RBSE Solutions Class 9 Maths Chapter 1 Vedic Mathematics More Ques

Mathematics Additional Questions

 

Question 1. (i) By Sutra Ekadhikena Purvena, add the following:
(a) 37995
68986
(b) km m
28 084
32 365
Answer:
(a) We can add these numbers using the Sutra Ekadhikena Purvena method. This method is used to add numbers by adjusting digits to avoid carrying. Here, we perform direct column-wise addition.
\( 37995 + 68986 = 106981 \)
(b) We add the kilometers and meters separately. This Vedic method helps simplify multi-digit addition by managing carries efficiently.

In simple words: For part (a), we just add the two numbers together to get the total. For part (b), we add the kilometers and meters separately, then combine the results.

🎯 Exam Tip: Always align numbers correctly by their place value before adding, especially when dealing with units like km and m, to ensure accurate calculations.

 

Question 1. (ii) By vedic method, subtract the following:
(a) 5652
2834
(b) hr. min. sec.
14 11 11
4 09 12
Answer:
(a) We subtract these numbers using the Vedic subtraction method, which can involve complements or direct subtraction.

(b) We subtract the time units (hours, minutes, seconds) separately, borrowing from higher units when necessary. Remember that 1 minute has 60 seconds and 1 hour has 60 minutes.

In simple words: For (a), we subtract the bottom number from the top number. For (b), we subtract hours from hours, minutes from minutes, and seconds from seconds, borrowing from the left if a number is too small.

🎯 Exam Tip: When subtracting time, always ensure you convert units correctly when borrowing (e.g., 1 minute = 60 seconds) to avoid errors.

 

Question 1. (iii) By Sutra Ekadhikena Purvena, multiply the following:
(a) 586 x 514
(b) 3993 x 3007
Answer:
(a) We use Sutra Ekadhikena Purvena, where the first digits are the same and the sum of the last digits is a power of 10. Here, the first digits are 5, and the sum of the units digits (6+4) is 10.
\( 586 \times 514 \)
The left part is \( 5 \times (5+1) = 5 \times 6 = 30 \).
The right part is \( 86 \times 14 = 1204 \).
Since our base is 100, the right part should have 2 digits. The product of the full numbers is obtained by combining these parts directly.
\( = 30 / 1204 \)
\( = 301204 \)
(b) We apply the same Sutra Ekadhikena Purvena. Here, the first digit is 3, and the sum of the last two parts (993 + 7) is 1000.
\( 3993 \times 3007 \)
The left part is \( 3 \times (3+1) = 3 \times 4 = 12 \).
The right part is \( 993 \times 007 = 6951 \).
Since our base is 1000, the right part should have 3 digits. We need to ensure the right side has enough zeros if the product is small.
\( = 12 / 006951 \)
\( = 12006951 \)
In simple words: We multiply numbers where the first parts are the same and the last parts add up to 10 or 100 or 1000. We multiply the first part by one more than itself, and then multiply the last parts together.

🎯 Exam Tip: This Sutra applies when the sum of the last digits (or groups of digits) of the numbers equals a power of 10 (10, 100, 1000, etc.) and the preceding digits are identical. Ensure the right part has the correct number of digits based on the power of 10 used.

 

Question 2. (i) Multiply with the help of Sutra Ekanyunena Purvena:
(a) 8567 x 9999
(b) 512 x 99
Answer:
(a) The Sutra Ekanyunena Purvena method is used when one number consists entirely of nines. We subtract 1 from the multiplicand to get the left part of the answer, and then subtract this result from the sequence of nines to get the right part.
\( 8567 \times 9999 \)
Left Hand Side (L.H.S.) = \( 8567 - 1 = 8566 \)
Right Hand Side (R.H.S.) = \( 9999 - 8566 = 1433 \)
So, \( 8567 \times 9999 = 85661433 \)
(b) We apply the same Sutra Ekanyunena Purvena for this multiplication.
\( 512 \times 99 \)
Left Hand Side (L.H.S.) = \( 512 - 1 = 511 \)
Right Hand Side (R.H.S.) = \( 99 - 511 \)
When the multiplicand (512) has more digits than the multiplier (99), we adjust by writing the multiplicand followed by as many zeros as there are nines in the multiplier, then subtract the multiplicand itself from this.
\( = 51100 + (99 - 511) \)
This can be thought of as \( 512 \times (100-1) = 51200 - 512 \).
\( = 51199 - 511 \)
\( = 50688 \)
In simple words: When multiplying by nines (like 99, 999), we take one less than the other number for the first part of the answer. Then, we subtract this first part from the nines for the second part.

🎯 Exam Tip: Remember to handle cases where the multiplicand has more digits than the number of nines correctly by using the 'all from 9, last from 10' concept or by simple subtraction from a base power.

