RBSE Solutions Class 10 Maths Chapter 1 Vedic Mathematics Exercise 1.2

By the method of upsutra yavaduham Tavaduram find the square:

Question 1. 93
Answer: To find the square of 93, we use the Upsutra Yavaduham Tavaduram method. This involves setting a base, determining the deviation, and then performing specific multiplication and addition steps. With a base of 10 and a deviation of +3, the number is first adjusted by the deviation. Then, we combine the product of the number, the deviation, and the square of the deviation.
\( 93^2 = 9(93 + 3)/(3)^2 \)
\( = 9(96)/9 \)
\( = 864/9 \)
Here, the base is 10, and the sub-base is \( 10 \times 9 = 90 \). The calculation combines the product of the sub-base with the sum of the number and deviation, along with the square of the deviation, to form the final result.
\( = 8649 \)
Thus, \( 93^2 = 8649 \).
In simple words: We use a special math trick to square 93. We break down the number, calculate its parts, and combine them. The answer is 8649.

🎯 Exam Tip: Remember to clearly identify the base and deviation for Upsutra Yavaduham Tavaduram method, as these form the foundation of your calculation.

Question 2. 206
Answer: To find the square of 206 using the Upsutra Yavaduham Tavaduram method, we choose a base of 100 and a sub-base digit of 2 (since \( 100 \times 2 = 200 \)). The deviation from the sub-base is +06. This Vedic method simplifies squaring numbers close to a multiple of a power of 10.
\( 206^2 = 2(206 + 06)/(06)^2 \)
\( = 2(212)/36 \)
\( = 424/36 \)
In this step, the right part (36) remains as is because the base is 100 (allowing two digits).
Thus, \( 206^2 = 42436 \).
In simple words: We use a special Vedic math rule to find the square of 206. We follow the steps, using 100 as the base, to get the answer 42436.

🎯 Exam Tip: Pay close attention to the number of digits allowed in each section of the answer, especially when working with bases like 10 or 100, to ensure correct carry-overs.

Question 3. 211
Answer: For 211 squared, we apply the same Vedic math method, Upsutra Yavaduham Tavaduram. With a base of 100 and a sub-base digit of 2, the deviation is +11. This technique efficiently calculates squares of numbers near a base, avoiding lengthy direct multiplication.
\( 211^2 = 2(211 + 11)/(11)^2 \)
\( = 2(222)/121 \)
\( = 444/121 \)
Since the base is 100, the rightmost section should have two digits. We keep '21' and carry over '1' to the middle section.
\( = 444+1/21 \)
\( = 445/21 \)
Thus, \( 211^2 = 44521 \).
In simple words: We use the same quick squaring trick for 211. Following the steps with 100 as the base and carrying over digits, we get 44521 as the square.

🎯 Exam Tip: When the squared deviation results in more digits than allowed by the base, remember to carry over the excess digits to the left side of the calculation.

Question 4. 405
Answer: The solution steps for Question 4 are not provided in the source material. Therefore, we are unable to present a worked solution for this question.
In simple words: The steps to solve this question are not available in the given information.

🎯 Exam Tip: Always double-check that all parts of a question, including instructions and solution steps, are present before attempting to solve it.

Question 5. 16
Answer: To find the square of 16 using a visual and columnar method, we split the number into tens and units digits. This method uses a structured way to perform the multiplication, often simplifying mental arithmetic. We effectively calculate \( (10+6)^2 \).
\( 16^2 = (1)^2 \quad \quad 1 \times 6 \quad \quad (6)^2 \)
\( \quad \quad \quad \quad \quad \quad +6 \) (This indicates doubling the middle product)
\( \quad = 1 \quad \quad 12 \quad \quad 36 \)
Now, we carry over digits from right to left.
\( = 1 \quad \quad (12+3) \quad \quad 6 \) (Carry 3 from 36)
\( = 1 \quad \quad 15 \quad \quad 6 \)
\( = (1+1) \quad \quad 5 \quad \quad 6 \) (Carry 1 from 15)
\( = 2 \quad \quad 5 \quad \quad 6 \)
Thus, \( 16^2 = 256 \).
In simple words: To square 16, we break it into parts, square each part, and combine them carefully, carrying over numbers. The answer is 256.

🎯 Exam Tip: When using columnar methods, always ensure you carry over digits correctly from right to left to avoid calculation errors.

Question 6. 31
Answer: To calculate the square of 31, we follow a similar columnar method as before. We effectively expand \( (30+1)^2 \). This visual technique helps in organizing the intermediate products and simplifies the final summation.
\( 31^2 = (3)^2 \quad \quad 3 \times 1 \quad \quad (1)^2 \)
\( \quad \quad \quad \quad \quad \quad +3 \) (This indicates doubling the middle product)
\( \quad = 9 \quad \quad 6 \quad \quad 1 \)
There are no carry-overs needed in this case, as each section is a single digit.
Thus, \( 31^2 = 961 \).
In simple words: Squaring 31 means multiplying it by itself. Using a step-by-step way, we find that \( 31^2 \) is 961.

