RBSE Solutions Class 8 Maths Chapter 5 Vedic Mathematics More Ques

Question 1. Multiplying the following by Urdhwtirgbhyaam Method
(i) 15 x 12
(ii) 60 x 18
(iii) 71 x 8
(iv) 122 x 4
(v) 706 x 56
(vi) 497 x 173
Answer:
(i) To multiply \(15 \times 12\) using the Urdhwtirgbhyaam method:
The numbers are 15 and 12.
Combine the numbers in three parts for the calculation:
\( \text{Part 1 (leftmost): } 1 \times 1 = 1 \)
\( \text{Part 2 (middle): } (1 \times 2) + (5 \times 1) = 2 + 5 = 7 \)
\( \text{Part 3 (rightmost): } 5 \times 2 = 10 \)
Now, combine these parts: \( 1 / 7 / 10 \).
Since our base is 10 (one digit per part), we process carries from right to left.
From \(10\), we write \(0\) and carry \(1\) to the middle part.
The middle part becomes \(7 + 1 = 8\).
The leftmost part remains \(1\).
So, the product is \(180\). This method is useful for quick mental calculations.
In simple words: To multiply 15 by 12 using this method, we multiply digits vertically and crosswise. The steps are: (1x1) / (1x2 + 5x1) / (5x2). This gives 1 / 7 / 10. We then carry over the extra numbers, so 10 becomes 0 with a carry of 1, making 7 into 8. The final answer is 180.

🎯 Exam Tip: Remember to always carry over any extra digits from the rightmost sections to the left sections, especially when the base is 10 and digits exceed 9 in any section.

Answer:
(ii) To multiply \(60 \times 18\) using the Urdhwtirgbhyaam method:
The numbers are 60 and 18.
Combine the numbers in three parts for the calculation:
\( \text{Part 1 (leftmost): } 6 \times 1 = 6 \)
\( \text{Part 2 (middle): } (6 \times 8) + (0 \times 1) = 48 + 0 = 48 \)
\( \text{Part 3 (rightmost): } 0 \times 8 = 0 \)
Now, combine these parts: \( 6 / 48 / 0 \).
Since our base is 10, we process carries from right to left.
From \(0\), we write \(0\). There is no carry.
From \(48\), we write \(8\) and carry \(4\) to the leftmost part.
The leftmost part becomes \(6 + 4 = 10\).
So, the product is \(1080\). This structured approach helps ensure accuracy.
In simple words: For 60 multiplied by 18, we get three sections: (6x1) / (6x8 + 0x1) / (0x8), which simplifies to 6 / 48 / 0. Then, we take the rightmost digit of each section and carry the tens part to the left. The 48 becomes 8 with a carry of 4, making the 6 into 10. The answer is 1080.

🎯 Exam Tip: Always make sure to account for any zero digits in the multiplication, as they can sometimes simplify sections but also hide potential errors if not handled carefully.

Answer:
(iii) To multiply \(71 \times 8\) using the Urdhwtirgbhyaam method, we treat 8 as 08 to apply the 2x2 digit rule:
The numbers are 71 and 08.
Combine the numbers in three parts for the calculation:
\( \text{Part 1 (leftmost): } 7 \times 0 = 0 \)
\( \text{Part 2 (middle): } (7 \times 8) + (1 \times 0) = 56 + 0 = 56 \)
\( \text{Part 3 (rightmost): } 1 \times 8 = 8 \)
Now, combine these parts: \( 0 / 56 / 8 \).
Since our base is 10, we process carries from right to left.
From \(8\), we write \(8\). There is no carry.
From \(56\), we write \(6\) and carry \(5\) to the leftmost part.
The leftmost part becomes \(0 + 5 = 5\).
So, the product is \(568\). This technique adapts to different digit counts by adding leading zeros.
In simple words: To multiply 71 by 8, we can think of 8 as 08. The parts are (7x0) / (7x8 + 1x0) / (1x8), which gives 0 / 56 / 8. We carry the 5 from 56 to the 0, making it 5, so the final answer is 568.

🎯 Exam Tip: When multiplying numbers with different digit counts using Urdhwatiryak, add leading zeros to the smaller number to make both numbers have the same number of digits for easier application of the method.

