RBSE Solutions Class 9 Maths Chapter 1 Vedic Mathematics Exercise 1.2

Mathematics Ex 1.2

Change the Following into Vinculum Numbers:

Question 1. 89
Answer: To convert 89 into a vinculum number using the Nikhilam Sutra (All from 9 and the last from 10) method, we look at the digits from right to left. For the last digit 9, we subtract it from 10 to get 1, which becomes \( \bar{1} \). We then increase the previous digit (8) by 1, making it 9. For this new digit 9, we subtract it from 10 to get 1, which becomes \( \bar{1} \). We then increase the digit before it (which is 0) by 1, making it 1. Thus, 89 can be written as \( 1\bar{1}\bar{1} \). This form represents 100 - 10 - 1, which equals 89.
\( 89 = 1\bar{1}\bar{1} \)
In simple words: To change 89 into a special "vinculum" number, you replace digits like 8 and 9 by subtracting them from 10 (or 9) and putting a bar over the new digit. You also add one to the digit before it. For 89, it becomes one-bar-one-bar-one, which means 100 minus 10 minus 1.

🎯 Exam Tip: Remember that a vinculum digit \( \bar{x} \) represents \( (10-x) \) in the last position or \( (9-x) \) in other positions, while also adding 1 to the digit immediately to its left.

Question 2. 878
Answer: To convert 878 into a vinculum number:
From right to left, for 8, we subtract from 10 to get \( \bar{2} \). Increment the previous digit (7) by 1, making it 8.
For this 8, we subtract from 10 to get \( \bar{2} \). Increment the previous digit (8) by 1, making it 9.
For this 9, we subtract from 10 to get \( \bar{1} \). Increment the previous digit (0) by 1, making it 1.
So, 878 can be written as \( 1\bar{1}\bar{2}\bar{2} \). This representation can be calculated as \( 1000 - 100 - 20 - 2 = 878 \).
\( 878 = 1\bar{1}\bar{2}\bar{2} \)
In simple words: We change the digits of 878 into vinculum form. You subtract the digits from 10 and add 1 to the digit before it. For example, 878 becomes one-bar-one-bar-two-bar-two.

🎯 Exam Tip: When converting to vinculum, identify digits 5 or larger from the right and apply the "All from 9, last from 10" rule, increasing the digit to the left of the group. Smaller digits (0-4) typically remain as they are or are part of a larger group that got changed.

Question 3. 9687
Answer: To convert 9687 into a vinculum number:
For 7, we subtract from 10 to get \( \bar{3} \). Increment 8 to 9.
For 9, we subtract from 10 to get \( \bar{1} \). Increment 6 to 7.
For 7, we subtract from 10 to get \( \bar{3} \). Increment 9 to 10.
So, 9687 can be written as \( 10\bar{3}\bar{1}\bar{3} \). This represents \( 10000 - 300 - 10 - 3 = 9687 \).
\( 9687 = 10\bar{3}\bar{1}\bar{3} \)
In simple words: We change the number 9687 into a special number with bars over some digits. You subtract the digits from 10 (or 9) and add one to the digit before them. In the end, 9687 becomes ten-bar-three-bar-one-bar-three.

🎯 Exam Tip: Practice converting numbers with different lengths to vinculum form to master the pattern of subtraction and incrementing digits.

Question 5. \( 3\bar{2}1 \)
Answer: To convert the vinculum number \( 3\bar{2}1 \) to a regular number, we break it down by place value. The `3` is in the hundreds place, the `\bar{2}` is in the tens place, and `1` is in the units place.
The digit `\(\bar{2}\)` means \( -2 \).
So, \( 3\bar{2}1 = 3 \times 100 + (-2) \times 10 + 1 \times 1 \)
\( = 300 - 20 + 1 \)
\( = 281 \)
This method helps convert numbers with negative digits into their standard decimal form.
In simple words: The bar over a number means it is negative. So, \( 3\bar{2}1 \) means 300, then subtract 20, then add 1. This gives us 281.

🎯 Exam Tip: Remember that a vinculum digit \( \bar{x} \) represents \( -x \) when converting back to a normal number, making sure to apply the negative value at its correct place value.