🎯 Exam Tip: Practice makes perfect with these columnar methods; visually separating the parts can reduce mistakes in multiplication and addition.

Question 7. 24
Answer: For the square of 24, we again use the columnar method of splitting the number. We are essentially calculating \( (20+4)^2 \). This method breaks down the two-digit multiplication into simpler steps.
\( 24^2 = (2)^2 \quad \quad 2 \times 4 \quad \quad (4)^2 \)
\( \quad \quad \quad \quad \quad \quad +8 \) (This indicates doubling the middle product)
\( \quad = 4 \quad \quad 16 \quad \quad 16 \)
Now, we carry over digits from right to left.
\( = 4 \quad \quad (16+1) \quad \quad 6 \) (Carry 1 from 16)
\( = 4 \quad \quad 17 \quad \quad 6 \)
\( = (4+1) \quad \quad 7 \quad \quad 6 \) (Carry 1 from 17)
\( = 5 \quad \quad 7 \quad \quad 6 \)
Thus, \( 24^2 = 576 \).
In simple words: To get \( 24^2 \), we follow a simple process of squaring parts of the number and combining them. The result is 576.

🎯 Exam Tip: Carefully manage carry-overs, especially when intermediate products are two-digit numbers, to ensure accuracy in the final result.

Question 8. 56
Answer: The complete solution steps for Question 8 are not provided in the source material. However, the final answer is given as \( 56^2 = 3136 \).
In simple words: The steps to solve this question are not available, but the answer for \( 56^2 \) is 3136.

🎯 Exam Tip: Even if only the final answer is provided, try to mentally or quickly work through the steps using the learned Vedic methods to verify it.

By Sutra Ekadhikena Poorvena find the square of the following:

Question 9. 45
Answer: To find the square of 45 using the Sutra Ekadhikena Poorvena method, we use a special rule for numbers ending in 5. This Vedic math trick makes squaring numbers ending in 5 very fast and easy.
We take the digit before 5 (which is 4) and multiply it by the next consecutive digit (4+1=5). This gives the left part of the answer.
\( 4 \times 5 = 20 \)
The right part is always the square of 5.
\( 5 \times 5 = 25 \)
Combining these two parts:
\( 45^2 = 20 / 25 = 2025 \).
In simple words: For numbers ending in 5, like 45, we use a simple trick. Multiply the first digit (4) by the next number (5), which is 20, and put 25 at the end. So, \( 45^2 \) is 2025.

🎯 Exam Tip: This method is highly efficient for numbers ending in 5; always ensure the first part is the product of the first digit and its successor, and the second part is always 25.

Question 10. 85
Answer: Using the same Sutra Ekadhikena Poorvena method for 85 squared, we apply the rule for numbers ending in 5. This principle leverages number patterns for quick calculations.
We multiply the tens digit (8) by its successor (9).
\( 8 \times 9 = 72 \)
Then, we append the square of 5.
\( 5 \times 5 = 25 \)
Combining the parts:
\( 85^2 = 72 / 25 = 7225 \).
In simple words: To square 85, we multiply 8 by 9 (which is 72) and add 25 at the end. The answer is 7225.

🎯 Exam Tip: This Vedic math sutra significantly reduces calculation time for specific types of numbers, making it valuable for competitive exams.

Question 11. 115
Answer: Applying the Ekadhikena Poorvena Sutra to 115, we consider the digits before 5 as '11'. This method works for any number whose last digit is 5, regardless of how many digits precede it.
We multiply 11 by the next consecutive number (12).
\( 11 \times 12 = 132 \)
We then append the square of 5.
\( 5 \times 5 = 25 \)
Combining the parts:
\( 115^2 = 132 / 25 = 13225 \).
In simple words: For \( 115^2 \), we multiply 11 by 12 (to get 132) and put 25 after it. So, \( 115^2 \) is 13225.

🎯 Exam Tip: Practice quickly multiplying consecutive numbers, as this is the core skill required for the first part of the Ekadhikena Poorvena method.

Question 12. 125
Answer: To find the square of 125 using the Ekadhikena Poorvena method, we take the number 12, which is before the 5. This method demonstrates how Vedic mathematics can simplify complex calculations into intuitive steps.
We multiply 12 by the next consecutive number, 13.
\( 12 \times 13 = 156 \)
Then, we simply add 25 (the square of 5) to the end of this result.
\( 5 \times 5 = 25 \)
Combining the parts:
\( 125^2 = 156 / 25 = 15625 \).
In simple words: To square 125, we multiply 12 by 13 (which is 156) and then add 25 to the end. The answer is 15625.

🎯 Exam Tip: Always double-check your multiplication of the leading number with its successor, as this is the largest component of the final answer.