 

Question 2. (ii) By vedic method, multiply the following:
(a) \( 11\frac {1}{3} \times 11\frac {2}{3} \)
(b) \( 1.15 \times 1.85 \)
Answer:
(a) We use the Nikhilam Sutra here, as the integral parts are the same (11) and the fractional parts sum to 1. This method simplifies the multiplication of mixed numbers or decimals.
\( 11\frac {1}{3} \times 11\frac {2}{3} \)
First, we multiply the integer part by one more than itself: \( 11 \times (11+1) = 11 \times 12 = 132 \).
Next, we multiply the fractional parts: \( \frac {1}{3} \times \frac {2}{3} = \frac {2}{9} \).
We combine these results. Notice that the sum of the fractional parts is \( \frac {1}{3} + \frac {2}{3} = 1 \).
\( = 132\frac {2}{9} \)
(b) We use a similar Vedic method for decimal multiplication where the sum of the decimal parts is a multiple of 10 or 100, and the integral parts are specific. Here, the integral parts are both 1, and the decimal parts sum to 1.
\( 1.15 \times 1.85 \)
I. Sum of the decimal parts = \( 0.15 + 0.85 = 1 \)
II. Multiply the integral parts, adding one to the first: \( 1 \times (1+1) = 2 \). This forms the left part of the answer.
III. Multiply the decimal parts: \( 0.15 \times 0.85 = 0.1275 \). This forms the right part.
Combining these, we get the product.
\( = 2.1275 \)
In simple words: For numbers like \( 11 \frac{1}{3} \), if the whole numbers are the same and the fractions add up to 1, you multiply the whole number by one more than itself, and then multiply the fractions. For decimals, if the whole numbers are the same and the decimal parts add up to 1, you do a similar process.

🎯 Exam Tip: Always check if the sum of the fractional or decimal parts equals 1 (or 10, or 100) and if the integral parts are identical; this indicates the applicability of specific Vedic multiplication Sutras.

 

Question 3. (i) Change the following into vinculum number:
(a) 898
(b) 18469
Answer:
(a) A vinculum number uses negative digits (represented by a bar over the digit) to express numbers more compactly or for easier calculation. We convert digits greater than 5 to their complement from 10, and carry 1 to the left.
\( 898 \)
Starting from the right, \( 8 = 10 - \overline{2} \), so we write \( \overline{2} \) and carry 1 to the 9.
\( 9+1 = 10 = 10 - \overline{0} \), so we write \( \overline{0} \) and carry 1 to the 8.
\( 8+1 = 9 \).
So, \( 898 = 9\overline{0}\overline{2} \)
(b) We apply the same vinculum conversion rules to this number.
\( 18469 \)
From right: \( 9 = 10 - \overline{1} \), carry 1 to 6. So \( 6+1=7 \). (No bar)
\( 7 \) (from 6+1) is \( 10 - \overline{3} \), carry 1 to 4. So \( 4+1=5 \). (No bar)
\( 5 \) is \( 10 - \overline{5} \), carry 1 to 8. So \( 8+1=9 \).
\( 9 \) is \( 10 - \overline{1} \), carry 1 to 1. So \( 1+1=2 \).
Thus, \( 18469 = 2\overline{1}5\overline{3}\overline{1} \)
In simple words: To change a number into vinculum form, we look for digits 5 or larger. We change them to "10 minus that digit" and put a bar on it, then add 1 to the digit on its left.

🎯 Exam Tip: When converting to vinculum numbers, remember to start from the rightmost digit and convert any digit greater than 5 by subtracting it from 10 and carrying over 1 to the next digit, placing a bar over the new digit.

 

Question 3. (ii) Change the following into ordinary number:
(a) \( 6\overline{2}45\overline{3}\overline{2} \)
(b) \( 4\overline{1}2\overline{3}\overline{3} \)
Answer:
(a) To convert a vinculum number to an ordinary number, we perform subtraction for digits with a bar and addition for normal digits, moving from right to left.
\( 6\overline{2}45\overline{3}\overline{2} \)
Starting from the rightmost digit: \( \overline{2} \) means \( 10-2=8 \). This means the digit to its left (3) reduces by 1, making it 2.
Next, \( \overline{3} \) (now 2) means \( 10-2=8 \). This means the digit to its left (5) reduces by 1, making it 4.
Next, 5 (now 4) remains 4.
Next, 4 remains 4.
Next, \( \overline{2} \) means \( 10-2=8 \). This means the digit to its left (6) reduces by 1, making it 5.
So, the number is \( 584488 \).
(The source shows \( 576468 \), which indicates a possible different interpretation of the vinculum placement or a calculation error in the source example.)
(b) We apply the same method to convert this vinculum number to its ordinary form.
\( 4\overline{1}2\overline{3}\overline{3} \)
Starting from the right: \( \overline{3} \) means \( 10-3=7 \). The digit to its left (3) reduces by 1, making it 2.
Next, \( \overline{3} \) (now 2) means \( 10-2=8 \). The digit to its left (2) reduces by 1, making it 1.
Next, 2 (now 1) remains 1.
Next, \( \overline{1} \) means \( 10-1=9 \). The digit to its left (4) reduces by 1, making it 3.
So, the number is \( 39187 \).
(The source shows \( 38767 \), again possibly indicating a different interpretation of the vinculum placement or a calculation error.)
In simple words: To change a vinculum number back to a normal number, we read from right to left. If a digit has a bar, we subtract it from 10, and then subtract 1 from the digit on its left. If it has no bar, we keep it as is.

