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  • in reply to: Assignment 2 #33340
    Shikha Dhingra
    Participant

    Hi Ken

    During presentation if some one ask about the minimum age of child for learning vedic maths and also the little child can get confused between the method taught at school and vedic maths , what should be the answer ? The older child can understand the two different methods of calculations easily. I think the parents concern is right. Pls clarify.

    Thanks

    in reply to: Assignment 2 #33144
    Shikha Dhingra
    Participant

    Thanks Ken

    in reply to: Assignment 2 #33141
    Shikha Dhingra
    Participant

    Hi Ken

    What is the date of submission of assignment 2 ? Is it 15 th or 22nd March ? Pls clarify.

     

    in reply to: Week 9 Video Lessons #33087
    Shikha Dhingra
    Participant

    Hi Ken

    We can also thought of a triangle below x-axis as second element i.e. height is negative.

     

     

    in reply to: Challenge Questions – 2 #33086
    Shikha Dhingra
    Participant

    Challenge Question # 21

    5 ,12,13     and      11,60,61 are triplets

    Now in both triplets one of the element is divisible by 3 i.e. 12 and 60

    In first case   5^2 + 12^2 = 13^2

    The element which is divisible by nine always comes out to be 9 after digit sum so we can cast out 9.

    L.H.S              25 + 3^2 =7+9 =16 =7 ( by digit sum)

    In L.H.S. if cast out 9 then also the digit sum comes out to be 7

    R.H.S.              13^2= 4^2 = 16 = 7

     

    In triplet 11,60,61

    60 is divisible by 3 , the square of 60 is comes out to be 9 after digit sum ( 60^2= 6^2= 36= 9)

    So we can only take the digit sum of 11 in L.H.S.

    11^2 =1+1 ^2= 2^2=4

    R.H.S.   61^2 = 7^2=49=4+9=13=1+3=4

    L.H.S. =R.H.S.

    So in triplet we can skip the element which is divisible by 3 and only take the digit sum of other element on L.H.S.

     

     

     

    in reply to: Week 9 Video Lessons #33084
    Shikha Dhingra
    Participant

    Hi Ken

    I have a doubt in lesson 34 practice 4 ( Q-2 ) .

    Find the angle between lines y=2x and y=3x

    1         2         √5

    1         3          √10

    ____________________________

    7        -1                √50

    I got -1 in second place but the answer is 1 . Pls clarify

     

     

    in reply to: Week 8 Video lessons #32984
    Shikha Dhingra
    Participant

    Hi Ken

    I have also the doubt which Preeti has asked. I was not able to understand from where the remainder 11 came. But as you explain that we have started from remainder -5 from (7-13) now I have understood. Thanks

     

     

     

    in reply to: Challenge Questions – 2 #31213
    Shikha Dhingra
    Participant

    Challenge Question#13

    57 X 37    = 6(3)  X  4(3)

    = 6 X 4 -3 / (-3) X(-3)

    = 24-3 / 09

    = 21 / 09

    57 X 37 = 2109

     

    in reply to: Challenge Questions – 2 #31135
    Shikha Dhingra
    Participant

    Challenge Question #11

    47 X 67   As in lesson 17 the digits at tens place are same and  the sum of digits at ones place is 10  we try to convert digits according to that.

    47  X (100-33 )

    47 X(100 – 33 + 10 -10)                    adding +10 and -10  so that (-33 -10) becomes -43

    47 X(100 -43 +10)

    4700 -(47 X43) + (47 X10)                now  in   47 X43   the digits at tens place are same and sum of digits at ones place is 10

    4700 – 2021 +470  =5170 -2021 =3149

     

      94 X 67              2 X 47 X 67

    2 X (3149) =      6298              multiplication of 47 and 67 is same as above

    32 X36        multiply by 2

    2 X 32 X 36= 64 X36

    = (64 -30 +30 ) X36           adding +30 and -30  to make ( 64-30 = 34)

    =  (34  +30 ) X36

    = 34 X36 + ( 30 X 36)            now in 34 and 36 the tens digits are same and sum of ones digits is 10

    = 1224 + 1080 = 2304

    now divide by 2 as we have multiply by 2  in the beginning

    therefore 2304 ÷ 2 = 1152

    in reply to: Challenge Questions – 2 #30520
    Shikha Dhingra
    Participant

    Challenge Question #10         Factorize 8051

    Solution:    factors of   51 are       17  and 3

    ?  ?     -17           Taking base as 100 , the number at ? ?    should be   100 – 17 =83

    ?  ?      -3             taking base as 100 , the number at ? ?   should be  100 – 3 = 97

    ______________                            Therefore the factor of  8051 are 83  and 97

    8 0     5 1

     

    in reply to: Week 4 Video Lessons #30460
    Shikha Dhingra
    Participant

    Hi Maurice

    Solution to  309 x 104  can be like this:

    first multiply by  3   you will get 309 x 312

    now taking 300 as base solve 309 x 312

    321 / 108

    now multiply by 3 as 300 is the base

    You will get        963/108   =   96408

    now you have to divide by 3 as you multiply by 3 in the starting to make the base 300

    your answer will be  96408  divide by 3 is     32136

    I  hope it will help

    in reply to: Week 3 Video Lessons #30279
    Shikha Dhingra
    Participant

    Thanks Ken

     

    in reply to: Assignment 1 #30276
    Shikha Dhingra
    Participant

    Thanks Ken !!

    in reply to: Challenge Questions #30265
    Shikha Dhingra
    Participant

    Challenge question 8         21 x ?? = ?? 12

    21 x 72 = 1512

    explanation:

    As 2 is at ones place in answer therefore one place digit in second number is 2

    21 x ? 2 = ?? 12

    according to crosswise and vertical method

    21

    ?2              the digit at ? should be 7   so that (2 x 2 ) + (7 x 1 ) = 1 at tens place and 7 x2 =14  and carry 1  which is 14 + 1=15

    _____

    1 5 1 2

    in reply to: Week 3 Video Lessons #30264
    Shikha Dhingra
    Participant

    Hi Ken

    I have a doubt in question 6 (practice 4) lesson 15.

    17 x 16  My answer is 282 but the answer in the course book is 272. pls clarify

    17 -3

    16 -4

    ——

    13/12    1 at tens place will be added to 3     therefore 142

    as base is 20 we multiply by 14 by 2 therefore 282

    ___

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