By Sutra Sankaiana - Viyavkalan find the square of the following:

Question 13. 23
Answer: To find the square of 23 using the Sutra Sankalana-Vyavakalan (also known as the "by addition and subtraction" method), we choose a convenient 'base' or auxiliary number close to 23, usually a multiple of 10. Here, we can choose 20 or 30. Let's use 20. The deviation (d) would be 3 (since 23 = 20 + 3). The formula is \( N^2 = (N+d)(N-d) + d^2 \).
We select a deviation of 3.
\( 23^2 = (23 + 3)(23 - 3) + 3^2 \)
\( = (26)(20) + 9 \)
\( = 520 + 9 \)
\( = 529 \)
Thus, \( 23^2 = 529 \). This method effectively uses the algebraic identity \( (a+b)(a-b) = a^2 - b^2 \) in a modified form to simplify squaring.
In simple words: To square 23, we add 3 and subtract 3, then multiply those numbers. Then we add the square of 3. So, we calculate \( 26 \times 20 + 9 = 520 + 9 = 529 \).

🎯 Exam Tip: Choose your deviation carefully; selecting a deviation that makes one of the factors a multiple of 10 simplifies the multiplication step significantly.

Question 15. 69
Answer: To find the square of 69 using the Sutra Sankalana-Vyavakalan method, we choose a convenient deviation. In this case, choosing 9 makes 69 close to 60 and 70 (69-9=60, 69+9=78). This method simplifies squaring by converting it into multiplication of two numbers and adding a square.
We select a deviation of 9.
\( 69^2 = (69 + 9)(69 - 9) + 9^2 \)
\( = (78)(60) + 81 \)
\( = 4680 + 81 \)
\( = 4761 \)
Thus, \( 69^2 = 4761 \). This Vedic principle is useful for numbers near a multiple of 10 or another easy-to-calculate value.
In simple words: To square 69, we use a trick: add 9 and subtract 9 from 69, then multiply the results. Add the square of 9 to that product. The final answer is 4761.

🎯 Exam Tip: Always remember to add the square of the chosen deviation at the end; this is a critical step in the Sankalana-Vyavakalan method.

Question 16. 89
Answer: For \( 89^2 \), we employ the same Sutra Sankalana-Vyavakalan technique. We choose 9 as our deviation to make the numbers easy to multiply (89-9=80, 89+9=98). This method makes calculations more efficient, especially for numbers close to the end of a decade.
We select a deviation of 9.
\( 89^2 = (89 + 9)(89 - 9) + 9^2 \)
\( = (98)(80) + 81 \)
\( = 7840 + 81 \)
\( = 7921 \)
Thus, \( 89^2 = 7921 \). The power of this method lies in transforming a square calculation into a simpler multiplication and addition.
In simple words: To find \( 89^2 \), we add and subtract 9 from 89, multiply those numbers, and add 81. The answer is 7921.

🎯 Exam Tip: Ensure that your chosen deviation simplifies the multiplication of the two factors (N+d) and (N-d) as much as possible for mental calculation ease.

By Dwandwa Yog find the Square:

Question 17. 362
Answer: To find the square of 362 using the Dwandwa Yog (Duplex) method, we break the number into groups of digits and apply the duplex operation. This ancient technique is very efficient for squaring multi-digit numbers.
The groups of digits for 362 are: 3, 36, 362, 62, 2.
We calculate the duplex (D) for each group:
\( D(3) = 3^2 = 9 \)
\( D(36) = 2 \times 3 \times 6 = 36 \)
\( D(362) = (2 \times 3 \times 2) + 6^2 = 12 + 36 = 48 \)
\( D(62) = 2 \times 6 \times 2 = 24 \)
\( D(2) = 2^2 = 4 \)
Combine these duplexes in sequence: \( 9 | 36 | 48 | 24 | 4 \)
Now, we adjust from right to left, carrying over digits based on a base of 10 (one digit per section):
\( \quad \quad 4 \)
\( \quad 24 \implies 4, \text{carry } 2 \)
\( 48+2 = 50 \implies 0, \text{carry } 5 \)
\( 36+5 = 41 \implies 1, \text{carry } 4 \)
\( 9+4 = 13 \)
Thus, \( 362^2 = 131044 \). Each carry-over step simplifies the section and ensures the final answer is correctly formed.
In simple words: To square 362, we use the Dwandwa Yog method. We find special values for different parts of the number and then add them up, carrying over numbers as needed. The final answer is 131044.

🎯 Exam Tip: Be meticulous with your carry-overs in the Dwandwa Yog method; a single mistake can lead to an incorrect final result for multi-digit squares.