🎯 Exam Tip: Pay close attention to how the vinculum bar is placed over specific digits, as this affects which digits are subtracted from 10 and which preceding digits are decremented. Practice this conversion carefully.

 

Question 4. (i) Multiply the following by Sutra Nikhilam (base method):
(a) 92 x 87
(b) 1007 x 1012
(c) 103 x 105 x 106
(d) 12 x 13 x 15
Answer:
(a) We use Sutra Nikhilam, taking 100 as the base. Deviations are \( 92 - 100 = -8 \) and \( 87 - 100 = -13 \).
\( 92 \times 87 \)
Left part: Cross-subtract one deviation from the other number (e.g., \( 92 - 13 = 79 \)).
Right part: Multiply the deviations (\( -8 \times -13 = 104 \)).
Since the base (100) has two zeros, the right part should have two digits. We carry over any extra digits.
\( = 79 / 104 \)
\( = (79+1) / 04 \)
\( = 8004 \)
(b) For 1007 x 1012, the base is 1000. Deviations are \( 1007 - 1000 = +007 \) and \( 1012 - 1000 = +012 \).
\( 1007 \times 1012 \)
Left part: Add one deviation to the other number (e.g., \( 1007 + 012 = 1019 \)).
Right part: Multiply the deviations (\( 007 \times 012 = 084 \)).
Since the base (1000) has three zeros, the right part should have three digits.
\( = 1019 / 084 \)
\( = 1019084 \)
(c) For 103 x 105 x 106, the base is 100. Deviations are \( +3, +5, +6 \). The calculation for three numbers is an extension of the two-number method.
\( 103 \times 105 \times 106 \)
Left part: \( (103 + 5 + 6) = 114 \).
Middle part: Sum of products of deviations taken two at a time: \( (3 \times 5) + (5 \times 6) + (6 \times 3) = 15 + 30 + 18 = 63 \).
Right part: Product of all deviations: \( (3 \times 5 \times 6) = 90 \).
Since the base (100) has two zeros, each part should have two digits.
\( = 114 / 63 / 90 \)
\( = 1146390 \)
(d) For 12 x 13 x 15, the base is 10. Deviations are \( +2, +3, +5 \). This also uses the three-number multiplication formula.
\( 12 \times 13 \times 15 \)
Left part: \( (12 + 3 + 5) = 20 \).
Middle part: Sum of products of deviations taken two at a time: \( (2 \times 3) + (3 \times 5) + (5 \times 2) = 6 + 15 + 10 = 31 \).
Right part: Product of all deviations: \( (2 \times 3 \times 5) = 30 \).
Since the base (10) has one zero, each part should have one digit. We carry over any extra digits.
\( = 20 / 31 / 30 \)
\( = 20 / (31+3) / 0 \) (carry 3 from 30)
\( = (20+3) / 4 / 0 \) (carry 3 from 34)
\( = 2340 \)
In simple words: When multiplying numbers close to a base (like 10, 100, 1000), we find how much they are over or under the base. Then we add/subtract these differences and multiply them to get the answer.

🎯 Exam Tip: When using Nikhilam Sutra, ensure the number of digits in the right-hand part of the answer matches the number of zeros in the chosen base; carry over any excess digits to the left-hand part.

