RBSE Solutions Class 6 Maths Chapter 7 Vedic Mathematics Important Questions

Multiple Choice Questions

Question 1. Meaning of ekadhiken is :
(i) one less
(ii) one more
(iii) equal
(iv) zero.
Answer: (ii) one more
In simple words: The term "ekadhiken" in Vedic Mathematics means "by one more than the previous one". It tells us to increase a number or digit by one.

🎯 Exam Tip: Remember that Vedic Math terms like 'ekadhiken' have specific meanings; knowing these definitions is key to solving problems.

Question 2. Ekadhiken number of 7 will be :
(i) 6
(ii) 7
(iii) 8
(iv) 9
Answer: (iii) 8
In simple words: To find the ekadhiken of the number 7, we simply add one to it. So, 7 plus 1 equals 8.

🎯 Exam Tip: Ekadhiken of a digit or number simply means adding 1 to it. It's a fundamental operation in Vedic Mathematics.

Question 3. In 289, ekadhiken of 2 will be :
(i) 389
(ii) 489
(iii) 189
(iv) 589
Answer: (i) 389
In simple words: When we are asked for the "ekadhiken of 2" in the number 289, we replace the digit 2 with "2 plus 1". So, 2 becomes 3, changing 289 to 389.

🎯 Exam Tip: When applying 'ekadhiken' to a specific digit within a number, only that digit is increased by one, leaving other digits as they are.

Question 4. In 46, ekadhiken poorven of 4 will be :
(i) 56
(ii) 36
Answer: (ii) 36
In simple words: The term "ekadhiken poorven" means to make the preceding digit one more. For example, if we consider a digit, we look at the digit before it and increase that digit by one.

🎯 Exam Tip: Always understand if the operation means 'one more' (ekadhiken) or 'one less' (eknunen), and how 'poorven' (previous) affects which digit is changed.

Question 6. Sign of vinkulum is :
(i) +
(ii) –
(iii) x
(iv) +
Answer: (ii) –
In simple words: The sign of vinkulum is a small horizontal bar placed over a digit or number. This bar means the number underneath it is negative.

🎯 Exam Tip: The vinkulum bar is crucial in Vedic Mathematics for representing negative numbers without using a minus sign in front, simplifying mental calculations.

Question 7. In 46, eknunen poorven of 6 will be :
(i) 26
(ii) 46
(iii) 76
(iv) 36
Answer: (iv) 36
In simple words: "Eknunen poorven of 6" means we decrease the digit that comes *before* 6 by one. In the number 46, the digit before 6 is 4. So, we make 4 one less, which becomes 3, resulting in 36.

🎯 Exam Tip: Eknunen poorven involves reducing the digit immediately preceding the specified digit by one, a key concept for subtractions.

Question 8. Parammitra digit of 4 will be :
(i) 2
(ii) 3
(iii) 6
(iv) 7
Answer: (iii) 6
In simple words: A "Parammitra digit" is a complementary digit. For any single digit, its Parammitra digit is the number that adds up to 10 with it. For the digit 4, its Parammitra digit is 6 because 4 plus 6 equals 10.

🎯 Exam Tip: Parammitra digits are also called friend numbers or complements to 10. They are very helpful in quick subtractions and other Vedic math calculations.

Very Short Answer Type Questions

Question 1. Write ekadhiken poorven of digit 1 in 125.
Answer: To find the ekadhiken poorven of digit 1 in 125, we place an Ekadhiken mark (a dot) over the digit immediately preceding 1. Since 1 is the first digit, we consider a conceptual 0 before it. So, we place the dot over this implied 0.
Ekadhiken poorven of digit 1 in 125 = \( \overset { \text{.} }{ 0 } 125 \)
\( \implies \) This means the digit before 1 (which is 0) becomes one more.
\( \implies 0+1 = 1 \)
So, the resulting number is 1125.
In simple words: When we say "ekadhiken poorven" of a digit, it means we add one to the digit just before it. For 1 in 125, we think of a zero before it, then we add 1 to that zero, making it 1125.

🎯 Exam Tip: Remember that for the first digit of a number, the 'poorven' (previous) digit is considered to be zero when applying operations like ekadhiken poorven.

Question 2. Write eknunen poorven of digit 6 in 2675.
Answer: To find the eknunen poorven of digit 6 in 2675, we place an Eknunen mark (a small dot below) over the digit immediately preceding 6. The digit before 6 is 2.
Eknunen poorven of digit 6 in 2675 = \( 2\underset { \text{.} }{ 6 } 75 \)
\( \implies \) This means the digit 2 (poorven of 6) becomes one less.
\( \implies 2-1 = 1 \)
So, the resulting number is 1675.
In simple words: To find "eknunen poorven of 6" in 2675, we take the digit before 6, which is 2, and reduce it by one. So, 2 becomes 1, making the number 1675.

🎯 Exam Tip: The 'eknunen poorven' operation is typically used in subtraction. It helps simplify the borrowing process by reducing the previous digit.

