# Week 6 Video Lessons

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• #28844

• #31217

Hi Ken,

I am in Texas. Because of the winter storms, we don’t have power and water.  I don’t know whether I am able to finish this week’s lessons and quiz on time. Can I Please have some extra time this week?

Thanks

• #31260

Sorry to hear about your situation Anupama. We are following it on the news.

I have sunbitted a ticket on your behalf.

• #31259

Hi Anupama,

We are praying for mother nature to be cooperative and hope power and normal life is set right soon.

After finishing lesson 23 and the quizzes, I am intrigued with deep knowledge Bharati Krishniji had  and i humbly bow down to him.

I am stuck at a silly step of square root of 256, trying to follow the steps –

marking off – 2/56 then 2 is close to sq. no. 1, so first figure must be 1.

From this 2, there is 1 leftover, with 15 and divide by twice the first figure will be 15/2 and this gives 7 with 1 left over. Should’t this step give me 6?

I am not thinking right somewhere.

I can square 16 well. 16^2 is square the 6=36 put the 6 with 3 carry , twice the product of 1 and 6 is 12, 12+3=15, put the 5 as the middle  figure. Now square if 1 =1 + carry 1 =2 as 100’s figure. =256

thank you

• #31261

Hi Savita,

When you get 7 as the 2nd figure in the square root of 256 you can check if it is right, as we did in the video, by comparing the square of that 7 with the 16 at the end of 256.

Since 16 and 49 are not the same we can go back and, instead of saying 15/2 = 7 remainder 1, we can say 15/2 = 6 remainder 3.

Doing that you then see that you have 36 at the end of your 256 now and that is the same as 6 squared.

So the answer is 16. Writing 6 rem 3 instead of 7 rem 1 is something we will be using in lesson 25 next week, so you have anticipated that.

The full square root process comes up in lesson 29 where the above process is extended.

• #31270

Are we going to deal with Square Roots with Irrational Numbers in this course?

• #31271

Yes we will Maurice – lesson 29.

• #31286

Thank you Mr. Ken.

Have a good weekend to the whole group.

• #31611

Dear Ken,

I was doing the practice problems of on square roots. I have a doubt.

while finding the 5229 square root, actually with the method of guessing the the last digit. The answer appears to be 73. But actually 73^2 is 5329. So from the method of guessing from the last digit. There is chance of error unless you are sure that the number is definitely a perfect square. Am  I correct in this? Please guide.

and if i try from the first method  to find out the answer, for 5229, from 52 …first number must be 7…7*7 = 49 the remainder is 32. and in 32, 14 (2*7 (first number)) goes in 2 times. So 32-28 = 4. Now 4 & 9 are remaining. Does it mean 49 is the remainder?

But 72^2 + 49 is not equal to 5229. Can you please elaborate?

• #31612

Hi Amara,

I’m not sure where you got the question from (if it’s from the free manual I think you may have copied it wrongly). Lesson 22 is about finding square roots of exact squares, so the method given will just give the first two figures of the answer, which for 5229 is 72.

49 is not the remainder until you have subtracted the duplex of 2 (the last figure of 72) from it. So 49-4=45: 45 is the remainder.

• #32842

Hi Ken,

Thank you for your reply. In lesson 22, while trying to find the square roots. I found this problem. Attached is the practice problem I was talking about. Instead of using the digit sum method for checking,  I tried to use the last figure method (attached). As you suggested, I can use the last figure method only when I know that it is a perfect square.

Can you please take a number which is not a perfect square and explain the remainder concept?

And one more request Ken, after the quiz, if we can know which question we have done wrong and what is the correct way, that would be really helpful. Because we are completing the quizzes, but we are not knowing which problem we have done wrongly and what is the correct answer for the same and the reasoning behind it. Thank you.

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• #32846

Hi Amara,

Actually no-one talks about remainders as far a square roots are concerned. I was trying to answer your question as best as I could. So I have explained about the remainders for 5229 in my reply above.