 

Question 4. (ii) Multiply the following by Sutra Nikhilam (sub-base method):
(a) 54 x 56
(b) 206 x 212
(c) 21 x 24 x 25
(d) 502 x 503 x 504
Answer:
(a) For 54 x 56, we use a sub-base of 50 (Base 10, Sub-base multiplier 5). Deviations are \( 54 - 50 = +4 \) and \( 56 - 50 = +6 \). This method is useful when numbers are close to multiples of a base.
\( 54 \times 56 \)
Left part: \( \text{Sub-base multiplier} \times (\text{Number} + \text{other deviation}) = 5 \times (54 + 6) = 5 \times 60 = 300 \).
Right part: Multiply deviations \( 4 \times 6 = 24 \).
Since the base (10) has one zero, the right part should have one digit. We carry over any extra digits.
\( = 300 / 24 \)
\( = (300+2) / 4 \) (carry 2 from 24)
\( = 3024 \)
(b) For 206 x 212, the sub-base is 200 (Base 100, Sub-base multiplier 2). Deviations are \( 206 - 200 = +06 \) and \( 212 - 200 = +12 \).
\( 206 \times 212 \)
Left part: \( \text{Sub-base multiplier} \times (\text{Number} + \text{other deviation}) = 2 \times (206 + 12) = 2 \times 218 = 436 \).
Right part: Multiply deviations \( 06 \times 12 = 72 \).
Since the base (100) has two zeros, the right part should have two digits.
\( = 436 / 72 \)
\( = 43672 \)
(c) For 21 x 24 x 25, the sub-base is 20 (Base 10, Sub-base multiplier 2). Deviations are \( +1, +4, +5 \). This requires extending the sub-base method for three numbers.
\( 21 \times 24 \times 25 \)
Left part: \( (\text{Sub-base multiplier})^2 \times (\text{Number} + \text{sum of other deviations}) = 2^2 \times (21 + 4 + 5) = 4 \times 30 = 120 \).
Middle part: \( \text{Sub-base multiplier} \times (\text{sum of products of deviations taken two at a time}) = 2 \times ((1 \times 4) + (4 \times 5) + (5 \times 1)) = 2 \times (4 + 20 + 5) = 2 \times 29 = 58 \).
Right part: Product of all deviations: \( (1 \times 4 \times 5) = 20 \).
Since the base (10) has one zero, each part should have one digit. We carry over any extra digits.
\( = 120 / 58 / 20 \)
\( = 120 / (58+2) / 0 \) (carry 2 from 20)
\( = (120+6) / 0 / 0 \) (carry 6 from 60)
\( = 12600 \)
(d) For 502 x 503 x 504, the sub-base is 500 (Base 100, Sub-base multiplier 5). Deviations are \( +02, +03, +04 \). This uses the extended sub-base method for three numbers.
\( 502 \times 503 \times 504 \)
Left part: \( (\text{Sub-base multiplier})^2 \times (\text{Number} + \text{sum of other deviations}) = 5^2 \times (502 + 03 + 04) = 25 \times 509 = 12725 \).
Middle part: \( \text{Sub-base multiplier} \times (\text{sum of products of deviations taken two at a time}) = 5 \times ((2 \times 3) + (3 \times 4) + (4 \times 2)) = 5 \times (6 + 12 + 8) = 5 \times 26 = 130 \).
Right part: Product of all deviations: \( (2 \times 3 \times 4) = 24 \).
Since the base (100) has two zeros, each part should have two digits. We carry over any extra digits.
\( = 12725 / 130 / 24 \)
\( = 12725 / (130) / 24 \) (no carry from 24)
\( = (12725+1) / 30 / 24 \) (carry 1 from 130 after 30 remains)
\( = 127263024 \)
In simple words: If numbers are near a multiple of a base (like 50, 200), we pick that multiple as a sub-base. Then we adjust the numbers by how far they are from the sub-base, multiply, and combine the parts, remembering to apply the sub-base multiplier to the first part.

🎯 Exam Tip: Clearly identify the base and the sub-base multiplier. For three numbers, remember the square of the sub-base multiplier for the leftmost part and the sub-base multiplier for the middle part.

 

Question 5. (i) Find the square of the following by Sutra Nikhilam (base method):
(a) 17
(b) 104
Answer:
(a) To find the square of 17 using Sutra Nikhilam, we use 10 as the base. The deviation is \( 17 - 10 = +7 \). This method is a quick way to square numbers close to a base.
\( 17^2 = (\text{Number} + \text{Deviation}) / (\text{Deviation})^2 \)
\( = (17 + 7) / 7^2 \)
\( = 24 / 49 \)
Since the base (10) has one zero, the right part (49) should have one digit. We carry over the extra digits.
\( = (24+4) / 9 \) (carry 4 from 49)
\( = 289 \)
(b) To find the square of 104, we use 100 as the base. The deviation is \( 104 - 100 = +04 \). This method is very efficient for numbers slightly above or below a convenient base.
\( 104^2 = (\text{Number} + \text{Deviation}) / (\text{Deviation})^2 \)
\( = (104 + 04) / (04)^2 \)
\( = 108 / 16 \)
Since the base (100) has two zeros, the right part (16) should have two digits. No carry is needed here.
\( = 10816 \)
In simple words: To square a number near a base (like 10 or 100), we add its difference from the base to itself for the first part. For the second part, we square that difference. Then we combine these two parts, carrying over digits if needed.

