Vedic Mathematics: Solved Examples | Improve Your Calculations: Vedic Maths (English) - Class 6 PDF Download

Addition and Subtraction

Example 1: Suppose we have to add 689 and 95.

Sol: We know 95 is nearer to 100, by keeping this in mind we can add 100 to 689 and subtract 5 later.
Hence 689 + 100 = 789 - 5 = 784, which is the required answer.

Take some other examples:
Example 2. 67 + 693 =?

Sol: It can be solved like 67 + 700 = 767 - 7 = 760.

Example 3. 454 + 27 =?

Sol: It can be solved like 454 + 30 = 484 - 3 = 481.

Similarly, subtraction operations can be solved using these tips.
Let’s take some examples:
Example 3: 367 - 37 =?

Sol: It can be solved like, 367 - 40 = 327 + 3 = 330.

Example 4:  289 - 58 = ?

Sol: It can be solved like, 289 - 60 = 229 + 2 = 231.

Sum of Series

For computing the sum of a series like

1. Consecutive numbers:

1, 2, 3, 4, 5 etc; or 12, 13, 14...

2. Consecutive EVEN numbers:
2, 4, 6, 8, etc or 12, 14, 16...

3. Consecutive ODD numbers:
3, 5, 7, 9, etc or 13, 15, 17,

We use the following formula
S = (F + L) (N) / 2
Where,
S = Sum of all numbers
F = First number in sequence
L = Last number in sequence
N = Number of Terms in sequence

Multiplication

Methods For Multiplication Of Number By Multiple Of 10 (I.e. By 10, 100, 1000 Etc.)

• This is quite simple just put the same number of zeroes behind the number as behind 1.
• Example: 23 × 100
• Here there are two zeroes behind 1, hence by putting two zeroes behind 23. We will get the answer.
• Therefore 23 × 100 = 2300
  Take another example e.g. 45 × 1000 = 45000

Methods For Multiplication Of Number By 5

• Let n is an number which is to be multiplied by 5.
  i.e. n x 5 = ?
• Now n could be either even or odd.
• Now if n is even, just half the number and put zero behind that.

Example: 44 x 5
Here 44 is an even number, now half of the 44 is 22 and by putting ‘0’ it become 220. Hence answer is 220.

• Now if n is odd, subtract one from n, and half that number (i.e. n - 1) and put five behind that.
  For example 47 x 5
• Here 45 is an odd number, now 47 - 1 = 46, half of this (46 / 2) is 23 and by putting ‘5’ behind that it become 235.
•  Hence, the answer is 235.

 Methods For Multiplication By 25

We know that 25 = 100 / 4, hence to ease the computation, multiply the number by 100 (it is very simple just put two zeroes at the end of the number) and then divide the number by 4.

Lets take one example

• 76 x 25 = ?
• Now first multiply 76 by 100 i.e. 76 x 100 = 7600
• Now divide 7600 by 4 i.e.    = 1900, hence the answer.

Methods For Multiplication By 25

As 50 =   , hence to ease the computation, multiply the number by 100 and then divide it by 2.

For example :

• 88 x 50 = ?
• First multiply 88 by 100, i.e. 88 x 100 = 8800.
• Now divide it by 2, i.e. 8800 by 2 i.e.   4400, hence the answer.

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Methods For Multiplication By 125

We know that 125 =   , hence to ease the computation, multiply the number by 1000 (it is very simple just put three zeroes at the end of the number) and then divide the number by 8.

Lets take one example
48 x 125 = ?

• Now first multiply 48 by 1000 i.e. 48 x 1000 = 48000
• Now divide 48000 by 8 i.e.   = 6000, hence the answer.

Methods For Multiplication By Numbers From 11 To 19

Multiplication By 11

Rule:

1. Prefix a zero to the multiplicand
2. Write down the answer one figure at a time, from right to left as in any multiplication.
The figures of the answer are obtained by adding to each successive digit of the multiplicand its
right neighbour. Remember the right neighbour is the right, (i.e., the correct) neighbour to be
added.
Example : 123 × 11 =?

