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VEDIC MATHEMATICS

VEDIC MATHEMATICS

The basic root of vedic mathematics is in the vedas.Here you will get an idea about the power of vedic mathematics.

Ekadhikena Purvena (By one more than the previous one)

Case1:Square of number ending in 5

As you know the square of 5 is 25, the right most two digits of a square of number ending in 5 is 25, that is the square of a number ending in 5 always contains 25 as the last 2 digits. We have to find out the other digits. For this purpose part the given number as given below.

Left side = 6 Right side = 5
The right hand side contains only '5'. The left side contains other digits in the given number. In the above example 6 in the left hand side.

• So our key number is 6
• Add one to this number is 7
• Multiply these two to get the result, i.e. 42
• There fore, the required result is 4225
• Considers another examples 35
• Draw a line to part the number 3/5
• Key number is 3
• Add one to this number 3+1=4
• Multiply these two 3*4 = 12
• So the answer is 12/25, i.e. 1225

We can write this procedures as below
35 = (3*(3+1))/(5*5) =3*4/25 =1225

Another example

45 = 4/5 = (4*(4+1))/(5+5)= 4*5/25 =2025

Case 2: When the sum of last digits is 10 and the previous parts are the same.

In this method multiply the last digits of the given numbers to find out the last 2 digits in the result.
Use the case 1 method to calculate the other digits in the result.
Eg.43*47
The last two digits of given numbers are 3 and 7ie; the last two digits in the result is 7*3 = 21
The other results in the digits is 4*(4+1) = 20 the required result = 2021

We can represent this procedure as follows
43*47 = (4*(4+1))/ (3*7) = (4*5)/ (21) =2021
Eg.2.
32*38 = (3*(3+1)) / (2*8)
=3*4 /16 =1216
If the last digits of the given numbers produce a single digit value, put ‘0’ to its left side to make it as a two digit number.
Eg.
21*28 =(2*(2+1)) / (1*8) =(2*3) / (8) =68
it is wrong.
Put zero to the left side of 8
Ie, 21*28 = (2*(2+1)) / (1*8) = (2*3)/ (8) = 6 / 08 =608

case 3: 1 divided by 19,29,39,…..
Consider 1 / 19
Take the left digit of 19 ie 1. Add one to this to get the key number
1+1 = 2
divide 1 / 2, result is 0 and remainder = 1 put the result to the right side of decimal point, this is the first digit in the required answer.

Combine the remainder and result to get a new number (10)
Again divide this number by 2. result = 5 remainder = 0 put 5 to the right most side of answer. Now answer becomes 0.05
Again combine the result and remainder to get a new number (5).
Divides this by 2.
Result = 2 remainder = 1
Do the above steps to find out the other digits. The answer is 0.052631578947368421

In 1 / 29 the keynumber is 3
In 1 / 39 the key number is 4

Nikhilam Navatashcaramam Dashalah (All from nine and the last from ten)
Case: 1 computing square of 8
Power of 10 nearest to 8 is 10, so take 10 as base, 8 is 2 less than 10. So 8-2=6 is the left side of the answer.Right hand side of the answer is square of 2. So answer is 64.
We can represent this as follows
Square of 8 = (8-2) / (2*2) =64
Square of 9 = (9-1) / (1*1) =81
If the number is greater than 10 take the surplus
Eg.
Square of 12 = (12+2) / (2*2) =14/4 = 144
Square of 13 = (13+3) / (3*3) =16/9 = 169

Square of 14 = (14+4) / (4*4) = 18 / 16
In the above case , the right side produces 2 digit number (16).. Take 6 as the right side digit and take 1 as carry. Add this carry to the left hand side number, ie add 1 with 18
So our answer is (18+1)/6 = 196

Square of 16 = (16+6) / (6*6) = 22 / 36 =(22+3)/6 = 256

Case 2
Square of 98

Power of 10 nearest to 10 is 100. So take 100 as base of 98. 100 is 2 less than 100. so the left hand side number is 98-2=96.since we are using 100 as base , the right hand side must be a digit number.
Ie square of 98 = (98-2) / (2 * 2) = 96 / 04 = 9604

Square of 99 = (99-1) / (1 * 1) = 98/01 = 9801

Square of 96 = (96-4) / (4 * 4) = 92/16 = 9216

Square of 92 = (92-8)/(8 * 8)= 84/64 = 8464

Square of 104= (104+4) / (4 * 4) = 108 / 16 = 10816
Square of 110 = (110 + 10 ) / ( 10 * 10 ) = 120 / 100 = (120 + 1 ) / 00 = 12100

Multiplying a number by 11
To multiply any 2 digit number by 11 , we just put the total of the 2 figures between 2 digits

36 * 11 = 3 / (3 + 6 ) / 6 = 396
74 * 11 = 7 / ( 7 + 4) / 4 = 7 / 11/ 4 = (7+1) / 1 / 4 = 814
234*11=2 / (2+3) (3+4) / 4 =2574

Multiply 8*7
The Power of 10 nearest to 8 and 7 is 10, so take 10 as base

Given numbers         Difference from 10
8                       - 2
7                       - 3
Multiply the elements 2and 3 to get the right most digits of required result I.e. 2 * 3= 6
To find out the other digits add diagonally either (8,-3) or (7,-2), Addition gives is the value 5.
The result is = 56

Multiplying 12 and 13
Given numbers         Difference from 10
12                         +  2
13                         +  3

Multiply the elements 2and 3 to get the right most digits of required result. I.e. 2 * 3= 6
To find out the other digits add diagonally either (12,3) or (13,2) .Addition gives is the value 15.
The result is = 156

Multiply 92 and 93

Given number         Difference from 100
92                 -8
93                 -7

Multiply the elements 8and 7 to get the right most digits of required result. I.e. 8 * 7= 56
To find out the other digits add diagonally either (92,-7) or (93,-8) .Addition gives is the value 85.
The result is = 8556

Multiplying with 99,999,9999,……………..
(Ekanyunena Purvena “By one less than the previous one)

E.g. 1. 77 multiplied by 99
One less than the multipliceny 77 is 76.
This is the left part of the answer.

the compliment of 76 from 99 is 23 (99-76).
This is the right part of the answer.

So result is 7623

E.g: 2. 666 multiplied by 999
One less than the multiplicant 666 is 665
the compliment of 665 form 999 is 334 (999-665)
So result is 6665334

E.g: 3. 123456789 x 999999999
One less then the multiplicand 123456789 is 123456788
the compliment of 123456788 from 99999999 is 876543211
So Result is 123456788876543211

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