take π = 3.14).
Solution:
Given radius of circular garden (r) = 15 m
Circumference of the circular garden = 2πr
= 2 × 3.14 × 15
= 94.2 m
So, length of the wire to make three rounds offense
= 3 × 94.2
= 282.6 cm
Cost of wire= ₹ 5 × 282.6
= ₹ 1413

6. Which of the following has larger area and by how much ?
(a) Rectangle with length 15 cm and breadth 5.4 cm
(b) Circle of diameter 5.6 cm.
Solution:
(a) Given length of rectangle = 15 cm
breadth = 5.4 cm
Area of rectangle = length × breadth
= 15 × 5.4
= 81 cm2

(b) Given diameter of circle (d) = 5.6 cm
radius (r) = \(\frac{d}{2}=\frac{5.6}{2}\)
= 2.8 cm
Area of the circle = πr²
= \(\frac {22}{7}\) × (2.8)²
= 24.64 cm²
Hence, Rectangle has more area = 81 – 24.64
= 56.36 cm²

7. From a rectangular sheet of length 15 cm and breadth 12 cm a circle of radius 3.5 cm is removed. Find the area of remaining sheet.
Solution:
Given length of rectangular sheet = 15 cm
Breadth of rectangle sheet = 12 cm
Area of rectangular sheet = length × breadth
= 15 × 12
= 180 cm²
Given radius of circle (r) = 3.5 cm
Area of circle = πr²
= \(\frac {22}{7}\) × (3.5)²
= 38.5 cm²
Since circle is removed from rectangular sheet.
So, area of remaining sheet = Area of rectangular Sheet – Area of circle
= 180 – 38.5
= 141.5 cm2

8. From a circular sheet of radius 7 cm, a circle of radius 2.1 cm is removed, find the area of remaining sheet.
Solution:
Radius of the circular sheet = 7 cm
Area of the circular sheet = 1 cm
= πr2 = \(\frac {22}{7}\) × 7 × 7 cm²
=154 cm²
Radius of the circle = 2.1 cm
Area of the circle
\(\frac {22}{7}\) × 2.1 × 1.1 = \(\frac{22}{7} \times \frac{21}{10} \times \frac{21}{10}\)
= \(\frac {1386}{100}\)
= 13.86 cm²
Area of the remaining sheet = 154 cm² – 13.86 cm²
= 140.14 cm²

9. Smeep took a wire of length 88 cm and bent it into the shape of a circle, find the radius and area of the circle. If the same wire is bent into a square, what will be the side of the square ? Which figure encloses more area ?
Solution:
Given length of wire = 88 cm
The wire is bent into the shape of circle.
Circumference of circle = length of the
2πr = 88
r = \(\frac{88}{2 \pi}=\frac{44}{\pi} \mathrm{cm}\)
= 14 cm
Area of the circle = πr²
= π × (14)²
= \(\frac {22}{7}\) × 14 × 14
= 616 cm²
If the same wire is bent into the square
Let side of the square = a
Perimeter of square = length of the wire
4 × a = 88
a = \(\frac {88}{4}\)
= 22 cm
Area of the square = (side)²
= (22)²
= 484 cm²
Hence circle enclosed more area.

10. A garden is 120 m long and 85 m broad. Inside the garden, there is a circular pit of diameter 14 m. Find the cost of planting the remaining part of the garden at the rate of ₹ 5.50 per square meter.
Solution:
Given length of garden = 120 m
Breadth of garden = 85 m
Area of garden = length × breadth
= 120 × 85
= 10200 m²
Given diameter of circular pit (d) = 14 m
radius (r) = \(\frac{d}{2}=\frac{14}{2}\)
= 7 m
Area of circular pit = πr²
= \(\frac {22}{7}\) × 7 × 7
= 154 m²
Remaining part of garden = Area of garden for planting – Area of circular pit
= 10200 – 154
= 10046 m²
Cost of planting the remaining part of the garden
= ₹ 5.50 × 10046
= ₹ 55243

11. In the figure PQ = QR and PR = 56 cm. The radius of inscribed circle is 7 cm. Q is centre of semicircle. What is the area of shaded region ?

Solution:
Given PQ = QR
PR = 56 cm
Radius of inscribed circle = 7 cm
So PR = PQ + QR
= PQ + PQ = 2PQ
Hence, PQ = \(\frac{\mathrm{PR}}{2}=\frac{56}{2}\)
= 28 cm
So QR = PQ = 28 cm
Area of shaded region = Area of semicircle of diameter PR – Area of semicircle of diameter PQ – Area of semicircle of diameter QR – Area of inscribe circle

12. The minute hand of a circular clock is 18 cm long. How far does the tip of minute hand move in one hour ?
Solution:
Given minute hand of a circular clock = 18 cm
Distance covered by minute hand in 1 hour = 2πr
= 2 × 3.14 × 18
= 2 × \(\frac {314}{100}\) × 18
= \(\frac {11304}{100}\)
= 113.04 cm.

