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RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3

March 6, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Exercise 2.3.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 7
Subject Maths
Chapter Chapter 2
Chapter Name भिन्न एवं दशमलव संख्याएँ
Exercise Ex 2.3
Number of Questions 6
Category RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3

प्रश्न 1
ज्ञात कीजिए
RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3
हल:
RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3

RBSE Solutions

प्रश्न 2
निम्नलिखित भिन्नों में से(RBSESolutions.com)प्रत्येक का व्युत्क्रम ज्ञात कीजिए।
RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3 img 1
हल:
RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3 img 2

RBSE Solutions

प्रश्न 3
ज्ञात कीजिए
RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3
हल:
RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3

RBSE Solutions

प्रश्न 4
ज्ञात कीजिए
RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3
हल:
RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3

RBSE Solutions

प्रश्न 5
6 रोटियों में से प्रत्येक रोटी को के \(\frac { 1 }{ 4 }\) टुकड़ों में बाँटने पर रोटी के \(\frac { 1 }{ 4 }\) भागों की संख्या कितनी होगी?
हल:
कुल रोटियाँ = 6, प्रत्येक रोटी को \(\frac { 1 }{ 4 }\) के टुकड़ों(RBSESolutions.com)में बाँटना है।
अत: \(\frac { 1 }{ 4 }\) भागों की संख्या = 6 ÷ \(\frac { 1 }{ 4 }\) = 6 x \(\frac { 4 }{ 1 }\) = 24 भाग

RBSE Solutions

प्रश्न 6
11\(\frac { 1 }{ 2 }\) सेमी. लम्बी रिबन में से \(\frac { 1 }{ 2 }\) सेमी. लम्बे कितने टुकड़े काटे जा सकते हैं?
हल:
रिबन की लम्बाई = 11\(\frac { 1 }{ 2 }\) सेमी. = \(\frac { 23 }{ 2 }\)सेमी.
अत: रिबन में से काटे गए (RBSESolutions.com)की संख्या = \(\frac { 23 }{ 2 }\) ÷ \(\frac { 1 }{ 2 }\)
= \(\frac { 23 }{ 2 }\) x \(\frac { 2 }{ 1 }\) = \(\frac { 23\times 1 }{ 1\times 1 }\) टुकड़े।

RBSE Solutions

We hope the RBSE Solutions for Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Ex 2.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 2 भिन्न एवं दशमलव संख्याएँ Exercise 2.3, drop a comment below and we will get back to you at the earliest.

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