Answer:
(iv) To multiply \(122 \times 4\) using the Urdhwtirgbhyaam method, we treat 4 as 004 to apply the 3x3 digit rule:
The numbers are 122 and 004.
Combine the numbers in five parts for the calculation:
\( \text{Part 1 (leftmost): } 1 \times 0 = 0 \)
\( \text{Part 2: } (1 \times 0) + (2 \times 0) = 0 + 0 = 0 \)
\( \text{Part 3 (middle): } (1 \times 4) + (2 \times 0) + (2 \times 0) = 4 + 0 + 0 = 4 \)
\( \text{Part 4: } (2 \times 4) + (2 \times 0) = 8 + 0 = 8 \)
\( \text{Part 5 (rightmost): } 2 \times 4 = 8 \)
Now, combine these parts: \( 0 / 0 / 4 / 8 / 8 \).
Since each part has a single digit, there are no carries needed.
So, the product is \(488\). This illustrates how the method expands for more digits.
In simple words: For 122 multiplied by 4, we use 004 for 4. We calculate five parts: (1x0) / (1x0+2x0) / (1x4+2x0+2x0) / (2x4+2x0) / (2x4). This gives 0 / 0 / 4 / 8 / 8. Since no part has more than one digit, there are no carries. The answer is 488.

🎯 Exam Tip: When multiplying numbers with three digits using Urdhwatiryak, remember there will be five sections in the intermediate result before carrying, following the pattern: product of first digits, sum of cross-products of first two pairs, sum of cross-products of all three digits, sum of cross-products of last two pairs, product of last digits.

Answer:
(v) To multiply \(706 \times 56\) using the Urdhwtirgbhyaam method, we treat 56 as 056 to apply the 3x3 digit rule:
The numbers are 706 and 056.
Combine the numbers in five parts for the calculation:
\( \text{Part 1 (leftmost): } 7 \times 0 = 0 \)
\( \text{Part 2: } (7 \times 5) + (0 \times 0) = 35 + 0 = 35 \)
\( \text{Part 3 (middle): } (7 \times 6) + (0 \times 5) + (6 \times 0) = 42 + 0 + 0 = 42 \)
\( \text{Part 4: } (0 \times 6) + (6 \times 5) = 0 + 30 = 30 \)
\( \text{Part 5 (rightmost): } 6 \times 6 = 36 \)
Now, combine these parts: \( 0 / 35 / 42 / 30 / 36 \).
Since our base is 10, we process carries from right to left.
From \(36\), we write \(6\) and carry \(3\) to Part 4.
Part 4 becomes \(30 + 3 = 33\). We write \(3\) and carry \(3\) to Part 3.
Part 3 becomes \(42 + 3 = 45\). We write \(5\) and carry \(4\) to Part 2.
Part 2 becomes \(35 + 4 = 39\). We write \(9\) and carry \(3\) to Part 1.
Part 1 becomes \(0 + 3 = 3\).
So, the product is \(39536\). This step-by-step carrying is crucial for the final answer.
In simple words: For 706 multiplied by 56 (as 056), we get five parts: (7x0) / (7x5+0x0) / (7x6+0x5+6x0) / (0x6+6x5) / (6x6). This gives 0 / 35 / 42 / 30 / 36. We then carry over numbers from right to left. Starting with 36, we write 6 and carry 3. This changes 30 to 33, then 42 to 45, then 35 to 39, and 0 to 3. The final answer is 39536.

🎯 Exam Tip: Pay close attention to the carrying process, especially when multiple digits need to be carried from one section to the next. It's often helpful to write down the carried digits explicitly.

Answer:
(vi) To multiply \(497 \times 173\) using the Urdhwtirgbhyaam method:
The numbers are 497 and 173.
Combine the numbers in five parts for the calculation:
\( \text{Part 1 (leftmost): } 4 \times 1 = 4 \)
\( \text{Part 2: } (4 \times 7) + (9 \times 1) = 28 + 9 = 37 \)
\( \text{Part 3 (middle): } (4 \times 3) + (9 \times 7) + (7 \times 1) = 12 + 63 + 7 = 82 \)
\( \text{Part 4: } (9 \times 3) + (7 \times 7) = 27 + 49 = 76 \)
\( \text{Part 5 (rightmost): } 7 \times 3 = 21 \)
Now, combine these parts: \( 4 / 37 / 82 / 76 / 21 \).
Since our base is 10, we process carries from right to left.
From \(21\), we write \(1\) and carry \(2\) to Part 4.
Part 4 becomes \(76 + 2 = 78\). We write \(8\) and carry \(7\) to Part 3.
Part 3 becomes \(82 + 7 = 89\). We write \(9\) and carry \(8\) to Part 2.
Part 2 becomes \(37 + 8 = 45\). We write \(5\) and carry \(4\) to Part 1.
Part 1 becomes \(4 + 4 = 8\).
So, the product is \(85981\). This method makes complex multiplications manageable.
In simple words: To multiply 497 by 173, we break it into five parts using the Urdhwatiryak method: (4x1) / (4x7+9x1) / (4x3+9x7+7x1) / (9x3+7x7) / (7x3). This gives 4 / 37 / 82 / 76 / 21. We then carry numbers from right to left. The 21 becomes 1 with a carry of 2. This continues until the leftmost part. The final answer is 85981.