Question 6. \( 2\bar{4}\bar{3}\bar{2} \)
Answer: To convert the vinculum number \( 2\bar{4}\bar{3}\bar{2} \) to a regular number:
This number can be understood as 2000 minus 400 minus 30 minus 2. Each barred digit represents a negative value at its place.
\( 2\bar{4}\bar{3}\bar{2} = 2 \times 1000 + (-4) \times 100 + (-3) \times 10 + (-2) \times 1 \)
\( = 2000 - 400 - 30 - 2 \)
\( = 1600 - 30 - 2 \)
\( = 1570 - 2 \)
\( = 1568 \)
Understanding the positional value of each barred digit is key to accurate conversion.
In simple words: We have 2, and then 4, 3, 2 all with bars. This means 2000 minus 400, then minus 30, and finally minus 2. When you do all these subtractions, you get 1568.

🎯 Exam Tip: When multiple digits are barred in a sequence, subtract each barred digit from its corresponding place value. For example, \( \bar{4}\bar{3}\bar{2} \) means \( -400 - 30 - 2 \).

Question 7. \( 4\bar{3}0\bar{2} \)
Answer: To convert the vinculum number \( 4\bar{3}0\bar{2} \) to a regular number:
We interpret the `\(\bar{3}\)` as \( -3 \) in the hundreds place and `\(\bar{2}\)` as \( -2 \) in the units place. The `0` remains as zero.
\( 4\bar{3}0\bar{2} = 4 \times 1000 + (-3) \times 100 + 0 \times 10 + (-2) \times 1 \)
\( = 4000 - 300 + 0 - 2 \)
\( = 3700 - 2 \)
\( = 3698 \)
This calculation shows how to correctly handle zeros and negative digits in vinculum numbers.
In simple words: This number has 4, a barred 3, a 0, and a barred 2. So, it means 4000 minus 300, then plus zero, then minus 2. The final answer is 3698.

🎯 Exam Tip: Pay close attention to the position of each barred digit as it indicates which place value needs to be subtracted. A '0' digit between barred numbers simply means no value is added or subtracted at that position.

Question 8. \( 450\bar{4}\bar{9} \)
Answer: To convert the vinculum number \( 450\bar{4}\bar{9} \) to a regular number:
Here, `\(\bar{4}\)` is in the tens place, and `\(\bar{9}\)` is in the units place. The `450` forms the thousands part.
\( 450\bar{4}\bar{9} = 450 \times 100 + (-4) \times 10 + (-9) \times 1 \)
\( = 45000 - 40 - 9 \)
\( = 44960 - 9 \)
\( = 44951 \)
This process applies the understanding of negative digits to larger numbers for accurate conversion.
In simple words: The number is 450, then a barred 4, then a barred 9. This means you start with 45000, then take away 40, and then take away 9. The final number you get is 44951.

🎯 Exam Tip: When converting complex vinculum numbers, it is helpful to first identify the positive and negative digit groups and then perform the arithmetic operations systematically from left to right.

Question 10. \( 94 \times 92 \)
Answer: To multiply \( 94 \times 92 \) using the Nikhilam Sutra (sub-base method):
Take the Base as 10 and Sub-base as \( 10 \times 9 = 90 \).
The deviations from the sub-base 90 are:
\( 94 - 90 = +4 \)
\( 92 - 90 = +2 \)
The formula is: \( \text{Sub-base Multiplier} \times (\text{Number}_1 + \text{Deviation}_2) | (\text{Deviation}_1 \times \text{Deviation}_2) \)
\( = 9 \times (94 + 2) | (4 \times 2) \)
\( = 9 \times 96 | 8 \)
\( = 864 | 8 \)
Since the base is 10, the right-hand side can hold one digit.
So, the product is \( 8648 \). This Vedic method simplifies multiplication by working with deviations from a base.
In simple words: We multiply 94 by 92 using a special trick. We use 90 as a helper number. 94 is 4 more than 90, and 92 is 2 more than 90. We then do some calculations with these small extra numbers, multiply by 9, and join the results to get 8648.

🎯 Exam Tip: When using the sub-base method, remember to multiply the left-hand side (sum of a number and deviation) by the sub-base multiplier. The number of digits on the RHS depends on the base (e.g., 1 for base 10, 2 for base 100).

