RBSE Solutions Class 6 Maths Chapter 7 Vedic Mathematics Exercise 7.6

Question 1. Find out the difference using vinkulum.
(i) \( \begin{array}{r} 96 \\ -49 \\ \hline \end{array} \)
(ii) \( \begin{array}{r} 932 \\ -245 \\ \hline \end{array} \)
(iii) \( \begin{array}{r} 952 \\ -788 \\ \hline \end{array} \)
(iv) \( \begin{array}{r} 834 \\ -547 \\ \hline \end{array} \)
Answer:
(i) To find \( 96 - 49 \):
1. Change the subtraction of 49 to adding its vinkulum number. The vinkulum digits for 49 are \( \overline{4} \) (for 4) and \( \overline{9} \) (for 9). So, we calculate \( 96 + \overline{4}\overline{9} \).
2. In the units place, add 6 and \( \overline{9} \): \( 6 + (-9) = -3 \). We write this as \( \overline{3} \).
3. In the tens place, add 9 and \( \overline{4} \): \( 9 + (-4) = 5 \).
4. This gives the intermediate vinkulum number \( 5\overline{3} \).
5. Now, convert \( 5\overline{3} \) to a general positive number.
- The units digit is \( \overline{3} \). Its Parammitra (complement from 10) is 7. We write 7 in the units place.
- Put a 'one less' mark on the preceding digit 5. So, 5 becomes \( 5-1=4 \).
6. Thus, \( 96 - 49 = 47 \).
(ii) To find \( 932 - 245 \):
1. Change the subtraction of 245 to adding its vinkulum number. The vinkulum digits for 245 are \( \overline{2} \), \( \overline{4} \), \( \overline{5} \). So, we calculate \( 932 + \overline{2}\overline{4}\overline{5} \).
2. In the units place, add 2 and \( \overline{5} \): \( 2 + (-5) = -3 \). We write this as \( \overline{3} \).
3. In the tens place, add 3 and \( \overline{4} \): \( 3 + (-4) = -1 \). We write this as \( \overline{1} \).
4. In the hundreds place, add 9 and \( \overline{2} \): \( 9 + (-2) = 7 \).
5. This gives the intermediate vinkulum number \( 7\overline{1}\overline{3} \).
6. Now, convert \( 7\overline{1}\overline{3} \) to a general positive number, starting from the right.
- The units digit is \( \overline{3} \). Its Parammitra (complement from 10) is 7. We write 7.
- Put a 'one less' mark on the preceding digit \( \overline{1} \). So, \( \overline{1} \) becomes \( (\overline{1}-1) = \overline{2} \). Wait, the source says parammitra of 1 (tens digit) i.e. 9 and put one less sign on poorven of `\overline{1}` i.e. 7. Let's follow the source's exact logic for conversion.
- For \( \overline{3} \) (units digit), its Parammitra is 7. So, the units digit of the answer is 7.
- Put a 'one less' mark on the digit before \( \overline{3} \), which is \( \overline{1} \). The source says this \( \overline{1} \) (with a dot below) becomes \( (\overline{1} - 1) \). The source states 'Put one less sign on the poorven digit of \( \overline{3} \) i.e., 9. Write \( \underset { . }{ 9 } =8 \)'. This implies 9 is the digit before \( \overline{3} \). The `7\overline{1}\overline{3}` means 7 in hundreds, `\overline{1}` in tens, `\overline{3}` in units.
Let's re-evaluate the conversion steps from the source for \( 7\overline{1}\overline{3} \):
- (f) Parammitra digit of 1 (tens digit) i.e., 9. (This is `10-1=9`, for the positive equivalent of `\overline{1}`)
- (g) Put one less sign on the poorven digit of \( \overline{1} \) i.e., 7. (This means 7 becomes 6)
- (h) Write \( \underset { . }{ 7 } =6 \).
- (i) Parammitra digit of \( \overline{3} \) (units digit) i.e., 7.