🎯 Exam Tip: For longer multiplications, keep your working neat and systematic. Each section's calculation should be done first, then the carrying handled in a clear, step-by-step manner.

Question 2. Find the product of the following
(i) 11 x 15
(ii) 12 x 18
(iii) 19 x 17
(iv) 28 x 22
(v) 51 x 49
(vi) 99 x 96
Answer:
(i) To find the product of \(11 \times 15\) using the Nikhilam method:
We choose Base \( = 10 \).
The numbers are 11 and 15.
Deviations from the base are:
For \(11\): \(11 - 10 = +1\)
For \(15\): \(15 - 10 = +5\)
Now, we form two parts of the answer:
\( \text{Left part (either number + other deviation): } 11 + 5 = 16 \text{ or } 15 + 1 = 16 \)
\( \text{Right part (product of deviations): } (+1) \times (+5) = 5 \)
Combine the parts: \( 16 / 5 \).
Since the right part (5) has only one digit and our base (10) also has one zero, no carrying is needed. This method is quick for numbers near a base.
Therefore, the product is \(165\).
In simple words: For 11 multiplied by 15, we use 10 as the base. The deviations are +1 and +5. We add a number to the other deviation (11+5=16) for the left part. We multiply the deviations (1x5=5) for the right part. Since 5 is a single digit, the answer is 165.

🎯 Exam Tip: When using the Nikhilam method, selecting an appropriate base (like 10, 100, 1000) close to the numbers is key to simplifying calculations. Also, correctly calculating the deviations is crucial.

Answer:
(ii) To find the product of \(12 \times 18\) using the Nikhilam method:
We choose Base \( = 10 \).
The numbers are 12 and 18.
Deviations from the base are:
For \(12\): \(12 - 10 = +2\)
For \(18\): \(18 - 10 = +8\)
Now, we form two parts of the answer:
\( \text{Left part (either number + other deviation): } 12 + 8 = 20 \text{ or } 18 + 2 = 20 \)
\( \text{Right part (product of deviations): } (+2) \times (+8) = 16 \)
Combine the parts: \( 20 / 16 \).
Since the right part (16) has two digits and our base (10) has only one zero, we need to carry. We keep the last digit \(6\) in the right part and carry the first digit \(1\) to the left part.
The left part becomes \(20 + 1 = 21\).
So, the product is \(216\). This method simplifies multiplications around a base number.
In simple words: For 12 multiplied by 18, with base 10, the deviations are +2 and +8. The left part is 12+8=20. The right part is 2x8=16. Since 16 has two digits and our base is 10, we keep 6 and carry 1 to the left part, making it 20+1=21. The final answer is 216.

🎯 Exam Tip: If the number of digits in the right part of the answer is more than the number of zeros in the base, carry the excess digits to the left part. If it's less, add leading zeros.

Answer:
(iii) To find the product of \(19 \times 17\) using the Nikhilam method:
We choose Base \( = 10 \).
The numbers are 19 and 17.
Deviations from the base are:
For \(19\): \(19 - 10 = +9\)
For \(17\): \(17 - 10 = +7\)
Now, we form two parts of the answer:
\( \text{Left part (either number + other deviation): } 19 + 7 = 26 \text{ or } 17 + 9 = 26 \)
\( \text{Right part (product of deviations): } (+9) \times (+7) = 63 \)
Combine the parts: \( 26 / 63 \).
Since the right part (63) has two digits and our base (10) has only one zero, we need to carry. We keep the last digit \(3\) in the right part and carry the first digit \(6\) to the left part.
The left part becomes \(26 + 6 = 32\).
So, the product is \(323\). This method makes it easy to multiply numbers close to a power of ten.
In simple words: For 19 multiplied by 17, with base 10, the deviations are +9 and +7. The left part is 19+7=26. The right part is 9x7=63. We carry the 6 from 63 to the left part, making it 26+6=32. The final answer is 323.

