logging in or signing up Vedic Mathematics Techniques Santoshmehta Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: Embed: Flash iPad Copy Does not support media & animations WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 27415 Category: Education License: All Rights Reserved Like it (41) Dislike it (5) Added: May 07, 2007 This Presentation is Public Favorites: 21 Presentation Description The presentation elaborates Vedic Maths Techniques for multiplication, squares, cubes, as discussed in the WiZiQ's public session on Vedic Mathematics Comments Posting comment... By: adikulal (19 month(s) ago) plz send this ppt to aditya.kulal@gmail.com! Saving..... Post Reply Close Saving..... 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See all Premium member Presentation Transcript Vedic Mathematics: Vedic Mathematics Session Topics Multiplication with common base Squares Cubes Two digit multiplication Three digit multiplication By Munish KumarSlide2: Multiplication with common base # Multiply 104 by 103 1 0 4 1 0 3 x 3 1 2 1 0 4 x x 0 0 0 x 1 0 7 1 2 Magic of Vedic 1 0 4 1 0 3 Here Common base is 100 4 3 1 2 1 0 7 (1 0 4 + 3 ) or (1 0 3 + 4) 4 x 3Slide3: Some problems to practice 105 x 107 102 x 108 107 x 109 102 x 103 104 x 112Slide4: Let us do some more multiplications 1 0 0 1 1 0 1 9 01 19 1 9 1 0 2 0 (1001 + 19 ) or (1 019 + 01) 01 x 19 Since the base is 1000, we take 3 digits on the right side 1,0 2 0,0 1 9 x 0 1 9 Slide5: Some practice problems 1005 x 1007 1002 x 1008 1004 x 1012 1013 x 1003Slide6: Multiplication with less than base 9 7 8 9 - 3 - 11 3 3 8 6 (9 7 - 11 ) or (89 - 03) -11 x -3 8 6 3 3 xSlide7: Let us do some practice 95 x 97 92 x 98 97 x 89 92 x 93Slide8: Squares Squares from 101 to 125 (101)2 (109)2 (112)2 (117)2 (123)2 Slide9: Squares from 75 to 99 Squares (97)2 (93)2 (89)2 (85)2 (77)2Slide10: Squares of numbers ending with 5 Squares Formula used (a5)2 = a(a+1) Ι 25 Where, a = Digits in the number other than 5 (25)2 (45)2 (95)2 (995)2 (1035)2 (25)2 =2(2+1) Ι 25 = 6 2 5 Here a = 2 (95)2 = 9(9+1) Ι 25 = 9025 Here a= 9 (45)2 = 4(4+1) Ι 25 = 2 0 2 5 Here a= 4 (995)2 = 99(99 + 1) Ι 25 = 990025 Here a = 99 (1035)2 = 103(103 + 1) Ι 25 = 1071225 Here a = 103Slide11: Squares Squares from 25 to 49 (47)2 (43)2 (39)2 (35)2 (27)2 Formula used N2 = ( 25 – X ) Ι X2 Where, X = By how much a number less than 50 (47)2 = (25 – 3) Ι 32 = 2 2 0 9 Here X = 3 (39)2 = (25 – 11) Ι 112 = 14 Ι 121 = 1 5 2 1 Here X = 11 (43)2 = (25 – 7) Ι 72 = 1 8 4 9 Here X = 7 (35)2 = (25 – 15) Ι 152 = 10 Ι 225 = 1 2 2 5 Here X = 15 (27)2 = (25 – 23) Ι 232 = 02 Ι 529 = 7 2 9 Here X = 23Slide12: Squares Squares from 51 to 74 (52)2 (58)2 (63)2 (65)2 (72)2 Formula used N2 = ( 25 + X ) Ι X2 Where, X = By how much a number is more than 50 (52)2 = (25 + 2) Ι 22 = 2 7 0 4 Here X = 2 (63)2 = (25 + 13) Ι 132 = 3 8 Ι 169 = 3 9 6 9 Here X = 13 (58)2 = (25 + 8) Ι 82 = 3 3 6 4 Here X = 8 (65)2 = (25 + 15) Ι 152 = 40 Ι 225 = 4 2 2 5 Here X = 15 (72)2 = (25 + 22) Ι 222 = 47 Ι 484 = 5 1 8 4 Here X = 22Slide13: Ending with 5 and difference 10 25 x 35 = 45 x 55 = 115 x 125 = 195 x 205 = 505 x 495 = 8 7 5 ALWAYS (32 – 1) = 8 2,475 14,375 39,975 2,49,975 What is the similarity in the followingSlide14: Base same and unit digits sum 10 53 x 57 = 26 x 24 = 117 x 113 = 109 x 101 = 55 x 55 = 30 21 7 x 3 5 x 6 624 13,221 11,009 3,025 What is the similarity in the followingSlide15: Cubes Formula used (N)3 = b + 3x Ι 3x2 Ι x3 Where, x = Difference from the base & b = base (104)3 (12)3 (95)3 (995)3 (1007)3 (104)3 = (100 + 3 x 4) Ι 3 x 42 Ι 43 = (100 +12) Ι 3 x 16 Ι 64 = 1 1 2 4 8 6 4 Here b = 100 & x = 4 (12)3 = (10 + 3 x 2) Ι 3 x 22 Ι 23 = (10 +6 ) Ι 3 x 4 Ι 8 = 16 Ι 12 Ι 8 = 1 7 2 8 Here b = 10 & x = 2 (95)3 = (100 + 3 x -5) Ι 3 x (-5)2 Ι (-5)3 = (100 -15) Ι 75 Ι -125 = 85 Ι 73 Ι 200 -125 = 8 5 7 3 7 5 Here, b = 100 & x = -5 Since the base is 10 . We will take only on digit in each block and carry forward the extra to consecutive next block in the left. Since we have a negative term in the extreme right block. So have to make it positive by borrowing 2 carry from consecutive left block and add to -125. (995)3 = (1000 + 3 x -5) Ι 3 x (-5)2 Ι (-5)3 = (1000 -15) Ι 75 Ι -125 = 985 Ι 073 Ι 200 -125 = 9 8 5, 0 7 3, 0 7 5 Here, b = 1000 & x = -5 As the base here is 1000, so we will take three digits in each block after making negative term in the extreme right block positive (1007)3 = (1000 + 3 x 7) Ι 3 x (7)2 Ι (7)3 = (1000 +21) Ι 147 Ι 343 = 1021 Ι 147 Ι 343 = 1 0, 2 1 1, 4 7, 3 4 3 Here, b = 1000 & x = 7Slide16: Three step multiplication # Multiply 36 by 94 3 6 9 4 x 6 x 4 = 24 3 x 4 + 9 x 6 = 66 9 x 3 = 27 2 7 6 6 2 4 Slide17: Let us do some more multiplications 1 1 1 1 1 3 11 13 1 4 3 1 2 4 (1 1 1 + 13 ) or (1 1 3 + 11) 11 x 13 This 1 is carried forward to the other side 1 2 5 4 3 xSlide18: Multiplication in different mood 9 7 1 0 9 - 3 + 9 -2 7 1 0 6 (9 7 + 9 ) or (109 - 03) 9 x -3 x Since we have a negative term on the right side of it So we will make it positive . For that we will borrow carry from left side, which is equal to 100 and add -27 to it 100 - 27 = 73 106 – 1 = 105 7 3 105 1 0,5 7 3 Slide19: Thank you Best of luck for your future You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.
