RBSE Solutions Class 8 Maths Chapter 5 Vedic Mathematics Important Questions

Rajasthan Board RBSE Class 8 Maths Chapter 5 Vedic Mathematics Additional Questions

I. Objective Type Questions

Question 1. How many groups makes when two digit number is multiplied by two digit number
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: When you multiply two numbers that each have two digits, the Vedic mathematics method divides the process into three main parts or groups to make it easier to solve. This helps keep the calculation organized.

🎯 Exam Tip: Remember that the number of groups in Vedic multiplication depends on the number of digits involved, simplifying the process into manageable parts.

Question 2. How many groups makes when three digit number is multiplied by three digit number
(a) 3
(b) 5
(c) 6
(d) 9
Answer: (b) 5
In simple words: For multiplying two numbers, each with three digits, the Vedic mathematics technique usually forms five different groups of calculations. This structure helps manage the larger number of digits involved.

🎯 Exam Tip: Understanding how many groups are formed for different digit multiplications is key to applying the Urdhva Tiryagbhyam (vertical and crosswise) method effectively.

Question 3. If base is 10 and sub-base is 30 then what sub-base digit will be
(a) 3
(b) 2
(c) 1
(d) 43
Answer: (a) 3
In simple words: To find the sub-base digit, you divide the sub-base by the base. Here, \( 30 \div 10 \) gives 3, so 3 is the sub-base digit. This digit is used to adjust the calculations when using a sub-base method.

🎯 Exam Tip: The sub-base digit is always found by dividing the sub-base by the chosen base, and it plays a crucial role in adjusting the left-hand side of the answer in Nikhilam multiplication.

Question 4. Sub-base digit is calculated as
(a) Base + Sub-base
(b) Sub-base - Base
(c) Sub-base / Base
Answer: (c) Sub-base / Base
In simple words: The sub-base digit is found by dividing the sub-base by the base. This number is very important in Vedic mathematics to adjust results when you are multiplying numbers that are close to a multiple of a base (like 20, 30, etc.).

🎯 Exam Tip: Always remember that the sub-base digit is the quotient obtained by dividing the sub-base by the base, which helps in scaling the results accurately.

Question 6. If the number is 22 and the base is 20, then the deviation is
(a) 10
(b) 20
(c) 2
(d) - 2
Answer: (c) 2
In simple words: Deviation means how far a number is from a chosen base. To find it, subtract the base from the number. Here, \( 22 - 20 = 2 \), so the deviation is 2.

🎯 Exam Tip: Always calculate deviation by subtracting the base from the number; a positive deviation means the number is greater than the base, and a negative deviation means it's smaller.

II. Short Answer Type Questions

Question 1. Find the product of following
(i) \( 98 \times 98 \)
(ii) \( 102 \times 102 \)
Answer:
(i) To find \( 98 \times 98 \):
Let's use Base = 10, Sub-base = \( 9 \times 10 = 90 \).
The sub-base digit is \( 90 \div 10 = 9 \).
The deviation from the sub-base for 98 is \( 98 - 90 = +8 \).
Step 1: Write the numbers and their deviations.
Number    Deviation
98          +8
98          +8
Step 2: Calculate the right-hand part (RHS) by multiplying the deviations: \( +8 \times +8 = 64 \).
Step 3: Calculate the left-hand part (LHS). Add one number to the other's deviation: \( 98 + 8 = 106 \). Then multiply this by the sub-base digit: \( 9 \times 106 = 954 \).
Step 4: Combine the LHS and RHS. Since our base is 10 and we have 2 digits in the RHS (64), we carry over the '6' to the LHS.
So, \( 954 / 64 \) becomes \( (954 + 6) / 4 = 960 / 4 \).
Combining these, we get 9604.
Therefore, the required product of \( 98 \times 98 \) is 9604.

(ii) To find \( 102 \times 102 \):
Let's use Base = 10, Sub-base = \( 10 \times 10 = 100 \).
The sub-base digit is \( 100 \div 10 = 10 \).
The deviation from the sub-base for 102 is \( 102 - 100 = +2 \).
Step 1: Write the numbers and their deviations.
Number    Deviation
102        +2
102        +2
Step 2: Calculate the right-hand part (RHS) by multiplying the deviations: \( +2 \times +2 = 4 \). Since the base is 100 (sub-base), we need two digits, so write it as 04.
Step 3: Calculate the left-hand part (LHS). Add one number to the other's deviation: \( 102 + 2 = 104 \). Then multiply this by the sub-base digit: \( 10 \times 104 = 1040 \).
Step 4: Combine the LHS and RHS. \( 1040 / 04 \).
Therefore, the required product of \( 102 \times 102 \) is 10404.
In simple words: We used a method where we first pick a base and a sub-base. Then we find how much each number is different from the sub-base. We multiply these differences for one part of the answer and add a deviation to the number (then multiply by sub-base digit) for the other part. Finally, we combine these parts, carrying over digits if needed, to get the final answer.