🎯 Exam Tip: Practice quickly identifying the correct base and deviations. This speeds up the Nikhilam method, especially when numbers are just slightly above or below a base.

Answer:
(iv) To find the product of \(28 \times 22\) using the Nikhilam method:
We choose Base \( = 10 \).
Since the numbers are close to 20, we select a Sub-base \( = 2 \times 10 = 20 \).
The Sub-base digit is \(20 \div 10 = 2\).
Deviations from the Sub-base are:
For \(28\): \(28 - 20 = +8\)
For \(22\): \(22 - 20 = +2\)
Now, we form two parts of the answer:
\( \text{Left part (either number + other deviation, multiplied by Sub-base digit): } (28 + 2) \times 2 = 30 \times 2 = 60 \text{ or } (22 + 8) \times 2 = 30 \times 2 = 60 \)
\( \text{Right part (product of deviations): } (+8) \times (+2) = 16 \)
Combine the parts: \( 60 / 16 \).
Since the right part (16) has two digits and our base (10) has only one zero, we need to carry. We keep the last digit \(6\) in the right part and carry the first digit \(1\) to the left part.
The left part becomes \(60 + 1 = 61\).
So, the product is \(616\). Using a sub-base makes this method versatile for wider number ranges.
In simple words: For 28 multiplied by 22, we use base 10 and sub-base 20. The deviations are +8 and +2. The left part is (28+2) multiplied by the sub-base digit (2), so (30)x2=60. The right part is 8x2=16. We carry the 1 from 16 to 60, making it 61. The final answer is 616.

🎯 Exam Tip: When using a sub-base, remember to multiply the left part of the answer by the sub-base digit before performing any carries. The carrying rules for the right part still depend on the main base.

Answer:
(v) To find the product of \(51 \times 49\) using the Nikhilam method:
We choose Base \( = 10 \).
Since the numbers are close to 50, we select a Sub-base \( = 5 \times 10 = 50 \).
The Sub-base digit is \(50 \div 10 = 5\).
Deviations from the Sub-base are:
For \(51\): \(51 - 50 = +1\)
For \(49\): \(49 - 50 = -1\)
Now, we form two parts of the answer:
\( \text{Left part (either number + other deviation, multiplied by Sub-base digit): } (51 + (-1)) \times 5 = 50 \times 5 = 250 \text{ or } (49 + 1) \times 5 = 50 \times 5 = 250 \)
\( \text{Right part (product of deviations): } (+1) \times (-1) = -1 \)
Combine the parts: \( 250 / -1 \).
Since the right part is negative, we need to adjust it. We borrow 1 from the left part (250 becomes 249). The borrowed 1 is equivalent to the Base (10) when moved to the right part.
So, the right part becomes \(10 + (-1) = 9\).
The left part is \(249\).
So, the product is \(2499\). Handling negative deviations properly is key for correct results.
In simple words: For 51 multiplied by 49, we use base 10 and sub-base 50. Deviations are +1 and -1. The left part is (51-1)x5=250. The right part is 1x(-1)=-1. Since the right part is negative, we borrow 1 from 250, making it 249, and add the base (10) to -1. So, -1 becomes 9. The final answer is 2499.

🎯 Exam Tip: When the product of deviations is negative, always convert it to a positive value by borrowing from the left part. The amount borrowed from the left part should be equal to the base of the Nikhilam calculation.

Answer:
(vi) To find the product of \(99 \times 96\) using the Nikhilam method:
We choose Base \( = 10 \).
Since the numbers are close to 90, we select a Sub-base \( = 9 \times 10 = 90 \).
The Sub-base digit is \(90 \div 10 = 9\).
Deviations from the Sub-base are:
For \(99\): \(99 - 90 = +9\)
For \(96\): \(96 - 90 = +6\)
Now, we form two parts of the answer:
\( \text{Left part (either number + other deviation, multiplied by Sub-base digit): } (99 + 6) \times 9 = 105 \times 9 = 945 \text{ or } (96 + 9) \times 9 = 105 \times 9 = 945 \)
\( \text{Right part (product of deviations): } (+9) \times (+6) = 54 \)
Combine the parts: \( 945 / 54 \).
Since the right part (54) has two digits and our base (10) has only one zero, we need to carry. We keep the last digit \(4\) in the right part and carry the first digit \(5\) to the left part.
The left part becomes \(945 + 5 = 950\).
So, the product is \(9504\). This method is particularly efficient for numbers near common multiples of ten.
In simple words: For 99 multiplied by 96, we use base 10 and sub-base 90. The deviations are +9 and +6. The left part is (99+6)x9=945. The right part is 9x6=54. We carry the 5 from 54 to the left part, making it 945+5=950. The final answer is 9504.

