vedic  

Posted by Your Friend

            127 x 123 As antyayor dasakepi works, we apply ekadhikena 127 x 123 = 12 x 13 / 7 x 3 12* 13 = (12 + 3)15 6(3*2)= 15621                 16    6                 13    3                 19   18 208                 88    12                 98    2                 86    24 8624 46*43 = take 50 as base you will gate 39 19.5 28 = 1978 76 + 86 15 12 = 162 373 + 474 7 14 7 8 4 7 divided by 9 23/9 = 2 is quotient and remainder is 5 ( 2+3) 134/9 = 14 remainder 8 1 is just the first figure 4 is the sum of first two figure 8 is the total of all figure 842/9 = 8 12 = 92 remainder is 14 now we will write it as 93 remainder is 5 73 * 6 = 42 18 = 438 282 * 4 = 8 32 8 = 1128 21 * 23 = 4 8 3  three steps multiply vertically on the left multiply crosswise and add multiply vertically on the right 61 * 31 = 18 9 1 21 * 26 = 4 14 6 = 546 234 * 11 = 2   5 7    4 = 2574 6x2 LOG 10 BANEGA 10000 AGAR HOGI GHAT 4

 
Example 2: 5x – 3y = 11 6x – 5y = 09 Now Nr. of x is (-3) (9) – (5) (11) = - 27 + 55 = 28 Dr. of x is (-3) (6) – (5) (-5) = - 18 + 25 = 07 x = Nr ÷ Dr = 28 ÷ 7 = 4 and for y, Nr is (11) (6) – (9)(5) = 66 – 45 = 21 Dr is 7 Hence y = 21 ÷ 7 = 3. Example 3: solve 3x + y = 5 4x – y = 9 Now we can straight away write the values as follows: (1)(9) – (-1)(5) 9 + 5 14 x = _____________ = _____ = ___ = 2 (1)(4) – (3)(-1) 4 + 3 7 (5)(4) – (9)(3) 20 – 27 -7 y = ____________ = _______ = ___ = -1 (1)(4) – (3)(-1) 4 + 3 7 Hence x = 2 and y = -1 is the solution. VARGAÑCA YOJAYET The meaning of the Sutra is 'what ever the deficiency subtract that deficit from the number and write along side the square of that deficit'. This Sutra can be applicable to obtain squares of numbers close to bases of powers of 10. Method-1 : Numbers near and less than the bases of powers of 10. Eg 1: 92 Here base is 10. The answer is separated in to two parts by a’/’ Note that deficit is 10 - 9 = 1 Multiply the deficit by itself or square it 12 = 1. As the deficiency is 1, subtract it from the number i.e., 9–1 = 8. Now put 8 on the left and 1 on the right side of the vertical line or slash i.e., 8/1. Hence 81 is answer. Method. 2 : Numbers near and greater than the bases of powers of 10. Eg.(1): 132 . Instead of subtracting the deficiency from the number we add and proceed as in Method-1. for 132 , base is 10, surplus is 3. Surplus added to the number = 13 + 3 = 16. Square of surplus = 32 = 9 Answer is 16 / 9 = 169. Cubing of Numbers: Example : Find the cube of the number 106. We proceed as follows: i) For 106, Base is 100. The surplus is 6. Here we add double of the surplus i.e. 106+12 = 118. (Recall in squaring, we directly add the surplus) This makes the left-hand -most part of the answer. i.e. answer proceeds like 118 / - - - - - ii) Put down the new surplus i.e. 118-100=18 multiplied by the initial surplus i.e. 6=108. Since base is 100, we write 108 in carried over form 108 i.e. . As this is middle portion of the answer, the answer proceeds like 118 / 108 /.... iii) Write down the cube of initial surplus i.e. 63 = 216 as the last portion i.e. right hand side last portion of the answer. Since base is 100, write 216 as 216 as 2 is to be carried over. Answer is 118 / 108 / 216 Now proceeding from right to left and adjusting the carried over, we get the answer 119 / 10 / 16 = 1191016. Eg.