🎯 Exam Tip: When using the Nikhilam method with a sub-base, always remember to multiply the left-hand side of the answer by the sub-base digit, and ensure the right-hand side has the correct number of digits as per the chosen base (usually 1 or 2).

Question 3. Find the cube of 96.
Answer:
To find the cube of 96, we use the Nikhilam formula for cubing numbers near a base.
Let Base = 100.
The deviation for 96 is \( 96 - 100 = -04 \).
The formula for \( N^3 \) using Nikhilam is: \( N + 2d / 3d^2 / d^3 \), where \( N \) is the number and \( d \) is the deviation.

\( (96)^3 = 96 + 2 \times (-04) / 3 \times (-04)^2 / (-04)^3 \)

Calculate each part:
Part 1: \( 96 + (2 \times -04) = 96 - 08 = 88 \)
Part 2: \( 3 \times (-04)^2 = 3 \times 16 = 48 \)
Part 3: \( (-04)^3 = -64 \)

So, we have: \( 88 / 48 / -64 \)

Now, we need to adjust for the negative part. Since the base is 100, we borrow from the left side.
\( 88 / 48 / -64 \)

From the middle part (48), borrow 1. This 1 becomes 100 in the rightmost part. So, \( 48 - 1 = 47 \).
The rightmost part becomes \( 100 - 64 = 36 \).
The result is \( 88 / 47 / 36 \).
Therefore, the cube of 96 is 884736.
In simple words: To find the cube of 96, we use a special math trick. We first find how much 96 is different from 100. Then we use a simple formula that gives us three parts. We put these parts together, carefully handling any negative numbers by borrowing from the left side, until we get the final answer.

🎯 Exam Tip: When cubing numbers using the Nikhilam method, pay close attention to the adjustments needed for negative deviations, ensuring each section has the correct number of digits (usually two for a base of 100) before combining.

Question 4. Find the cube of 15.
Answer:
To find the cube of 15, we use the Nikhilam formula for cubing numbers near a base.
Let Base = 10.
The deviation for 15 is \( 15 - 10 = +5 \).
The formula for \( N^3 \) using Nikhilam is: \( N + 2d / 3d^2 / d^3 \), where \( N \) is the number and \( d \) is the deviation.

\( (15)^3 = 15 + 2 \times 5 / 3 \times (5)^2 / (5)^3 \)

Calculate each part:
Part 1: \( 15 + (2 \times 5) = 15 + 10 = 25 \)
Part 2: \( 3 \times (5)^2 = 3 \times 25 = 75 \)
Part 3: \( (5)^3 = 125 \)

So, we have: \( 25 / 75 / 125 \)

Now, we need to adjust these parts, as our base is 10, meaning each part should ideally have one digit (after any carry-overs).
From 125, carry 12 to the left: \( 75 + 12 = 87 \). The rightmost digit remains 5.
So, now we have: \( 25 / 87 / 5 \)

From 87, carry 8 to the left: \( 25 + 8 = 33 \). The middle digit remains 7.
So, now we have: \( 33 / 7 / 5 \)
Therefore, the cube of 15 is 3375.
In simple words: To find \( 15^3 \), we use a Vedic math trick by picking 10 as our base and finding how much 15 is different from it. We then use a formula to get three parts of the answer. Because our base is 10, we make sure each part has only one digit by carrying any extra digits to the left side until we have the final number.

🎯 Exam Tip: When cubing with Nikhilam, always ensure to handle carry-overs from right to left according to the base (e.g., if base is 10, carry tens; if base is 100, carry hundreds) to get the correct final product.