🎯 Exam Tip: When the numbers are larger but still close to a multiple of a base (like 90 in this case), using a sub-base simplifies the calculation much more than trying to use a base of 100.

Question 3. Multiply the following three numbers by formula Nikhilam
(i) 11 x 12 x 13
(ii) 8 x 9 x 10
(iii) 6 x 7 x 8
(iv) 27 x 28 x 29
(v) 98 x 99 x 99
(vi) 51 x 52 x 53
Answer:
(i) To multiply \(11 \times 12 \times 13\) using the Nikhilam method:
We choose Base \( = 10 \).
Deviations from the base are:
For \(11\): \(11 - 10 = +1\)
For \(12\): \(12 - 10 = +2\)
For \(13\): \(13 - 10 = +3\)
Now, we form three parts of the answer:
\( \text{Part 1 (leftmost): (Number + Sum of other two deviations) } = 11 + 2 + 3 = 16 \text{ (or } 12+1+3=16 \text{ or } 13+1+2=16 \text{)} \)
\( \text{Part 2 (middle): (Sum of products of deviations taken two at a time) } = (1 \times 2) + (2 \times 3) + (1 \times 3) = 2 + 6 + 3 = 11 \)
\( \text{Part 3 (rightmost): (Product of all deviations) } = 1 \times 2 \times 3 = 6 \)
Combine the parts: \( 16 / 11 / 6 \).
Since our base is 10 (one digit per part), we process carries from right to left.
From Part 3 (\(6\)), we write \(6\). There is no carry.
From Part 2 (\(11\)), we write \(1\) and carry \(1\) to Part 1.
Part 1 becomes \(16 + 1 = 17\).
So, the product is \(1716\). This method makes multiplying three numbers less complicated.
In simple words: For 11x12x13 with base 10, deviations are +1, +2, +3. We get three parts: (11+2+3) / (1x2+2x3+1x3) / (1x2x3). This gives 16 / 11 / 6. We then carry from right to left. The 11 becomes 1 with a carry of 1 to 16, making it 17. The final answer is 1716.

🎯 Exam Tip: When applying Nikhilam for three numbers, remember the three sections: one number plus other deviations, sum of pairwise products of deviations, and product of all deviations. Master these definitions to avoid confusion.

Answer:
(ii) To multiply \(8 \times 9 \times 10\) using the Nikhilam method:
We choose Base \( = 10 \).
Deviations from the base are:
For \(8\): \(8 - 10 = -2\)
For \(9\): \(9 - 10 = -1\)
For \(10\): \(10 - 10 = 0\)
Now, we form three parts of the answer:
\( \text{Part 1 (leftmost): (Number + Sum of other two deviations) } = 8 + (-1) + 0 = 7 \text{ (or } 9+(-2)+0=7 \text{ or } 10+(-2)+(-1)=7 \text{)} \)
\( \text{Part 2 (middle): (Sum of products of deviations taken two at a time) } = (-2 \times -1) + (-1 \times 0) + (0 \times -2) = 2 + 0 + 0 = 2 \)
\( \text{Part 3 (rightmost): (Product of all deviations) } = (-2) \times (-1) \times 0 = 0 \)
Combine the parts: \( 7 / 2 / 0 \).
Since all parts have single digits and the base is 10, no carrying is needed.
So, the product is \(720\). The presence of zero deviations can simplify calculations significantly.
In simple words: For 8x9x10 with base 10, deviations are -2, -1, 0. The three parts are: (8-1+0) / (-2x-1 + -1x0 + 0x-2) / (-2x-1x0). This gives 7 / 2 / 0. No carries are needed. The final answer is 720.