13. Multiple choice questions :

Question (i).
The circumference of a circle of diameter 10 cm is :
(a) 31.4 cm
(b) 3.14 cm
(c) 314 cm
(d) 35.4 cm
Answer:
(a) 31.4 cm

Question (ii).
The circumference of a circle with radius 14 cm is :
(a) 88 cm
(b) 44 cm
(c) 22 cm
(d) 85 cm
Answer:
(a) 88 cm

Question (iii).
What is the area of the circle of radius 7 cm ?
(a) 49 cm
(b) 22 cm²
(c) 154 cm²
(d) 308 cm²
Answer:
(c) 154 cm²

Question (iv).
Find the diameter of a circle whose area is 154 cm² ?
(a) 4 cm
(b) 6 cm
(c) 14 cm
(d) 12 cm
Answer:
(c) 14 cm

Question (v).
A circle has area 100 times the area of another circle. What is the ratio of their circumferences ?
(a) 10 : 1
(b) 1 : 10
(c) 1 : 1
(d) 2 : 1
Answer:
(a) 10 : 1

Question (vi).
Diameter of a circular garden is 9.8 cm. Which of the following is its area ?
(a) 75.46 cm²
(b) 76.46 cm²
(c) 74.4 cm²
(d) 76.4 cm²
Answer:
(a) 75.46 cm²

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4

1. An rectangular park is 80 m long and 65 in wide. A path of 5 m width is constructed outside the park. Find the area of path.
Solution:
Let ABCD be a rectangular park.
Length of the park = 80 m
Breadth of the park = 65 m

Area of the rectangular park
ABCD = Length × Breadth
= 80 m × 65 m
= 5200 m²
Length of rectangular garden EFGH (including park)
= 80 + 5 + 5
= 90 m
Breadth = 65 + 5 + 5
= 75 m
Area of rectangular path EFGH = 90 × 75
= 6750 m²
Area of the path = Area of rectangular park EFGH – Area of rectangle ABCD
= 6750 – 5200
= 1550 m²

2. A rectangular garden is 110 m long and 72 m broad. A path of uniform width 8 m has to be constructed around it. Find the cost of gravelling the path at ₹ 11.50 per m².
Solution:
Let ABCD represents the rectangular garden and the shaded region represents the path of width 8 m around the garden.
Length of rectangular garden l = 110 m

Breadth of rectangular garden b = 72 m
Area of rectangular garden ABCD = (110 × 72) m²
= 7920 m²
Length of rectangular garden including path = 110 m + (8m + 8m)
= 126 m
Breadth of rectangular garden including path = 72m + (8m + 8m) = 88 m
Area of garden including path = (126 × 88) m²
= 11088 m²
Area of path = Area of garden including path – Area of garden
Area of path = (11088 – 7920) m²
= 3168 m²
Cost of gravelling 1 m² of path = ₹ 11.50
Cost of gravelling 2928 m2 of path = ₹ 3168 × 11.50
= ₹ 36432

3. A room is 12 m long and 8 m broad. It is surrounded by a verandah, which is 3 m wide all around it. Find the cost of flooring the verandah with marble at ₹ 275 per m².
Solution:
Let ABCD. represents the rectangular floor of room and shaded region represents the verandah 3 m wide all along the outside of a room.

PQ = (3 + 12 + 3) m
= 18 m
PS = (3 + 8 + 3) m
= 14 m
Area of rectangle ABCD = 1 × b
= AB × AD
= 12 m × 8m
= 96 m²
Area of recangle PQRS = 1 × b
= PQ × PS
= 18 m × 14 m
= 252 m²
Area of verandah = [Area of rectangle PQRS] – [Area of rectangle ABCD]
= (252 – 96) m²
= 156 m²
(Rate of flooting the verandha with marble verandah = ₹ 275 per m²)
Cost of flooring verandah with moble.
= ₹ (156 × 275)
= ₹ 42900.

4. A sheet of paper measures 30 cm × 24 cm. A strip of 4 cm width is cut from it, all around. Find the area of remaining sheet and also the area of cut out strip.
Solution:
Let ABCD represent the sheet of 30 cm × 24 cm and shaded region represents the 4 cm width to be cut

PQ = (30 – 4 – 4) cm
= 22 cm
PS = (24 – 4 – 4) cm
= 16 cm

(i) remaining sheet
[Area of rectangle ABCD] – [Area of rectangle PQRS]
= (30 × 24 – 22 × 16)
= (720 – 352 = 368) cm²
Area of the cut our strip i.e. area of rectangle PQRS = 22 × 16 cm²
= 352 cm²

5. A path of 2 m wide is built along the border inside a square garden of side 40 m. Find :

Question (i).
The Area of path.
Solution:
Let ABCD be the square park of side 40 m and the shaded region represents the path 2 m wide
EF = 40 m – (2 + 2) m
= 36 m

Area of square park ABCD = (Side)²
= 40 × 40
= 1600 m²
Area of EFGH = (Side)²
= 36 × 36
= 1296 m²
Area of path = Area of square park ABCD – Area of EFGH
= (1600 – 1296) m²
= 304 m²

Question (ii).
The cost of planting grass in the remaining portion of the garden at the rate of ₹ 50 per m².
Solution:
Cost of planting grass = 50 per m²
Cost of planting grass 1m² = ₹ 50
Cost of 1296 m² = 1296 × 50
= ₹ 64800

6. A nursery school play ground is 150 m long and 75 m wide. A portion of 75 m × 75 m is kept for see-saw slides and other park equipments. In the remaining portion 3 m wide path parallel to its width and parallel to remaining length (as shown in fig). The remaining area is covered by grass. Find the area covered by grass.

