The Power of Vedic Maths
Format: PDF / Kindle (mobi) / ePub
2nd REVISED & UPDATED EDITION
Vedic mathematics is gaining widespread popularity among the student community as well as maths lovers. The absence of a book, explaining the techniques in a simple language, has been felt acutely for a long time. This book has been written using a step-by-step approach, and attempts to fill the existing void. It includes several solved problems in addition to 1000 practice problems with answers. It also includes a special chapter which shows the application of the techniques to problems set in competitive exams like CAT, CET etc.
People from all walks of life including school and college students, teachers, parents and also those from non-mathematical areas of study will discover the joys of solving mathematical problems using the wonderful set of techniques called Vedic Maths.
will subtract 5 from the GD 8 to get 3 (ND) Divide 3 by 6 to get the quotient as 0 and remainder as 3 Carry the remainder 3 to the next digit 5 and get 35 (GD) Now, the previous two quotient digits are 0 and 1. We pad them with one zero and multiply by the flag 524 as follows: We will subtract 2 from the GD 35 to get 33 (ND). Divide 33 by 6 to get the quotient as 5 and remainder as 3 Carry the remainder 3 to the next digit 1 and get 31 (GD) Now, the previous three quotient digits are
taken as the next base dividend as explained above Since the last digit of the divisor is 8 which has a difference of 1 from 9, we multiply the quotient by 1 and add to the base dividend at each stage to compute the gross dividend. Example 1: Let us now see the details for 5 / 28 Steps Add 2 to the divisor to get 30 The modified division is now 5 / 30 Remove the zero from the denominator by placing a decimal point in the numerator giving the modified division as 0.5 / 3 We will not
before. It will be seen that there is no need to make any adjustment at any column and the procedure can progress very fast. The final result is 329 with remainder 49, or in decimal form it is 329.6363. g) Application in squares In order to get the square of any number, we need to compute duplex values of a lot of numbers. If some of the digits are above 5, the computation of the duplex becomes tedious. Consider the square of 897. The computation of the duplex is quite time-consuming
(Subtract 500 from each number) Consider the steps We will scale up this intermediate result by multiplying 517 by 5. Since 517 × 5 = 2585 So the answer becomes 2585 / (–480). Since the primary base is 100, we will carry 5 from the left which is equal to a carry of 500 (primary base 100 * 5) and subtract 480 from 500 giving 20. So the final answer becomes 2580/020 = 258020. Or, it can be considered as 258500 - 480 giving 258020. G) Primary base 1000 and secondary base 500 for 532 ×
consider the following example The final answer is 9073792. Steps: Digit (a) : Multiply the right most digits 2 and 6 to get 12. Write down 2 and show the carry as 1. Digit (b) : Carry out a 2-digit cross multiplication between 3 and 6 (18), and 2 and 5 (10) and add them to get 28. Add the carry digit 1 and get 29. Write down 9 and show the carry as 2. Digit (c) : Carry out a 3-digit cross multiplication between 1 and 6 (6), 2 and 2 (4) and 3 and 5 (15) and add them to get 25. Add the