## Forum Replies Created

Viewing 15 posts - 1 through 15 (of 21 total)
• Author
Posts
• 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

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

___

Viewing 15 posts - 1 through 15 (of 21 total)
error:
Scroll to Top