🎯 Exam Tip: When one of the deviations is zero, it often makes the product of deviations zero, and simplifies some of the pairwise products, speeding up the calculation.

Answer:
(iii) To multiply \(6 \times 7 \times 8\) using the Nikhilam method:
We choose Base \( = 10 \).
Deviations from the base are:
For \(6\): \(6 - 10 = -4\)
For \(7\): \(7 - 10 = -3\)
For \(8\): \(8 - 10 = -2\)
Now, we form three parts of the answer:
\( \text{Part 1 (leftmost): (Number + Sum of other two deviations) } = 6 + (-3) + (-2) = 1 \)
\( \text{Part 2 (middle): (Sum of products of deviations taken two at a time) } = (-4 \times -3) + (-3 \times -2) + (-2 \times -4) = 12 + 6 + 8 = 26 \)
\( \text{Part 3 (rightmost): (Product of all deviations) } = (-4) \times (-3) \times (-2) = -24 \)
Combine the parts: \( 1 / 26 / -24 \).
Since Part 3 (\(-24\)) is negative, we need to adjust it. We borrow 3 from Part 2 (\(26\)), making Part 2 become \(23\). The borrowed 3 is multiplied by the Base (10), making it \(30\).
Part 3 becomes \(30 + (-24) = 6\).
Now we have: \( 1 / 23 / 6 \).
Part 2 (\(23\)) has two digits, and our base is 10, so we carry \(2\) to Part 1.
Part 1 becomes \(1 + 2 = 3\).
So, the product is \(336\). This process of adjusting negative parts is a common step in Vedic math.
In simple words: For 6x7x8 with base 10, deviations are -4, -3, -2. The parts are: (6-3-2) / (-4x-3 + -3x-2 + -2x-4) / (-4x-3x-2). This gives 1 / 26 / -24. We change -24 to positive by borrowing 3 from 26 (becomes 23), then 3x10-24=6. So now 1 / 23 / 6. Then we carry 2 from 23 to 1, making it 3. The final answer is 336.

🎯 Exam Tip: When dealing with negative products of deviations, remember to borrow from the left section, multiplying the borrowed amount by the base to make the rightmost section positive. Always adjust from right to left.

Answer:
(iv) To multiply \(27 \times 28 \times 29\) using the Nikhilam method:
We choose Base \( = 10 \).
Since the numbers are close to 30, we select a Sub-base \( = 3 \times 10 = 30 \).
The Sub-base digit is \(30 \div 10 = 3\). (Note: The source implies a sub-base of 20 with a sub-base digit of 2 for its calculations, which aligns with 27-20=7. Let's follow the source's implied sub-base of 20).
Let's use Sub-base \( = 2 \times 10 = 20 \).
The Sub-base digit is \(20 \div 10 = 2\).
Deviations from the Sub-base are:
For \(27\): \(27 - 20 = +7\)
For \(28\): \(28 - 20 = +8\)
For \(29\): \(29 - 20 = +9\)
Now, we form three parts of the answer:
\( \text{Part 1 (leftmost): (Number + Sum of other two deviations) } = 27 + 8 + 9 = 44 \text{ (This is multiplied by } \text{Sub-base digit}^2 = 2^2 = 4 \text{)} \)
\( \text{So, Part 1 } = 44 \times 4 = 176 \)
\( \text{Part 2 (middle): (Sum of products of deviations taken two at a time) } = (7 \times 8) + (8 \times 9) + (9 \times 7) = 56 + 72 + 63 = 191 \text{ (This is multiplied by } \text{Sub-base digit} = 2 \text{)} \)
\( \text{So, Part 2 } = 191 \times 2 = 382 \)
\( \text{Part 3 (rightmost): (Product of all deviations) } = 7 \times 8 \times 9 = 504 \)
Combine the parts: \( 176 / 382 / 504 \).
Since our base is 10 (one digit per section for the product of deviations), we process carries from right to left.
From Part 3 (\(504\)), we write the last digit \(4\). We carry \(50\) to Part 2.
Part 2 becomes \(382 + 50 = 432\). We write the last digit \(2\). We carry \(43\) to Part 1.
Part 1 becomes \(176 + 43 = 219\).
So, the product is \(21924\). This is an excellent example of using a sub-base with three numbers.
In simple words: For 27x28x29, we use base 10 and sub-base 20 (sub-base digit 2). Deviations are +7, +8, +9. The three parts are: (27+8+9)x\(2^2\) / (7x8+8x9+9x7)x2 / (7x8x9). This gives 176 / 382 / 504. We then carry over from right to left. From 504, we write 4 and carry 50. From 382+50=432, we write 2 and carry 43. From 176+43=219, we write 219. The final answer is 21924.

