RBSE Solutions Class 5 Maths Chapter 4 Vedic Mathematics Exercise 4.4

Rajasthan Board RBSE Class 5 Maths Chapter 4 Vedic Mathematics Ex 4.4

Question 1. Multiply by Nikhilam sutra on base 10-
(1) 12 x 9
(2) 15 x 12
(3) 13 x 17
(4) 8 x 9
(5) 14 x 11
(6) 9 x 16
(7) 12 x 13
(8) 13 x 10
Answer:
(1) \( 12 \times 9 \)
We use the Nikhilam method with base 10.
The numbers are 12 and 9. Their deviations from base 10 are \( +2 \) and \( -1 \) respectively.
\[ \begin{array}{ccc} 12 & & +2 \\ 9 & & -1 \\ \hline 12-1 & | & (+2) \times (-1) \\ 11 & | & -2 \end{array} \]
Since the right side is a negative number, we borrow 1 from the left side. This '1' represents 10 units in the right column because our base is 10.
So, the left side becomes \( 11 - 1 = 10 \).
The right side becomes \( 10 - 2 = 8 \).
Combining these parts, we get \( 10 \text{ | } 8 \).
Thus, \( 12 \times 9 = 108 \).
In simple words: We multiply 12 by 9 using the Nikhilam method. First, we find the deviations from the base 10, which are +2 and -1. Then, we multiply these deviations to get -2. For the left side, we cross-add (12 - 1 = 11). Since we can't have a negative number on the right, we borrow 1 from the left side (making it 10) and add it to the right side (10 - 2 = 8). This gives us 108. This method makes multiplication easier by breaking it down into smaller steps.

🎯 Exam Tip: Remember to adjust the left side when borrowing to make the right side positive; this is crucial for accurate Vedic multiplication using Nikhilam sutra.

Answer:
(2) \( 15 \times 12 \)
We use the Nikhilam method with base 10.
The numbers are 15 and 12. Their deviations from base 10 are \( +5 \) and \( +2 \) respectively.
\[ \begin{array}{ccc} 15 & & +5 \\ 12 & & +2 \\ \hline 15+2 & | & 5 \times 2 \\ 17 & | & 10 \end{array} \]
Since our base is 10, only one digit should be on the right side. We carry over the '1' from '10' to the left side.
So, the left side becomes \( 17 + 1 = 18 \).
The right side becomes the remaining digit, which is \( 0 \).
Combining these parts, we get \( 18 \text{ | } 0 \).
Thus, \( 15 \times 12 = 180 \).
In simple words: For \( 15 \times 12 \), we use base 10. The deviations are +5 and +2. Multiplying them gives 10. For the left side, we add 15 + 2 to get 17. Because our base (10) has only one zero, we keep only one digit (0) on the right side and carry over the '1' to the left side. So, 17 + 1 becomes 18. Combining the left and right sides gives us 180. This carrying over ensures the number correctly reflects the product.

🎯 Exam Tip: When the product of deviations has more digits than the number of zeros in the base, always carry over the extra digits to the left side.

Answer:
(3) \( 13 \times 17 \)
We use the Nikhilam method with base 10.
The numbers are 13 and 17. Their deviations from base 10 are \( +3 \) and \( +7 \) respectively.
\[ \begin{array}{ccc} 13 & & +3 \\ 17 & & +7 \\ \hline 13+7 & | & 3 \times 7 \\ 20 & | & 21 \end{array} \]
Since our base is 10, only one digit should be on the right side. We carry over the '2' from '21' to the left side.
So, the left side becomes \( 20 + 2 = 22 \).
The right side becomes the remaining digit, which is \( 1 \).
Combining these parts, we get \( 22 \text{ | } 1 \).
Thus, \( 13 \times 17 = 221 \).
In simple words: For \( 13 \times 17 \), using base 10, the deviations are +3 and +7. Their product is 21. The left side is 13 + 7, which equals 20. As the base is 10, we keep '1' on the right side and carry '2' to the left side, making it 20 + 2 = 22. So, the final product is 221. This method helps simplify larger multiplications.

🎯 Exam Tip: Practice quickly identifying the deviations from the chosen base, as this is the first and most crucial step in the Nikhilam method.

Answer:
(4) \( 8 \times 9 \)
We use the Nikhilam method with base 10.
The numbers are 8 and 9. Their deviations from base 10 are \( -2 \) and \( -1 \) respectively.
\[ \begin{array}{ccc} 8 & & -2 \\ 9 & & -1 \\ \hline 8-1 & | & (-2) \times (-1) \\ 7 & | & 2 \end{array} \]
Since the product of deviations is positive and a single digit, no borrowing or carrying is needed.
Combining the left and right parts, we get \( 7 \text{ | } 2 \).
Thus, \( 8 \times 9 = 72 \).
In simple words: To multiply \( 8 \times 9 \) using base 10, the deviations are -2 and -1. Multiplying these gives +2. For the left side, we subtract (8 - 1 = 7). Since the right side is positive, we combine 7 and 2 directly to get 72. This shows how negative deviations work in the Nikhilam method.