🎯 Exam Tip: For squares using Nikhilam Sutra, ensure the number of digits in the right-hand part of the answer matches the number of zeros in the chosen base.

 

Question 5. (ii) Find the square of the following by Sutra Nikhilam (sub-base method):
(a) 64
(b) 308
Answer:
(a) To find the square of 64 using the sub-base method, we choose a sub-base of 60 (Base 10, Sub-base multiplier 6). The deviation is \( 64 - 60 = +4 \). This allows squaring numbers that are multiples of a base.
\( 64^2 = (\text{Sub-base multiplier} \times (\text{Number} + \text{Deviation})) / (\text{Deviation})^2 \)
\( = (6 \times (64 + 4)) / 4^2 \)
\( = (6 \times 68) / 16 \)
\( = 408 / 16 \)
Since the base (10) has one zero, the right part (16) should have one digit. We carry over the extra digits.
\( = (408+1) / 6 \) (carry 1 from 16)
\( = 4096 \)
(b) To find the square of 308, we use a sub-base of 300 (Base 100, Sub-base multiplier 3). The deviation is \( 308 - 300 = +08 \). This method is efficient for squaring numbers near multiples of 100.
\( 308^2 = (\text{Sub-base multiplier} \times (\text{Number} + \text{Deviation})) / (\text{Deviation})^2 \)
\( = (3 \times (308 + 08)) / (08)^2 \)
\( = (3 \times 316) / 64 \)
\( = 948 / 64 \)
Since the base (100) has two zeros, the right part (64) should have two digits. No carry is needed here.
\( = 94864 \)
In simple words: For squaring numbers near a multiple of 10 (like 60 or 300), we first find the difference from that multiple. Then, we multiply the number plus this difference by the multiple itself for the first part, and square the difference for the second part.

🎯 Exam Tip: For the sub-base method, carefully determine the correct sub-base multiplier and apply it only to the left-hand side calculation. Ensure the right side maintains the correct number of digits relative to the primary base.

 

Question 6. (i) Find the cube of the following by Sutra Nikhilam (base method):
(a) 12
(b) 105
Answer:
(a) To find the cube of 12 using Sutra Nikhilam, we use 10 as the base. The deviation is \( 12 - 10 = +2 \). The method for cubing involves three parts, based on the number and its deviation.
\( 12^3 = (\text{Number} + 2 \times \text{Deviation}) / (3 \times (\text{Deviation})^2) / ((\text{Deviation})^3) \)
\( = (12 + 2 \times 2) / (3 \times 2^2) / 2^3 \)
\( = (12 + 4) / (3 \times 4) / 8 \)
\( = 16 / 12 / 8 \)
Since the base (10) has one zero, each part should have one digit. We carry over any extra digits.
\( = 16 / (12+0) / 8 \) (no carry from 8)
\( = (16+1) / 2 / 8 \) (carry 1 from 12)
\( = 1728 \)
(b) To find the cube of 105, we use 100 as the base. The deviation is \( 105 - 100 = +05 \). This is a fast way to cube numbers close to 100.
\( 105^3 = (\text{Number} + 2 \times \text{Deviation}) / (3 \times (\text{Deviation})^2) / ((\text{Deviation})^3) \)
\( = (105 + 2 \times 05) / (3 \times (05)^2) / (05)^3 \)
\( = (105 + 10) / (3 \times 25) / 125 \)
\( = 115 / 75 / 125 \)
Since the base (100) has two zeros, each part should have two digits. We carry over any extra digits.
\( = 115 / (75+1) / 25 \) (carry 1 from 125)
\( = 115 / 76 / 25 \)
\( = 1157625 \)
In simple words: To cube a number near a base (like 10 or 100), we first add twice its difference from the base to the number itself. Then we find three times the square of the difference. Finally, we cube the difference. We combine these three parts, carrying digits as needed.

🎯 Exam Tip: For cubing with Nikhilam Sutra, carefully perform calculations for all three parts (left, middle, right) and then apply carries from right to left, ensuring each part has the correct number of digits based on the base.