Question 18. 453
Answer: To find the square of 453 using the Dwandwa Yog method, we arrange the digits into groups and apply the duplex formula to each group. This systematic approach breaks down large squaring problems into manageable parts.
The groups for 453 are: 4, 45, 453, 53, 3.
We calculate the duplex (D) for each:
\( D(4) = 4^2 = 16 \)
\( D(45) = 2 \times 4 \times 5 = 40 \)
\( D(453) = (2 \times 4 \times 3) + 5^2 = 24 + 25 = 49 \)
\( D(53) = 2 \times 5 \times 3 = 30 \)
\( D(3) = 3^2 = 9 \)
Combine these duplexes in sequence: \( 16 | 40 | 49 | 30 | 9 \)
Now, we adjust from right to left, carrying over digits (base 10, one digit per section):
\( \quad \quad \quad \quad 9 \)
\( \quad \quad \quad 30 \implies 0, \text{carry } 3 \)
\( \quad \quad 49+3 = 52 \implies 2, \text{carry } 5 \)
\( \quad 40+5 = 45 \implies 5, \text{carry } 4 \)
\( 16+4 = 20 \)
Thus, \( 453^2 = 205209 \). The Dwandwa Yog method effectively handles the complexities of squaring larger numbers.
In simple words: We square 453 using the Dwandwa Yog method. We break the number into parts and follow special multiplication and addition rules. After carrying over numbers, the final square is 205209.

🎯 Exam Tip: For longer numbers, clearly writing down each duplex and then systematically performing carry-overs is crucial to avoid errors.

Question 19. 4312
Answer: To find the square of 4312 using the Dwandwa Yog method, we create groups of digits and apply the duplex operation to each. This method efficiently calculates squares by breaking down the calculation based on digit positions.
The seven digit groups for 4312 are: 4, 43, 431, 4312, 312, 12, 2.
We calculate the duplex (D) for each:
\( D(4) = 4^2 = 16 \)
\( D(43) = 2 \times 4 \times 3 = 24 \)
\( D(431) = (2 \times 4 \times 1) + 3^2 = 8 + 9 = 17 \)
\( D(4312) = (2 \times 4 \times 2) + (2 \times 3 \times 1) = 16 + 6 = 22 \)
\( D(312) = (2 \times 3 \times 2) + 1^2 = 12 + 1 = 13 \)
\( D(12) = 2 \times 1 \times 2 = 4 \)
\( D(2) = 2^2 = 4 \)
Combine these duplexes in sequence: \( 16 | 24 | 17 | 22 | 13 | 4 | 4 \)
Now, we adjust from right to left, carrying over digits (base 10, one digit per section):
\( \quad \quad \quad \quad \quad \quad 4 \)
\( \quad \quad \quad \quad \quad 4 \)
\( \quad \quad \quad \quad 13 \implies 3, \text{carry } 1 \)
\( \quad \quad \quad 22+1 = 23 \implies 3, \text{carry } 2 \)
\( \quad \quad 17+2 = 19 \implies 9, \text{carry } 1 \)
\( \quad 24+1 = 25 \implies 5, \text{carry } 2 \)
\( 16+2 = 18 \)
Thus, \( 4312^2 = 18593344 \). This systematic breakdown is a key feature of Vedic mathematics, simplifying complex calculations.
In simple words: To find \( 4312^2 \), we use Dwandwa Yog by splitting the number into groups, finding their special duplex values, and combining them with carry-overs. The answer is 18593344.

🎯 Exam Tip: When dealing with numbers with an even number of digits for Dwandwa Yog, ensure the middle duplex is calculated as twice the product of the two innermost digits.

Question 20. 2456
Answer: To find the square of 2456 using the Dwandwa Yog method, we group the digits and calculate the duplex for each. This method allows for structured and rapid calculation of squares of large numbers.
The seven digit groups for 2456 are: 2, 24, 245, 2456, 456, 56, 6.
We calculate the duplex (D) for each:
\( D(2) = 2^2 = 4 \)
\( D(24) = 2 \times 2 \times 4 = 16 \)
\( D(245) = (2 \times 2 \times 5) + 4^2 = 20 + 16 = 36 \)
\( D(2456) = (2 \times 2 \times 6) + (2 \times 4 \times 5) = 24 + 40 = 64 \)
\( D(456) = (2 \times 4 \times 6) + 5^2 = 48 + 25 = 73 \)
\( D(56) = 2 \times 5 \times 6 = 60 \)
\( D(6) = 6^2 = 36 \)
Combine these duplexes in sequence: \( 4 | 16 | 36 | 64 | 73 | 60 | 36 \)
Now, we adjust from right to left, carrying over digits (base 10, one digit per section):
\( \quad \quad \quad \quad \quad \quad 36 \implies 6, \text{carry } 3 \)
\( \quad \quad \quad \quad \quad 60+3 = 63 \implies 3, \text{carry } 6 \)
\( \quad \quad \quad \quad 73+6 = 79 \implies 9, \text{carry } 7 \)
\( \quad \quad \quad 64+7 = 71 \implies 1, \text{carry } 7 \)
\( \quad \quad 36+7 = 43 \implies 3, \text{carry } 4 \)
\( \quad 16+4 = 20 \implies 0, \text{carry } 2 \)
\( 4+2 = 6 \)
Thus, \( 2456^2 = 6031936 \). Understanding the duplex function for single digits, pairs, and triplets is key to mastering this method.
In simple words: To square 2456, we use Dwandwa Yog by finding special values for each group of digits and then adding them up with careful carry-overs. The final answer is 6031936.