Step 1: Prefix a zero to the multiplicand so that it reads 0123.

Step 2:

Therefore, 123 × 11 = 1353 (which you can easily verify by a conventional multiplication).

Multiplication By 12

The method is exactly the same as in the case of 11 except that you double each number before
adding the right neighbour.

Example: 13 × 12 = ?

Step 1: Prefix a zero to the multiplicand so that it reads 013.

Step 2:

Therefore, 13 × 12 = 156 (which you can again verify by a conventional multiplication).

Multiplication From 13 To 19

The reason why the rule is different for multiplication by 11 and by 12 is obviously because the right digits are different.

• The right digit, we could call the Parent Index Number (PIN). Thus in 11, the PIN is 1, and in 12 it is
  2. (In 13, it is 3; in 14, it is 4 etc.) When the PIN is 1, we are simply taking each figure of the multiplication and (we could call this figure the Parent Figure - PF in short) as such and adding the right neighbour. 
• When the PIN is 2, we are doubling the PF and then adding the right neighbour.
• Obviously, if the PIN is 3 (as in 3), we would treble the PF and then add the right neighbour. 
• If the PIN is 4 (as in 14), we would quadruple (i.e. multiply by 4) the PF and then add the right neighbour. 
• If PIN is 9 (as in 19), we would multiply the PF by 9 and then add the right neighbour, what is the advantage of the method?
  We need know the tables only up to 9 and still multiply by a simple process of addition. Let us see some examples below:

Multiplication of Big Numbers

• Consider the conventional multiplication of two 2-digit numbers 12 and 23 shown below:
  
  It is obvious from the above that
• The right digit 6 of the answer is the product obtained by the "vertical" multiplication of the right digit of
  the multiplicand and of the multiplier.
• The left digit 2 of the answer is the product obtained by the "vertical" multiplication of the left digit of the
  multiplicand and of the multiplier.
• The middle digit 7 of the answer is the sum of 3 and 4. The 3 is the product of the left digit of the
  multiplicand and the right digit of the multiplier; the 4 is the product of the right digit of the multiplicand
  and the left digit of the multiplier. This means that, to obtain the middle digit, one has to multiply
  "across" and add the two products (in our example 1 × 3 + 2 × 2).
• The working in our above example can therefore be depicted as
  
  and can be summarized as
  
• When the units figure is "one" in both the numbers being multiplied, the process of multiplication is
  simplified further. Consider the following multiplication:
  
  You will notice that the middle digit of the answer is 2 × 1 + 1 × 3 i.e. (2 + 3) × 1. So, instead of multiplying
  "across" for the middle term, you could simply add the tens digit of the two numbers.
  Therefore, 31 × 21 = 6 / (2 + 3) / 1 = 651.
• Similarly, in 81 × 91, you could obtain the middle term as 17 by merely adding 8 and 9.
• Like wise, when the tens figure is "one" in both the numbers being multiplied, you could obtain the middle
  term by simply adding the units digit of the two numbers. For instance, the middle term in 12 × 17 is 2 + 7, i.e.
  9, in 8 × 12 it is 8 + 2 i.e. 10 etc.
  
  If the units figure or the tens figure is the same in the two numbers, the process of multiplication could be
  simplified as shown in the following examples:
  1. 

1. Multiplication of 2 Three-Digit Numbers

• Let us consider the multiplicand to be ABC and the multiplier to be DEF, as shown below:

• The extreme right digit of the answer is obtained (as before) by vertical multiplication as C × F.
• The extreme left digit is also obtained (as before) by vertical multiplication as A × B.
• The "middle" digits are obtained (as before) by multiplying across. Progressing one step at a time to the left, the "middle" digits are successively
  B × F + C × E
  A × F + C × D + B × E
  A × E + B × D
  The process is set out in detail for following examples below.
  

2. Multiplication Of Numbers Of Different Lengths

In the examples we saw above, both the multiplicand and the multiplier contained the same number of digits.

But what if the two numbers were to contain a different number of digits; for instance, how would we multiply 286 and 78?