Solution:
Area of school ground
= 150 m × 75 m = 11250 m²
Area kept for see-saw slides and other equipments
= 75 × 75
= 5625 m²
Area of path parallel to width of ground = 75 × 3
= 225 m²
Area common to both paths = 3 × 3
= 9 m²
Total area covered by path
= (225 + 225 – 9)
= 441 m²
Area covered by grass = Area of ground – (Area kept for see-saw slides + area covered by paths)
= 11250 – (5625 + 441)
= (11250 – 6066) m²
= 5184 m²

7. Two cross roads each of width 8 m cut at right angle through the centre of a rectangular park of length 480 m and breadth 250 m and parallel to its sides. Find the area of roads. Also, find the area of park excluding cross roads.
Solution:

ABCD represent the rectangular park of length AB = 480 m and breadth BC = 250 m. Area of shaded portion i.e. area of rectangle EFGH and PQRS represent the area of cross roads, but the area of square KLMN is taken twice, So it will be subtracted.
Now EF = 480, FG = 8 m, PQ = 250 m, QR = 8 m, KL = 8 m.
Area covered by roads = Area of rectangle EFGH + area of rectangle PQRS – Area of square KLMN
= (EF × FG) + (PQ × QR) – (KL)²
= (480 × 8) + (250 × 8) – (8 × 8)
= 3840 + 2000 – 64
Area of the road = 5776 m²
Area of park excluding cross roads = 250 × 480 – (250 × 8 + 480 × 8 – 8 × 8)
= 114224 m²

8. In a rectangular field of length 92 m and breadth 70 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of field. If the width of each road is 4 m, find.
(i) The area covered by roads.
(ii) The cost of constructing the roads at the rate of ₹ 150 per m².
Solution:
Let ABCD represents the rectangular field of length ; AB = 92 m and breadth; AD = 70 m. Let the area of shaded portion
i. e. area of the rectangle PQRS and the area of rectangle EFGH represents the area of cross roads.

But in doing this, area of square KLMN is taken twice which is to be subtracted.
Now PQ = 4 m, PS = 70 m
and EH = 4 m, EF = 92 m
and KL = 4 m, KN = 4 m

= PQ × PS + EF × EH – KL × KN
= [(4 × 70) + (92 × 4) – (4 × 4)] m²
= (280 + 368 – 16) m²
= (648 – 16) m²
= 632 m²

(ii) Cost of constructing 1 m² of roads = ₹ 150
Therefore cost of constructing 632 m² of roads = ₹ (150 × 632)
= ₹ 94800.

9. Find the area of shaded region in each of the following figures.

Question (i).

Solution:
Length of rectangle ABDC
= 3m + 15m + 3m
= 21 m
Breadth of rectangle ABDC
= 2m + 12m + 2m
= 16 m
Area of rectangle ABDC
= length × breadth
= 21 × 16 m²
= 336 m²
Length of rectangle PQRS = 15 m
Breadth of rectangle PQRS = 12 m
Area of rectangle PQRS = 15 × 12 m²
= 180 m²
Area of shaded region = Area of rectangle ABCD – Area of rectangle PQRS
= 336 m² – 180 m²
= 156 m²

Question (ii).

Solution:

SR = PQ = 2.5 m
EH = FG = 4 m
KL = 2.5 m
LM = 4 m
Area of shaded region = [Area of rectangle PQRS] + [Area of rectangle EFGH] – Area of rectangle KLMN
= 40 × 2.5 + 80 × 4 – 2.5 × 4
= 100 + 320 – 10
= 420 – 10
= 410 m²

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.4

1. Which of the following can be the sides of a triangle ?
(a) 8 cm, 10 cm, 18 cm
(b) 6 cm, 4 cm, 8 cm
(c) 35 cm, 38 cm, 40 cm
(d) 3 cm, 4 cm, 10 cm
Solutions:
Drawing of a triangle with given three sides is possible if sum of lengths of any two sides of a triangle is greater than the third side.
(a) Since 8 + 10 = 18
So, 8 cm, 10 cm, 18 cm cannot be the lengths of the sides of the triangle

(b) Since 6 + 4 > 8
4 + 8 >6
8 + 6 > 4
So, 6 cm, 4 cm, 8 cm can be lengths of the sides of a triangle.

(c) Since 35 + 38 > 40
38 + 40 > 35
40 + 35 > 38
So, 35 cm, 38 cm, 40 cm can be lengths of the sides of a triangle.

(d) Since 3 + 4 < 10
So, 3 cm, 4 cm, 10 cm cannot be the lengths of the sides of a triangle.

2. A point O is in interior of a ΔABC use symbols >, < or = to make the following statements true.

Answer:
(a) OA + OB  >  AB
(b) OB + OC  >  BC
(c) OA + OC  >  AC

3. ABCD is a quadrilateral
Is AB + BC + CD + DA > AC + BD ?

Solution:
Yes ; AB + BC + CD + DA > AC + BD
Proof : In ΔABC ; AB + BC > AC
[∵ sum of lengths of any two sides of a triangle is always greater than the third side] …. (i)
In ΔADC ; CD + DA > AC
[using same reason as above] …. (ii)
In ΔABD ; AB + DA > BD
[using same reason as above] …. (iii)
In ΔBCD ; BC + CD > BD
[using same reason as above] …. (iv)
Adding (i), (ii), (iii) and (iv), we get :
(AB + BC) + (CD + DA) + (AB + DA) + (BC + CD] > AC + AC + BD + BD (AB + AB) + (BC + BC) + (CD + CD) + (DA + DA) > 2AC + 2BD
2AB + 2BC + 2CD + 2DA > 2AC + 2BD
or 2 (AB + BC + CD + DA) > 2 (AC + BD)
or AB + BC + CD + DA > AC + BD
Proved.

4. AD is a median of < ABC
Is AB + BC + CA > 2AD ?

Solution:
In ΔABD
AB + BD > AD …(1)
{∵ Sum of the lengths of any two sides of a triangle is greater than the third side}
In ΔACD
CA + DC > AD …(2) (using same reason as above)
Adding (1) and (2), we get
AB + BD + CA + DC > AD + AD
AB + (BD + DC) + CA > 2AD
[D is mid point of BC BD + DC = BC]
Hence AB + BC + CA > 2AD.