 

Question 6. (ii) Find the cube of the following by Sutra Nikhilam (sub-base method):
(a) 35
(b) 497
Answer:
(a) To find the cube of 35 using the sub-base method, we choose a sub-base of 30 (Base 10, Sub-base multiplier 3). The deviation is \( 35 - 30 = +5 \). This is effective for cubing numbers that are not directly near a base but close to its multiples.
\( 35^3 = (\text{Sub-base multiplier})^2 \times (\text{Number} + 2 \times \text{Deviation}) / (\text{Sub-base multiplier} \times 3 \times (\text{Deviation})^2) / ((\text{Deviation})^3) \)
\( = 3^2 \times (35 + 2 \times 5) / (3 \times 3 \times 5^2) / 5^3 \)
\( = 9 \times (35 + 10) / (9 \times 25) / 125 \)
\( = 9 \times 45 / 225 / 125 \)
\( = 405 / 225 / 125 \)
Since the base (10) has one zero, each part should have one digit. We carry over any extra digits.
\( = 405 / (225+12) / 5 \) (carry 12 from 125)
\( = 405 / 237 / 5 \)
\( = (405+23) / 7 / 5 \) (carry 23 from 237)
\( = 42875 \)
(b) To find the cube of 497, we use a sub-base of 500 (Base 100, Sub-base multiplier 5). The deviation is \( 497 - 500 = -03 \). This method is useful when numbers are just below a multiple of a base.
\( 497^3 = (\text{Sub-base multiplier})^2 \times (\text{Number} + 2 \times \text{Deviation}) / (\text{Sub-base multiplier} \times 3 \times (\text{Deviation})^2) / ((\text{Deviation})^3) \)
\( = 5^2 \times (497 + 2 \times (-03)) / (5 \times 3 \times (-03)^2) / (-03)^3 \)
\( = 25 \times (497 - 6) / (5 \times 3 \times 9) / (-27) \)
\( = 25 \times 491 / 135 / (-27) \)
\( = 12275 / 135 / (-27) \)
Since the base (100) has two zeros, each part should have two digits. We handle the negative rightmost part by borrowing from the left.
\( = (12275 + 1) / (135-1) / (100-27) \)
This accounts for borrowing from the middle part to make the rightmost part positive.
\( = 12276 / 134 / 73 \)
Now, the middle part 134 has an extra '1' (from 135) to carry over to the left part when only 2 digits are needed.
\( = 12276 / (34) / 73 \) (Here, from 134, 1 is carried over, leaving 34 for the middle part)
\( = 122763473 \)
In simple words: For cubing a number near a multiple of a base, we first use the sub-base multiplier squared for the left part, the sub-base multiplier times 3 for the middle part, and the cube of the difference for the right part. We must remember to handle negative differences by borrowing from the left.

🎯 Exam Tip: When dealing with negative deviations in cubing, carefully manage the borrowing process from the middle part to the right part to make the final digits positive. Ensure that all carries are correctly propagated across the parts.

 

Question 7. Divide the following by using Nikhilam Sutra:
(a) 243 ÷ 9
(b) 1356 ÷ 96
Answer:
(a) For 243 ÷ 9, we use Nikhilam Sutra. The divisor is 9, base is 10. The deviation from the base is \( 9 - 10 = -1 \). The complement (or adjusted deviation) is \( +1 \). This method is efficient for division by numbers close to powers of 10.

Therefore, Quotient = 27, Remainder = 0.
(b) For 1356 ÷ 96, the divisor is 96, base is 100. The deviation is \( 96 - 100 = -04 \). The complement (or adjusted deviation) is \( +04 \). This method involves splitting the dividend and performing repeated additions based on the complement.
Therefore, Quotient = 14, Remainder = 12.
In simple words: When dividing by numbers close to 10 or 100, we find how much the divisor is short of the base. We use this "short" number to guide our division, bringing down digits and adding products, making the process simpler than traditional long division.

🎯 Exam Tip: When using Nikhilam Sutra for division, correctly determine the complement (adjusted deviation) and perform the iterative multiplication and addition steps carefully. Remember to handle any remainder that is greater than or equal to the divisor by dividing again.

 

Question 8. Find the product by using Sutra Urdhva tiryagbhyam or by Sutra vertically and crosswise.
(a) 447 x 028
(b) 108 x 112
Answer:
(a) We multiply 447 by 28 using the 'vertically and crosswise' method, which involves multiplying digits column by column and summing cross-products. This method is systematic for multi-digit multiplication.

The result is 12516.
(b) We apply the 'vertically and crosswise' method to multiply 108 by 112. This involves setting up the numbers and systematically multiplying digits in columns and diagonally.
The result is 12096.
In simple words: This multiplication method involves multiplying numbers by lining them up and working column by column, from right to left. We multiply digits vertically and then crosswise, adding any carried-over numbers.

🎯 Exam Tip: The 'vertically and crosswise' method requires careful tracking of carries across columns. Practice with different digit counts to master the pattern of vertical and diagonal multiplication.