Obviously we could prefix a zero to 78 (so that it becomes 078, a 3 digit number) and proceed as in any multiplication of two 3 digit numbers.

The following examples will clarify the procedure:

3. Multiplication of 9 and 7

The closest base to the two numbers in this case is 10
Therefore

9 - 1 (The remainder after subtracting the number from 10)
7 - 3 (The remainder after subtracting the number from 10)

The right hand side of the answer will be 1 × 3 = 3
The left hand side can be computed either by subtracting 3 from 9 or 1 from 7 which is 6. Therefore, the
answer is 63.

4. Numbers, which are close to 100

The closest base in this case will be 100
Therefore

Here, 6 in the first row is the difference 100 and 94 and the 13 in the second row is the difference between 100 and 87. The right hand side of the answer is obtained by the multiplication of 6 and 13 which is 78 and the left hand side is obtained by the difference between either 87 and 6 or 94 and 13, both of which give the answer 81.

5. Numbers which are Greater than the Closest Power of 10

Find the product of 108 and 112
The closest base is 100 in this case as well
Therefore

The procedure is the same with only difference being that instead of subtracting the remainder of one number from the other number, we add in this case as the number were marginally larger than the nearest power of 10. When the number of digits of the product of the remainders is greater than the power of 10 closest to the two numbers
E.g.

As the product of 16 and 8 is 128 which is a three digit number as against 2 being the power of 10 in100, we carry forward the digits on the left more than 2 digits (in this case) and add to 76, the left hand side of the answer.
Example:  When one of the number is lesser than the closest power of 10 and the other greater than the
 closest power of 10 like 88 and 106.

Sol: 
1. Identify the closest power of 10:

• The closest power of 10 to both numbers is 100.

2. Find the differences from the base (100):

• For 88, the difference from 100 is 88 - 100 = -12 (88 is 12 less than 100).

• For 106, the difference from 100 is 106 - 100 = 6 (106 is 6 more than 100).

3. Apply the Formula
$A-x\left)\right(B-x)=(A-B)+x2$(A−x)(B−x)=(A−B)+x2

Where:

• A = 88

• B = 106

• x = 100 (the base power of 10)

4. First part of the result (multiply the differences):

• Multiply the differences: (-12) × 6 = -72.

5. Second part (multiply the base by the sum of the differences):

• Sum of the differences: -12 + 6 = -6.

• Multiply this sum by the base (100): 100 × (-6) = -600.

6. Final step (combine the results):

• The first part: Base² = 100 × 100 = 10000.

• Add the result from step 4: 10000 + (-600) = 9400.

• Finally, add the result from step 5: 9400 + (-72) = 9328.

6.  Multiplication of Numbers which are not Close to the Nearest Power of 10

• Let us take the case of multiplication of 41 by 43. Going by the earlier method we have the nearest power of 10 as 100 or 10. 
• In the former case, the remainders are 59 and 57, the multiplication of which will be as tedious as the multiplication of these two numbers. 
• In the latter case, The remainders will be 31 and 33 which will be equally difficult 

Therefore, we need to look at an alternative method. In this case, we can take 50, which is a sub multiple of 100 or a multiple of 10 and proceed

Method 1: Take 50 as the base which is half of 100

Sol: 

since 50 = 100/2. We divide the left hand side number also by 2 while retaining the right hand
side. Therefore the answer will be 1763.

Method 2: We can start off instead of using 50 as the base, we can use 40 as the base

Sol:  and since 40 is 4 times 10, we multiply 44 by 4 to yield by 176 and join the right hand side to
yield 1763 - the same answer.

