RBSE Solutions Class 6 Maths Chapter 7 Vedic Mathematics Exercise 7.5

Question 1. Find the sum of vinkulum numbers.
(i) \( \begin{array}{r} 63 \\ +43 \\ \hline \end{array} \)
(ii) \( \begin{array}{r} 73 \\ +42 \\ \hline \end{array} \)
(iii) \( \begin{array}{r} 82 \\ +55 \\ \hline \end{array} \)
(iv) \( \begin{array}{r} 89 \\ +78 \\ \hline \end{array} \)
(v) \( \begin{array}{r} 53 \\ +21 \\ \hline \end{array} \)
Answer:
(i) To find the sum of \( 6\overline{3} \) and \( 4\overline{3} \):
First, add the unit digits: \( \overline{3} + \overline{3} = \overline{6} \).
Next, add the tens digits: \( 6 + \overline{4} = 2 \).
So, the sum in vinkulum form is \( 2\overline{6} \).
To convert \( 2\overline{6} \) into a general number, we use the Parammitra digit of \( \overline{6} \), which is \( 4 \). We also put one less sign on the poorven digit (the digit to the left) of \( 6 \), which is \( 2 \). This means \( 2 \) becomes \( 1 \).
Thus, \( 2\overline{6} = 14 \).
(ii) To find the sum of \( 73 \) and \( 4\overline{2} \):
First, add the unit digits: \( 3 + \overline{2} = 1 \).
Next, add the tens digits: \( 7 + \overline{4} = 3 \).
So, the sum is \( 31 \).
(iii) To find the sum of \( 8\overline{2} \) and \( 5\overline{5} \):
First, add the unit digits: \( \overline{2} + \overline{5} = \overline{7} \).
Next, add the tens digits: \( 8 + \overline{5} = 3 \).
So, the sum in vinkulum form is \( 3\overline{7} \).
To convert \( 3\overline{7} \) into a general number, we use the Parammitra digit of \( \overline{7} \), which is \( 3 \). We also put one less sign on the poorven digit of \( 7 \), which is \( 3 \). This means \( 3 \) becomes \( 2 \).
Thus, \( 3\overline{7} = 23 \).
(iv) To find the sum of \( 8\overline{9} \) and \( \overline{7}8 \):
First, add the unit digits: \( \overline{9} + 8 = \overline{1} \). This means \( -9 + 8 = -1 \).
Next, add the tens digits: \( 8 + \overline{7} = 1 \). This means \( 8 - 7 = 1 \).
So, the sum in vinkulum form is \( 1\overline{1} \).
To convert \( 1\overline{1} \) into a general number, we use the Parammitra digit of \( \overline{1} \), which is \( 9 \). We put one less sign on the poorven digit of \( 1 \), which is \( 1 \). This means \( 1 \) becomes \( 0 \).
Thus, \( 1\overline{1} = 09 \), or simply \( 9 \).
(v) To find the sum of \( 5\overline{3} \) and \( \overline{2}1 \):
First, add the unit digits: \( \overline{3} + 1 = \overline{2} \). This means \( -3 + 1 = -2 \).
Next, add the tens digits: \( 5 + \overline{2} = 3 \). This means \( 5 - 2 = 3 \).
So, the sum in vinkulum form is \( 3\overline{2} \).
To convert \( 3\overline{2} \) into a general number, we use the Parammitra digit of \( \overline{2} \), which is \( 8 \). We put one less sign on the poorven digit of \( 2 \), which is \( 3 \). This means \( 3 \) becomes \( 2 \).
Thus, \( 3\overline{2} = 28 \).
In simple words: Vinkulum numbers have negative digits represented with a bar. To add them, we add each place value separately, then combine the results. If a digit has a bar, we convert it to a regular number by subtracting it from 10 and reducing the digit to its left by one.
🎯 Exam Tip: Always pay close attention to which digits have a vinkulum bar, as it changes how you perform the addition and conversion to a regular number.