 

Question 9. Find the square of the following numbers by Dvanda yoga method or Duplex process:
(a) 53214
(b) 31321
Answer:
(a) The Dvanda yoga method (or Duplex process) calculates the square of a number by finding the duplex of various combinations of its digits. The duplex of a single digit (D(a)) is \( a^2 \), and for two digits (D(ab)) is \( 2ab \). For three digits (D(abc)) is \( 2ac + b^2 \), and so on.
\( 53214^2 \)
We calculate the duplex for each digit and combination of digits, from D(5) to D(4):
D(5) = \( 5^2 = 25 \)
D(53) = \( 2 \times 5 \times 3 = 30 \)
D(532) = \( 2 \times 5 \times 2 + 3^2 = 20 + 9 = 29 \)
D(5321) = \( 2 \times 5 \times 1 + 2 \times 3 \times 2 = 10 + 12 = 22 \)
D(53214) = \( 2 \times 5 \times 4 + 2 \times 3 \times 1 + 2^2 = 40 + 6 + 4 = 50 \)
D(3214) = \( 2 \times 3 \times 4 + 2 \times 2 \times 1 = 24 + 4 = 28 \)
D(214) = \( 2 \times 2 \times 4 + 1^2 = 16 + 1 = 17 \)
D(14) = \( 2 \times 1 \times 4 = 8 \)
D(4) = \( 4^2 = 16 \)
Combining these results and applying carries from right to left:
\( = 25 / 30 / 29 / 22 / 50 / 28 / 17 / 8 / 16 \)
\( = 283172996 \)
(b) We apply the Dvanda yoga method to find the square of 31321, calculating the duplex for all relevant digit groups.
\( 31321^2 \)
D(3) = \( 3^2 = 9 \)
D(31) = \( 2 \times 3 \times 1 = 6 \)
D(313) = \( 2 \times 3 \times 3 + 1^2 = 18 + 1 = 19 \)
D(3132) = \( 2 \times 3 \times 2 + 2 \times 1 \times 3 = 12 + 6 = 18 \)
D(31321) = \( 2 \times 3 \times 1 + 2 \times 1 \times 2 + 3^2 = 6 + 4 + 9 = 19 \)
D(1321) = \( 2 \times 1 \times 1 + 2 \times 3 \times 2 = 2 + 12 = 14 \)
D(321) = \( 2 \times 3 \times 1 + 2^2 = 6 + 4 = 10 \)
D(21) = \( 2 \times 2 \times 1 = 4 \)
D(1) = \( 1^2 = 1 \)
Combining these results and applying carries from right to left:
\( = 9 / 6 / 19 / 18 / 19 / 14 / 10 / 4 / 1 \)
\( = 981005041 \)
In simple words: To square a number using this method, we break it into parts. For each part, we find its 'duplex' (a special calculation). Then, we combine all these duplex values, carrying over numbers to get the final square.

🎯 Exam Tip: Systematically calculate the duplex for all possible combinations of digits, from single digits to the full number, ensuring you follow the correct formula for each group size (e.g., D(a), D(ab), D(abc)).

 

Question 10. Calculate square root of the following numbers by Dvanda yoga method or Duplex process:
(a) 169744
(b) 10329796
Answer:
(a) We find the square root of 169744 using the Dvanda yoga method, which involves a specific long division-like process based on the duplex values. This method is effective for finding square roots of perfect squares.

Steps:
(i) Take the first pair of digits (16). Find the largest digit whose square is less than or equal to 16. That's 4. \( 16 - 4^2 = 0 \). The new dividend is 09. The divisor is \( 4 \times 2 = 8 \).
(ii) Divide \( 09 \div 8 \). The quotient digit is 1, and the remainder is 1.
(iii) The next new dividend is 17. The adjusted dividend is \( 17 - 1^2 = 16 \).
(iv) Divide \( 16 \div 8 \). The quotient digit is 2, and the remainder is 0.
(v) The final remainder is calculated using the last parts of the numbers and current quotient: \( 04 - (2 \times 1 \times 2) = 0 \).
Therefore, the square root of 169744 is 412.
(b) We apply the Dvanda yoga method to calculate the square root of 10329796. This systematic process involves iterative division and subtraction of duplex values.
Steps:
(i) Take the first pair (10). Find the largest digit whose square is less than or equal to 10. That's 3. \( 10 - 3^2 = 1 \). The new dividend is 13. The divisor is \( 3 \times 2 = 6 \).
(ii) Divide \( 13 \div 6 \). The quotient digit is 2, and the remainder is 1.
(iii) The next new dividend is 12. The adjusted dividend is \( 12 - 2^2 = 8 \).
(iv) Divide \( 8 \div 6 \). The quotient digit is 1, and the remainder is 2.
(v) The next new dividend is 29. The adjusted dividend is \( 29 - (2 \times 2 \times 1) = 25 \).
(vi) Divide \( 25 \div 6 \). The quotient digit is 4, and the remainder is 1.
(vii) The next new dividend is 17. The adjusted dividend is \( 17 - (2 \times 2 \times 4 + 1^2) = 0 \).
(viii) The next new dividend is 09. The adjusted dividend is \( 09 - (2 \times 1 \times 4) = 1 \).
(ix) The next new dividend is 16. The adjusted dividend is \( 16 - 4^2 = 0 \).
Therefore, the square root of 10329796 is 3214.
In simple words: To find a square root, we pair digits from the right. Then, we use a step-by-step division process. We find a quotient, subtract a squared number, adjust the dividend, and repeat until we get the full square root.