Question 11. \( 71 \times 73 \)
Answer: To multiply \( 71 \times 73 \) using the Nikhilam Sutra (sub-base method):
Take the Base as 10 and Sub-base as \( 10 \times 7 = 70 \).
The deviations from the sub-base 70 are:
\( 71 - 70 = +1 \)
\( 73 - 70 = +3 \)
The formula is: \( \text{Sub-base Multiplier} \times (\text{Number}_1 + \text{Deviation}_2) | (\text{Deviation}_1 \times \text{Deviation}_2) \)
\( = 7 \times (71 + 3) | (1 \times 3) \)
\( = 7 \times 74 | 3 \)
\( = 518 | 3 \)
Since the base is 10, the right-hand side can hold one digit.
So, the product is \( 5183 \). This method makes multiplying numbers near a common base or sub-base much faster.
In simple words: To multiply 71 by 73, we use 70 as our helper number. 71 is 1 more than 70, and 73 is 3 more than 70. We cross-add one number with the other's extra part, multiply by 7, and then join with the product of the extra parts to get 5183.

🎯 Exam Tip: Always double-check your deviations from the chosen sub-base. A small error here will lead to an incorrect final product.

Question 13. \( 11 \times 12 \times 13 \)
Answer: To multiply \( 11 \times 12 \times 13 \) using Vedic Mathematics (Nikhilam method for three numbers):
Take the Base as 10.
The deviations from the base 10 are:
\( 11 - 10 = +1 \)
\( 12 - 10 = +2 \)
\( 13 - 10 = +3 \)
The formula for \( (N+d_1)(N+d_2)(N+d_3) \) is:
\( N^2 + (d_1+d_2+d_3)N | (d_1d_2 + d_2d_3 + d_3d_1) | (d_1d_2d_3) \)
Since N=10, the formula parts become:
First Part (LHS): \( (\text{Number}_1 + \text{Deviation}_2 + \text{Deviation}_3) \)
\( = 11 + 2 + 3 = 16 \)
Second Part (Middle): \( (\text{Deviation}_1 \times \text{Deviation}_2) + (\text{Deviation}_2 \times \text{Deviation}_3) + (\text{Deviation}_3 \times \text{Deviation}_1) \)
\( = (1 \times 2) + (2 \times 3) + (3 \times 1) \)
\( = 2 + 6 + 3 = 11 \)
Third Part (RHS): \( (\text{Deviation}_1 \times \text{Deviation}_2 \times \text{Deviation}_3) \)
\( = 1 \times 2 \times 3 = 6 \)
So, we have \( 16 | 11 | 6 \).
Since the base is 10, each section on the right-hand side can hold only one digit. We carry over any tens.
RHS: \( 6 \)
Middle: \( 11 \implies \) write \( 1 \), carry over \( 1 \) to LHS.
LHS: \( 16 + 1 (\text{carry}) = 17 \)
So, the product is \( 1716 \). This method makes multiplying numbers slightly above a base simple.
In simple words: To multiply 11, 12, and 13 together, we use 10 as our main helper number. Each number is a little bit more than 10 (1, 2, and 3 more). We add these small extra numbers in a special way and also multiply them together. Then we combine these parts, carrying over tens when needed, to get the answer 1716.

🎯 Exam Tip: For three numbers, remember the three sections: (Number + sum of other two deviations) | (sum of products of two deviations at a time) | (product of all three deviations). Carry over digits from right to left based on the base power.

Question 14. \( 97 \times 98 \times 99 \)
Answer: To multiply \( 97 \times 98 \times 99 \) using Vedic Mathematics (Nikhilam method for three numbers):
Take the Base as 100.
The deviations from the base 100 are:
\( 97 - 100 = -3 \)
\( 98 - 100 = -2 \)
\( 99 - 100 = -1 \)
Using the formula for \( (N+d_1)(N+d_2)(N+d_3) \):
First Part (LHS): \( (\text{Number}_1 + \text{Deviation}_2 + \text{Deviation}_3) \)
\( = 97 + (-2) + (-1) = 97 - 2 - 1 = 94 \)
Second Part (Middle): \( (d_1 \times d_2) + (d_2 \times d_3) + (d_3 \times d_1) \)
\( = (-3 \times -2) + (-2 \times -1) + (-1 \times -3) \)
\( = 6 + 2 + 3 = 11 \)
Third Part (RHS): \( (d_1 \times d_2 \times d_3) \)
\( = (-3 \times -2 \times -1) = -6 \)
So, we have \( 94 | 11 | -6 \).
Since the base is 100, each section on the right-hand side must hold two digits. The rightmost part is negative, so we need to adjust by borrowing.
From Middle part (11), borrow 1. This 1 becomes 100 for the RHS. So, middle part becomes \( 11 - 1 = 10 \).
RHS: \( 100 - 6 = 94 \).
Now the parts are \( 94 | 10 | 94 \).
All parts have two digits (or are LHS), so we combine them.
The product is \( 941094 \). This systematic approach helps manage negative deviations efficiently.
In simple words: We multiply 97, 98, and 99. We use 100 as the helper number because they are all close to it. They are 3, 2, and 1 less than 100. We add and multiply these small "less than" numbers in a special way. Because one part is negative, we "borrow" from the middle part. After all the adjustments, we get 941094.