Question 5. Find the value of \( 101 \times 102 \times 103 \) by applying 'nikhilam' formula.
Answer:
To multiply \( 101 \times 102 \times 103 \) using the Nikhilam formula:
Let Base = 100.
The numbers are 101, 102, 103.
The deviations from the base 100 are:
\( d_1 = 101 - 100 = +1 \)
\( d_2 = 102 - 100 = +2 \)
\( d_3 = 103 - 100 = +3 \)

The formula for multiplying three numbers using Nikhilam is:
\( (N + d_1 + d_2 + d_3) / (d_1 d_2 + d_1 d_3 + d_2 d_3) / (d_1 d_2 d_3) \)

Calculate each part:
Part 1 (Left-hand part): Sum of one number and the other two deviations:
\( 100 + (+1) + (+2) + (+3) = 100 + 6 = 106 \)

Part 2 (Middle part): Sum of the products of deviations taken two at a time:
\( (1 \times 2) + (1 \times 3) + (2 \times 3) = 2 + 3 + 6 = 11 \)

Part 3 (Right-hand part): Product of all deviations:
\( 1 \times 2 \times 3 = 6 \)

Combine the parts: \( 106 / 11 / 06 \). Since the base is 100, each right-hand section (middle and rightmost) should have two digits. 11 already has two digits, and 6 needs to be written as 06.

So, the final product is 1061106.
In simple words: We are multiplying three numbers that are close to 100. First, we find how much each number is more than 100. Then we use a special Vedic math formula that puts these numbers together in three parts. We then combine these parts, making sure that the middle and right parts have two digits each (because our base is 100), to get the final answer.

🎯 Exam Tip: When multiplying three numbers near a base using Nikhilam, ensure the deviations are correctly calculated and that each section of the combined result (middle and right) has the appropriate number of digits corresponding to the base (e.g., two digits for a base of 100).

Question 6. Multiply by Urdhwtirygbhyaam method \( 426 \times 351 \)
Answer:
To multiply \( 426 \times 351 \) using the Urdhva Tiryagbhyam method (Vertical and Crosswise method):
This method involves multiplying digits vertically and crosswise in a structured way.

Let the numbers be \( ABC \) and \( PQR \).
The steps are:
Step I (Units digit): \( C \times R \)
Step II (Tens digit): \( B \times R + C \times Q \)
Step III (Hundreds digit): \( A \times R + C \times P + B \times Q \)
Step IV (Thousands digit): \( A \times Q + B \times P \)
Step V (Ten Thousands digit): \( A \times P \)

For \( 426 \times 351 \): \( A=4, B=2, C=6 \) and \( P=3, Q=5, R=1 \).

Step 1: Vertical multiplication of units digits:
\( 6 \times 1 = 6 \)

Step 2: Crosswise multiplication of units and tens digits (and add):
\( (2 \times 1) + (6 \times 5) = 2 + 30 = 32 \)

Step 3: Crosswise multiplication of all three digits (extreme vertical, middle crosswise, then add):
\( (4 \times 1) + (6 \times 3) + (2 \times 5) = 4 + 18 + 10 = 32 \)

Step 4: Crosswise multiplication of tens and hundreds digits (and add):
\( (4 \times 5) + (2 \times 3) = 20 + 6 = 26 \)

Step 5: Vertical multiplication of hundreds digits:
\( 4 \times 3 = 12 \)

Combining these results from left to right, with carry-overs:
\( 12 / 26 / 32 / 32 / 6 \)

Now, we process from right to left, carrying over digits.
Starting with 6 (units digit).
Next part is 32. Keep 2, carry over 3: \( 32 + 3 = 35 \). Keep 5, carry over 3.
Next part is 32. Keep 2, carry over 3: \( 32 + 3 = 35 \). Keep 5, carry over 3.
Next part is 26. Keep 6, carry over 2: \( 26 + 3 = 29 \). Keep 9, carry over 2.
Next part is 12. Add carry over 2: \( 12 + 2 = 14 \). Keep 14.

So, the sequence of digits is 149526.
Hence, the required product \( 426 \times 351 \) is 149526.
In simple words: We multiply two three-digit numbers using a special criss-cross method. We break down the multiplication into five steps, each involving different digits. We then add up these parts, carrying over extra numbers from right to left, just like in normal addition, to get our final answer. This method makes big multiplications easier.

🎯 Exam Tip: The Urdhva Tiryagbhyam method is highly efficient for multi-digit multiplication; ensure you correctly identify the vertical and crosswise products for each 'group' and carefully manage all carry-overs from right to left to avoid errors.

