Mathematics

Vedic Math Examples

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12 November 2020

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Introduction

We learnt about the origin of Vedic Math, the sutras and sub-sutras contained in it and how they help tremendously in reducing the effort and time needed to do arithmetic calculations.

In this article, let us see more examples and illustrations based on the various sutras discussed in the previous article.

 

Also read:


What are Sutras? with Examples

In this section, we will see sample problems along with step by step solutions based on the sutras presented in Vedic Math.

Ekadhikena Purvena

This Sutra helps in finding out the square of a number ending with 5.

Problem 1

 

 

Find the square of 45

Solution: 

Step 1: The last part of the answer has to be 25

Step 2: Multiply the remaining number (excluding 5) in the original number by its successor. So, in this case, we multiply 4 and 5. This gives us 20.

Step 3: Join the numbers obtained in step 2 and step 1, in that order.

Answer: 2025
Problem 2

 

 

Find the square of 115

Solution: 

Step 1: The last part of the answer has to be 25

Step 2: Multiply the remaining number (excluding 5) in the original number by its successor. In this case, we multiply 11 and 12. This gives us 132.

Step 3: Join the numbers obtained in step 2 and step 1, in that order.

Answer: 13225

Nikhilam Navatashcaramam

This Sutra helps in finding the product of two numbers closer to 10 or to any power of 10.

Problem 1

 

 

Find the product of 95 and 96

Solution: 

Step 1: Find the difference between the numbers and their closest multiple of 10.

We get, 

100 – 95 = 5

100 – 96 = 4

Step 2: Multiply the 2 numbers obtained in step 1. 

5 × 4 = 20

Step 3: Find the difference between one of the given numbers and the result obtained in step1 for the second number.

We get, 

95 – 4 = 91

Or

96 – 5 = 91

Step 4: Join the numbers obtained in step 3 and step 2, in that order.

Answer: 9120
Problem 2

 

 

Find the product of 103 and 105

Solution: 

Step 1: Find the difference between the numbers and their closest multiple of 10.

We get, 

100 – 103 = -3

100 – 105 = -5

Step 2: Multiply the 2 numbers obtained in step 1. 

-3 × -5 = 15

Step 3: Find the difference between one of the given numbers and the result obtained in step1 for the second number.

We get, 

103 - (-5) = 108

Or

105 - (-3) = 108

Step 4: Join the numbers obtained in step 3 and step 2, in that order.

Answer: 10815
Problem 3

 

 

Find the product of 103 and 98

Solution: 

Step 1: Find the difference between the numbers and their closest multiple of 10.

We get, 

100 – 103 = -3

100 – 98 = 2

Step 2: Multiply the 2 numbers obtained in step 1. 

-3 × 2 = -06

This can also be written as 06  (By Vinculum method)

Step 3: Find the difference between one of the given numbers and the result obtained in step1 for the second number.

We get, 

103 - (2) = 101

Or

98 - (-3) = 101

Step 4: Join the numbers obtained in step 3 and step 2, in that order.

We get, \(\begin{align}101\overline {06} \end{align}\)

Since we have a negative component in the answer (with a bar in the top), we expand the number according to their place values and add them.

\(\begin{align}101\overline {06} \end{align}\)  = 10000 + 0 + 100 - 0 - 6

Answer: 10094

Ekayunena Purvena

This Sutra helps in finding the product of two numbers when one of the numbers is entirely made of 9

Problem 1

 

 

Find the product of 52 and 99

Solution:

In this example, the multiplicand 52 and the multiplier 9 have the same number of digits.

Step 1: Find the predecessor of the number (which is not entirely made of 9). In this case we get 52 -1 = 51

Step 2: Subtract the same number from the multiple of 10 which is closest to the number entirely made up of 9

100 - 52 = 48

Step 3: Join the numbers obtained in step 1 and step 2, in that order.

Answer: 5148
Problem 2

 

 

Find the product of 316 and 999

Solution:

In this example, the multiplicand 316 and the multiplier 999 have the same number of digits.