🎯 Exam Tip: Remember that multiplying two negative deviations results in a positive product, which simplifies the combining step.

Answer:
(5) \( 14 \times 11 \)
We use the Nikhilam method with base 10.
The numbers are 14 and 11. Their deviations from base 10 are \( +4 \) and \( +1 \) respectively.
\[ \begin{array}{ccc} 14 & & +4 \\ 11 & & +1 \\ \hline 14+1 & | & 4 \times 1 \\ 15 & | & 4 \end{array} \]
Since the product of deviations is positive and a single digit, no borrowing or carrying is needed.
Combining the left and right parts, we get \( 15 \text{ | } 4 \).
Thus, \( 14 \times 11 = 154 \).
In simple words: For \( 14 \times 11 \) with base 10, the deviations are +4 and +1. Their product is 4. The left side is 14 + 1, which is 15. Since the product of deviations (4) is a single digit and positive, we simply combine the left and right parts to get 154. This method helps quickly multiply numbers close to a base.

🎯 Exam Tip: For numbers above the base, cross-addition can be done using either number plus the other's deviation (e.g., \( 14+1 \) or \( 11+4 \)); both will give the same result for the left part.

Answer:
(6) \( 9 \times 16 \)
We use the Nikhilam method with base 10.
The numbers are 9 and 16. Their deviations from base 10 are \( -1 \) and \( +6 \) respectively.
\[ \begin{array}{ccc} 9 & & -1 \\ 16 & & +6 \\ \hline 9+6 & | & (-1) \times 6 \\ 15 & | & -6 \end{array} \]
Since the right side is a negative number, we borrow 1 from the left side. This '1' represents 10 units in the right column because our base is 10.
So, the left side becomes \( 15 - 1 = 14 \).
The right side becomes \( 10 - 6 = 4 \).
Combining these parts, we get \( 14 \text{ | } 4 \).
Thus, \( 9 \times 16 = 144 \).
In simple words: To multiply \( 9 \times 16 \) using base 10, the deviations are -1 and +6. Multiplying these gives -6. For the left side, we add 9 + 6 to get 15. As we cannot have a negative number on the right, we borrow 1 from the left side (making it 14) and add 10 to the right side (10 - 6 = 4). This results in 144. This technique is important for handling mixed deviations.

🎯 Exam Tip: Be extra careful when one deviation is positive and the other is negative, as their product will be negative, requiring a borrowing step.

Answer:
(7) \( 12 \times 13 \)
We use the Nikhilam method with base 10.
The numbers are 12 and 13. Their deviations from base 10 are \( +2 \) and \( +3 \) respectively.
\[ \begin{array}{ccc} 12 & & +2 \\ 13 & & +3 \\ \hline 12+3 & | & 2 \times 3 \\ 15 & | & 6 \end{array} \]
Since the product of deviations is positive and a single digit, no borrowing or carrying is needed.
Combining the left and right parts, we get \( 15 \text{ | } 6 \).
Thus, \( 12 \times 13 = 156 \).
In simple words: For \( 12 \times 13 \) with base 10, the deviations are +2 and +3. Their product is 6. The left side is 12 + 3, which is 15. Since the product of deviations is a single digit and positive, we simply combine the left and right parts to get 156. This is a straightforward application of the Nikhilam method.

🎯 Exam Tip: This method works best for numbers that are relatively close to a chosen base, making calculations simpler and faster.

Answer:
(8) \( 13 \times 10 \)
We use the Nikhilam method with base 10.
The numbers are 13 and 10. Their deviations from base 10 are \( +3 \) and \( +0 \) respectively.
\[ \begin{array}{ccc} 13 & & +3 \\ 10 & & +0 \\ \hline 13+0 & | & 3 \times 0 \\ 13 & | & 0 \end{array} \]
Since the product of deviations is zero, and the left side is 13, no borrowing or carrying is needed.
Combining the left and right parts, we get \( 13 \text{ | } 0 \).
Thus, \( 13 \times 10 = 130 \).
In simple words: To multiply \( 13 \times 10 \) using base 10, the deviations are +3 and 0. Their product is 0. The left side is 13 + 0, which is 13. Combining these gives 130. This shows that the Nikhilam method works even when one number is the base itself.

🎯 Exam Tip: Any number multiplied by the base (like 10) can be solved quickly by simply adding a zero to the end of the number.