🎯 Exam Tip: When dealing with negative deviations, remember to borrow from the left section to make the rightmost section positive. Each unit borrowed from a left section equals the base (e.g., 100 for base 100) when carried to the right.

Question 15. \( 102 \times 103 \times 104 \)
Answer: To multiply \( 102 \times 103 \times 104 \) using Vedic Mathematics (Nikhilam method for three numbers):
Take the Base as 100.
The deviations from the base 100 are:
\( 102 - 100 = +2 \)
\( 103 - 100 = +3 \)
\( 104 - 100 = +4 \)
Using the formula for \( (N+d_1)(N+d_2)(N+d_3) \):
First Part (LHS): \( (\text{Number}_1 + \text{Deviation}_2 + \text{Deviation}_3) \)
\( = 102 + 3 + 4 = 109 \)
Second Part (Middle): \( (d_1 \times d_2) + (d_2 \times d_3) + (d_3 \times d_1) \)
\( = (2 \times 3) + (3 \times 4) + (4 \times 2) \)
\( = 6 + 12 + 8 = 26 \)
Third Part (RHS): \( (d_1 \times d_2 \times d_3) \)
\( = 2 \times 3 \times 4 = 24 \)
So, we have \( 109 | 26 | 24 \).
Since the base is 100, each section on the right-hand side can hold two digits. All parts fit this rule.
Combine the parts directly.
The product is \( 1092624 \). This method is particularly efficient for numbers close to a base.
In simple words: To multiply 102, 103, and 104, we use 100 as our helper number. These numbers are 2, 3, and 4 more than 100. We add one number to the other two "extra" parts. We also multiply the extra parts in pairs and all three together. We then put these parts together to get 1092624.

🎯 Exam Tip: When all deviations are positive, the combination of the parts is straightforward, as there are no negative values or borrowings involved. Ensure each segment holds the correct number of digits as per the base.

Question 16. \( 99 \times 101 \times 103 \)
Answer: To multiply \( 99 \times 101 \times 103 \) using Vedic Mathematics (Nikhilam method for three numbers):
Take the Base as 100.
The deviations from the base 100 are:
\( 99 - 100 = -1 \)
\( 101 - 100 = +1 \)
\( 103 - 100 = +3 \)
Using the formula for \( (N+d_1)(N+d_2)(N+d_3) \):
First Part (LHS): \( (\text{Number}_1 + \text{Deviation}_2 + \text{Deviation}_3) \)
\( = 99 + 1 + 3 = 103 \)
Second Part (Middle): \( (d_1 \times d_2) + (d_2 \times d_3) + (d_3 \times d_1) \)
\( = (-1 \times 1) + (1 \times 3) + (3 \times -1) \)
\( = -1 + 3 - 3 = -1 \)
Third Part (RHS): \( (d_1 \times d_2 \times d_3) \)
\( = (-1 \times 1 \times 3) = -3 \)
So, we have \( 103 | -1 | -3 \).
Since the base is 100, each section on the right-hand side must hold two digits. Both middle and RHS parts are negative, requiring adjustments.
First, adjust RHS: From middle part (-1), borrow a unit. This is tricky because the middle part is also negative. Instead, we carry over from the LHS (103).
Take 1 from \( 103 \), so LHS becomes \( 102 \). This 1 becomes 100 for the middle part.
Middle part: \( 100 + (-1) = 99 \).
Now the parts are \( 102 | 99 | -3 \).
Next, adjust RHS again: From middle part (99), borrow 1. This 1 becomes 100 for the RHS. So, middle part becomes \( 99 - 1 = 98 \).
RHS: \( 100 + (-3) = 97 \).
Now the parts are \( 102 | 98 | 97 \).
All parts have two digits (or are LHS), so we combine them.
The product is \( 1029897 \). Handling mixed positive and negative deviations requires careful adjustments.
In simple words: To multiply 99, 101, and 103, we use 100 as our helper number. Their "extra" parts are -1, +1, and +3. We add and multiply these parts in a special way. Because some parts end up negative, we have to "borrow" from the parts on the left until all parts are positive. After these steps, the final answer is 1029897.