Question 7. Multiply by using Nikhilam formula : \( 22 \times 23 \)
Answer:
To multiply \( 22 \times 23 \) using the Nikhilam formula with a sub-base:
Let Base = 10.
Choose Sub-base = \( 2 \times 10 = 20 \). (Since both 22 and 23 are close to 20).
The sub-base digit is \( 20 \div 10 = 2 \).

Deviations from the sub-base 20:
For 22: \( d_1 = 22 - 20 = +2 \)
For 23: \( d_2 = 23 - 20 = +3 \)

The Nikhilam formula for two numbers near a sub-base is:
\( (\text{Sub-base digit} \times (N_1 + d_2)) / (d_1 \times d_2) \)
or \( (\text{Sub-base digit} \times (N_2 + d_1)) / (d_1 \times d_2) \)

Calculate the parts:
Right-hand part (RHS): Product of deviations
\( (+2) \times (+3) = 6 \)

Left-hand part (LHS): Sub-base digit multiplied by (one number + other deviation)
\( 2 \times (22 + 3) = 2 \times 25 = 50 \)
Alternatively: \( 2 \times (23 + 2) = 2 \times 25 = 50 \)

Combine the parts: \( 50 / 6 \)
Since the base is 10, the right-hand part (6) should have only one digit. It does.
So, the combined product is 506.
Therefore, the required product of \( 22 \times 23 \) is 506.
In simple words: We multiply 22 by 23 using a Vedic math shortcut called Nikhilam. We choose 20 as our working number (sub-base) and find how much 22 and 23 are different from 20. Then we multiply these differences for one part of the answer. For the other part, we add one number to the deviation of the other, then multiply by the sub-base digit. Finally, we put these two parts together to get the answer.

🎯 Exam Tip: When using the Nikhilam method with a sub-base, remember to multiply the sum of (number + deviation) by the sub-base digit to correctly form the left-hand part of the product.

Question 8. Multiply by using Nikhilam formula \( 103 \times 103 \times 103 \)
Answer:
To multiply \( 103 \times 103 \times 103 \) using the Nikhilam formula:
Let Base = 100.
The number is 103.
The deviation from the base 100 is \( d = 103 - 100 = +03 \).

The formula for cubing a number using Nikhilam (which is essentially multiplying three identical numbers) is:
\( (N + 2d) / (3d^2) / (d^3) \)
This is equivalent to the generalized three-number multiplication formula where \( d_1=d_2=d_3=d \):
\( (N + d + d + d) / (d \cdot d + d \cdot d + d \cdot d) / (d \cdot d \cdot d) \)
\( (N + 3d) / (3d^2) / (d^3) \)

Here, \( N = 100 \) (conceptually, or starting number), and \( d = +03 \).

Part 1 (Left-hand part): \( 103 + 3 \times (+03) = 103 + 9 = 109 \)
Part 2 (Middle part): \( 3 \times (+03)^2 = 3 \times 9 = 27 \)
Part 3 (Right-hand part): \( (+03)^3 = 27 \)

Combine the parts: \( 109 / 27 / 27 \).
Since the base is 100, each section should have two digits. All parts already have two digits.

So, the final product is 1092727.
In simple words: We are finding the cube of 103 by using a Vedic math trick. We first see that 103 is 3 more than 100 (our base). Then, we use a special formula that creates three parts of the answer using this difference. We combine these parts, making sure each part has two digits because our base is 100, to get the final result.

🎯 Exam Tip: When cubing a number using Nikhilam, ensure the deviations are correct and that the intermediate parts are adjusted for the base (e.g., two digits for a base of 100) before combining them for the final answer.

Question 9. Divide by using Dhwajank Method \( 18542 \div 52 \)
Answer:
To divide \( 18542 \div 52 \) using the Dhwajank Method:
In this method, the divisor is split into a main digit and a flag digit.
Here, divisor = 52. Let the main digit be 5 and the flag digit be 2.

Arrange the division as follows:
The flag digit (2) is placed above the main digit (5). The dividend (18542) is written, with the last digit (2) separated by a vertical line, as the flag digit has one digit.
\[ \begin{array}{c|c c c|c} \multicolumn{2}{r}{3} & 5 & 6 & \text{Quotient} \\ \cline{2-5} 5^2 & 1 & 8 & 5 & 4^2 \\ \multicolumn{2}{r}{} & & & -6 \\ \multicolumn{2}{r}{} & & & \quad 42 \\ \multicolumn{2}{r}{} & & & -12 \\ \cline{5-5} \multicolumn{2}{r}{} & & & \quad 30 \\ \multicolumn{2}{r}{3} & 4 & 4 & \\ \multicolumn{2}{r}{3} & 5 & 6 & \\ \end{array} \]
(Note: The visual layout is simplified here, but the steps below explain the process.)

