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in Vedic Mathematics by (49.9k points)
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Convert the general numbers into its vinkulum. 

(i) 8

(ii) 27 

(iii) 82 

(iv) 78 

(v) 96

1 Answer

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by (48.6k points)
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Best answer

(i)

Hints : 

(a) Put vinkulum line on the parammitra digit of 8 i.e.,2 = bar 2 

(b) Put one more sign on poorven digit of 8 i.e., 0 = bar \(\dot{0}\)2

(c) Write \(\dot{0}\) = 1. 

(d) Thus, we got the vinkulum of 8 = bar 2

(ii) 27 

Hints : 

= bar 87 

(a) Digit 7 would be as it is and 

\(\dot{0}\) 8 bar 7 = 7 vinkulum line would be on param 

= 187 mitra digit of 2 i.e., 8. 

(b) One more sign on the poorven of 8 i.e., 0. 

(c) Write \(\dot{0}\) = 1, 

(d) Thus, we got vinkulum of 27 

(iii) 82 

Hint : 

= 2 bar 2 

(a) Digit 2 would be as it is and vinkulum line would be on 

= \(\dot{0}\) 2 bar 2 parammitra digit of 8 i.e., 2. 

= 1 2 bar 2 

(b) One more sign on the poorven digit of 8 ie., 0. 

(c) Write \(\dot{0}\) = 1 

(d) Thus, we get vinkulum of 82. 

(iv) 78 

Hint : 

= 3 bar 8 (a) Digit 8 would be as it is and vinkulum line would be on 

= \(\dot{0}\) 3 bar 8 parammitra digit of 7 i.e., 3. 

= 1 3 bar 8 

(b) One more sign on the poorven digit of 7 ie., 0

(c) Write \(\dot{0}\) = 1. 

(d) Thus, we got vinkulum of 78. 

(v) 96 

Hint : 

= 1 bar 6 (a) Digit 6 would be as it is and 

= \(\dot{0}\) 1 bar 6 vinkulum line would be on parammitra digit of 9 i.e., 1. 

= 1 1 bar 6 (b) One more sign on the poorven digit of 9 Le., 0. 

(c) Write \(\dot{0}\) = 1. 

(d) Thus, we got vinkulum of 96

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