🎯 Exam Tip: When both the middle and rightmost sections are negative, perform adjustments sequentially from left to right. First, adjust the middle section using the LHS, then adjust the rightmost section using the new middle section value.

Find the Square of the Following Numbers

Question 17. 103
Answer: To find the square of 103 using the Vedic method (Nikhilam Sutra):
Take the Base as 100.
The deviation from the base 100 is:
\( 103 - 100 = +3 \)
The formula for squaring numbers near a base is: \( (\text{Number} + \text{Deviation}) | (\text{Deviation})^2 \)
First Part (LHS): \( 103 + 3 = 106 \)
Second Part (RHS): \( (+3)^2 = 9 \)
So, we have \( 106 | 9 \).
Since the base is 100, the right-hand side must hold two digits. So, 9 should be written as 09.
\( 106 | 09 \)
Combine the parts.
The square of 103 is \( 10609 \). This method provides a quick way to find squares for numbers near a base.
In simple words: To square 103, we use 100 as our helper number. 103 is 3 more than 100. We add this 3 to 103, and then we square the 3. We get 106 and 9. Because our helper number (100) has two zeros, the 9 needs two digits, so we write 09. Putting them together gives 10609.

🎯 Exam Tip: When squaring numbers using the Nikhilam Sutra, ensure the number of digits in the right-hand side part matches the number of zeros in the base. If fewer, add leading zeros; if more, carry over.

Question 18. 95
Answer: To find the square of 95 using the Vedic method (Nikhilam Sutra):
Take the Base as 100.
The deviation from the base 100 is:
\( 95 - 100 = -5 \)
The formula for squaring numbers near a base is: \( (\text{Number} + \text{Deviation}) | (\text{Deviation})^2 \)
First Part (LHS): \( 95 + (-5) = 95 - 5 = 90 \)
Second Part (RHS): \( (-5)^2 = 25 \)
So, we have \( 90 | 25 \).
Since the base is 100, the right-hand side must hold two digits. 25 already has two digits.
Combine the parts.
The square of 95 is \( 9025 \). This technique effectively squares numbers slightly below a base.
In simple words: To square 95, we use 100 as our helper number. 95 is 5 less than 100. We subtract this 5 from 95, and then we square the 5. We get 90 and 25. Putting them together gives 9025.

🎯 Exam Tip: A negative deviation squared will always result in a positive number. Be careful to calculate \( (-d)^2 \) correctly, for example, \( (-5)^2 = 25 \), not -25.

Question 19. 204
Answer: To find the square of 204 using the Vedic method (Upa-Sutra "Anurupyena" or sub-base method):
Take the Base as 100 and Sub-base as \( 100 \times 2 = 200 \).
The deviation from the sub-base 200 is:
\( 204 - 200 = +4 \)
The formula for squaring numbers near a sub-base is: \( \text{Sub-base Multiplier} \times (\text{Number} + \text{Deviation}) | (\text{Deviation})^2 \)
First Part (LHS): \( 2 \times (204 + 4) = 2 \times 208 = 416 \)
Second Part (RHS): \( (+4)^2 = 16 \)
So, we have \( 416 | 16 \).
Since the base is 100, the right-hand side must hold two digits. 16 already has two digits.
Combine the parts.
The square of 204 is \( 41616 \). This method expands the base method to numbers near multiples of the base.
In simple words: To square 204, we use 200 as our helper number. 204 is 4 more than 200. We add this 4 to 204, then multiply that by 2 (because 200 is 2 times 100), and we also square the 4. This gives us 416 and 16. Putting them together gives 41616.

🎯 Exam Tip: In the sub-base method for squaring, do not forget to multiply the left-hand side part by the sub-base multiplier before combining it with the right-hand side.