🎯 Exam Tip: For the Dwandwa Yog method, correctly identifying the digit groups and applying the appropriate duplex formula for each (single digit, two digits, three digits, etc.) is the most critical step.

Find cube by Sutra Nikhilam

Question 21. 14
Answer: To find the cube of 14 using the Sutra Nikhilam method, we choose a convenient base, which is 10 in this case. The deviation (d) from the base is +4. The Nikhilam formula for cubes involves three parts: \( (N + 2d) \), \( (3d^2) \), and \( (d^3) \). This method simplifies cubing numbers close to a base.
Here, Base \( = 10 \), Deviation \( d = +4 \).
Part 1: \( N + 2d = 14 + 2(4) = 14 + 8 = 22 \)
Part 2: \( 3d^2 = 3(4^2) = 3(16) = 48 \)
Part 3: \( d^3 = 4^3 = 64 \)
Combine these parts: \( 22 | 48 | 64 \)
Now, adjust from right to left (since the base is 10, each section should ideally have one digit):
\( \quad \quad 64 \implies 4, \text{carry } 6 \)
\( \quad 48+6 = 54 \implies 4, \text{carry } 5 \)
\( 22+5 = 27 \)
Thus, \( 14^3 = 2744 \). This systematic adjustment is crucial to arrive at the correct final answer.
In simple words: To cube 14, we use the Nikhilam method with a base of 10. We calculate three parts using the deviation and then combine them, carrying over numbers to get 2744.

🎯 Exam Tip: For the Nikhilam method, ensure the number of digits carried over or borrowed from each section corresponds to the chosen base (e.g., one digit for base 10, two for base 100).

Question 22. 97
Answer: To find the cube of 97 using the Sutra Nikhilam method, we choose a base of 100. The deviation (d) from the base is \( 97 - 100 = -3 \). The Nikhilam formula for cubes provides a streamlined way to cube numbers near a base.
Here, Base \( = 100 \), Deviation \( d = -03 \).
Part 1: \( N + 2d = 97 + 2(-3) = 97 - 6 = 91 \)
Part 2: \( 3d^2 = 3(-3)^2 = 3(9) = 27 \)
Part 3: \( d^3 = (-3)^3 = -27 \)
Combine these parts: \( 91 | 27 | -27 \)
Now, adjust from right to left (since the base is 100, each section should ideally have two digits):
For the last part, \( -27 \), we borrow from the middle part. We need to add 27 to make it zero, so we borrow 1 from the middle (which represents 100 units).
\( \quad 27-1 | (100 - 27) \)
\( \quad = 26 | 73 \)
Combining all parts: \( 91 | 26 | 73 \)
Thus, \( 97^3 = 912673 \). This efficient method for cubing is particularly helpful for numbers close to powers of ten.
In simple words: To cube 97, we use the Nikhilam method with 100 as the base. We find the deviation, calculate three parts, and then adjust by borrowing from the middle part to handle the negative number. The final answer is 912673.

🎯 Exam Tip: When the deviation cube (\( d^3 \)) is negative, remember to borrow a multiple of the base from the middle section to make it positive, ensuring the correct number of digits per section.

Question 23. 27
Answer: To find the cube of 27, we use the Sutra Nikhilam method with a sub-base.
Here, Base \( = 10 \), Sub-base \( = 20 \) (so sub-base digit \( = 2 \)), Deviation \( d = +7 \).
The formula for cubes with a sub-base is: \( (\text{sub-base digit})^2 (N + 2d) / (\text{sub-base digit} \times 3d^2) / d^3 \)
Part 1: \( 2^2 (27 + 2 \times 7) = 4(27 + 14) = 4(41) = 164 \)
Part 2: \( 2 \times 3 \times 7^2 = 2 \times 3 \times 49 = 6 \times 49 = 294 \)
Part 3: \( 7^3 = 343 \)
Combine these parts: \( 164 | 294 | 343 \)
Now, adjust from right to left (since the base is 10, each section should ideally have one digit, but due to sub-base, we carry over appropriately):
\( \quad \quad \quad 343 \implies 3, \text{carry } 34 \)
\( \quad \quad 294 + 34 = 328 \implies 8, \text{carry } 32 \)
\( 164 + 32 = 196 \)
Thus, \( 27^3 = 19683 \). This method is powerful for cubing numbers that are not direct multiples of 10.
In simple words: To cube 27, we use the Nikhilam method with a sub-base of 20. We calculate three parts using the deviation +7 and the sub-base digit, then combine them by carrying over. The answer is 19683.

🎯 Exam Tip: When using a sub-base, remember to multiply the first part by the square of the sub-base digit and the second part by the sub-base digit itself before combining the sections.