Vedic Mathematics Techniques Santoshmehta Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: Embed: Flash iPad Copy Does not support media & animations WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 27415 Category: Education License: All Rights Reserved Like it (41) Dislike it (5) Added: May 07, 2007 This Presentation is Public Favorites: 21 Presentation Description The presentation elaborates Vedic Maths Techniques for multiplication, squares, cubes, as discussed in the WiZiQ's public session on Vedic Mathematics Comments Posting comment... By: adikulal (19 month(s) ago) plz send this ppt to aditya.kulal@gmail.com! Saving..... Post Reply Close Saving..... Edit Comment Close By: sakthi888 (21 month(s) ago) pl send this ppt mail id: sakthivishal20@gmail.com Saving..... Post Reply Close Saving..... Edit Comment Close By: amarifam (23 month(s) ago) Dear Sir, This presentation is just great. Is it possible to allow me to have the presentation so that i can teach my students. my e-mail address is amarifam2@yahoo.com Saving..... Post Reply Close Saving..... Edit Comment Close By: msnraju09 (25 month(s) ago) pl send to my mail add, msnraju09@gmail.com.. Saving..... Post Reply Close Saving..... Edit Comment Close By: mihir_kanani777 (27 month(s) ago) hey the ppt is too good if you cud just help me bye sending the ppt to my id : mihir_kanani777@yahoo.com and even helping me out wid the other subtracting and division methods.. it would b of great he;p.. thanking you in anticipation.. Saving..... Post Reply Close Saving..... Edit Comment Close loading.... See all Premium member Presentation Transcript Vedic Mathematics: Vedic Mathematics Session Topics Multiplication with common base Squares Cubes Two digit multiplication Three digit multiplication By Munish KumarSlide2: Multiplication with common base # Multiply 104 by 103 1 0 4 1 0 3 x 3 1 2 1 0 4 x x 0 0 0 x 1 0 7 1 2 Magic of Vedic 1 0 4 1 0 3 Here Common base is 100 4 3 1 2 1 0 7 (1 0 4 + 3 ) or (1 0 3 + 4) 4 x 3Slide3: Some problems to practice 105 x 107 102 x 108 107 x 109 102 x 103 104 x 112Slide4: Let us do some more multiplications 1 0 0 1 1 0 1 9 01 19 1 9 1 0 2 0 (1001 + 19 ) or (1 019 + 01) 01 x 19 Since the base is 1000, we take 3 digits on the right side 1,0 2 0,0 1 9 x 0 1 9 Slide5: Some practice problems 1005 x 1007 1002 x 1008 1004 x 1012 1013 x 1003Slide6: Multiplication with less than base 9 7 8 9 - 3 - 11 3 3 8 6 (9 7 - 11 ) or (89 - 03) -11 x -3 8 6 3 3 xSlide7: Let us do some practice 95 x 97 92 x 98 97 x 89 92 x 93Slide8: Squares Squares from 101 to 125 (101)2 (109)2 (112)2 (117)2 (123)2 Slide9: Squares from 75 to 99 Squares (97)2 (93)2 (89)2 (85)2 (77)2Slide10: Squares of numbers ending with 5 Squares Formula used (a5)2 = a(a+1) Ι 25 Where, a = Digits in the number other than 5 (25)2 (45)2 (95)2 (995)2 (1035)2 (25)2 =2(2+1) Ι 25 = 6 2 5 Here a = 2 (95)2 = 9(9+1) Ι 25 = 9025 Here a= 9 (45)2 = 4(4+1) Ι 25 = 2 0 2 5 Here a= 4 (995)2 = 99(99 + 1) Ι 25 = 990025 Here a = 99 (1035)2 = 103(103 + 1) Ι 25 = 1071225 Here a = 103Slide11: Squares Squares from 25 to 49 (47)2 (43)2 (39)2 (35)2 (27)2 Formula used N2 = ( 25 – X ) Ι X2 Where, X = By how much a number less than 50 (47)2 = (25 – 3) Ι 32 = 2 2 0 9 Here X = 3 (39)2 = (25 – 11) Ι 112 = 14 Ι 121 = 1 5 2 1 Here X = 11 (43)2 = (25 – 7) Ι 72 = 1 8 4 9 Here X = 7 (35)2 = (25 – 15) Ι 152 = 10 Ι 225 = 1 2 2 5 Here X = 15 (27)2 = (25 – 23) Ι 232 = 02 Ι 529 = 7 2 9 Here X = 23Slide12: Squares Squares from 51 to 74 (52)2 (58)2 (63)2 (65)2 (72)2 Formula used N2 = ( 25 + X ) Ι X2 Where, X = By how much a number is more than 50 (52)2 = (25 + 2) Ι 22 = 2 7 0 4 Here X = 2 (63)2 = (25 + 13) Ι 132 = 3 8 Ι 169 = 3 9 6 9 Here X = 13 (58)2 = (25 + 8) Ι 82 = 3 3 6 4 Here X = 8 (65)2 = (25 + 15) Ι 152 = 40 Ι 225 = 4 2 2 5 Here X = 15 (72)2 = (25 + 22) Ι 222 = 47 Ι 484 = 5 1 8 4 Here X = 22Slide13: Ending with 5 and difference 10 25 x 35 = 45 x 55 = 115 x 125 = 195 x 205 = 505 x 495 = 8 7 5 ALWAYS (32 – 1) = 8 2,475 14,375 39,975 2,49,975 What is the similarity in the followingSlide14: Base same and unit digits sum 10 53 x 57 = 26 x 24 = 117 x 113 = 109 x 101 = 55 x 55 = 30 21 7 x 3 5 x 6 624 13,221 11,009 3,025 What is the similarity in the followingSlide15: Cubes Formula used (N)3 = b + 3x Ι 3x2 Ι x3 Where, x = Difference from the base & b = base (104)3 (12)3 (95)3 (995)3 (1007)3 (104)3 = (100 + 3 x 4) Ι 3 x 42 Ι 43 = (100 +12) Ι 3 x 16 Ι 64 = 1 1 2 4 8 6 4 Here b = 100 & x = 4 (12)3 = (10 + 3 x 2) Ι 3 x 22 Ι 23 = (10 +6 ) Ι 3 x 4 Ι 8 = 16 Ι 12 Ι 8 = 1 7 2 8 Here b = 10 & x = 2 (95)3 = (100 + 3 x -5) Ι 3 x (-5)2 Ι (-5)3 = (100 -15) Ι 75 Ι -125 = 85 Ι 73 Ι 200 -125 = 8 5 7 3 7 5 Here, b = 100 & x = -5 Since the base is 10 . We will take only on digit in each block and carry forward the extra to consecutive next block in the left. Since we have a negative term in the extreme right block. So have to make it positive by borrowing 2 carry from consecutive left block and add to -125. (995)3 = (1000 + 3 x -5) Ι 3 x (-5)2 Ι (-5)3 = (1000 -15) Ι 75 Ι -125 = 985 Ι 073 Ι 200 -125 = 9 8 5, 0 7 3, 0 7 5 Here, b = 1000 & x = -5 As the base here is 1000, so we will take three digits in each block after making negative term in the extreme right block positive (1007)3 = (1000 + 3 x 7) Ι 3 x (7)2 Ι (7)3 = (1000 +21) Ι 147 Ι 343 = 1021 Ι 147 Ι 343 = 1 0, 2 1 1, 4 7, 3 4 3 Here, b = 1000 & x = 7Slide16: Three step multiplication # Multiply 36 by 94 3 6 9 4 x 6 x 4 = 24 3 x 4 + 9 x 6 = 66 9 x 3 = 27 2 7 6 6 2 4 Slide17: Let us do some more multiplications 1 1 1 1 1 3 11 13 1 4 3 1 2 4 (1 1 1 + 13 ) or (1 1 3 + 11) 11 x 13 This 1 is carried forward to the other side 1 2 5 4 3 xSlide18: Multiplication in different mood 9 7 1 0 9 - 3 + 9 -2 7 1 0 6 (9 7 + 9 ) or (109 - 03) 9 x -3 x Since we have a negative term on the right side of it So we will make it positive . For that we will borrow carry from left side, which is equal to 100 and add -27 to it 100 - 27 = 73 106 – 1 = 105 7 3 105 1 0,5 7 3 Slide19: Thank you Best of luck for your future