Question 20. 225
Answer: To find the square of 225 using the Vedic method (Upa-Sutra "Anurupyena" or sub-base method):
Take the Base as 100 and Sub-base as \( 100 \times 2 = 200 \).
The deviation from the sub-base 200 is:
\( 225 - 200 = +25 \)
The formula for squaring numbers near a sub-base is: \( \text{Sub-base Multiplier} \times (\text{Number} + \text{Deviation}) | (\text{Deviation})^2 \)
First Part (LHS): \( 2 \times (225 + 25) = 2 \times 250 = 500 \)
Second Part (RHS): \( (+25)^2 = 625 \)
So, we have \( 500 | 625 \).
Since the base is 100, the right-hand side must hold two digits. Here, 625 has three digits.
We keep the last two digits (25) for the RHS and carry over the remaining digit (6) to the LHS.
LHS: \( 500 + 6 = 506 \)
RHS: \( 25 \)
Combine the parts.
The square of 225 is \( 50625 \). This method simplifies calculations for numbers that are further from the base but close to a sub-base.
In simple words: To square 225, we use 200 as our helper number. 225 is 25 more than 200. We add this 25 to 225 and multiply by 2 (since 200 is 2 times 100). We also square the 25. This gives us 500 and 625. Since 625 is too big for the right side, we take the extra '6' and add it to the 500. So we get 506 and 25, which makes 50625.

🎯 Exam Tip: If the right-hand side part has more digits than allowed by the base, always carry over the extra digits to the left-hand side, just like in standard addition.

Question 22. 91
Answer: To find the cube of 91 using the Vedic method (Nikhilam Sutra):
Take the Base as 100.
The deviation from the base 100 is:
\( 91 - 100 = -9 \)
The formula for cubing numbers near a base is: \( (\text{Number} + 2 \times \text{Deviation}) | (3 \times \text{Deviation}^2) | (\text{Deviation}^3) \)
First Part (LHS): \( 91 + 2 \times (-9) = 91 - 18 = 73 \)
Second Part (Middle): \( 3 \times (-9)^2 = 3 \times 81 = 243 \)
Third Part (RHS): \( (-9)^3 = -729 \)
So, we have \( 73 | 243 | -729 \).
Since the base is 100, each part on the right-hand side must hold two digits. The RHS is negative, and the Middle part has more than two digits.
Step 1: Adjust RHS. To make -729 positive, we need to add 800 (8 times the base 100). So, borrow 8 from the Middle part.
Middle part becomes \( 243 - 8 = 235 \).
RHS becomes \( 800 - 729 = 71 \).
Now the parts are: \( 73 | 235 | 71 \).
Step 2: Adjust Middle part. 235 has three digits, but it should only have two. Carry over the leading digit (2) to the LHS.
LHS becomes \( 73 + 2 = 75 \).
Middle part becomes \( 35 \).
Now the parts are: \( 75 | 35 | 71 \).
All parts now have the correct number of digits and are positive.
Combine the parts.
The cube of 91 is \( 753571 \). Careful adjustments are essential when dealing with negative deviations and carry-overs in cubing.
In simple words: To cube 91, we use 100 as our helper number. 91 is 9 less than 100. We do three steps: 1) subtract 9 twice from 91, 2) multiply 3 by 9 squared, and 3) cube the -9. This gives us 73, 243, and -729. Because of the negative parts and too many digits, we borrow from the left side and adjust until all numbers are positive and have two digits each. The final answer is 753571.

🎯 Exam Tip: When cubing numbers with negative deviations, perform adjustments systematically from the rightmost section. Borrowing from a section means adding the base value (or a multiple of it) to the right and subtracting units from the left.

Question 23. 32
Answer: To find the cube of 32 using the Vedic method (Upa-Sutra "Anurupyena" or sub-base method):
Take the Base as 10 and Sub-base as \( 10 \times 3 = 30 \).
The deviation from the sub-base 30 is:
\( 32 - 30 = +2 \)
The formula for cubing numbers near a sub-base is: \( \text{Multiplier}^2 \times (\text{Number} + 2 \times \text{Deviation}) | \text{Multiplier} \times (3 \times \text{Deviation}^2) | (\text{Deviation}^3) \)
Here, the Multiplier is 3.
First Part (LHS): \( 3^2 \times (32 + 2 \times 2) = 9 \times (32 + 4) = 9 \times 36 = 324 \)
Second Part (Middle): \( 3 \times (3 \times 2^2) = 3 \times (3 \times 4) = 3 \times 12 = 36 \)
Third Part (RHS): \( 2^3 = 8 \)
So, we have \( 324 | 36 | 8 \).
Since the base is 10, each part on the right-hand side must hold one digit.
RHS: \( 8 \) (one digit, correct)
Middle: \( 36 \). Keep \( 6 \), carry over \( 3 \) to LHS.
LHS: \( 324 + 3 = 327 \).
Combine the parts.
The cube of 32 is \( 32768 \). This method is powerful for quickly cubing numbers that are multiples of a base plus a small deviation.
In simple words: To cube 32, we use 30 as our helper number. 32 is 2 more than 30. We use a special formula that involves multiplying by 3 (the helper multiplier) and its square. This creates three parts: 324, 36, and 8. Because our helper number (10) has one zero, we need to make sure each part has only one digit on the right. We carry over any extra digits from right to left to get the final answer 32768.