Question 24. Find the cube of 395.
Answer: To find the cube of 395, we use the Sutra Nikhilam method with a sub-base.
Here, Base \( = 100 \), Sub-base \( = 400 \) (so sub-base digit \( = 4 \)), Deviation \( d = 395 - 400 = -5 \).
The formula for cubes with a sub-base is: \( (\text{sub-base digit})^2 (N + 2d) / (\text{sub-base digit} \times 3d^2) / d^3 \)
Part 1: \( 4^2 (395 + 2 \times (-5)) = 16(395 - 10) = 16(385) = 6160 \)
Part 2: \( 4 \times 3 \times (-5)^2 = 4 \times 3 \times 25 = 300 \)
Part 3: \( (-5)^3 = -125 \)
Combine these parts: \( 6160 | 300 | -125 \)
Now, adjust from right to left (since the base is 100, each section should ideally have two digits). For the last part, \( -125 \), we borrow from the middle part. We need to add 125 to make it positive. We borrow 2 from the middle (which represents \( 2 \times 100 = 200 \) units).
\( 6160 | (300 - 2) | (200 - 125) \)
\( = 6160 | 298 | 75 \)
Finally, adjust for any carry-overs to the first part. Here, 298 means 2 'hundreds' are carried to the first part.
\( = (6160 + 2) | 98 | 75 \)
\( = 6162 | 98 | 75 \)
Thus, \( 395^3 = 61629875 \). The use of a negative deviation and sub-base demonstrates the flexibility of Nikhilam Sutra.
In simple words: To cube 395, we use the Nikhilam method with a base of 100 and a deviation of -5. We calculate three parts, then adjust them by borrowing from the middle part and carrying over. The answer is 61629875.

🎯 Exam Tip: When dealing with negative deviations in the Nikhilam method, carefully manage the borrowing process from the preceding sections to ensure all parts are positive before final combination.

By Upsutra Anurupyena find the cube of the following:

Question 25. 16
Answer: To find the cube of 16 using the Upsutra Anurupyena method, we consider the number as \( (1 \quad 6) \). This method involves setting up four columns based on the digits of the number. The columns are formed by \( a^3 | a^2b | ab^2 | b^3 \), where a is the tens digit and b is the units digit.
For 16 (a=1, b=6):
Column I: \( a^3 = 1^3 = 1 \)
Column II: \( a^2b = 1^2 \times 6 = 6 \)
Column III: \( ab^2 = 1 \times 6^2 = 36 \)
Column IV: \( b^3 = 6^3 = 216 \)
Initial combination: \( 1 | 6 | 36 | 216 \)
Now, we double the middle terms (Column II and Column III) and add them to themselves:
\( 6 \times 2 = 12 \)
\( 36 \times 2 = 72 \)
New combination: \( 1 | (6+12) | (36+72) | 216 \)
\( = 1 | 18 | 108 | 216 \)
Finally, adjust from right to left (base 10, one digit per section):
\( \quad \quad \quad 216 \implies 6, \text{carry } 21 \)
\( \quad \quad 108+21 = 129 \implies 9, \text{carry } 12 \)
\( \quad 18+12 = 30 \implies 0, \text{carry } 3 \)
\( 1+3 = 4 \)
Thus, \( 16^3 = 4096 \). This structured approach makes cubing two-digit numbers systematic.
In simple words: To cube 16, we use the Anurupyena method. We break the number into parts, calculate each column, double the middle parts, and then add them up with proper carrying over. The answer is 4096.

🎯 Exam Tip: Ensure that you correctly double only the middle two terms (\( a^2b \) and \( ab^2 \)) and not the extreme terms (\( a^3 \) and \( b^3 \)) in the Anurupyena method.

Question 26. Find the cube of 33.
Answer: To find the cube of 33 using the Upsutra Anurupyena method, we treat the number as having digits \( a=3 \) and \( b=3 \). This method is especially straightforward when the digits of the number are the same.
The four columns are based on \( a^3 | a^2b | ab^2 | b^3 \):
Column I: \( a^3 = 3^3 = 27 \)
Column II: \( a^2b = 3^2 \times 3 = 27 \)
Column III: \( ab^2 = 3 \times 3^2 = 27 \)
Column IV: \( b^3 = 3^3 = 27 \)
Initial combination: \( 27 | 27 | 27 | 27 \)
Now, we double the middle terms (Column II and Column III) and add them to themselves:
\( 27 \times 2 = 54 \)
New combination: \( 27 | (27+54) | (27+54) | 27 \)
\( = 27 | 81 | 81 | 27 \)
Finally, adjust from right to left (base 10, one digit per section):
\( \quad \quad \quad 27 \implies 7, \text{carry } 2 \)
\( \quad \quad 81+2 = 83 \implies 3, \text{carry } 8 \)
\( \quad 81+8 = 89 \implies 9, \text{carry } 8 \)
\( 27+8 = 35 \)
Thus, \( 33^3 = 35937 \). This method is an elegant way to cube two-digit numbers without direct multiplication.
In simple words: To cube 33, we use the Anurupyena method. We set up four parts (each based on 3 cubed or its combinations), double the two middle parts, and then carry over numbers to get 35937.

🎯 Exam Tip: When both digits of the number are the same, the four initial parts of the Anurupyena method will be identical, simplifying the setup process.