(1): 1023 = (102 + 4) / 6 X 2 / 23 FALSE 1061208 Observe initial surplus = 2, next surplus =6 and base = 100. Sunyam Samyasamuccaye SŨNYAM SĀMYASAMUCCAYE The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that Samuccaya is Zero.' i.e., it should be equated to zero. The term 'Samuccaya' has several meanings under different contexts. i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all the terms concerned and proceed as follows. Example 1: The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all its terms. Hence by the sutra it is zero, i.e., x = 0. Example 2: 5(x+1) = 3(x+1) No need to proceed in the usual procedure like 5x + 5 = 3x + 3 5x – 3x = 3 – 5 2x = -2 or x = -2 ÷ 2 = -1 Simply think of the contextual meaning of ‘ Samuccaya ‘ Now Samuccaya is ( x + 1 ) x + 1 = 0 gives x = -1 ii) Now we interpret ‘Samuccaya ‘as product of independent terms in expressions like (x+a) (x+b) Example 3: ( x + 3 ) ( x + 4) = ( x – 2) ( x – 6 ) Here Samuccaya is 3 x 4 = 12 = -2 x -6 Since it is same , we derive x = 0 This example, we have already dealt in type ( ii ) of Paravartya in solving simple equations. iii) We interpret ‘ Samuccaya ‘ as the sum of the denominators of two fractions having the same numerical numerator. Consider the example. 1 1 ____ + ____ = 0 3x-2 2x-1 Instead of this, we can directly put the Samuccaya i.e., sum of the denominators i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0 giving 5x = 3 x = 3 / 5 It is true and applicable for all problems of the type m m ____ + _____ = 0 ax+b cx+d iii) We now interpret ‘Samuccaya’ as combination or total. If the sum of the numerators and the sum of the denominators be the same, then that sum = 0. Consider examples of type ax + b ax + c _____ = ______ ax + c ax + b As per Samuccaya (ax+b) + (ax+c) = 0 2ax+b+c = 0 2ax = -b-c -(c+b) x = ______ 2a Hence the statement. Consider the examples of the type, where N1 + N2 = K ( D1 + D2 ), where K is a numerical constant, then also by removing the numerical constant K, we can proceed as above. Example 6: 2x + 3 x + 1 _____ = ______ 4x + 5 2x + 3 Here N1 + N2 = 2x + 3 + x + 1 = 3x + 4 D1 + D2 = 4x + 5 + 2x + 3 = 6x + 8 = 2 ( 3x + 4 ) Removing the numerical factor 2, we get 3x + 4 on both sides. 3x + 4 = 0 3x = -4 x = - 4 / 3. v) ‘Samuccaya‘ with the same meaning as above, i.e., case (iv), we solve the problems leading to quadratic equations. In this context, we take the problems as follows; If N1 + N2 = D1 + D2 and also the differences N1 ~ D1 = N2 ~ D2 then both the things are equated to zero, the solution gives the two values for x. Now ‘Samuccaya’ sutra comes to help us in a beautiful way as follows : Observe N1 + N2 = 3x + 2 + 2x + 5 = 5x + 7 D1 + D2 = 2x + 5 + 3x + 2 = 5x + 7 Further N1 ~ D1 = ( 3x + 2 ) – ( 2x + 5 ) = x – 3 N2 ~ D2 = ( 2x + 5) – ( 3x + 2 ) = - x + 3 = - ( x – 3 ) Hence 5x + 7 = 0 , x – 3 = 0 5x = -7 , x = 3 i.e., x = -7 / 5 , x = 3 vi)‘Samuccaya’ with the same sense but with a different context and application . Example 9: 1 1 1 1 ____ + _____ = ____ + ____ x - 4 x – 6 x - 2 x - 8 Now ‘Samuccaya’ sutra, tell us that, if other elements being equal, the sum-total of the denominators on the L.H.S. and their total on the R.H.S. be the same, that total is zero. Now D1 + D2 = x – 4 + x – 6 = 2x – 10, and D3 + D4 = x – 2 + x – 8 = 2x – 10 By Samuccaya, 2x – 10 gives 2x = 10 10 x = __ = 5 2 Sunyam Samya Samuccaye in Certain Cubes: Consider the problem ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3. For the solution by the traditional method we follow the steps as given below: But once again observe the problem in the vedic sense We have ( x – 4 ) + ( x – 6 ) = 2x – 10. Taking out the numerical factor 2, we have ( x – 5 ) = 0, which is the factor under the cube on R.H.S. In such a case “Sunyam samya Samuccaye” formula gives that x – 5 = 0. Hence x = 5 Think of solving the problem (x–249)3 + (x+247)3 = 2(x–1)3 Think of solving the problem (x–249)3 + (x+247)3 = 2(x–1)3 Mathematics - Why Vedic Mathematics (x + 2)3  x + 1 ______ = _____ (x + 3)3 x + 4 Observe that ( N1 + D1 ) with in the cubes on L.H.S. is x + 2 + x + 3 = 2x + 5 and N2 + D2 on the right hand side is x + 1 + x + 4 = 2x + 5. By vedic formula we have 2x + 5 = 0 x = - 5 / 2.  

This entry was posted on 6:18 PM . You can leave a response and follow any responses to this entry through the Subscribe to: Post Comments (Atom) .

0 comments

Newer Post Older Post