Class 6

MP Board Class 6th Social Science Solutions Chapter 23 District Administration

MP Board Class 6th Social Science Chapter 23 Text Book Exercise

MP Board Class 6th Social Science Short Answer Type Questions

Question 1.
Question (a)
What is the head of the district administration called?
Answer:
The head of the District Administration is called the Collector or the District Magistrate. He/she is from the Indian Administrative Services.

Question (b)
Name two important departments in the district.
Answer:
The two important departments in the district are the Education Department, the Health Department.

Question (c)
Why has the state been divided into districts and tehsils?
Answer:
For an effective administration states are divided into districts and tehsils. A district is the most important unit. The district administration implements the government policies and maintains law and order in the district. The administration includes the collector, various officers and employees.

Question (d)
Write the name of your neighboring district.
Answer:
The name of neighboring district is Bhopal.

MP Board Class 6th Social Science Long Answer Type Questions 

Question 2.
Question (a)
Write any four functions of the collector.
Answer:
The four functions of the district administration are:

1. To maintain law and order.
2. To maintain land record and realize land revenue.
3. To provide civic amenities and the development of district in every sphere.
4. To supervises all functioning of all departments and offices of the government.

Question (b)
What are civil cases? Which court decided these case?
Answer:
The civil cases are related to property and money disputes. These are heard in Civil Courts (Civil Judge).

Question (c)
What are criminal cases? Which court decides these cases?
Answer:
The criminal cases are related to thefts, criminal assaults and murders. These are heard by the criminal courts (Session Judge).

Question (d)
What system has been developed in the districts for the monitoring of educational activities?
Answer:
For monitoring the educational activities Jan Shiksha Kendra, Janpad Shiksha Kendra and Zila Shiksha Kendra have been formed in all the districts. Jan Shikshaks, District Project Coordinator, Block Resource Center Coordinator and Principal, District Institute of Education and Training work in these institutions.

Question 3.
Match the following

Answer:

Project Work:

Question 1.
Look the chart given in this chapter and with the help of your teacher make a list of the persons working in the area you live along with their designation.
Answer:
Do with the help of your teacher.

MP Board Class 6th Social Science Solutions

MP Board Class 6th Maths Solutions Chapter 5 प्रारंभिक आकारों को समझना Ex 5.1

प्रश्न 1.
रेखाखण्ड की तुलना केवल देखकर करने से क्या हानि है ?
हल :
रेखाखण्ड की तुलना केवल देखकर करने पर अधिक त्रुटियाँ होने की सम्भावना है।

प्रश्न 2.
एक रेखाखण्ड की लम्बाई मापने के लिए रूलर की अपेक्षा डिवाइडर का प्रयोग करना क्यों अधिक अच्छा है ?
हल :
रूलर की अपेक्षा डिवाइडर का उपयोग करने से रेखाखण्ड की सही माप सम्भव है।

प्रश्न 3.
कोई रेखाखण्ड \(\overline{A B}\) खींचिए। A और B के बीच स्थित कोई बिन्दु C लीजिए। AB, BC और CA की लम्बाई मापिये। क्या AB = AC + CB है ?
(टिप्पणी: यदि किसी रेखा पर बिन्दु A, B, C इस प्रकार स्थित हों कि AC + CB = AB है, तो निश्चित रूप से बिन्दु C बिन्दु A और B के बीच स्थित होता है।)
हल :
रेखाखण्डों AB, BC और AC की लम्बाइयाँ निम्नलिखित हैं
\(\overline{A B}\) = 4.6 सेमी,
AC = 3.2 सेमी,
BC = 1.4 सेमी,

∵AC + BC = 3.2 सेमी + 1.4 सेमी
= 4.6 सेमी = AB
AB = AC + BC है।

प्रश्न 4.
एक रेखाखण्ड पर बिन्दु A, B और C इस प्रकार स्थित हैं कि AB = 5 सेमी, BC = 3 सेमी और AC = 8 सेमी है। इनमें से कौन-सा बिन्दु अन्य दोनों बिन्दुओं के बीच स्थित है ?
हल :
∵AB = 5 सेमी, BC = 3 सेमी
लेकिन AC = 8 सेमी
∴AB + BC = 5 सेमी + 3 सेमी
= 8 सेमी = AC

∴बिन्दु B, A और C के बीच में है।

प्रश्न 5.
जाँच कीजिए कि संलग्न आकृति में D रेखाखण्ड \(\overline{A G}\) का मध्य-बिन्दु है।

हल :
∴AD = AB + BC + CD = 3 इकाई
और DG = DE + EF + FG = 3 इकाई
∴AD = DG (प्रत्येक = 3 इकाई)
अत: D, \(\overline{A G}\) का मध्य-बिन्दु है।

प्रश्न 6.
B रेखाखण्ड \(\overline{A C}\) का मध्य-बिन्दु है, जहाँ A, B, C और D एक ही रेखा पर स्थित हैं। बताइए कि AB = CD क्यों है?