Question 3. What will be the deviation of number 7 and 93?
Answer: Deviation is found by subtracting the number from its nearest base (like 10, 100, 1000).
For number 7:
Base for 7 = 10
Deviation = \( 10 - 7 = 3 \)
For number 93:
Base for 93 = 100
Deviation = \( 100 - 93 = 7 \)
In simple words: Deviation tells us how far a number is from a base number like 10 or 100. For 7, it's 3 away from 10. For 93, it's 7 away from 100.

🎯 Exam Tip: Always choose the nearest power of 10 (10, 100, 1000, etc.) as the base to calculate deviation accurately in Vedic Multiplication methods like Nikhilam.

Short/Long Answer Type Questions

Question 1. Find the sum of 65 + 68, using the formula of ekadhiken poorven.
Answer: We will add 65 and 68 using the Ekadhiken Poorven method. This method helps manage carries in addition.
Let's write the numbers vertically:
\[\begin{array}{r} 65 \\ +\quad \overset{\bullet}{6}8 \\ \hline 133 \\ \end{array}\]
Step 1: Add the unit digits: \( 5 + 8 = 13 \). Write down 3 in the units place. Mark 'ekadhiken' (a dot) on the 'poorven' (previous) digit of the tens place (which is the 6 in 68).
Step 2: Add the tens digits along with the ekadhiken mark. The 6 in 65 plus the 6 in 68 (which now acts as 7 due to the ekadhiken mark). So, \( 6 + 7 = 13 \). Write down 3 in the tens place. Mark 'ekadhiken' on the 'poorven' digit of the hundreds place (which is conceptually 0 before 65). This conceptually makes the hundreds digit 1.
Step 3: Combine the hundreds digit (which became 1 due to the ekadhiken mark) with the result.
So, the sum is 133.
In simple words: To add 65 and 68 using this method, we add units first (5+8=13), keep 3 and put a dot on the 6 of 68 (making it 7). Then, we add the tens digits (6+7=13), keep 3 and put a dot on the number before 65 (making it 1). The final answer is 133.

🎯 Exam Tip: When using ekadhiken poorven for addition, remember to place the ekadhiken dot on the digit *preceding* the column where a carry-over occurs, effectively adding one to it.

Question 2. Find the subtraction 74 – 69, using the formula of eknunen poorven.
Answer: We will subtract 69 from 74 using the Eknunen Poorven method. This method helps simplify borrowing.
Let's write the numbers vertically:
\[\begin{array}{r} 74 \\ -\quad \underset{\text{•}}{6}9 \\ \hline 05 \\ \end{array}\]
Step 1: Subtract the unit digits: \( 4 - 9 \). Since 4 is smaller than 9, we cannot subtract directly. Instead, we use the Parammitra (complementary) digit of 9, which is \( 10 - 9 = 1 \). Add this 1 to 4: \( 4 + 1 = 5 \). Write 5 in the units place.
Step 2: When we use a Parammitra digit, we mark 'eknunen' (a dot below) on the 'poorven' (previous) digit of the tens place in the top number (which is 7). So, 7 becomes \( \underset { \text{.} }{ 7 } \) (meaning 6).
Step 3: Now, subtract the tens digits: \( 6 - 6 = 0 \). Write 0 in the tens place.
So, the difference is 05, or simply 5.
In simple words: To subtract 69 from 74, first we cannot do 4 minus 9. So, we find the "friend" of 9 (which is 1), and add it to 4 (4+1=5). Then, because we used a friend number, we make the digit before 7 (which is 6) one less. So, 7 becomes 6. Finally, we do 6 minus 6, which is 0. The answer is 05 or 5.

🎯 Exam Tip: Eknunen poorven simplifies subtraction by converting borrowing into reducing the previous digit and adding the complement of the subtrahend digit.

Question 4. Solve 5 x 8 by formula of Nikhilam.
Answer: We will solve \( 5 \times 8 \) using the Nikhilam Navatashcaramam Dashatah (All from 9 and the Last from 10) formula, taking base as 10.
Numbers and their deviations from base 10 are:
Number    Deviation
5               -5 (since \( 5 - 10 = -5 \))
8               -2 (since \( 8 - 10 = -2 \))
The calculation is done in two parts: Left Hand Side (LHS) and Right Hand Side (RHS).
LHS: Cross-subtract one deviation from the other number. Either \( 5 - 2 \) or \( 8 - 5 \). Both give 3.
RHS: Multiply the deviations: \( (-5) \times (-2) = 10 \).
So we have \( 3 | 10 \).
Since the base is 10, the RHS should only have one digit. We carry over the '1' from 10 to the LHS.
LHS becomes \( 3 + 1 = 4 \).
RHS becomes \( 0 \).
Combining these, the product is 40.
In simple words: To multiply 5 by 8 using the Nikhilam rule, we first see how much each number is different from 10 (our base). 5 is -5 from 10, and 8 is -2 from 10. We then add one number to the other's difference (5-2=3). And we multiply the differences (-5 x -2 = 10). This gives us "3 and 10". Since 10 has two digits, we move the '1' to the '3' (making 4), and leave '0'. So the answer is 40.

🎯 Exam Tip: The Nikhilam Sutra is perfect for multiplying numbers close to a base (like 10, 100). Remember to adjust the RHS to match the number of zeros in the base by carrying over if needed.