[Preeti Pathak]

Hi Ken,

This is related to week 7 lesson.
In practice 4, lesson 27, multiplying 7868 * 9 ,
I could do it with base=10000 (for 7868) and base=10(for 9).
I was trying with base=7*1000 and and base=10
base = 7*1000, 7 8 6 8 +868
base=10, 9 -1
—————
7* 6868/(868) (i.e. bar 868)
We need to multiply by 7 here as the base is 7000.How will we proceed after this step?. we need only 1 digit in the right hand side as the smaller base is 10.Please explain and let me know if I have made any mistake ?
Thank you

[Preeti Pathak]

Hi,

It was really interesting to learn the use of duplexes in squaring the algebraic expressions and it saves so much time as compared to calculating the same with the conventional method. With the conventional method , to find the square of ,suppose, say a trinomial, we write

(x+2y+5)(x+2y+5)=x(x+2y+5)+2y(x+2y+5)+5(x+2y+5) and then solve each bracket and finally add. But by using duplexes ,we can find the answer in one step.

At first, using the flag digit for division seemed somewhat difficult, but with practice it improved. Now ,with 3 figure divisor also,

it can be done easily. Also, very well understood the use of ‘proportionately’ in solving the division problems.

Thank you

 

 

[Preeti Pathak]

Hello,

Enjoyed week 2 videos lessons, digit sum concept, calculating from left to right .

In Vedic Maths ,we can calculate from left to right or right to left whichever we find  useful , is very helpful .Its flexible. In the beginning, calculating from  left to right seems difficult but after practicing ,we can do it in a better way.

Thanks

[Preeti Pathak]

Hello,

I learnt a lot from videos. mental addition, doubling and halving, dividing by 5,50,25 ,attempted the quiz. It was a sort of revision for me(as I did introductory course too).

But ,somehow couldn’t paticipate in forum discussions as didn’t have access. So posting it now.

 

 

 

[Preeti Pathak]

Sorry, there was one calculation mistake in earlier post..

47 *67=3149 ( not 3139)

then, using the “Proportionately ” Sutra, to find 94 *  67,

= 2 * (47 * 67)

= 2 * (3149) gives 6298

Thanks

[Preeti Pathak]

47 * 67

For the first part of the answer,

=Product of the digits in tens place  that add up to 10 +  digit in ones place

=(4*6) + 7=24+7= 31_____(1)

For the second part,

simply multiply the digits in one’s place, so 7 * 7=49________(2)

combining (1) & (2) gives

47 * 67 = 3139

To find 94 * 67

we can use the sutra “proportionately” here.

94= 2 * 47.

47 * 67= 3139 and double it ,so 94 * 67= 2 *3139=6278

To find 32 * 36

36 * 2=72

using the “proportionately” sutra, 7 and 3 add up to 10,

32 * 72 = 2304 and halve it ,so 32 *36=1152

Thanks

[Preeti Pathak]

Thank you.Got it. I did not cosider doubling or halving both the numbers.

[Preeti Pathak]

Hello,

Thank you.

In this case ,for  using “one more than the one before ” ,34 will have to be written as 39-5.

31  *   34

34=(39-5),will give

31 * (39-5)=(31 *39) -(31*5)

=1209-155

= 1054

Is this the way(explained above) ,we can do ?( “proportionately is not used here)

For using proportionately,

if we halve 34, will get 17 .If we double ,will get 68. can’t find 17 * 31 or 68*31.(using proportionately)

similarly,doubling 31 will give 62.Can’t find 62 * 34(using proportionately).

Please explain how it can be done “using proportionately”?

 

[Preeti Pathak]

Hi Ken,

In lesson 17, practice 3,

Q.7 )      46  *   54,    I did this way.

46 * (44 + 10)

= 46 * 44 +46*10

=2024 + 460(using “one more than before” method)= 2024+ 460=2484

and also, by “using the average”method, 46*54=(50*50)-(4*4)=2484

Q.8)   31 * 34 , we can get the answer by ” vertïcally and crosswise “method.  Can we use “one more than one before”method in this case?

Thanks

[Preeti Pathak]

okay, thank you.

[Preeti Pathak]

Hi Ken,
There are 3 quizzes in week 5.Is it necessary to complete only after lesson 19 or we can complete after each lesson (i.e lesson 17,18 and 19)?
Thanks

[Preeti Pathak]

Hi Maurice !

I tried this multiplication with base as 100. I would share.

base=100

309  * 104

if we take one-third of 309, we will get 103

now,

base=100

103         +3 (deficiency from 100)

*

104          +4(deficiency from 104)

_______________

(103+4)/ 12=107/12

and now, we will have to multiply by 3(since in the beginning,we took 1/3 of 309.

s0 3*(10712)=    32136 is the answer

Thanks

 

 

 

[Preeti Pathak]

Thanks Ken. Got it now where I was doing mistake. we have to do final halving at last.

[Preeti Pathak]

Hello,

In lesson 15, for multiplying 204  with 107 ,Iwe used base as 100. Can we do it with base as 200 by first multiplying 107 by 2.?I tried this way.

base =200    204     +4

*

214      +14

________________

218/56 and if we do half of it (since 107 was doubled)we will get 10928(not 10914).

But with base as 200,if we multiply by 2(i.e.218*2),we will get 43656 and then again need to half of it giving the answer to be 21828.(however if we still make it half,will get the answer). Please explain if we can do it with base 200 and where I am getting wrong?

Thanks

 

[Preeti Pathak]

Hello,

In lesson 14, multiplying near base 100(other bases also),the vertcally and crosswise method is really a faster way to get the answer as compared to the conventional method.And the added advantage is that it works for other bases (ex,200,200 etc) also.

The three proofs(Arithmetic,Geometric and Algebraic)are well explained to answer why the method(for multiplying numbers near base) works.

Extension of base multiplication method ,use of “Proportionately” in solving other multiplication problems is helpful.Also,it saves lot of time in doing the calculations.

Table patten,Sequence Pattern are intresting topics.Really enjoyed representing the sequence on the 9-point circle.I drew the table pattern(for 7 times table).

The use of “reversing” is explained in a simple manner in “Thinking of a number”.

Also, the use of equations in finding the “unknown number” is explained in a good way, especially when we are doing it mentally .

Had a great learning.

Thanks

 

 

g

 

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