7.  Multiplication of Three Digit Number with 101 

For example
349 × 101
Add 3 and 9. This is 12. 2 is the middle digit of the answer, 1 is carried forward.
Take the first two digits i.e. 34. Add carry forward, 1 to it. This is 35. These are the first two digits of the answer.
Take the last two digits of the number, 49. 49 is the last two digits of the answer.
Hence answer is 35249.
Lets take few more examples

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8. Multiply Three Digit Number with  7 × 11 × 13

• What's the answer?In this case , Simply write the number twice.
• For Example: 346 × 7 × 11 × 13 = 346346
  845 × 7 × 11 × 13 = 845845

9. Multiplication of Numbers Close to Each Other

• In this shortcut, we use the application of algebra. we know a² - b² = (a + b) (a - b)
• Now as 24 × 26 can be written like (25 + 1) × (25 - 1),
  Now using the above formula it is equal to 25² - 1² = 625 - 1 = 624, which is the required answer.
  Similarly
  87 × 93 = (90 + 3) x (90 - 3) = 90² - 3² = 8100 - 9 = 8091
  28 × 32 = (30 + 2) × (30 - 2) = 30² - 2² = 900 - 4 = 896

Squares of Numbers

1. Numbers Method

​
 

2. Some other Methods of Squaring

 i. Let's start with numbers 11 - 19.
Square the ones digit. This becomes your last digit.
Step 1: Multiply the ones digit by 2. This becomes your second digit.
Step 2: Retain the tens digit (i.e., 1) and append the above results.
Take 172, for example.
7² = 49. The 9 is your last digit. Carry the 4 to the left.
7 x 2 = 14. The 4 (plus the 4 remaining from the first step) is your second digit, 8. As before carry the 1
to the left.
ii. General Method of squaring numbers 51 to 100.
Step 1: Subtract the number you want to square from 100. (100 - x)
Step 2: Take this difference and subtract it from the number you want to square. (x - (100 - x)). This is the
first 2 digits of your answer.
Step 3:  Square the difference, the answer from step 1. This is the last 2 digits of your answer.
Take 96², for example.
100 - 96 = 4.
96 - 4 = 92. You're halfway there. 92__.
4² = 16. 

Thus the answer is : 9,216.

iii. If you like that, well check this out for numbers 50-59:
It's a simple rule: 
Step 1: Square the tens digit and add to that the ones digit of your original number. That's your first
two numbers. 
Step 2: Second, square your ones digit. That's your last two numbers.
Example,
1. 5² + 4 = 29. That's our first two numbers.
2. 4² = 16. That's our last two numbers. Thus, the answer 2916.

iv. Numbers greater than 100?
Here are the steps and an example.
1. Subtract 100 from the number you want to square (x - 100).
2. Take the difference and add it to the number you want to square. (x + (x - 100)). This is the first 3 digits of
your answer.
3. Square the difference, the answer from step 1. This is the last 2 digits of your answer.

4. Example: 112²
112 - 100 = 12
112 + 12 = 124. 124__.
12² = 144. Carry the 1 over the 4.
__144
124__
12,544

v. What about numbers around 1000?
Use the same rules above. Here are 2 examples:
996²
1000 - 996 = 4
996 - 4 = 992. 992___ First 3 digits.
4² = 16. Last 3 digits.
Answer: 992,016

3. Squaring Numbers Ending in 9

In this method, count the number of nines in the question.

• Let there are n = number of nines. 
  Then square of the number = (n - 1) nines | 8 | (n - 1) zeroes | 1
• (99)² = two nines i.e. n = 2, hence the answer is ⇒ (2 - 1) nines | 8 | (2 - 1) zeroes | 1 = 9801
• (999)² = here n = 3, hence the answer is ⇒ two nines | 8 | two zeroes | 1 = 998001
• (9999)² = here n = 4, hence the answer is ⇒ 99980001

4. Squaring Numbers between 101 and 199

​For example (119)² = ?

• Add last two digit of number to the number 119 + 19 = 138
•  Square last two digits 19 × 19 = 361
• Take last two digits, 61 as last two digit of answer. Rest of the numbers are carry forwards.
• Add carry forward, 3 to 138. 138 + 3 = 141
• These are first three digits of answer, So answer is 14161.

Take few more examples.
i. (145)² = ?
a) 145 + 45 = 190
b) 45 × 45 = 2025
c) 190 + 20 = 210
Hence the answer is, 21025.

ii. 106 × 106
a) 106 + 06 = 112
b) 06 × 06 = 36
c) 112 + 0 = 112
Hence the answer is, 11236.