हल :
∵ B, AC मध्य का मध्य-बिन्दु है,
∴AB = BC
इसी प्रकार C, BD का मध्य-बिन्दु है,
∴BC = CD
अतः AB = BC = CD
अर्थात् AB = CD

प्रश्न 7.
पाँच त्रिभुज खींचिए और इनकी भुजाओं को मापिए। प्रत्येक स्थिति में जाँच कीजिए कि किन्हीं दो भुजाओं की लम्बाइयों का योग तीसरी भुजा की लम्बाई से सदैव बड़ा है।
हल :
(1) ABC में,
AB = 1.7 सेमी, BC = 3 सेमी
और
AC = 2.3 सेमी

∴AB + BC = 1.7 सेमी + 3 सेमी = 4.7 सेमी
और BC + CA = 3 सेमी + 2.3 = 5.3 सेमी;
AC + AB = 2.3 सेमी + 1.7 सेमी = 4.0 सेमी
∵4.7 सेमी > 2.3 सेमी
∴ AB + BC > AC इसी प्रकार (BC + AC) > AB ; (AB + AC) > BC

(2) ∆PQR में,
PQ = 3 सेमी, QR = 3 सेमी और PR = 3 सेमी

∴PQ + QR= 3 सेमी + 3 सेमी = 6 सेमी
∵6 सेमी > 3 सेमी
∴(PQ + QR) > RP
इसी प्रकार, (QR + RP) > PQ
(RP + PQ) > QR

(3) ∆XYZ में,
XY = 3 सेमी, YZ = 4 सेमी और ZX = 5 सेमी

अब ∵ XY + YZ = 3 सेमी + 4 सेमी = 7 सेमी
YZ + ZX= 4 सेमी + 5 सेमी = 9 सेमी
ZX + XY = 5 सेमी + 3 सेमी = 8 सेमी
∴(XY + YZ) > ZX ,
(YZ + ZX) > XY
और (ZX + XY) > YZ

(4) ∆KLM में,
∵KL = 1.5 सेमी, LM = 2.5 सेमी, और MK = 2 सेमी
अब, ∵LK + LM = 1.5 सेमी + 2.5 सेमी = 4 सेमी

LM + MK = 2.5 सेमी + 2 सेमी = 4.5 सेमी
MK + KL = 2 सेमी + 1.5 सेमी = 3.5 सेमी
∴(KL + LM) > MK
(LM + MK) > KL
और (MK + KL) > LM

(5) ∆ABC में,
∵AB = 2.5 सेमी, BC = 2.8 सेमी और CA = 4.8 सेमी

अब, ∵AB + BC = 2.5 सेमी + 2.8 सेमी = 5.3 सेमी
BC + CA = 2.8 सेमी + 4.8 सेमी = 7.6 सेमी
CA + AB = 4.8 सेमी + 2.5 सेमी = 7.3 सेमी
∴(AB + BC) > CA
(BC + CA) > AB
और (CA + AB) > BC
उपर्युक्त सभी स्थितियों में त्रिभुज की किन्हीं दो भुजाओं की लम्बाइयों का योग तीसरी भुजा की लम्बाई से बड़ा है।

पाठ्य-पुस्तक पृष्ठ संख्या # 99

प्रयास कीजिए

प्रश्न 1.
आधे घूर्णन के लिए कोण का नाम क्या है?
उत्तर-
आधे घूर्णन के लिए कोण का नाम सरल कोण (दो समकोण) है। .

प्रश्न 2.
एक-चौथाई घूर्णन के लिए कोण का नाम क्या है?
उत्तर-
एक-चौथाई धूर्णन के लिए कोण का नाम एक समकोण है।

प्रश्न 3.
एक घड़ी पर आधे घूर्णन, एक-चौथाई घूर्णन और तीन-चौथाई घूर्णन के लिए पाँच अन्य स्थितियाँ दीजिए।
हल :
(1) आधे घूर्णन के लिए स्थितियाँ :
(i) 12 से 6
(ii) 3 से 9
(iii) 1 से 7
(iv) 2 से 8
(v) 4 से 10

(2) एक-चौथाई घूर्णन के लिए स्थितियाँ :
(i) 12 से 3
(ii) 1 से 4
(iii) 2 से 5
(iv) 9 से 12
(v) 7 से 10

(3) तीन-चौथाई घूर्णन के लिए स्थितियाँ :
(i) 2 से 11
(ii) 3 से 12
(iii) 4 से 1
(iv) 5 से 2
(v) 6 से 3

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 7 भिन्न Ex 7.1

पाठ्य-पुस्तक पृष्ठ संख्या # 147

प्रश्न 1.
छायांकित भाग को निरूपित करने वाली भिन्न लिखिएहल :


हल :
छायांकित भाग को निरूपित करने वाली भिन्नः

प्रश्न 2.
दी हुई भिन्न के अनुसार छायांकित कीजिए
हल :

प्रश्न 3.
निम्न में कोई गलती है, तो पहचानिएयह है।
यह \(\frac { 1 }{ 2 }\) है। यह \(\frac { 1 }{ 4 }\) है। यह \(\frac { 3 }{ 4 }\) है।

हल :
(i) ∵ भाग समान नहीं हैं।
∴ छायांकित भाग \(\frac { 1 }{ 2 }\) नहीं है।

(ii) ∵ भाग समान नहीं हैं।
∴ छायांकित भाग \(\frac { 1 }{ 4 }\) नहीं है।

(iii) ∵ भाग समान नहीं हैं।
∴ छायांकित भाग \(\frac { 3 }{ 4 }\) नहीं है।

प्रश्न 4.
8 घण्टे एक दिन की कौन-सी भिन्न है?
हल :
∵ 1 दिन में 24 घण्टे होते हैं।
∴ अभीष्ट भिन्न \(=\frac{8}{24}=\frac{1}{3}\)