🎯 Exam Tip: For three numbers with a sub-base, remember the multipliers for each section: the leftmost section is multiplied by (sub-base digit)^2, the middle by (sub-base digit), and the rightmost remains as is before carrying. This distinction is crucial.

Question 4. Division operation using Dhwajank method
(1) \( 1737 \div 21 \)
Answer: To divide 1737 by 21 using the Dhwajank method, we first split the divisor into a main part and a flag digit. Here, the main part is 2 and the flag digit is 1. We then set up the dividend, dividing it based on the number of flag digits. We perform the division in steps, adjusting the dividend based on the flag digit and the quotient digit from the previous step. The process continues until all digits of the dividend are used.
Quotient = 82
Remainder = 15
In simple words: When you divide 1737 by 21 using a special method called Dhwajank, you get 82 as the main answer and 15 left over.

🎯 Exam Tip: Remember to correctly identify the main part and the flag digit from the divisor, as this is crucial for setting up the Dhwajank division accurately.

Question 4. (2) \( 37941 \div 47 \)
Answer: Using the Dhwajank method to divide 37941 by 47, we designate 4 as the main divisor and 7 as the flag digit. The division involves a series of steps where each new dividend is adjusted by subtracting the product of the previous quotient digit and the flag digit. This systematic process helps in finding the correct quotient and remainder.
Quotient = 807
Remainder = 12
In simple words: When you divide 37941 by 47 using the Dhwajank method, you get 807 as the main answer and 12 is left over.

🎯 Exam Tip: Practice adjusting the dividend by subtracting the product of the flag digit and the newly found quotient digit; this is often where mistakes occur.

Question 4. (3) \( 23754 \div 74 \)
Answer: For the division of 23754 by 74 using the Dhwajank method, 7 is the main divisor and 4 is the flag digit. The calculations proceed by performing standard division with the main divisor and then making adjustments using the flag digit to determine the true remainder and quotient for each step. This ensures accuracy in the final result.
Quotient = 321
Remainder = 0
In simple words: Dividing 23754 by 74 with the Dhwajank method gives 321, with nothing left over (zero remainder).

🎯 Exam Tip: A zero remainder indicates that the divisor completely divides the dividend, which is an important checkpoint in division problems.

Question 4. (4) \( 3257 \div 74 \)
Answer: To divide 3257 by 74 using the Dhwajank method, 7 acts as the main divisor and 4 is the flag digit. The steps involve repeatedly dividing a part of the dividend by the main divisor and then subtracting the 'corrected' product from the dividend to find the next adjusted remainder. This method is efficient for multi-digit division.
Quotient = 44
Remainder = 1
In simple words: Using the Dhwajank method to divide 3257 by 74, we find the answer is 44 with 1 left behind.

🎯 Exam Tip: Pay close attention to carrying over numbers in the Dhwajank method, as incorrect carrying can lead to errors in subsequent steps.

Question 4. (5) \( 7453 \div 79 \)
Answer: For dividing 7453 by 79 using the Dhwajank method, the main divisor is 7 and the flag digit is 9. The method requires careful step-by-step calculation, where partial dividends are formed, divided, and then adjusted based on the flag digit's contribution to subtract from the new dividend. This ensures precision in finding the quotient and remainder.
Quotient = 94
Remainder = 27
In simple words: When 7453 is divided by 79 using the Dhwajank method, the result is 94 with a remainder of 27.

🎯 Exam Tip: Always double-check your subtraction steps, especially when dealing with negative adjustments from the flag digit's product.

Question 4. (6) \( 59241 \div 82 \)
Answer: To divide 59241 by 82 using the Dhwajank method, 8 is the main divisor and 2 is the flag digit. The division proceeds by finding trial quotients using the main divisor and then refining these by accounting for the flag digit's multiplication with the previous quotient digits. This structured approach helps manage complex divisions.
Quotient = 722
Remainder = 37
In simple words: Dividing 59241 by 82 using the Dhwajank method yields a quotient of 722 and a remainder of 37.

🎯 Exam Tip: The final remainder must always be less than the divisor; if it's not, recheck your last few steps.