I believe you can see your answers to the quizzes and which ones are wrong after the quiz is completed. For an explanation please post in the thread or contact me.

• #32971

Thank you Ken

• #31615

Hello,

I have a doubt on video lesson 24 (general squaring).

Into the video I can see this calculus: (2x+3y+5)^2=4x^2+12xy+9y^2+20x+30y+25.

I think it is correct, but I noticed that in the video a + sign is missing between 20x and 30y.

Thanks

• #31616

Hi Fabio,

Yes you are right. Thanks for letting us know.

• #31633

Hi Ken,

This is regarding Triples. I solved all the problems from chapter 9 (TM – IL) except c and d in Practice E. (picture attached)

I am not sure whether I didn’t understand the sums correctly or it was a printing error.

Regards,

Mallika

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• #31635

Hi Mallika,

In c you first find that a triple for angle CMB is 4,5,-
So you can get the required angle by doubling this and taking from the triple for 180 degrees.

I hope this helps – let me know if it does not.

The problem in d is similar but you will need to introduce a value for the square’s side length, say 2.

• #31672

Hi Ken,

Thanks for the explanation. This is how I solved it (attached).  Please let me know if any corrections are needed.

Triples subtraction is introduced in later videos I guess (Lesson 13 in the book).

Regards.

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• #31684

Hi Mallika,

c is good. You get take a triple from 180 degrees most easily by just changing the sign of the first element.
Actually I forgot that triple subtraction comes up later in the book (in VMTTC we come to it in lesson 34), so my advice was not the best. I should have said to drop a perpendicular from M onto AB, call that point P. Then angle PMA is clearly 5,4,- and the required angle is double that.

My advice would be to draw DB, then you can take 4,2,- from 1,1,- (the triple for 45 degrees), and then double.

• #31725

Hi Ken,

I have reworked the sums using your suggestions. This looks a lot easier for me now.

Thanks a ton for all the patience. 🙂

Attached the reworked sums.

Regards.

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• #31727

Yes, that looks great Mallika. Well done.

• #31730

Thanks Ken.

• #32845

Hi Mallika,

Very nice to see your practice. Can you please tell me which book you are doing these practice questions. what is TM-IL ? Please let me know. Thank you.

• #32851

Thank you Deepthi.

TM -IL means Teachers Manual – Intermediate Level   by Ken sir

• #32972

Thank you Mallika.

• #32000

Hi,

It was interesting to listen, understand, and solve the duplexes of 1, 2, 3, 4, and 5 digit numbers and finding squares of  1, 2, 3, 4 digit numbers. And it takes practice to really get a hang of straight division of dividing a no. with 2 and  3 digit nos. and +ve and – ve flag figures.  It is inspiring to see the use of bar figures in division to make the process easier. The examples to show variations in which there was a 7 digit no ( dividend) and a 6 digit divisor  seemed complicated at first but after couple of tries, I was glad to understand it fully. The proportionately section was captivating. There is so much to do and play with numbers.

I am so impressed that Mallika is able to work on the triples.

Thank you,

Thanks,

• #32442

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

• #32568

That’s right Preeti. It’s brilliant the way that the Duplex automatically collects up similar terms.

• #32662

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

• #32663

Hi Preeti,

You are mixing your bases: 7000 and 10.
You would need to multiply the LHS by 7 only if you use 7000 as base for both numbers.

Good try though.

• #32676

okay, understood. Don’t have to multiply the LHS  by  7.Can you please explain how to proceed further   with base=7000.

• #32681

It would be rather absurd to find 7868×9 using a base of 7000 as your deficiencies will both the large numbers, 868 and 6991.

• #32685

Okay. Actually I meant multiplying 7868(with base=7000) and 9(with base=10).

Thanks

• #32698

It gets a lot more complicated when you mix bases like this. When I looked into it I decided it was not worth the hassle.

I think there is an article or two on this though in the Online Journal in my website.

But I would suggest you use 8000 as base, not 7000, as then both numbers will be below a base  and your deficiency will be only 132.