प्रश्न 5.
40 मिनट एक घण्टे की कौन-सी भिन्न है?
हल :
∵ 1 घण्टे में 60 मिनट होते हैं।
∴ अभीष्ट भिन्न \(=\frac{40}{60}=\frac{2}{3}\)

प्रश्न 6.
आर्या, अभिमन्यु और विवेक एक साथ, बाँट कर खाना खाते हैं। आर्या दो सैंडविच लेकर आता है-एक सब्जी वाला और दूसरा जैम (Jam) वाला। अन्य दो लड़के अपना खाना लाना भूल गए। आर्या अपने सैंडविचों को उन दोनों के साथ बाँटकर खाने को तैयार हो जाता है, ताकि प्रत्येक व्यक्ति को प्रत्येक सैंडविच में से बराबर भाग मिले।
(a) आर्या अपनी सैंडविचों को किस प्रकार बाँटे कि प्रत्येक को बराबर भाग मिले ?
(b) प्रत्येक लड़के को एक सैंडविच का कौन-सा भाग मिलेगा?
हल :
(a) आर्या प्रत्येक सैंडविच को तीन बराबर भागों में बाँटेगा।
(b) प्रत्येक लड़के को सैंडविच का \(\frac { 1 }{ 3 }\) भाग मिलेगा।

प्रश्न 7.
कंचन ड्रेसों (Dresses) को रँगती है। उसे 30 ड्रेस रँगनी थीं। उसने अब तक 20 ड्रेस रंग ली हैं। उसने ड्रेसों की कितनी भिन्न रँग ली हैं ?
हल :
कंचन की ड्रेस रँगनी थीं = 30
उसने ड्रेस रंग ली = 20
∴ रँगी हुई ड्रेसों की अभीष्ट भिन्न = \(\frac { 1 }{ 2 }\)

प्रश्न 8.
2 से 12 तक की प्राकृत संख्याएँ लिखिए। अभाज्य संख्याएँ इनकी कौन-सी भिन्न हैं ?
हल :
2 से 12 तक की प्राकृत संख्याएँ हैं- 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
अभाज्य संख्याएँ हैं- 2, 3, 5, 7 और 11
दी हुई कुल संख्याएँ = 11,
अभाज्य कुल संख्याएँ = 5
∴ अभीष्ट भिन्न = \(\frac { 5 }{ 11 }\)

प्रश्न 9.
102 से 113 तक की प्राकृत संख्याएँ लिखिए। अभाज्य संख्याएँ इनकी कौन-सी भिन्न हैं ?
हल :
102 से 113 तक की प्राकृत संख्याएँ हैं
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113.
∴ दी हुई कुल प्राकृत संख्याएँ = 12
अभाज्य संख्याएँ = 103, 107, 109, 113
∴ कुल अभाज्य संख्याएँ = 4
अतः अतः अभीष्ट भिन्न = \(\frac { 4 }{ 12 }\)

प्रश्न 10.
इन वृत्तों की कौन-सी भिन्नों में × है?

हल :
वृत्तों की कुल संख्या = 8
× वाले वृत्तों की संख्या = 4
अभीष्ट भिन्न = \(\frac { 4 }{ 8 }\)

प्रश्न 11.
क्रिस्टिन अपने जन्म-दिन पर एक सीडी प्लेयर (CD Player) प्राप्त करती है। वह तब से सीडी इकट्ठी करना प्रारम्भ कर देती है। वह 3 सीडी खरीदती है और 5 सीडी उपहार के रूप में प्राप्त करती है। उसके द्वारा खरीदी गई सीडी की संख्या, कुल सीडी की संख्या की कौन-सी भिन्न है ?
हल :
बाजार से खरीदी गई सीडी की संख्या = 3
उपहार में प्राप्त सीडी की संख्या = 5
∴ सीडी की कुल संख्या = 3 + 5 = 8
∴ खरीदी गई सीडी की अभीष्ट भिन्न = \(\frac { 3 }{ 8 }\)
उपहार से प्राप्त सीडी की अभीष्ट भिन्न = \(\frac { 5 }{ 8 }\)

पाठ्य-पुस्तक पृष्ठ संख्या # 150

प्रयास कीजिए

प्रश्न 1.
संख्या रेखा पर \(\frac { 3 }{ 5 }\) को दर्शाइए।
हल :
\(\frac { 3 }{ 5 }\)

प्रश्न 2.
संख्या रेखा पर \(\frac{1}{10}, \frac{0}{10}, \frac{5}{10}\) और \(\frac { 10 }{ 10 }\) को दर्शाइए।
हल :

प्रश्न 3.
क्या आप 0 से 1 के बीच कोई अन्य भिन्न को दर्शा सकते हैं ? ऐसी पाँच भिन्नें और लिखिए जिन्हें आप दर्शा सकते हैं और उन्हें संख्या रेखा पर दर्शाइए।
हल :
हाँ, 0 और 1 के बीच में अन्य भिन्न को दर्शा सकते हैं, ऐसी 5 भिन्नै निम्न हैं