Step 1: Find the predecessor of the number (which is not entirely made of 9). In this case, we get 316 -1 = 315

Step 2: Subtract the same number from the multiple of 10, which is closest to the number entirely made up of 9

1000 - 316 = 684

Step 3: Join the numbers obtained in step 1 and step 2 in that order.

Answer: 315684
Problem 3

 

 

Find the product of 52 and 9

Solution:

In this example, the multiplicand 52 and the multiplier 9 do not have the same number of digits. The multiplier has a lesser number of digits than the multiplicand.

Step 1: Subtract the successor of the first digit of the multiplicand from the multiplicand itself.

Successor of 5 is 6.

So, we get 52 - 6 = 46

Step 2: Subtract the last digit of the multiplicand from 10

10 - 2 = 8

Step 3: Join the numbers obtained in step 1 and step 2 in that order.

Answer: 468
Problem 4

 

 

Find the product of 52 and 999

Solution:

In this example, the multiplicand 52 and the multiplier 999 do not have the same number of digits. Multiplier has more digits than the multiplicand.

Step 1: Find the predecessor of the multiplicand

So, we get 52 - 1 = 51

Step 2: The second part of the number is 9

Step 3: Find the compliment of the multiplicand

100 - 52 = 48

Step 3: Join the numbers obtained in step 1 and step 2 and step3, in that order.

Answer: 51948

Anurupyena

This Sutra helps in finding the product of two numbers that are close to each other.

Problem 1

 

 

Find the product of 37 and 35

Solution: 

Step 1: Choose the base as the nearest multiple of 10. In this case, we can choose the base as 40 (as opposed to choosing the nearest power of 10 as base in Nikhilam Navatashcaramam)

4 ×10 = 40. Here 4 is the factor.

Step 2: Find the difference between the numbers and the base

We get, 

40 – 37 = 3

40 – 35 = 5

Step 3: Multiply the 2 numbers obtained in step 2. 

3 × 5 = 15

Retain 5 and carry over 1

Step 4: Find the difference between one of the given numbers and the result obtained in step1 for the second number.

We get, 

37 – 5 = 32

Or

35 – 3 = 32

Step 5: Multiply/Divide (based on how the factor is arrived at, in step 1) the number obtained in step 4 by the factor identified in step 1

So, we get 32 × 4 = 128

Step 6: Add the number obtained in step 5 to the carryover (if any) from step 3

128 + 1 = 129

Step 7: Join the numbers obtained in step 6 and step 3, in that order.

Answer: 1295
Problem 2

 

 

Find the product of 377 and 355

Solution: 

Step 1: Choose the base as the nearest multiple of 10. In this case, we can choose the base as 400 (as opposed to choosing the nearest power of 10 as base in Nikhilam Navatashcaramam)

4 ×100 = 400. Here 4 is the factor.

Step 2: Find the difference between the numbers and the base

We get, 

400 – 377 = 23

400 – 355 = 45

Step 3: Multiply the 2 numbers obtained in step 2. 

23 × 45 = 1035

Retain 35 and carry over 10

Step 4: Find the difference between one of the given numbers and the result obtained in step1 for the second number.

We get, 

377 – 45 = 332

Or

355 – 23 = 332

Step 5: Multiply/Divide (based on how the factor is arrived at, in step 1) the number obtained in step 4 by the factor identified in step 1

So, we get 332 × 4 = 1328

Step 6: Add the number obtained in step 5 to the carry over (if any) from step 3

1328 + 10 = 1338

Step 7: Join the numbers obtained in step 6 and step 3, in that order.

Answer: 133835

Nikhilam

This Sutra helps in dividing a number by a divisor, which is closer to 10 or any power of 10 and is lesser than 10.

Problem 1

 

 

Divide 224 by 9

Solution: 

Step 1: Find the difference between the divisor and the closest power of 10

So, we get 10 - 9 = 1

Step 2: Dividend is to be split into Quotient and Remainder. The remainder should have the same number of digits as in the divisor (as read from the right end of the number). The remaining portion of the dividend is to be

considered as the Quotient.