Step-by-step calculation:
1. The main divisor is 5, and the flag digit is 2. The last digit of the dividend (2) is separated in the remainder column.
2. Take the first two digits of the dividend, 18. Divide 18 by 5. \( 18 \div 5 = 3 \) with a remainder of 3. Write 3 as the first digit of the quotient and place the remainder 3 before 5 to make it 35.
3. The new dividend is 35. Subtract the product of the last quotient digit and the flag digit: \( 35 - (3 \times 2) = 35 - 6 = 29 \).
4. Divide 29 by 5. \( 29 \div 5 = 5 \) with a remainder of 4. Write 5 as the next quotient digit and place the remainder 4 before 4 to make it 44.
5. The new dividend is 44. Subtract the product of the last quotient digit and the flag digit: \( 44 - (5 \times 2) = 44 - 10 = 34 \).
6. Divide 34 by 5. \( 34 \div 5 = 6 \) with a remainder of 4. Write 6 as the next quotient digit and place the remainder 4 before the last separated digit (2) to make it 42.
7. The final remainder is 42. Subtract the product of the last quotient digit and the flag digit: \( 42 - (6 \times 2) = 42 - 12 = 30 \).

Thus, the Quotient = 356 and Remainder = 30.
In simple words: This method for dividing big numbers uses a "flag" digit. We divide the main part of the number by a single digit, find a quotient, and then adjust it by subtracting a value found by multiplying the flag digit with the quotient digit. We repeat this until we find the final answer and any leftover amount.

🎯 Exam Tip: In the Dhwajank method, carefully manage the main divisor and flag digit. Remember to calculate the 'modified dividend' by subtracting the product of the flag digit and the previous quotient digit before performing the next division step.

Question 11. Solve \( 3732 \div 42 \) by 'Dhwajank' Sutra.
Answer:
To solve \( 3732 \div 42 \) using the Dhwajank Sutra:
Here, the divisor is 42. Let the main digit be 4 and the flag digit be 2.

Step 1: Set up the division. Place the flag digit (2) above the main digit (4). Separate the last digit of the dividend (2) in the remainder column.
\( 4^2 | 373 | 2 \)

Step 2: Take the first part of the dividend, 37. Divide 37 by the main digit 4.
\( 37 \div 4 = 8 \) (Quotient) with a remainder of 5. (Hint: \( 37 \div 4 \), first digit of quotient = 8, remainder = 5).
Write 8 as the first digit of the quotient. Place the remainder 5 before the next digit 3, making the new dividend 53.

Step 3: Calculate the modified dividend. Subtract the product of the previous quotient digit (8) and the flag digit (2) from the new dividend (53).
Modified dividend = \( 53 - (8 \times 2) = 53 - 16 = 37 \). (Hint: Modified dividend = \( 53 - 8 \times 2 = 37 \)).

Step 4: Divide the modified dividend (37) by the main digit 4.
\( 37 \div 4 = 8 \) (Quotient) with a remainder of 5. (Hint: \( 37 \div 4 \), second digit of quotient = 8, remainder = 5).
Write 8 as the next digit of the quotient. Place the remainder 5 before the last digit 2, making the new dividend 52.

Step 5: Calculate the final remainder. The new dividend is 52. Subtract the product of the last quotient digit (8) and the flag digit (2).
Modified dividend (final remainder) = \( 52 - (8 \times 2) = 52 - 16 = 36 \). (Hint: Modified dividend = \( 52 - 8 \times 2 = 36 \)).

The process is complete as there are no more digits in the dividend.
Thus, the Quotient = 88 and Remainder = 36.
In simple words: This division method works by splitting the divisor into two parts: a main number and a "flag" number. We divide normally using the main number, but then we adjust our leftover amount by subtracting a value from the "flag" number. This continues step-by-step until we find our final answer and any remaining amount.

🎯 Exam Tip: Carefully manage the 'new dividend' and 'modified dividend' at each step in the Dhwajank method; ensure you subtract the product of the last quotient digit and the flag digit from the new dividend before dividing by the main digit.