इन्हें संख्या रेखा पर क्रमश: A, B, C, D और E द्वारा दर्शाया गया है।

प्रश्न 4.
0 से 1 के बीच में कितनी भिन्नें स्थित हैं ? सोचिए, चर्चा कीजिए और अपने उत्तर को लिखिए।
हल :
0 और 1 के बीच में असंख्य भिन्न हैं।

पाठ्य-पुस्तक पृष्ठ संख्या # 151

प्रयास कीजिए

प्रश्न 1.
एक उचित भिन्न लिखिए
(a) जिसका अंश 5 और हर 7 है।
(b) जिसका हर 9 है और अंश 5 है।
(c) जिसके अंश और हर का योग 10 है। आप इस प्रकार की कितनी भिन्न लिख सकते हैं ?
(d) जिसका हर उसके अंश से 4 अधिक है।
(कोई पाँच भिन्न बनाइए। आप और कितनी भिन्न बना सकते हैं ?)
हल :
(a) अंश = 5, हर = 7.
∴अभीष्ट भिन्न = \(\frac { 5 }{ 7 }\)

(b) हर = 9, अंश = 5
∴अभीष्ट भिन्न = \(\frac { 5 }{ 9 }\)

(c) सम्भव युग्म जिनके अंश और हर का योग 10 है- 0, 10; 1, 9; 2, 8; 3, 7; 4, 6;
∴अभीष्ट भिन्ने है- \(\frac{0}{10}, \frac{1}{9}, \frac{2}{8}, \frac{3}{7}, \frac{4}{6}\)

(d) अभीष्ट भिन्ने है- \(\frac{1}{5}, \frac{2}{6}, \frac{3}{7}, \frac{4}{8}, \frac{5}{9}\) इत्यादि।
दी हुई परिस्थिति के अनुसार भिन्नों की संख्या असंख्य हो सकती है।

प्रश्न 2.
एक भिन्न दी हुई है। इसे देखकर आप कैसे बता सकते हैं कि यह भिन्न
(a) 1 से छोटी है?
(b) 1 के बराबर है?
हल :
(a) यदि अंश हर से छोटा है, तो भिन्न 1 से छोटी है।
(b) यदि अंश और हर बराबर हैं तो भिन्न 1 के बराबर है।

प्रश्न 3.
संकेत ‘>’, ‘<‘ या ‘=’ का प्रयोग करके रिक्त स्थानों को भरिए
हल :
(a) \(\frac { 1 }{ 2 }\) < 1
(b) \(\frac { 3 }{ 5 }\) < 1 (c) 1 > \(\frac { 7 }{ 8 }\)
(d) \(\frac { 4 }{ 4 }\) = 1
(e) \(\frac { 0 }{ 6 }\) < 1
(f) \(\frac { 2005 }{ 2005 }\) = 1

पाठ्य-पुस्तक पृष्ठ संख्या # 152

प्रश्न 1.
हर 7 वाली पाँच विषम भिन्नै लिखिए।
हल :
हर 7 वाली पाँच विषम भिन्नें हैं–
\(\frac{8}{7}, \frac{9}{7}, \frac{10}{7}, \frac{11}{7}, \frac{12}{7}\)

प्रश्न 2.
अंश 11 वाली पाँच विषम भिन्नै लिखिए।
हल :
अंश 11 वाली विषम भिन्नें हैं
\(\frac{11}{5}, \frac{11}{6}, \frac{11}{7}, \frac{11}{8}\) और \(\frac { 11 }{ 9 }\)

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 5 प्रारंभिक आकारों को समझना Intext Questions

पाठ्य-पुस्तक पृष्ठ संख्या # 96

सोचिए, चर्चा कीजिए एवं लिखिए

प्रश्न 1.
अन्य कौन-सी त्रुटियाँ और कठिनाइयाँ हमारे सम्मुख आ सकती हैं ?
उत्तर-

• प्रेक्षण लेने में त्रुटियाँ हो सकती हैं। हमारे प्रेक्षण सदैव सही नहीं होते।
• रेखाओं को ट्रेस करने में, तुलना करने में कठिनाई हो सकती है।
• पैमाने की मोटाई अधिक होने से लम्बाई नापने में त्रुटि हो सकती है। पैमाने को उचित चिह्न पर न रखने पर अथवा दूसरे कोने पर चिह्न को पढ़ने में त्रुटि हो सकती है।

इन त्रुटियों से बचने के लिए डिवाइडर का प्रयोग कर सकते हैं।

प्रश्न 2.
यदि रूलर पर अंकित चिन्हों को ठीक प्रकार से न पढ़ा जाए, तो किस प्रकार की त्रुटि हो सकती है ? इससे कैसे बचा जा सकता है ?
उत्तर-
रूलर से सही माप लेने के लिए आँख सही स्थान पर स्थित होनी चाहिए। आँख को चिह्न के ठीक ऊपर रखना चाहिए। आँख से तिरछा देखने पर चिह्न को पढ़ने में त्रुटि हो सकती है।

पाठ्य-पुस्तक पृष्ठ संख्या # 97

प्रयास कीजिए

प्रश्न 1.
एक पोस्टकार्ड लीजिए। उपर्युक्त तकनीक का प्रयोग करके, उसकी दो आसन्न भुजाओं को मापिए।
हल :
पोस्टकार्ड की दो आसन्न भुजाओं की लम्बाई क्रमशः 14 सेमी व 9 सेमी है।