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Cube Roots of Exact Cubes 

• The lowest cubes, i.e. the cubes of the first nine natural numbers are 1, 8, 27, 64, 125, 216, 343, 512
  and 729.
• Thus, they all have their own distinct endings; and there is no possibility of overlapping or doubt as in
  the case of squares. 
• Therefore, the last digit of the cube root of an exact cube is obvious:
  
• In other words,
  (i) 1, 4, 5, 6, 9 and 0 repeat themselves in the cube-ending; and
  (ii) 2, 3, 7 and 8 have an inter-play of complements from 10.
• The number of digits in a cube root is the same as the number of 3-digit groups in the original cube
  including a single digit or a double-digit group if there is any:
• The first digit of the cube-root will always be obvious from the first group in the cube.
• Thus, the number of digits, the first digit and the last digit of the cube root of an exact cube are the data
  with which we start, when we begin the work of extracting the cube root of an exact cube.
•  To instantly determine the cube root, follow these easy steps:
  Example: Find the cube root of 438976.
  Step 1:  Drop the last three digits and find the largest cube contained in 438. This is 7³ = 343, so the ten’s-digit
  is 7. (This is why you had to memorize the cubes of the digits 1 through 9).
  Step 2:  Now go back to the last three digits. Look at the last digit, 6. That's the same ending as 6³, so your
  unit’s digit is 6. So the cube root of 438976 is 76.

Comparison Of Fractions 

• We know that, if the denominator of all fractions is the same, then the fraction having a bigger numerator is bigger in value and having smaller numerator is smaller in value.
• Let’s take one example.
   the first fraction is the smallest and the last one is the biggest.
• Also if the numerator of all fractions is same, then the fraction having bigger denominator is smaller in value and vice versa.
  Let’s take one example.
   , the first fraction is the biggest and the last one is the smallest.
• What happens when if neither numerator nor denominator of the fractions is same. We know in this case, basic method is by LCM method, i.e. making the denominator same. But we don’t know that we can even make the numerator same. So we have to check that whose LCM is easy to find out, depending upon the value of the numerators or the denominators.
• Let’s understand it with the help of an example.
  
  Now we can see that finding out of the LCM of the 25, 35 and 30 is somewhat tedious. So we find out the LCM of the numerators, i.e. LCM of 1, 2 and 3. Which is 6. So our problem reduced to comparison of

1. Comparison of Fractions in Certain Fashion

• When the relationship between each of the numerators and denominators is the same, the difference between them is the same and with the same sign. It has two cases.

1. If each numerator is smaller than its denominator by the same value. 

• In this situation the fraction having bigger digits will be bigger in value and vice versa.
• Let’s take one example:
  
• Using above method we can easily compare all of them. 
   is the biggest,   is second biggest,
   is third biggest and   is the smallest.

2. If each numerator is greater than its denominator by same value. 

• In this situation the fraction having bigger digits will be smaller and vice versa.
• Let’s take one example:
  Compare
• Using above method we can easily say  is the biggest, is second biggest and  is the
  smallest.
  

2. Logical Comparison

• This shortcut more depends upon our logic. The basic requirement to use this method is, we should be able to calculate the approximate value of the fraction.
• Let’s take one example
  
• Now look at the first fraction, we see that denominator is 28 and half of the 28 is 14, 15 is greater than 14,
• Hence value of this fraction is slightly more than   
• Similarly in second fraction, denominator is 34 and half of 34 is 17, 16 is less than 17, hence value of this fraction is slightly less than .
• Combining these two results we can judge that   is greater than

Percentage

Basic Shortcut

The most basic tip for questions involving percentages is memorize the fractional equivalent of the

percentage. i.e.

Similarly, we can generate lot more.

This means when we have to calculate 25 % of 50, we can directly calculate one-fourth of 50, which is 12.5

Similarly, if we have to calculate 5 % of 400, we calculate one-twentieth (1/20) of the 400, which is 20.