5. The length of two sides of a triangle are 4 cm and 6 cm. Between what two measures should the length of the third side fall ?
Solution:
We know that the sum of two sides of a triangle is always greater than the third side.
Therefore, third side has to be less than the sum of the two sides.
The third is thus less than 4 cm + 6 cm = 10 cm
The third side cannot be less than the difference of the two sides.
Thus the third side has to be more than 6 cm – 4 cm = 2 cm the length of third side should be greater than 2 cm and less than 10 cm.
Hence the length of the third side should fall between 2 cm and 10 cm.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 8 Comparing Quantities MCQ Questions

Multiple Choice Questions

Question 1.
Find the ratio of ₹ 10 to 10 paise.
(a) 1 : 1
(b) 100 : 1
(c) 1000 : 1
(d) 1000 : 10
Answer:
(b) 100 : 1

Question 2.
The ratio of ₹ 5 to 50 Paise is :
(a) 5 : 50
(b) 1 : 10
(c) 10 : 1
(d) 50 : 5.
Answer:
(c) 10 : 1

Question 3.
The ratio of 15 kg to 210 g is :
(a) 15 : 210
(b) 15 : 21
(c) 500 : 7
(d) 7 : 500.
Answer:
(c) 500 : 7

Question 4.
The Percentage of \(\frac {12}{16}\) is :
(a) 25%
(b) 12%
(c) 75%
(d) 16%
Answer:
(c) 75%

Question 5.
Convert \(\frac {5}{4}\) into percent.
(a) 100%
(b) 125%
(c) 75%
(d) 16%
Answer:
(b) 125%

Question 6.
Convert 12.35 into percent.
(a) 12.35%
(b) 123.5%
(c) 1235%
(d) 1.235%
Answer:
(c) 1235%

Question 7.
What percent part of figure is shaded ?

(a) 30%
(b) 50%
(c) 60%
(d) 20%
Answer:
(c) 60%

Question 8.
15% of 250 is :
(a) 250
(b) 375
(c) 37.5
(d) 3750
Answer:
(c) 37.5

Fill in Blanks :

Question 1.
25% of 120 litres is ………….. litres.
Answer:
30

Question 2.
The ratio of 4 km to 300 m is …………..
Answer:
40

Question 3.
The price at which an article is purchased is called …………..
Answer:
Cost price

Question 4.
If the selling price of an article is less than to cost prices then there is …………..
Answer:
loss

Question 5.
The symbol ………….. stands for percent
Answer:
%

Write True or False

Question 1.
The Ratio 1 : 5 and 2 : 15 are equivalent. (True/False)
Answer:
False

Question 2.
A ratio remains unchanged, if both of its terms are multiplied or divided by the same number. (True/False)
Answer:
True

Question 3.
If the selling price of an article is more than its cost price then there is a profit. (True/False)
Answer:
True

Question 4.
If cost price and selling price both are equal then three is profit. (True/False)
Answer:
False

Question 5.
Profit loss percentage is calculated on the cost price. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

1. Find what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case.

Question (i).
Gardening shears bought for ₹ 250 and sold for ₹ 325
Solution:
C.P. of gardening shears = ₹ 250
S.P. of gardening shears = ₹ 325
Profit = S.P. – C.P.
= ₹ 325 – ₹ 250
= ₹ 75
Profit percentage = \(\left[\frac{\text { Profit }}{\text { Cost price }} \times 100\right] \%\)
= \(\left[\frac{75}{250} \times 100\right] \%\)
= 30%

Question (ii).
A refrigerater bought for ₹ 12,000 and sold at ₹ 13,500
Solution:
C.P. of refrigerator = ₹ 12,000
S.P. of refregerator = ₹ 13,500
Profit = S.P. – C.P.
= ₹ 13,500 – ₹ 12,000
= ₹ 1500
Profit percentage = \(\left(\frac{\text { Profit }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{1500}{12000} \times 100\right) \%\)
= 12.5%

Question (iii).
A cupboard bought for ₹ 2,500 and old at ₹ 3,000.
Solution:
C.P. of card board = ₹ 2,500
S.P. of card board = ₹ 3,000
Profit = S.P. – C.P.
= ₹ 3000 – ₹ 2500
= ₹ 500
Profit percentage = \(\left(\frac{\text { Profit }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{500}{2500} \times 100\right) \%\)
= 20%

Question (iv).
A shirt bought for ₹ 250 and sold at ₹ 150
Solution:
C.P. of shirt = ₹ 250
S.P. of shirt a ₹ 150
Since S.P. is less than C.P.
So, there will be a loss
Loss = C.P. – S.P.
= ₹ 250 – ₹ 150
= ₹ 100
Loss percentage = \(\left(\frac{\text { Loss }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{100}{250} \times 100\right) \%\)
= 40%

2. A shopkeeper buys an article for ₹ 735 and sold it for ₹ 850. Find his profit or loss.
Solution:
C.P. of an article = ₹ 735
S.P. of an article = ₹ 850
Profit = ₹ 850 – ₹ 735
= ₹ 115

3. Kirti bought a saree for ₹ 2500 and sold it for ₹ 2300. Find her loss and loss percent.
Solution:
C.P. of saree = ₹ 2500
S.P. of saree = ₹ 2300
Loss = C.P. – S.P.
= ₹ 2500 – ₹ 2300
= ₹ 200
Loss percentage = \(\left(\frac{\text { Loss }}{\text { C.P. }} \times 100\right) \%\)
= \(\frac {200}{2500}\) × 100
= 8%