प्रश्न 2.
कोई तीन वस्तुएँ चुनिए जिनके ऊपरी सिरे सपाट हों। डिवाइडर और रूलर का प्रयोग करते हुए, इन ऊपरी सिरों की सभी भुजाओं को मापिए।।
हल :
तीन वस्तुएँ-

• गणित की कक्षा 6 की पुस्तक,
• अभ्यास पुस्तिका,
• फोटो फ्रेम।

इनकी माप निम्नांकित है

• गणित की कक्षा 6 की पुस्तक : 27.5 सेमी, 21 सेमी, 27-5 सेमी, और 21 सेमी।
• अभ्यास पुस्तिका : 20 सेमी, 16 सेमी, 20 सेमी और 16 सेमी।
• फोटोफ्रेम : 25 सेमी, 20 सेमी, 25 सेमी और 20 सेमी।

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.2

पाठ्य-पुस्तक पृष्ठ संख्या # 84-85

प्रश्न 1.
पाठ्य-पुस्तक में दी हुई वक्रों को
(i) खुली या
(ii) बंद वक्रों के रूप में वर्गीकृत कीजिए :
हल :
(i) खुली वक्र – (a) व (c)
(ii) बन्द वक्र – (b), (b) व (e)

प्रश्न 2.
निम्न को स्पष्ट करने के लिए रफ आकृतियाँ बनाइए:
(a) खुला वक्र
(b) बन्द वक्र
हल :

प्रश्न 3.
कोई भी बहुभुज खींचिए और उसके अभ्यन्तर को छायांकित (Shade) कीजिए।
हल :
ABCDE एक बहुभुज है जिसके अभ्यंतर को छायांकित किया गया है।

प्रश्न 4.
संलग्न आकृति को देखकर निम्न प्रश्नों के उत्तर दीजिए :
(a) क्या यह एक वक्र है ?
(b) क्या यह बन्द है?

हल :
(a) हाँ, यह एक वक्र है।
(b) हाँ, यह बन्द वक्र है।

प्रश्न 5.
रफ आकृतियाँ बनाकर, यदि सम्भव हो, तो निम्न को स्पष्ट कीजिए :
(a) एक बन्द आकृति जो बहुभुज नहीं है।
(b) केवल रेखाखण्डों से बनी हुई खुली वक्र
(c) दो भुजाओं वाला एक बहुभुज।
हल :
(a) बन्द आकृति जो बहुभुज नहीं है।

(b) रेखाखण्डों से बनी हुई खुली वक्र

(c) दो भुजाओं वाला बहुभुज असम्भव है।

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 10 Mensuration Ex 10.3

Question 1.
Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Solution:
(a) Area of rectangle = length × breadth
= 3 cm × 4 cm = 12 cm²

(b) Area of rectangle = length × breadth
= 12 m × 21 m = 252 m²

(c) Area of rectangle = length × breadth
= 2 km × 3 km = 6 km²

(d) Area of rectangle = length × breadth
= 2 m × 70 cm = 2 m × 0.7 m = 1.4 m²

Question 2.
Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Solution:
(a) Area of square = side × side
= 10 cm × 10 cm = 100 cm²

(b) Area of square = side × side
= 14 cm × 14 cm = 196 cm²

(c) Area of square = side × side
= 5 m × 5 m = 25 m²

Question 3.
The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Solution:
(a) Area of rectangle = length × breadth
= 9m × 6m = 54m²

(b) Area of rectangle = length × breadth
= 17 m × 3 m = 51 m²

(c) Area of rectangle = length × breadth
= 4 m × 14 m = 56 m²
Thus, rectangle (c) has the largest area, i.e. 56 m² and rectangle (b) has the smallest area, i.e., 51 m².

Question 4.
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
SolutionL
Length of rectangle = 50 m
Area of rectangle = 300 m²
Since, area of rectangle = length × breadth
Therefore, breadth = \(\frac{\text { area of rectangle }}{\text { length }}\)
= \(\frac{300}{50}\) m = 6 m
Thus, the breadth of the garden is 6 m.

Question 5.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs. 8 per hundred sq m?
Solution:
Length of land = 500 m
Breadth of land = 200 m
Area of land = length × breadth
= 500 m × 200 m = 1,00,000 sq m
Cost of tiling 100 sq m of land = Rs. 8
∴ Cost of tiling 1,00,000 sq m of land
= Rs. \(\frac{8 \times 100000}{100}\) = Rs. 8000

Question 6.
A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:
Length of table-top = 2 m
Breadth of table-top = 1 m 50 cm = 1.50 m
∴ Area of table-top = length × breadth
= 2 m × 1.50 m = 3 m²

Question 7.
A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Solution:
Length of room = 4 m
And breadth of room = 3 m 50 cm = 3.50 m
∴ Area of carpet = length × breadth
= 4 m × 3.50 m = 14 m²

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution:
Length of floor = 5 m
And breadth of floor = 4 m
Area of floor = length × breadth
= 5m × 4m = 20m²
Now, side of square carpet = 3 m
Area of square carpet = side × side
= 3m × 3m = 9m²
∴ Area of floor that is not carpeted
= 20 m² – 9 m² = 11 m²

Question 9.
Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Solution:
Side of square flower bed = 1 m Area of square flower bed = side × side
= 1m × 1m = 1m²
∴ Area of 5 square flower beds = (1 × 5) m²
= 5 m²
Now, length of land = 5 m
And breadth of land = 4 m
∴ Area of land = length × breadth = 5m × 4m
= 20 m²
∴ Area of remaining part
= Area of land – Area of 5 flower beds
= 20 m² – 5 m² = 15 m²

Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimeters).

