RBSE Solutions Class 6 Maths Chapter 7 Vedic Mathematics Exercise 7.4

Step-by-Step Textbook Solutions for Class 6 Mathematics Chapter 07 Vedic Mathematics

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Question 1. Convert the vinkulum number into general number :
(i) \( 3\overline {5} \)
(ii) \( 5\overline {4} \)
(iii) \( 13\overline {2} \)
(iv) \( 5\overline {42} \)
(v) \( 6\overline {23} \)
Answer:
To convert a vinkulum number to a general number, we use the "All from 9, last from 10" rule for the barred digits and the "Ekanyunena Purvena" (one less than the previous) rule for the digit before the barred section. This method efficiently handles negative digits in Vedic math.
(i) For \( 3\overline {5} \):
• The vinkulum digit is 5. Its parammitra (complement from 10) is \( 10 - 5 = 5 \).
• The digit before the vinkulum is 3. We apply Ekanyunena Purvena, making it \( 3 - 1 = 2 \).
• So, the general number is 25.
(ii) For \( 5\overline {4} \):
• The vinkulum digit is 4. Its parammitra (complement from 10) is \( 10 - 4 = 6 \).
• The digit before the vinkulum is 5. We apply Ekanyunena Purvena, making it \( 5 - 1 = 4 \).
• So, the general number is 46.
(iii) For \( 13\overline {2} \):
• The vinkulum digit is 2. Its parammitra (complement from 10) is \( 10 - 2 = 8 \).
• The digit before the vinkulum is 3. We apply Ekanyunena Purvena, making it \( 3 - 1 = 2 \).
• The leftmost digit 1 remains unchanged.
• So, the general number is 128.
(iv) For \( 5\overline {42} \):
• The last vinkulum digit is 2. Its parammitra (complement from 10) is \( 10 - 2 = 8 \).
• The next vinkulum digit is 4. Its parammitra (complement from 9) is \( 9 - 4 = 5 \).
• The digit before the vinkulum block is 5. We apply Ekanyunena Purvena, making it \( 5 - 1 = 4 \).
• So, the general number is 458.
(v) For \( 6\overline {23} \):
• The last vinkulum digit is 3. Its parammitra (complement from 10) is \( 10 - 3 = 7 \).
• The next vinkulum digit is 2. Its parammitra (complement from 9) is \( 9 - 2 = 7 \).
• The digit before the vinkulum block is 6. We apply Ekanyunena Purvena, making it \( 6 - 1 = 5 \).
• So, the general number is 577.
In simple words: To change a vinkulum number to a regular one, you find the number that adds up to 10 for the last barred digit, and numbers that add up to 9 for any other barred digits. Then, you subtract one from the digit that is just before the barred part.
🎯 Exam Tip: Remember the two main rules for vinkulum conversion: "All from 9 and the Last from 10" for the barred digits, and "Ekanyunena Purvena" (one less) for the digit preceding the vinkulum part. Practice these steps with different numbers to become fluent.

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RBSE Solutions for Class 6 Mathematics Chapter 07 Vedic Mathematics

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