🎯 Exam Tip: When using the sub-base method for cubing, ensure the multiplier is correctly squared for the LHS part and applied normally for the middle part. Also, strictly adhere to the digit count per section as determined by the base.

Question 24. 208
Answer: To find the cube of 208 using the Vedic method (Upa-Sutra "Anurupyena" or sub-base method):
Take the Base as 100 and Sub-base as \( 100 \times 2 = 200 \).
The deviation from the sub-base 200 is:
\( 208 - 200 = +8 \)
The formula for cubing numbers near a sub-base is: \( \text{Multiplier}^2 \times (\text{Number} + 2 \times \text{Deviation}) | \text{Multiplier} \times (3 \times \text{Deviation}^2) | (\text{Deviation}^3) \)
Here, the Multiplier is 2.
First Part (LHS): \( 2^2 \times (208 + 2 \times 8) = 4 \times (208 + 16) = 4 \times 224 = 896 \)
Second Part (Middle): \( 2 \times (3 \times 8^2) = 2 \times (3 \times 64) = 2 \times 192 = 384 \)
Third Part (RHS): \( 8^3 = 512 \)
So, we have \( 896 | 384 | 512 \).
Since the base is 100, each part on the right-hand side must hold two digits.
RHS: \( 512 \). Keep \( 12 \), carry over \( 5 \) to the Middle part.
Middle: \( 384 + 5 = 389 \). Keep \( 89 \), carry over \( 3 \) to the LHS.
LHS: \( 896 + 3 = 899 \).
Combine the parts.
The cube of 208 is \( 8998912 \). This method showcases the structured approach of Vedic mathematics for complex calculations.
In simple words: To cube 208, we use 200 as our helper number. 208 is 8 more than 200. We use a formula that involves multiplying by 2 (the helper multiplier) and its square. This creates three parts: 896, 384, and 512. Because our helper number (100) has two zeros, each part on the right must have two digits. We carry over any extra digits from right to left. After carrying over the 5 from 512 and 3 from 389, we get 8998912.

🎯 Exam Tip: When performing cubing near a sub-base, remember that the multiplier affects the LHS and Middle sections differently. Always make sure to manage the carry-overs correctly based on the number of zeros in the base.

Divide the Following by Using Nikhilam Sutra

Question 25. \( 1245 \div 97 \)
Answer: To divide \( 1245 \) by \( 97 \) using the Nikhilam Sutra:
Divisor = 97.
Base = 100.
Deviation (Complement) from base = \( 100 - 97 = 03 \).
Since the base is 100 (two zeros), we separate the last two digits of the dividend for the remainder section.
Dividend: \( 12 | 45 \)

\[ \begin{array}{c|c c c} \text{Divisor } 97 & 1 & 2 & 45 \\ + \text{Complement } 03 & & 0 & 3 \\ & & & 0 & 6 \\ \hline & 1 & 2 & 81 \\ \end{array} \]
Steps:
1. Bring down the first digit of the dividend (1). This is the first digit of the quotient.
2. Multiply this quotient digit (1) by the complement (03). Write the result (03) starting under the next column (under 2, then 4).
3. Add the digits in the second column (2 + 0 = 2). This is the next digit of the quotient.
4. Multiply this new quotient digit (2) by the complement (03). Write the result (06) starting under the next column (under 4, then 5).
5. Add the digits in the remainder section (45 + 03 + 06 = 81).
The quotient is \( 12 \) and the remainder is \( 81 \).
This Vedic method simplifies division by working with complements from a base.
In simple words: To divide 1245 by 97, we use 100 as a helper. 97 is 3 less than 100, so our helper number is 03. We write down 1, then multiply it by 03 and place the numbers. Then we add the next column, write 2, and multiply it by 03. After adding up the last parts, we find that the answer is 12 with 81 left over.