Solution:
(a) We have,

Area of square HKLM = 3 × 3 cm² = 9 cm²
Area of rectangle I]CH = 1 × 2 cm² = 2 cm²
Area of square FEDG = 3 × 3 cm² = 9 cm²
Area of rectangle ABCD = 2 × 4 cm² = 8 cm²
∴ Total area of the figure = (9 + 2 + 9 + 8) cm² = 28 cm²

(b) We have,

Area of rectangle ABCD = 3 × 1 cm² = 3 cm²
Area of rectangle BJEF = 3 × 1 cm² = 3 cm²
Area of rectangle FGHI = 3 × 1 cm² = 3 cm²
∴ Total area of the figure = (3 + 3 + 3) cm² = 9 cm²

Question 11.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

Solution:
(a) We have,

Area of rectangle ABCD = 2 × 10 cm² = 20 cm²
Area of rectangle DEFG = 10 × 2 cm² = 20 cm²
∴ Total area of the figure = (20 + 20) cm²
= 40 cm²

(b) We have,

There are 5 squares each of side 7 cm.
Area of one square = 7 × 7 cm² = 49 cm²
∴ Area of 5 squares = 5 × 49 cm² = 245 cm²

(c) We have,

Area of rectangle ABCD = 5 × 1 cm² = 5 cm²
Area of rectangle EFGH = 4 × 1 cm² = 4 cm²
∴ Total area of the figure = (5 + 4) cm²
= 9 cm²

Question 12.
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm.
Solution:
(a) Area of rectangular region
= length × breadth = 100 cm × 144 cm = 14400 cm²
Area of one tile = 12 cm × 5 cm = 60 cm²

Thus, 240 tiles are required.

(b) Area of rectangular region
= length × breadth = 70 cm × 36 cm = 2520 cm²
Area of one tile = 12 cm × 5 cm = 60 cm²
∴ Number of tiles
= \(\frac{\text { Area of rectangular region }}{\text { Area of one tile }}=\frac{2520}{60}\) = 40
Thus, 42 tiles are required.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 9 Data Handling Ex 9.2

Question 1.
Total number of animals in five villages are as follows:
Village A : 80 Village B : 120
Village C : 90 Village D : 40
Village E : 60
Prepare a pictograph of these animals using one symbol to represent 10 animals and answer the following questions :
(a) How many symbols represent animals of village E?
(b) Which village has the maximum number of animals?
(c) Which village has more animals : village A or village C?
Solution:

(a) 6 symbols represent animals of village E.
(b) Village B has the maximum number of animals.
(c) Village C has more animals than village A.

Question 2.
Total number of students of a school in different years is shown in the following table.

A. Prepare a pictograph of students using one symbol img 2 to represent 100 students and answer the following questions:
(a) How many symbols represent total number of students in the year 2002?
(b) How many symbols represent total number of students for the year 1998?
B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?
Solution:
A.

(a) 6 symbols represent total number of students in the year 2002.
(b) Five completed and one incomplete symbols represent total number of students for the year 1998.
B.

Pictograph B is more informative than pictograph A.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 1.
Find the areas of the following figures by counting square:

Solution:
(a) Number of filled squares = 9
∴ Area covered by filled squares
= (9 × 1) sq units = 9 sq units

(b) Number of filled squares = 5
∴ Area covered by filled squares
= (5 × 1) sq units = 5 sq units

(c) Number of fully-filled squares = 2
Number of half-filled squares = 4
∴ Area covered by fully-filled squares
= (2 × 1) sq units = 2 sq units
Area covered by half-filled squares
= (4 × \(\frac{1}{2}\)) sq units = 2 sq units
∴ Total area = (2 + 2) sq units = 4 sq units

(d) Number of filled squares = 8
∴ Area covered by filled squares
= (8 × 1) sq units = 8 sq units

(e) Number of filled squares = 10
∴ Area covered by filled squares
= (10 × 1) sq units = 10 sq units

(f) Number of fully-filled squares = 2
Number of half-filled squares = 4
∴ Area covered by fully-filled squares
= (2 × 1) sq units = 2 sq units
Area covered by half-filled squares
= (4 × \(\frac{1}{2}\)) sq units = 2 sq units
∴ Total area = (2 + 2) sq units = 4 sq units

(g) Number of fully-filled squares = 4
Number of half-filled squares = 4
∴ Area covered by fully-filled squares
= (4 × 1) sq units = 4 sq units
Area covered by half-filled squares
= (4 × \(\frac{1}{2}\)) sq units = 2 sq units
∴ Total area = (4 + 2) sq units = 6 sq units

(h) Number of filled squares = 5 .
∴ Area covered by filled squares
= (5 × 1) sq units = 5 sq units

(i) Number of filled squares = 9
∴ Area covered by filled squares
= (9 × 1) sq units = 9 sq units

(j) Number of fully-filled squares = 2
Number of half-filled squares = 4
∴ Area covered by fully-filled squares
= (2 × 1) sq units = 2 sq units
Area covered by half-filled squares
= (4 × \(\frac{1}{2}\)) sq units = 2 sq units
∴ Total area = (2 + 2) sq units = 4 sq units