In this case, Dividend is 224. So, Remainder is 4 (as the divisor has only one digit), and 22 is the Quotient

Step 3: Multiply the one’s digit of the quotient with the number obtained in step 1. Add this number to the units digit of the quotient.

  • 2 × 1 = 2
  • 2 + 2 = 4

So quotient is 24

Step 3: Multiply the units digit of the number obtained in step 3 to the quotient obtained in step 1, and add it to the remainder

4 + 4 = 8

Step 4: The numbers obtained in step 3 and step 4 are the quotient and remainder respectively

Answer: Quotient: 24 | Remainder: 8

Paravartya 

This Sutra helps in dividing a number by a divisor which is closer to 10 (or any power of 10) and is greater than 10.

Problem 1

 

 

Divide 224 by 12

Solution: 

Step 1: Discard the first digit of the divisor and transpose the remaining digits

Discard 1 from 12

Transposing 2, we get \(\begin{align}\overline {2} \end{align}\)

Step 2: Dividend is to be split into Quotient and Remainder. Remainder should have the same number of digits as in the divisor (as read from the right end of the number) from step 1. The remaining portion of the dividend is to

be considered as the Quotient.

In this case, Dividend is 224. So, Remainder is 4 (as the divisor has 1 digit) and 22 is the Quotient

Step 3: Multiply the one’s digit of the quotient with the number obtained in step 1. Add this number to the units digit of the quotient.

2 × \(\begin{align}\overline {2} \end{align}\) = \(\begin{align}\overline {4} \end{align}\)

22 + \(\begin{align}\overline {4} \end{align}\) = 2\(\begin{align}\overline {2} \end{align}\)

Quotient is 2\(\begin{align}\overline {2} \end{align}\)=20 -2(By Vinculum method)

Quotient: 18

Step 3: Multiply the units digit of the number obtained in step 3 to the quotient obtained in step 1, and add it to the remainder

\(\begin{align}\overline {2} \end{align}\) × \(\begin{align}\overline {2} \end{align}\) = 4

4 + 4 = 8

Step 4: The numbers obtained in step 3 and step 4 are the quotient and remainder respectively

Answer: Quotient: 18 | Remainder: 8

Vedic Math Practice Worksheet

Using the examples shown above, try to identify the correct Sutra and solve the problems following the specified rules.

1 65 × 65
2 18 × 99
3 97 × 98
4 98 ×102
5 22 × 999
6 56 ×57
7 105 × 105
8 3456 ÷ 9
9 1003 ÷ 11
10 657 × 680
11 234 × 99
12 225 × 225
13 245 ÷ 32
14 3456 × 3600
15 1455 ÷ 78

Summary

All the Vedic Math techniques can be best understood by focusing on the patterns of the numbers in the given arithmetic problem. There is a definite relation between the numbers in the problems and the answers. 
Specific techniques may seem harder on the first go. But constant practice will help us to master these techniques. Once mastered, the results of employing these techniques are astounding. 
I request everyone to explore all the available sutras and sub-sutras and make the best use of this priceless gift that has been passed down to us by ancient Indians.

Written by Gayathri Sivasubramanian, Cuemath Teacaher


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FAQs

Why is the importance of Vedic Math?

Vedic Math has numerous techniques and rules that help us perform various arithmetic calculations in time. It reduces the calculating time from many minutes to very few seconds.

Who invented Vedic Math?

The techniques in Vedic Math Indian Mathematician are said to have been put together by Shri Bharati Krishna Tirthaji. It is said to have been obtained from Atharva Veda.

How do you find the square of a number using Vedic Math?

We can use this Vedic Math tricks to find the square of any number ending in 5:

Problem 1

 

 

Find the square of 45

Solution: 

Step 1: The last part of the answer has to be 25

Step 2: Multiply the remaining number (excluding 5) in the original number by its successor. So, in this case, we multiply 4 and 5. This gives us 20.

Step 3: Join the numbers obtained in step 2 and step 1, in that order.

Answer: 2025

External References

Wikipedia.org - Vedic Mathematics 


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