4. An article was sold for ₹ 252 with a profit of 5%. What was its cost price ?
Solution:
S.P. of an article = ₹ 252
Profit = 5%
Let C.P. of article = ₹ 100
Profit = 5% of ₹ 100
= ₹ 5
S.P. of article = ₹ 100 + ₹ 5
= ₹ 105
If S.P. is ₹ 105, then C.P. = ₹ 100
If S.P. is ₹ 1 then C.P. = ₹ \(\frac {100}{105}\)
If S.P. is ₹ 252, then C.P. = ₹ \(\frac {100}{105}\) × 252
= ₹ 240

5. Amrit buys a book for ₹ 275 and sells it at a loss of 15%. For how much does she sell it ?
Solution:
C.P. of book = ₹ 275
Loss = 15%
∴ Loss on ₹ 275 = ₹ \(\frac {15}{100}\) × 275
= ₹ 41.25
Thus, S.P. of book = ₹ 275 – ₹ 41.25
= ₹ 233.75

6. Juhi sells a washing machine for ₹ 13500. She losses 20% in the bargain. What was the price at which she bought it ?
Solution:
S.P. of washing machine = ₹ 13500
Let C.P. = ₹ 100
Loss = 20%
S.P. = ₹ 100 – ₹ 20
= ₹ 80
If S.P. of washing machine is ₹ 80 then its cost price = ₹ 100
If S.P. of washing machine is ₹ 1 then its cost price = ₹ \(\frac {100}{80}\)
If S.P. of washing machine is ₹ 13500
then its cost price = ₹ \(\frac {100}{80}\) × 13500
= ₹ 16875

7. Anita takes a loan of ₹ 5000 at 15% per year as rate of interest. Find the interest she has to pay at the end of one year.
Solution:
Here, Principal (P) = ₹ 5000
Rate (R) = 15% per year
Time (T) = 1 year
Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{5000 \times 15 \times 1}{100}\)
= ₹ 750

8. Find the amount to be paid at the end of 3 years in each case :

Question (i).
Principal = ₹ 1200 at 12% p.a.
Solution:
P = ₹ 1200, R = 12% p.a.
T = 3 years
Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{1200 \times 12 \times 3}{100}\)
= ₹432
Amount = Principal + Simple Interest
= ₹ 1200 + ₹ 432
= ₹ 1632

Question (ii).
Principal = ₹ 7500 at 5% p.a.
Solution:
P = ₹ 7500, R = 5% p.a. T = 3 years
SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹\(\frac{7500 \times 5 \times 3}{100}\)
= ₹ 1125
Amount = P + S.I. = ₹ 7500 + ₹ 1125
= ₹ 8625

9. Find the time when simple interest on ₹ 2500 at 6% p.a. is ₹ 450
Solution:
P = ₹ 2500, R = 6% p.a. Time = T = ?,
S.I. = ₹ 450
T = \(\frac{\mathrm{SI} \times 100}{\mathrm{P} \times \mathrm{R}}\)
T = \(\frac{450 \times 100}{2500 \times 6}\)
= 3 years

10. Find the rate of interest when simple interest on ₹ 1560 in 3 years is ₹ 585.
Solution:
Principal (P) = ₹ 1560
Time (T) = 3 years
S.I. = ₹ 585
R = \(\frac{\text { SI } \times 100}{\mathrm{P} \times \mathrm{T}}\)
= \(\frac{585 \times 100}{1560 \times 3}\)
= \(\frac {125}{10}\)
= 12.5
Thus Rate of interest is 12.5% p.a.

11. If Nakul gives an interest of ₹ 45 for, one year at 9% rate p.a. what is the sum he borrowed ?
Solution:
Here S.I.= ₹ 45, R = 9% p.a.
T = 1 year, P = ?
P = \(\frac{\text { S.I. } \times 100}{\mathrm{R} \times \mathrm{T}}\)
= \(\frac{45 \times 100}{9 \times 1}\)
= ₹ 500

12. If ₹ 14,000 is invested at 4% per annum simple interest, how long will it take for the amount to reach ₹ 16240 ?
Solution:
P = ₹ 14,000
R = 4% p.a.
T = ?
A = ₹ 16240
S.I. = A – P
= ₹ 16240 – ₹ 14,000
= ₹ 2240
T = \(\frac{\text { SI } \times 100}{\mathrm{R} \times \mathrm{P}}\)
= \(\frac{2240 \times 100}{14000 \times 4}\)
= 4 years

13. Multiple Choice Questions :

Question (i).
If a man buys an article for ₹ 80 and sells it for ₹ 100, then gain percentage is
(a) 20%
(b) 25%
(c) 40%
(d) 125%
Answer:
(b) 25%

Question (ii).
If a man buys an article for ₹ 120 and sells it for ₹ 100, then his loss percentage is
(a) 10%
(b) 20%
(c) 25%
(d) 16\(\frac {2}{3}\)%
Answer:
(d) 16\(\frac {2}{3}\)%

Question (iii).
The salary of a man is ₹ 24000 per month. If he gets an increase of 25% in the salary, then the new salary per month is
(a) ₹ 2,500
(b) ₹ 28,000
(c) ₹ 30,000
(d) ₹ 36,000
Answer:
(c) ₹ 30,000

Question (iv).
On selling an article for ₹ 100, Renu gains ₹ 20 Her gain percentage is
(a) 25%
(b) 20%
(c) 15%
(d) 40%
Answer:
(a) 25%