(k) Number of fully-filled squares = 4
Number of half-filled squares = 2
∴ Area covered by fully-filled squares
= (4 × 1) sq units = 4 sq units
Area covered by half-filled squares
= (2 × \(\frac{1}{2}\)) sq units = 1 sq units
∴ Total area = (4 + 1) sq units = 5 sq units

(l) Number of fully-filled squares = 3,
Number of half-filled squares = 2,
Number of more than half-filled squares = 4
and number of less than half-filled squares = 4.
Now, estimated area covered by
fully-filled squares = 3 sq units,
half-filled squares = (2 × \(\frac{1}{2}\)) sq units
= 1 sq unit,
more than half-filled squares = 4 sq units
and less than half-filled squares
= 0 sq unit
∴ Total area = (3 + 1 + 4 + 0) sq units
= 8 sq units.

(m) Number of fully-filled squares = 7,
Number of more than half-filled squares = 7
and number of less than half-filled squares = 5
Estimated area covered by
fully-filled squares = 7 sq units,
more than half-filled squares = 7 sq units
and less than half-filled squares = 0 sq unit
∴ Total area = (7 + 7 + 0) sq units = 14 sq units

(n) Number of fully-filled squares = 10,
Number of more than half-filled squares = 8
and number of less than half-filled squares = 5
Estimated area covered by
fully-filled squares = 10 sq units,
more than half-filled squares = 8 sq units
less than half-filled squares = 0 sq unit
∴ Total area = (10 + 8 + 0) sq units
= 18 sq units.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 9 Data Handling Ex 9.1

Question 1.
In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks
Solution:

(a) 5 + 4 + 3 = 12 students obtained marks equal to or more than 7.
(b) 2 + 3 + 3 = 8 students obtained marks below 4.

Question 2.
Following is the choice of sweets of 30 students of Class VI.
Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
(a) Arrange the names of sweets in a table using tally marks.
(b) Which sweet is preferred by most of the students?
Solution:

(b) Ladoo is preferred by most of the students.

Question 3.
Catherine threw a dice 40 times and noted the number appearing each time as shown below:

Make a table and enter the data using tally marks. Find the number that appeared
(a) The minimum number of times.
(b) The maximum number of times.
(c) Find those numbers that appear an equal number of times
Solution:

(a) 4 appeared minimum number of times.
(b) 5 appeared maximum number of times.
(c) 1 and 6 appeared equal number of times.

Question 4.
Following pictograph shows the number of tractors in five villages.

Observe the pictograph and answer the following questions.
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) Flow many more tractors village C has as compared to village B.
(iv) What is the total number of tractors in all the five villages?
Solution:
(i) Village D has the minimum number of tractors.
(ii) Village C has the maximum number of tractors.
(iii) Village C has 8 – 5 = 3 more tractors than village B.
(iv) Total number of tractors = 6 + 5 + 8 + 3 + 6 = 28

Question 5.
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:

Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in Class VI less than the number of girls in Class V?
(c) Flow many girls are there in Class VII?
Solution:

(a) Class VIII has the minimum number of girl students.
(b) No, the number of girls in Class VI is greater than the number of girls in Class V.
(c) There are 12 girls in Class VII.

Question 6.
The sale of electric bulbs on different days of a week is shown below:

Observe the pictograph and answer the following questions:
(a) How many bulbs were sold on Friday?
(b) On which day were the maximum number of bulbs sold?
(c) On which of the days same number of bulbs were sold?
(d) On which of the days minimum number of bulbs were sold?
(e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week ?
Solution:

(a) Number of bulbs sold on Friday is 14.
(b) Maximum number of bulbs were sold on Sunday.
(c) Same number of bulbs were sold on Wednesday and Saturday.
(d) Minimum number of bulbs were sold on Wednesday and Saturday.
(e) The total number of bulbs were sold in the given week = 86
Number of cartons required for 9 bulbs = 1
∴ Number of cartons required for 86 bulbs = 86 ÷ 9 = 9.55 = 10
Therefore, 10 cartons were needed in the given week.

Question 7.
In a village six fruit merchants sold the following number of fruit baskets in a particular season:

Observe this pictograph and answer the following questions:
(a) Which merchant sold the maximum number of baskets?
(b) How many fruit baskets were sold by Anwar?
(c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?
Solution:

(a) Martin sold the maximum number of baskets.
(b) 700 fruit baskets were sold by Anwar.
(c) Anwar, Martin and Ranjit Singh have sold more than 600 baskets.

MP Board Class 6th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 10 Mensuration Ex 10.1

Question 1.
Find the perimeter of each of the following figures:

Solution:
(a) Perimeter = Sum of all the sides
= 4 cm + 2 cm + 1 cm + 5 cm = 12 cm

(b) Perimeter = Sum of all the sides
= 23 cm + 35cm + 40 cm + 35cm = 133 cm

(c) Perimeter = Sum of all the sides
= 15 cm + 15 cm + 15 cm + 15 cm = 60 cm

(d) Perimeter = Sum of all the sides
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm

(e) Perimeter = Sum of all the sides
= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm

(f) Perimeter = Sum of all the sides
= 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm

Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution:
Total length of tape required
= Perimeter of rectangle
= 2 (length + breadth)
= 2 (40 +10) cm = 2 × 50 cm = 100 cm = 1 m

Thus, the total length of tape required is 100 cm or 1 m.

Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Solution:
Length of table-top = 2 m 25 cm = 2.25 m
Breadth of table-top = 1 m 50 cm = 1.50 m
Perimeter of table-top = 2 × (length + breadth)
= 2 × (2.25 + 1.50) m
= 2 × 3.75 m = 7.50 m
Thus, perimeter of table-top is 7.5 m.

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solution:
Length of wooden strip
= Perimeter of photograph
= 2 × (length + breadth)
= 2 (32 + 21) cm = 2 × 53 cm = 106 cm
Thus, the length of the wooden strip required is 106 cm.

Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution:
Since 4 rows of wires are needed. Therefore, the total length of wire is equal to 4 times the perimeter of land.
Perimeter of land = 2 × (length + breadth)
= 2 × (0.7 + 0.5) km = 2 × 1.2 km = 2.4 km
= 2.4 × 1000 m = 2400 m
Thus, the length of wire
= 4 × 2400 m = 9600 m = 9.6 km

Question 6.
Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Solution:
(a) Perimeter of ∆ABC
= AB + BC + CA
= 3 cm+ 5 cm+ 4 cm
= 12 cm

(b) Perimeter of equilateral ∆ABC
= 3 × side
= 3 × 9 cm
= 27 cm

(c) Perimeter of ∆ABC
= AB + BC + CA
= 8 cm + 6 cm + 8 cm
= 22 cm

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
Perimeter of triangle
= Sum of all three sides
= 10 cm + 14 cm + 15 cm = 39 cm
Thus, perimeter of triangle is 39 cm.

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:
Perimeter of a regular hexagon
= 6 × length of one side
= 6 × 8m
= 48m
Thus, the perimeter of the regular hexagon is 48 m.

Question 9.
Find the side of the square whose perimeter is 20 m.
Solution:
Perimeter of square = 4 × side
⇒ 20 m = 4 × side ⇒ side = \(\frac{20}{4}\) m = 5m
Thus, the side of square is 5 m.

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
Perimeter of regular pentagon = 5 × side
⇒ 100 cm = 5 × side ⇒ side = \(\frac{100}{5}\) cm = 20 cm
Thus, the side of regular pentagon is 20 cm.

Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Solution:
Length of string = Perimeter of each shape
(a) Perimeter of square = 4 × side
⇒ 30 cm = 4 × side ⇒ side = \(\frac{30}{4}\) cm = 7.5 cm
Thus, the length of each side of square will be 7.5 cm.

(b) Perimeter of equilateral triangle = 3 × side
⇒ 30 cm = 3 × side ⇒ side = \(\frac{30}{3}\) cm = 10 cm
Thus, the length of each side of equilateral triangle will be 10 cm.

(c) Perimeter of regular hexagon = 6 × side
⇒ 30 cm = 6 × side ⇒ side = \(\frac{30}{6}\) cm = 5 cm
Thus, the length of each side of regular hexagon will be 5 cm.

Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solution:
Let the length of third side be x cm. Length of other two sides are 12 cm and 14 cm.
Now, perimeter of triangle = 36 cm
⇒ 12 + 14 + x = 36 ⇒ 26 + x = 36
⇒ x = 36 – 26 ⇒ x = 10
Thus, the length of third side is 10 cm.

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of Rs. 20 per metre.
Solution:
Side of square park = 250 m
Perimeter of square park = 4 × side
= 4 × 250 m = 1000 m
Since, cost of fencing for 1 metre = Rs. 20
Therefore, cost of fencing for 1000 metres
= Rs. 20 × 1000 = Rs. 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs. 12 per metre.
Solution:
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park = 2 × (length + breadth)
= 2 × (175 + 125) m = 2 × 300 m = 600 m
Since, cost of fencing park for 1 metre = Rs. 12
Therefore, cost of fencing park for 600 m = Rs. 12 × 600 = Rs. 7,200

Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Solution:
Distance covered by Sweety
= Perimeter of square park = 4 × side
= 4 × 75 m = 300 m
Thus, distance covered by Sweety is 300 m.
Now, distance covered by Bulbul
= Perimeter of rectangular park
= 2 × (length + breadth)
= 2 × (60 + 45) m = 2 × 105 m = 210 m
Thus, Bulbul covers a distance of 210 m.
So, Bulbul covers less distance.

Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?

Solution:
(a) Perimeter of square = 4 × side
= 4 × 25 cm = 100 cm

(b) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (40 + 10) cm = 2 × 50 cm = 100 cm

(c) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (30 + 20) cm = 2 × 50 cm = 100 cm

(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm
= 100 cm
Thus, all the figures have same perimeter.

Question 17.
Avneet buys 9 square paving slabs, each with side of \(\frac{1}{2}\) m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [see fig. (i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [see fig. (ii)]?
(c) Which has a greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)

Solution:
(a) Side of one small square = \(\frac{1}{2}\) m
∴ Side of given square = \(\frac{3}{2}\) m
Perimeter of square = 4 × side
= 4 × \(\frac{3}{2}\) m = 6 m

(b) Perimeter of given figure
= Sum of all sides = 20 × \(\frac{1}{2}\) m = 10 m

(c) The arrangement cross has greater perimeter.
(d) It is not possible to determine the arrangement with perimeter greater than 10 m.

MP Board Class 6th Maths Solutions