Question 27. 41
Answer: To find the cube of 41 using the Upsutra Anurupyena method, we consider the number as \( (4 \quad 1) \). Let \( a=4 \) and \( b=1 \). This method structures the cubing process into distinct, manageable steps.
The four columns are based on \( a^3 | a^2b | ab^2 | b^3 \):
Column I: \( a^3 = 4^3 = 64 \)
Column II: \( a^2b = 4^2 \times 1 = 16 \)
Column III: \( ab^2 = 4 \times 1^2 = 4 \)
Column IV: \( b^3 = 1^3 = 1 \)
Initial combination: \( 64 | 16 | 4 | 1 \)
Now, we double the middle terms (Column II and Column III) and add them to themselves:
\( 16 \times 2 = 32 \)
\( 4 \times 2 = 8 \)
New combination: \( 64 | (16+32) | (4+8) | 1 \)
\( = 64 | 48 | 12 | 1 \)
Finally, adjust from right to left (base 10, one digit per section):
\( \quad \quad \quad 1 \)
\( \quad \quad 12 \implies 2, \text{carry } 1 \)
\( \quad 48+1 = 49 \implies 9, \text{carry } 4 \)
\( 64+4 = 68 \)
Thus, \( 41^3 = 68921 \). The Anurupyena method is an efficient way to cube two-digit numbers systematically.
In simple words: To cube 41, we use the Anurupyena method. We set up four parts based on the digits 4 and 1, double the middle parts, and then carry over numbers. The answer is 68921.

🎯 Exam Tip: When cubing numbers using Anurupyena, clearly distinguish between the "initial combination" and the "doubled middle terms" step to avoid confusion during calculation.

Question 28. 52
Answer: To find the cube of 52, we can use the Sutra Nikhilam method with a sub-base.
Here, we choose a Base \( = 50 \) (implicitly, the actual base is 10, and 50 is a sub-base), Sub-base digit \( = 5 \) (since \( 50 = 5 \times 10 \)), Deviation \( d = +2 \) (since \( 52 - 50 = 2 \)).
The formula for cubes with a sub-base is: \( (\text{sub-base digit})^2 (N + 2d) / (\text{sub-base digit} \times 3d^2) / d^3 \)
Part 1: \( 5^2 (52 + 2 \times 2) = 25(52 + 4) = 25(56) = 1400 \)
Part 2: \( 5 \times 3 \times 2^2 = 5 \times 3 \times 4 = 60 \)
Part 3: \( 2^3 = 8 \)
Combine these parts: \( 1400 | 60 | 8 \)
Now, adjust from right to left (since we are working with base 10 for the carry-overs, one digit per section for the smaller parts):
\( \quad \quad 8 \)
\( \quad 60 \implies 0, \text{carry } 6 \)
\( 1400+6 = 1406 \)
Thus, \( 52^3 = 140608 \). This method is an excellent alternative for cubing two-digit numbers whose units digit is small.
In simple words: To cube 52, we use the Nikhilam method with 50 as a special base. We calculate three parts using the deviation +2 and the sub-base digit 5. Then we combine these parts by carrying over numbers. The answer is 140608.

🎯 Exam Tip: Choosing the nearest convenient multiple of 10 (like 50 for 52, 20 for 24) as a sub-base is key to making Nikhilam calculations simpler and faster.

Find cube by sutra Ekadhikene purvena

Formula: \( (\text{Square of ten's digit}) \times (\text{number} + \text{deviation}) / (\text{ten's digit} \times 3 \times \text{deviation}^2) / (\text{deviation})^3 \)
where deviation \( = (\text{unit digit} \times 3) - 10 \)

Question 29. 45
Answer: To find the cube of 45 using the Sutra Ekadhikena Poorvena method for cubes, we identify the ten's digit as 4 and the unit's digit as 5. This method, specific to Vedic mathematics, simplifies the cubing process through a formula based on deviations.
First, calculate the deviation:
Deviation \( = (5 \times 3) - 10 = 15 - 10 = 5 \)
Now, apply the formula to find the three parts:
Part 1: \( (\text{Ten's digit})^2 \times (\text{Number} + \text{deviation}) = 4^2 \times (45 + 5) = 16 \times 50 = 800 \)
Part 2: \( \text{Ten's digit} \times 3 \times (\text{deviation})^2 = 4 \times 3 \times 5^2 = 4 \times 3 \times 25 = 300 \)
Part 3: \( (\text{deviation})^3 = 5^3 = 125 \)
Combine these parts: \( 800 | 300 | 125 \)
Now, adjust from right to left (as per the specific Ekadhikena Poorvena cube rule):
\( \quad \quad \quad 125 \implies 5, \text{carry } 12 \)
\( \quad \quad 300+12 = 312 \implies 2, \text{carry } 31 \)
\( 800+31 = 831 \)
Thus, \( 45^3 = 83125 \). The Ekadhikena Poorvena Sutra can effectively handle cubing through a structured approach.
In simple words: To cube 45, we use a special formula. We first find the deviation, then calculate three parts, and finally combine them by carrying over numbers. The answer is 83125.

