RBSE Solutions Class 7 Maths Chapter 6 वैदिक गणित Exercise 6.4

NCERT Solutions for Class 7 Mathematics: Chapter 06 वैदिक गणित

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Practice Class 7 Mathematics Solutions: Chapter 06 वैदिक गणित

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Question 1. योग कीजिए। (सूत्र विलोकनम् एवं तिर्यक गुणन से)
(i) \( \frac { 1 }{ 9 } + \frac { 4 }{ 9 } \)
(ii) \( \frac { 7 }{ 15 } + \frac { 2 }{ 15 } \)
(iii) \( \frac { 1 }{ 2 } + \frac { 3 }{ 5 } \)
(iv) \( \frac { 4 }{ 3 } + \frac { 2 }{ 5 } \)
(v) \( \frac { 1 }{ 3 } + \frac { 1 }{ 4 } + \frac { 1 }{ 5 } \)
(vi) \( \frac { 1 }{ 2 } + \frac { 2 }{ 3 } + \frac { 3 }{ 5 } \)
Answer:
(i) \( \frac { 1 }{ 9 } + \frac { 4 }{ 9 } \)
यहाँ दोनों भिन्नों के हर समान हैं, इसलिए अंशों को सीधे जोड़ देंगे।
\( \frac { 1 + 4 }{ 9 } = \frac { 5 }{ 9 } \)
(ii) \( \frac { 7 }{ 15 } + \frac { 2 }{ 15 } \)
हर समान होने के कारण, अंशों को जोड़कर वही हर रखा जाता है।
\( \frac { 7 + 2 }{ 15 } = \frac { 9 }{ 15 } \)
\( \implies \frac { 3 }{ 5 } \)
(iii) \( \frac { 1 }{ 2 } + \frac { 3 }{ 5 } \)
यहाँ हर अलग-अलग हैं, इसलिए हम भिन्नों को तिर्यक गुणा विधि से जोड़ेंगे।
\( \frac { 1 \times 5 + 3 \times 2 }{ 2 \times 5 } = \frac { 5 + 6 }{ 10 } \)
\( \implies \frac { 11 }{ 10 } \)
(iv) \( \frac { 4 }{ 3 } + \frac { 2 }{ 5 } \)
अलग-अलग हर होने पर तिर्यक गुणा विधि का उपयोग करें।
\( \frac { 4 \times 5 + 2 \times 3 }{ 3 \times 5 } = \frac { 20 + 6 }{ 15 } \)
\( \implies \frac { 26 }{ 15 } \)
(v) \( \frac { 1 }{ 3 } + \frac { 1 }{ 4 } + \frac { 1 }{ 5 } \)
तीन भिन्नों को जोड़ने के लिए, पहले हरों का लघुत्तम समापवर्त्य (LCM) निकालते हैं, फिर अंशों को जोड़ते हैं।
\( \frac { 1 \times 4 \times 5 + 1 \times 3 \times 5 + 1 \times 3 \times 4 }{ 3 \times 4 \times 5 } = \frac { 20 + 15 + 12 }{ 60 } \)
\( \implies \frac { 47 }{ 60 } \)
(vi) \( \frac { 1 }{ 2 } + \frac { 2 }{ 3 } + \frac { 3 }{ 5 } \)
यहाँ भी तीन भिन्नों को जोड़ने के लिए LCM विधि का प्रयोग किया जाएगा।
\( \frac { 1 \times 3 \times 5 + 2 \times 2 \times 5 + 3 \times 2 \times 3 }{ 2 \times 3 \times 5 } = \frac { 15 + 20 + 18 }{ 30 } \)
\( \implies \frac { 53 }{ 30 } \)
In simple words: जब भिन्नों को जोड़ना हो, तो सबसे पहले उनके हरों को देखें। अगर हर समान हैं, तो केवल अंशों को जोड़ें और हर वही रहने दें। अगर हर अलग-अलग हैं, तो तिर्यक गुणा या LCM (लघुत्तम समापवर्त्य) विधि का उपयोग करके हरों को बराबर बनाएं, फिर अंशों को जोड़ें। भिन्नों को जोड़ने का मतलब है कि आप बराबर हिस्सों को एक साथ मिला रहे हैं।

🎯 Exam Tip: भिन्नों का योग करते समय, हमेशा सुनिश्चित करें कि आप अंत में भिन्न को उसके सबसे सरल रूप में बदल दें।

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Mathematics Class 7 Curriculum Solutions: Chapter 06 वैदिक गणित

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Concept-Driven Answers for Class 7 Mathematics

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Are the Mathematics RBSE solutions for Class 7 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the RBSE Solutions Class 7 Maths Chapter 6 वैदिक गणित Exercise 6.4 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Mathematics concepts are applied in case-study and assertion-reasoning questions.

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