Question (v).
The simple interest on ₹ 6000 at 8% p.a. for one year is
(a) ₹ 600
(b) ₹ 480
(c) ₹ 400
(d) ₹ 240
Answer:
(b) ₹ 480

Question (vi).
If Rohini borrows ₹ 4800 at 5% p.a. simple interest, then the amount she has to return at the end of 2 years is.
(a) ₹ 480
(b) ₹ 5040
(c) ₹ 5280
(d) ₹ 5600
Answer:
(c) ₹ 5280

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

1. Convert the following fractions into percents

(i). \(\frac {1}{8}\)
Solution:
\(\frac {1}{8}\) = \(\frac {1}{8}\) × 100
= \(\frac {25}{2}\)
= 12.5
Thus, \(\frac {1}{8}\) = 12.5%

(ii). \(\frac {49}{50}\)
Solution:
\(\frac {49}{50}\) = \(\frac {49}{50}\) × 100
= 98
Thus, \(\frac {49}{50}\) = 98%

(iii). \(\frac {5}{4}\)
Solution:
\(\frac {5}{4}\) = \(\frac {5}{4}\) × 100
= 125
Thus, \(\frac {5}{4}\) = 125%

(iv). 1\(\frac {3}{8}\)
Solution:
1\(\frac {3}{8}\) = \(\frac {11}{8}\) × 100
= \(\frac {275}{2}\)
= 137\(\frac {1}{2}\)
Thus, 1\(\frac {3}{8}\) = 137\(\frac {1}{2}\)%

2. Convert the following percents into fractions in simplest form

(i). 25%
Solution:
25% = \(\frac {25}{100}\)
= \(\frac {1}{4}\)

(ii). 150%
Solution:
150% = \(\frac {150}{100}\)
= \(\frac {3}{2}\)

(iii). 7\(\frac {1}{2}\)
Solution:
7\(\frac {1}{2}\) = \(\frac {15}{2}\) × \(\frac {1}{100}\)
= \(\frac {3}{40}\)

3. (i) Anita secured 324 marks out of 400 marks. Find the percentage of marks secured by Anita.
(ii) Out of 32 students, 8 are absent from the class. What is the percentage of students who are absent ?
(iii) There are 120 voters, 90 out of them voted. What percent did not vote ?
Solution:
(i) Anita scored 324 marks out of 400 marks.
∴ Percentage of marks secured by Anita = \(\left(\frac{324}{400} \times 100\right) \%\) = 50%

(ii) Out of 32 students 8 students are absent
∴ Percentage of students who are absent = \(\frac {8}{32}\) × 100% = 25%

(iii) Total voters = 120
Voters who voted = 90
The voters who did not vote = 120 – 90
= 30
Percentage voters who did not vote = \(\frac {30}{120}\) × 100%
= 25%

4. Estimate the part of figure which is shaded and hence find the percentage of the part which is shaded.

Solution:
(i) Shaded part = \(\frac {2}{4}\) = \(\frac {1}{2}\)
Percentage of shaded part = \(\left(\frac{1}{2} \times 100\right) \%\)
= 50%

(ii) Shaded part = \(\frac {2}{6}\) = \(\frac {1}{3}\)
Percentage of shaded part = \(\left(\frac{1}{3} \times 100\right) \%\)
= 33\(\frac {1}{3}\)%

(iii) Shaded part = \(\frac {5}{8}\)
Percentage of shaded part = \(\left(\frac{5}{8} \times 100\right) \%\)
= \(\frac {125}{2}\)%
= 62.5%

5. Convert the following percentages into ratios in simplest form :

(i). 14%
Solution:
14% = 14 × \(\frac {1}{100}\)
= \(\frac {7}{50}\)
= 7 : 50

(ii). 1\(\frac {3}{4}\)%
Solution:
1\(\frac {3}{4}\)% = \(\frac {7}{4}\) × \(\frac {1}{100}\)
= \(\frac {7}{400}\)
= 7 : 400

(iii). 33\(\frac {1}{3}\)%
Solution:
33\(\frac {1}{3}\)% = \(\frac {100}{3}\) × \(\frac {1}{100}\)
= \(\frac {1}{3}\)
= 1 : 3

6. Express the following ratios as percentages :

(i). 5 : 4
Solution:
5:4 = \(\frac {5}{4}\) × 100
= 125%

(ii). 1 : 1
Solution:
1 : 1 = \(\frac {1}{1}\) × 100
= 100%

(iii). 2 : 3
Solution:
2 : 3 = \(\frac {2}{3}\) × 100
= \(\frac {200}{3}\)%
= 66\(\frac {2}{3}\)%

(iv). 9 : 16
Solution:
9 : 16 = \(\frac {9}{16}\) × 100
= \(\frac {225}{4}\)%
= 56\(\frac {1}{4}\)%

7. Chalk contains calcium, carbon and sand in the ratio 12 : 3 : 10. Find the percentage of carbon in the chalk.
Solution:
Calcium : Carbon : sand
= 12 : 3 : 10
Total of ratios = 12 + 3 + 10
= 25
Percentage of carbon is chalk
= \(\frac {3}{25}\) × 100
= 12%

8. Convert each part of the following ratios into percentage :

(i). 3 : 1
Solution:
Total of ratios = 3 + 1 = 4
Percentage of first part = \(\frac {3}{4}\) × 100
= 75%
Percentage of second part = \(\frac {1}{4}\) × 100
= 25%

(ii). 1 : 4
Solution:
Total of ratios = 1 + 4 = 5
Percentage of first part = \(\frac {1}{5}\) × 100
= 20%
Percentage of second part = \(\frac {4}{5}\) × 100
= 80%