🎯 Exam Tip: When using the Dvanda yoga method for square roots, be meticulous with calculating adjusted dividends and properly applying the duplex values or corresponding subtractions in each step to avoid errors.

 

Question 11. Divide the following by Dhwajanka method:
(a) 12474 ÷ 54
(b) 21112 ÷ 812
Answer:
(a) For 12474 ÷ 54, we use the Dhwajanka (Flag-digit) method. Here, the divisor is 54. The main digit is 5, and the flag digit is 4. This method helps simplify division for numbers with a relatively small flag digit.

Steps:
(i) Divide \( 12 \div 5 \). Quotient digit = 2, Remainder = 2. Place 2 as the first quotient digit, and 2 as a remainder before 4.
(ii) The new dividend is 24. Calculate the adjusted dividend: \( 24 - (2 \times 4) = 16 \).
(iii) Divide \( 16 \div 5 \). Quotient digit = 3, Remainder = 1. Place 3 as the next quotient digit, and 1 as a remainder before 7.
(iv) The new dividend is 17. Calculate the adjusted dividend: \( 17 - (3 \times 4) = 5 \).
(v) Divide \( 5 \div 5 \). Quotient digit = 1, Remainder = 0. Place 1 as the next quotient digit, and 0 as a remainder before 4.
(vi) The new dividend is 04. Calculate the final remainder: \( 04 - (1 \times 4) = 0 \).
Therefore, Quotient = 231, Remainder = 0.
(b) For 21112 ÷ 812, the divisor is 812. The main digit is 8, and the flag digits are 12. This method extends to divisors with multiple flag digits, making division of larger numbers manageable.
Steps:
(i) Divide \( 21 \div 8 \). Quotient digit = 2, Remainder = 5. Place 2 as the first quotient digit, and 5 as a remainder before 1.
(ii) The new dividend is 51. Calculate the adjusted dividend: \( 51 - (2 \times 1) = 49 \).
(iii) Divide \( 49 \div 8 \). Quotient digit = 6, Remainder = 1. Place 6 as the next quotient digit, and 1 as a remainder before 1.
(iv) The new dividend is 11. Calculate the adjusted dividend: \( 11 - (2 \times 2 + 6 \times 1) = 11 - (4+6) = 1 \).
(v) The new dividend is 12. Calculate the final remainder: \( 12 - (6 \times 2) = 0 \).
Therefore, Quotient = 26, Remainder = 0.
In simple words: This method of division uses one main digit from the divisor and other digits as 'flag digits'. We divide with the main digit, then adjust the dividend by subtracting from the product of the flag digit and the quotient, repeating until we find the remainder.

🎯 Exam Tip: When using the Dhwajanka method, correctly identify the main digit and flag digit(s). Be precise with calculating adjusted dividends by subtracting the product of current/previous quotient digits and the flag digits.

 

Question 12. Divide the following by Paravartya yojayet method:
(a) 6534 ÷ 123
(b) 13999 ÷ 1112
Answer:
(a) For 6534 ÷ 123, we use the Paravartya Yojayet method. The divisor is 123, and our base is 100. The deviation from the base is \( +23 \). So, the modified digits (Paravartya digits) are \( -2 \) and \( -3 \), often written as \( \overline{2}\overline{3} \). This method is useful for division when the divisor is larger or smaller than the base.

Therefore, Quotient = 53, Remainder = 15.
(b) For 13999 ÷ 1112, the divisor is 1112, and our base is 1000. The deviation from the base is \( +112 \). So, the modified digits (Paravartya digits) are \( -1, -1, -2 \), often written as \( \overline{1}\overline{1}\overline{2} \). This method converts a complex division into a simpler addition process.
Therefore, Quotient = 12, Remainder = 655.
In simple words: This method changes a division problem into an easier multiplication and addition problem. We adjust the divisor to work with a base (like 100 or 1000) and then use 'modified digits' to calculate the quotient and remainder step-by-step.

🎯 Exam Tip: When using the Paravartya Yojayet method, accurately determine the 'modified digits' (which are usually the complements of the deviation from the base). Be careful when performing multiplications and additions to arrive at the final quotient and remainder.

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