🎯 Exam Tip: Accurately calculating the 'deviation' using the given sub-formula is the initial and crucial step for the Ekadhikena Poorvena method for cubes.

Question 30. 73
Answer: To find the cube of 73, we use the Sutra Nikhilam method with a sub-base.
Here, Base \( = 100 \), Sub-base \( = 100 \) (so sub-base digit \( = 1 \)), Deviation \( d = 73 - 100 = -27 \).
The formula for cubes with a sub-base is: \( (\text{sub-base digit})^2 (N + 2d) / (\text{sub-base digit} \times 3d^2) / d^3 \)
Part 1: \( 1^2 (73 + 2 \times (-27)) = 1(73 - 54) = 19 \)
Part 2: \( 1 \times 3 \times (-27)^2 = 3 \times 729 = 2187 \)
Part 3: \( (-27)^3 = -19683 \)
Combine these parts: \( 19 | 2187 | -19683 \)
Now, adjust from right to left (since the base is 100, each section should ideally have two digits). For \( -19683 \), we borrow from the middle part. We need to add 19683 to make it positive, so we borrow 197 from the middle (which represents \( 197 \times 100 = 19700 \) units).
\( 19 | (2187 - 197) | (19700 - 19683) \)
\( = 19 | 1990 | 17 \)
Adjust again for carry-overs to the first part. Here, 1990 means 19 'hundreds' are carried to the first part.
\( = (19 + 19) | 90 | 17 \)
\( = 38 | 90 | 17 \)
Thus, \( 73^3 = 389017 \). This method is highly efficient for numbers close to a base, including those requiring negative deviations.
In simple words: To cube 73, we use the Nikhilam method with 100 as the base and a deviation of -27. We calculate three parts, borrow from the middle to handle the negative cube, and carry over digits to get 389017.

🎯 Exam Tip: When using Nikhilam with a large negative deviation, be very careful with the borrowing calculations in the middle and last sections to avoid sign and magnitude errors.

Question 31. 24
Answer: To find the cube of 24 using the Sutra Nikhilam method with a sub-base.
Here, Base \( = 10 \), Sub-base \( = 20 \) (so sub-base digit \( = 2 \)), Deviation \( d = +4 \) (since \( 24 - 20 = 4 \)).
The formula for cubes with a sub-base is: \( (\text{sub-base digit})^2 (N + 2d) / (\text{sub-base digit} \times 3d^2) / d^3 \)
Part 1: \( 2^2 (24 + 2 \times 4) = 4(24 + 8) = 4(32) = 128 \)
Part 2: \( 2 \times 3 \times 4^2 = 2 \times 3 \times 16 = 96 \)
Part 3: \( 4^3 = 64 \)
Combine these parts: \( 128 | 96 | 64 \)
Now, adjust from right to left (since the base is 10, each section should ideally have one digit):
\( \quad \quad \quad 64 \implies 4, \text{carry } 6 \)
\( \quad \quad 96+6 = 102 \implies 2, \text{carry } 10 \)
\( 128+10 = 138 \)
Thus, \( 24^3 = 13824 \). This method is especially useful for numbers slightly above a multiple of 10.
In simple words: To cube 24, we use the Nikhilam method with a sub-base of 20. We calculate three parts using the deviation +4 and the sub-base digit 2. Then, we combine these parts by carrying over numbers. The answer is 13824.

🎯 Exam Tip: Remember to correctly multiply the middle section by the sub-base digit and the first section by the square of the sub-base digit before carrying over.

Question 32. 106
Answer: To find the cube of 106 using the Sutra Nikhilam method, we choose a base of 100. The deviation (d) from the base is \( +6 \). The Nikhilam formula for cubes provides a straightforward way to cube numbers close to a base.
Here, Base \( = 100 \), Deviation \( d = +6 \).
Part 1: \( N + 2d = 106 + 2 \times 6 = 106 + 12 = 118 \)
Part 2: \( 3d^2 = 3 \times 6^2 = 3 \times 36 = 108 \)
Part 3: \( d^3 = 6^3 = 216 \)
Combine these parts: \( 118 | 108 | 216 \)
Now, adjust from right to left (since the base is 100, each section should ideally have two digits):
\( \quad \quad \quad 216 \implies 16, \text{carry } 2 \)
\( \quad \quad 108+2 = 110 \implies 10, \text{carry } 1 \)
\( 118+1 = 119 \)
Thus, \( 106^3 = 1191016 \). This method highlights the efficiency of Vedic mathematics for calculations involving numbers near powers of ten.
In simple words: To cube 106, we use the Nikhilam method with 100 as the base. We find the deviation, calculate three parts, and then combine them, carrying over numbers while keeping two digits in each section. The answer is 1191016.

🎯 Exam Tip: Ensure that the intermediate sections (Part 2 and Part 3) are always represented with the correct number of digits corresponding to the chosen base (e.g., two digits for base 100) before final adjustment.