(iii). 4 : 5 : 6.
Solution:
Total of ratios = 4 + 5 + 6 = 15
Percentage of first part = \(\frac {4}{15}\) × 100
= \(\frac {8}{3}\)%
= 26\(\frac {2}{3}\)%
Percentage of second part = \(\frac {5}{15}\) × 100
= \(\frac {100}{3}\)%
= 33\(\frac {1}{3}\)%
Percentage of third part = \(\frac {6}{15}\) × 100
= 40%

9. Convert the following percentages to decimals :

(i). 28%
Solution:
28% = \(\frac {28}{100}\)
0.28

(ii). 3%
Solution:
3% = \(\frac {3}{100}\)
= 0.03

(iii). 37\(\frac {1}{2}\)%
Solution:
37\(\frac {1}{2}\)%
= \(\frac {75}{2}\) × \(\frac {1}{100}\)=
= \(\frac {3.75}{100}\)
= 0.375

10. Convert the following decimals to percentage :

(i). 0.65
Solution:
0.65 = (0.65 × 100)%
= \(\left(\frac{65}{100} \times 100\right) \%\)
= 65%

(ii). 0.9
Solution:
0.9 = (0.9 × 100)%
= \(\left(\frac{9}{10} \times 100\right) \%\)
= 90%

(iii). 2.1
Solution:
2.1 = (2.1 × 100)%
= \(\left(\frac{21}{10} \times 100\right) \%\)
= 210%

11. (i) If 65% of students in a class have a bicycle, then what percent of the students do not have a bicycle ?
(ii) We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what percent are mangoes ?
Solution:
(i) Percentage of student having a bicycle = 65%
Percentage of students do not have a bicycle = (100 – 65)%
= 35%

(ii) Percentage of apples = 50%
Percentage of oranges = 30%
Percentage Of mangoes = (100 – (50% + 30%)
= (100 – 80)%
= 20%

12. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Original population = 25000
Decreased population = 24,500
Decrease in population = (25000 – 24500)
= 500
Percentage decrease = \(\frac{\text { Decrease in population }}{\text { Original population }} \times 100 \%\)
= \(\frac {500}{25000}\) × 100%
= 2%

13. Arun bought a plot for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase ?
Solution:
Original price of plot = ₹ 3,50,000
The increased price of plot = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – ₹ 3,50,000
= ₹ 20,000.
Percentage of price increase
= \(\left(\frac{\text { Increase in price }}{\text { Original price }} \times 100\right)\)
= \(\frac {20,000}{350000}\) × 100%
= \(\frac {40}{7}\)%
= 5\(\frac {5}{7}\)

14. Find :

(i). 15% of 250
Solution:
15% of 250 = \(\frac {15}{100}\) × 250
= \(\frac {375}{10}\)
= 37.5

(ii). 25% of 120 litres
Solution:
25% of 120 litres = \(\frac {25}{100}\) × 20 litres
= 30 liters.

(iii). 4% of 12.5
Solution:
4% of 12.5 = \(\frac {4}{100}\) × \(\frac {125}{10}\)
= \(\frac {5}{10}\)
= 0.5

(iv). 12% of ₹ 250.
Solution:
12% of ₹ 250 = ₹\(\frac {12}{100}\) × 250
= ₹300

15. Multiple Choice Questions :

Question (i).
The ratio 2 : 3 expressed as percentage is
(a) 40%
(b) 60%
(c) 66\(\frac {2}{3}\)%
(d) 33\(\frac {1}{3}\)%
Answer:
(c) 66\(\frac {2}{3}\)%

Question (ii).
If 30% of x is 72, then x is equal to
(a) 120
(b) 240
(c) 360
(d) 480
Answer:
(b) 240

Question (iii).
0.025 when expressed as a percent is
(a) 250%
(b) 25%
(c) 4%
(d) 2.5%
Answer:
(d) 2.5%

Question (iv).
In a class, 45% of students are girls. If there are 22 boys in the class, then the total number of students in the class is
(a) 30
(b) 36
(c) 40
(d) 44
Answer:
(c) 40

Question (v).
What percent of \(\frac {1}{7}\) is \(\frac {2}{35}\) ?
(a) 20%
(b) 25%
(c) 30%
(d) 40%
Answer:
(d) 40%

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

1. Find the ratio of :

Question (i).
₹ 5 to 50 paise
Solution:
To find ratio of ₹ 5 to 50 paise.
Firstly convert both the quantities into same units.
₹ 1 = 100 paise
₹ 5 = 5 × 100 paise = 500 paise.
So, ratio of 500 paise to 50 paise
= \(\frac{500}{50}=\frac{10}{1}\)
= 10 : 1
Hence, required ratio is 10 : 1

Question (ii).
15 kg to 210 g
Solution:
To find ratio of 15 kg to 210 g
Firstly, convert both the quantities into same unit
1 kg = 1000 g
15 kg = 15 × 1000 g = 15000 g.
So, ratio of 15000 to 210 g
= \(\frac{15000}{210}=\frac{500}{7}\)
= 500 : 7
Hence, required ratio is 500 : 7

Question (iii).
4 m to 400 cm
Solution:
To find ratio of 4 m to 400 cm
Firstly convert both the quantities into same unit.
1 m = 100 cm
4 m = 4 × 100 cm = 400 cm
So, ratio of 400 cm to 400 cm
= \(\frac{400}{400}=\frac{1}{1}\)
= 1 : 1
Hence, required ratio is 1 : 1