- (j) Put one less sign on the poorven digit of \( \overline{3} \) i.e., 9. (This means 9 becomes 8)
- (k) Write \( \underset { . }{ 9 } =8 \).
- Combining these, the process is: Units `\overline{3}` becomes 7. Tens `\overline{1}` becomes 9. The digit before `\overline{1}` (which is 7 in `7\overline{1}\overline{3}`) becomes `7-1=6`. The digit before `\overline{3}` (which is `\overline{1}`) becomes 9, and the previous digit of that (which is 7) becomes 6.
Let's follow the standard process: \( 7\overline{1}\overline{3} \)
- Convert \( \overline{3} \): \( 10-3 = 7 \). Reduce the preceding digit \( \overline{1} \) by 1, so it becomes \( \overline{2} \). The number is now \( 7\overline{2}7 \).
- Convert \( \overline{2} \): \( 10-2 = 8 \). Reduce the preceding digit 7 by 1, so it becomes 6. The number is now \( 687 \).
- Thus, \( 932 - 245 = 687 \).
(iii) To find \( 952 - 788 \):
1. Change the subtraction of 788 to adding its vinkulum number: \( \overline{7} \), \( \overline{8} \), \( \overline{8} \). So, we calculate \( 952 + \overline{7}\overline{8}\overline{8} \).
2. In the units place, add 2 and \( \overline{8} \): \( 2 + (-8) = -6 \). We write this as \( \overline{6} \).
3. In the tens place, add 5 and \( \overline{8} \): \( 5 + (-8) = -3 \). We write this as \( \overline{3} \).
4. In the hundreds place, add 9 and \( \overline{7} \): \( 9 + (-7) = 2 \).
5. This gives the intermediate vinkulum number \( 2\overline{3}\overline{6} \).
6. Now, convert \( 2\overline{3}\overline{6} \) to a general positive number, starting from the right.
- Convert \( \overline{6} \): \( 10-6 = 4 \). Reduce the preceding digit \( \overline{3} \) by 1, so it becomes \( \overline{4} \). The number is now \( 2\overline{4}4 \).
- Convert \( \overline{4} \): \( 10-4 = 6 \). Reduce the preceding digit 2 by 1, so it becomes 1. The number is now \( 164 \).
- Thus, \( 952 - 788 = 164 \).
(iv) To find \( 834 - 547 \):
1. Change the subtraction of 547 to adding its vinkulum number: \( \overline{5} \), \( \overline{4} \), \( \overline{7} \). So, we calculate \( 834 + \overline{5}\overline{4}\overline{7} \).
2. In the units place, add 4 and \( \overline{7} \): \( 4 + (-7) = -3 \). We write this as \( \overline{3} \).
3. In the tens place, add 3 and \( \overline{4} \): \( 3 + (-4) = -1 \). We write this as \( \overline{1} \).
4. In the hundreds place, add 8 and \( \overline{5} \): \( 8 + (-5) = 3 \).
5. This gives the intermediate vinkulum number \( 3\overline{1}\overline{3} \).
6. Now, convert \( 3\overline{1}\overline{3} \) to a general positive number, starting from the right.
- Convert \( \overline{3} \): \( 10-3 = 7 \). Reduce the preceding digit \( \overline{1} \) by 1, so it becomes \( \overline{2} \). The number is now \( 3\overline{2}7 \).
- Convert \( \overline{2} \): \( 10-2 = 8 \). Reduce the preceding digit 3 by 1, so it becomes 2. The number is now \( 287 \).
- Thus, \( 834 - 547 = 287 \).
In simple words: To subtract using vinkulum, change the second number into its vinkulum form (negative digits). Then add the numbers. If you get negative digits in the answer, convert them to positive numbers by taking their complement from 10 and reducing the digit before them by one. Repeat this until all digits are positive.

🎯 Exam Tip: Remember to apply the 'one less than the previous digit' rule (poorven) correctly whenever you convert a vinkulum digit to its positive form. This is crucial for accuracy.