🎯 Exam Tip: When applying Nikhilam division, correctly identifying the base and its complement is crucial. Also, ensure the remainder section has the same number of digits as the zeros in the base (e.g., two digits for base 100).

Question 26. \( 311 \div 8 \)
Answer: To divide \( 311 \) by \( 8 \) using the Nikhilam Sutra:
Divisor = 8.
Base = 10.
Deviation (Complement) from base = \( 10 - 8 = 2 \).
Since the base is 10 (one zero), we separate the last one digit of the dividend for the remainder section.
Dividend: \( 31 | 1 \)

\[ \begin{array}{c|c c c} \text{Divisor } 8 & 3 & 1 & 1 \\ + \text{Complement } 2 & & 6 & \\ & & & 14 \\ \hline & 3 & 7 & 15 \\ \end{array} \]
Steps:
1. Bring down the first digit of the dividend (3). This is the first digit of the quotient.
2. Multiply this quotient digit (3) by the complement (2). Write the result (6) under the next column (under 1).
3. Add the digits in the second column (1 + 6 = 7). This is the next digit of the quotient.
4. Multiply this new quotient digit (7) by the complement (2). Write the result (14) under the next column (under 1).
5. Add the digits in the remainder section (1 + 14 = 15).
So, initially, the quotient is \( 37 \) and the remainder is \( 15 \).
However, the remainder (15) is greater than the divisor (8). We need to correct this.
Corrected Quotient: Add \( 1 \) to the quotient: \( 37 + 1 = 38 \).
Corrected Remainder: Subtract the divisor from the remainder: \( 15 - 8 = 7 \).
The final quotient is \( 38 \) and the remainder is \( 7 \). This correction step ensures the remainder is always less than the divisor.
In simple words: To divide 311 by 8, we use 10 as our helper. 8 is 2 less than 10, so our helper is 2. We divide the number 311 using a special method, which first gives us an answer of 37 with 15 left over. Since 15 is bigger than 8, we have to adjust: we add 1 to the answer to make it 38, and we subtract 8 from the leftover part to make it 7. So, the final answer is 38 with 7 left over.

🎯 Exam Tip: Always check if the calculated remainder is less than the divisor. If not, perform the correction steps: add 1 to the quotient and subtract the divisor from the remainder until the remainder is smaller than the divisor.

Question 27. \( 1013 \div 88 \)
Answer: To divide \( 1013 \) by \( 88 \) using the Nikhilam Sutra:
Divisor = 88.
Base = 100.
Deviation (Complement) from base = \( 100 - 88 = 12 \).
Since the base is 100 (two zeros), we separate the last two digits of the dividend for the remainder section.
Dividend: \( 10 | 13 \)

\[ \begin{array}{c|c c c} \text{Divisor } 88 & 1 & 0 & 13 \\ + \text{Complement } 12 & & 1 & 2 \\ & & & 1 & 2 \\ \hline & 1 & 1 & 45 \\ \end{array} \]
Steps:
1. Bring down the first digit of the dividend (1). This is the first digit of the quotient.
2. Multiply this quotient digit (1) by the complement (12). Write the result (12) starting under the next column (1 under 0, 2 under 1).
3. Add the digits in the second column (0 + 1 = 1). This is the next digit of the quotient.
4. Multiply this new quotient digit (1) by the complement (12). Write the result (12) starting under the next column (1 under 1, 2 under 3).
5. Add the digits in the remainder section (13 + 2 + 1 + 2 = 45).
The quotient is \( 11 \) and the remainder is \( 45 \).
Since the remainder (45) is less than the divisor (88), no further correction is needed. This elegant method simplifies long division into a series of additions and multiplications.
In simple words: To divide 1013 by 88, we use 100 as our helper number. 88 is 12 less than 100, so our helper is 12. We write down the first digit 1, multiply it by 12, and place the numbers below. Then we add the next column (0+1=1), multiply this new 1 by 12, and place these numbers. Finally, we add up the last section to get the remainder. The answer is 11 with 45 left over.

🎯 Exam Tip: In Nikhilam division, ensure correct placement of multiplied complement digits, shifting one column to the right for each subsequent quotient digit. Be careful with additions, especially in the remainder section.