Question (iv).
30 days to 36 hours
Solution:
To find ratio of 30 days to 36 hours
1 day = 24 hours
30 days = 30 × 24 hours
= 720 hours
So, ratio of 720 hours to 36 hours
= \(\frac{720}{36}=\frac{20}{1}\)
= 20 : 1
Hence required ratio is 20 : 1

2. Are the ratios 1: 2 and 2 :3 equivalent ?
Solution:
To check this we need to know whether
1 : 2 and 2 : 3 are equal
First convert the ratios into fractions
1 : 2 is written as \(\frac {1}{2}\)
2 : 3 is written as \(\frac {2}{3}\)
Now to change these fraction into like fraction.
Make denominator of both the fraction same.
\(\frac{1}{2}=\frac{1}{2} \times \frac{3}{3}=\frac{3}{6}\) and \(\frac{2}{3}=\frac{2}{3} \times \frac{2}{2}=\frac{4}{6}\)
4 > 3
\(\frac{4}{6}>\frac{3}{6}\)
Hence 1 : 2 and 2 : 1 are not equivalent

3. If the cost of 6 toys is ₹ 240, find the cost of 21 toys.
Solution:
The more the number of toys one purchases, the more is the amount to be paid.
Therefore, there is a direct proportion between the number of toys and the amount to be paid.
Let x be number of toys to be purchased
∴ 6 : 240 : : 21 : x
\(\frac{6}{240}=\frac{21}{x}\)
x = \(\frac{21 \times 240}{6}\) = ₹ 840
Thus cost of 21 toys is ₹ 840.

4. The car that I own can go 150 km with 25 litres of petrol. How far can it go with 30 litres of petrol ?
Solution:
More km ↔ More petrol
So, the consumption of petrol and the distance travelled by a car is a case of direct proportion.
Let the distance travelled be x km.
∴ 150 : 25 : : x : 30
\(\frac{150}{25}=\frac{x}{30}\)
x = \(\frac{150 \times 30}{25}\)
x = 180
Thus, it can go 180 km is 30 litres of petrol.

5. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students ?
Solution:
The more the number of students, the more is the number of computer needed.
More students ↔ More will be the computed needed.
There is a direct proportion between the number of students and the number of computers needed. Let computer needed will be x.
6 : 3 : : 24 : x
\(\frac{6}{3}=\frac{24}{x}\)
6 × x = 24 × 3
x = \(\frac{24 \times 3}{6}\) = 12
Thus, 12 computers will be needed.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.5

1. Which of the following are meaningless:

Question (a)
IC
Solution:
Since I can be subtracted only from V and X not from C
So IC is meaningless.

Question (b)
VD
Solution:
VD, since V cannot be subtracted
So VD is meaningless.

Question (c)
XCVII
Solution:
XCVII = XC + VII = 90 + 7 = 97
So XCVII is meaningful.

Question (d)
IVC
Solution:
Since IV cannot be subtracted from C.
So IVC is meaningless.

Question (e)
XM.
Solution:
Since X can be subtracted only from L and C not from M.
So XM is meaningless.

2. Write the following in Hindu Arabic Numerals:

Question (a)
XXV
Solution:
XXV = X + X + V
= 10 + 10 + 5 = 25

Question (b)
XLV
Solution:
XLV = XL + V = 40 + 5 = 45

Question (c)
LXXIX
Solution:
LXXIX = L + X + X + IX
= 50 + 10 + 10 + 9 = 79

Question (d)
XCIX
Solution:
XCIX = XC + IX = 90 + 9 = 99

Question (e)
CLXIX
Solution:
CLXIX = CL + X + IX = 40 + 10 + 9 = 59

Question (f)
DCLXII
Solution:
DCLXII = D + C + L + X + II
= 500 + 100 + 50 + 10 + 2 = 662

Question (g)
DLXIX
Solution:
DLXIX = D + L + X + IX
= 500 + 50 + 10 + 9
= 569

Question (h)
DCCLXVI
Solution:
DCCLXVI = D + CC + L + X + VI
= 500 + 200 + 50 + 10 + 6 = 766

Question (i)
CDXXXVIII
Solution:
CDXXXVIII = CD + XXX + VIII
= 400 + 30 + 8 = 438

Question (j)
MCCXLVI = M + CC + XL + VI
= 1000 + 200 + 40 + 6
= 1246

3. Express each of the following as Roman numerals:

Question (i)
(a) 29
(b) 63
(c) 94
(d) 99
(e) 156
(f) 293
(g) 472
(h) 638
(i) 1458
(j) 948
(k) 199
(l) 499
(m) 699
(n) 299
(o) 999
(p) 1000
Solution:
(a) 29 = 20 + 9 = XXIX
(b) 63 = 60 + 3 = LXIII
(c) 94 = 90 + 4 = XCIV
(d) 99 = 90 + 9 = XCIX
(e) 156 = 100 + 50 + 6 = CLVI
(f) 293 = 200 + 90 + 3 = CCXCffl
(g) 472 = 400 + 70 + 2 = CDLXXII
(h) 638 = 600 + 30 + 8 = DCXXXVIII
(i) 1458 = 1000 + 400 + 50 + 8 = MCDLVIII
(j) 948 = 900 + 40 + 8 = CMXLVIH
(k) 199 = 100 + 90 + 9 = CXCIX
(l) 499 = 400 + 90 + 9 = CDXCIX
(m) 699 = 600 + 90 + 9 = DCXCIX
(n) 299 = 200 + 90 + 9 = CCXCIX
(o) 999 = 900 + 90 + 9 = CMXCIX
(p) 1000 = M.