Number System UPSC CSAT | CSAT Maths Syllabus Questions | CSAT Maths Question and Answers PDF EduTap

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[@shahidias1884]

Respected sir, please continue with this series. This is better than any other paid classes.

[@vishnudutt7889]

Thank you dear.
Series will countinue.

[@vishalnaik3423]

​@vishnudutt7889 will it cover complete csat for free ? Is there any difference between video lectures with free content VS paid content? Please reply

[@MarriageBureauHaryanaRishtey]

Sir it's humble request to you
Please aap poore CSAT ko you tube lectures mein iss tarah cover karwa den ki students ko kisi bhi paid courses ki jarurat na pade taaki jo *students Paid courses afford nahi kar sakte vo bhi prepare kar paayen*🙂🥲
Jai shree Ram
Jai Hind 🧡🤍💚

[@gungunsingh6475]

Finally, I've solved H.W!!!!-Q(6):ans- Op(b)-6
Explanation:
Stp(i) : Break the given number 3798125P369 in pairs of 3-3 digits from our R---->L
i.e. 37 ,981 ,25P ,369
Stp(ii) : Now giving ( - , + ) alternative sign from our L ----> R in above broken pairs...
i.e. -37+981-25P+369
=369+981-37-(25P)
=369+981-37-(250+P) [0+P=P
=369+981-37-250-P
=1350-287-P
=1063-P
When we divide (1063-P) by 7, we get (6-P) as a remainder(r) and we know that r>=0 in a=bq+r
i.e. 6-P=0 => P=6 answer.
Thanks a lot Sir to motivating me through this initiative...

[@vishnudutt7889]

Your answer is correct

[@AnIndian142]

Hello dear
Can you explain me beyond
369+981-37-25P
Please🙏😊

[@jagriti...]

I solve it ...by traditional method.

[@Mysterious-o5s]

I have confusion on your step 2

[@rinkukharbas774]

Yes how step 2 used sir did not told this

[@Hqr7]

Last year was able to clear csat just becoz of your class. Thank you one of the best teachers for csat.

[@EduTapUPSC]

Great
Also share your result and experience with us at [e-mail]

[@so_ul_fighter]

Super 🎉

[@iasraviranjan4082]

answer for the question i.e. homework is 6
taking 3 - 3 pairs
369-25P + 981 - 37
369 - ( 250 + P ) + 981 - 37
(1063 - P )
when dividing 1063 by 7 we get 6 as a remainder and if we subtract 6 from 1063 it is divisible by 7
hence , P = 6 ans

[@vishnudutt7889]

Your answer is correct✅

[@AnIndian142]

How to write 25P =250+P

[@iasraviranjan4082]

@@AnIndian142 for example if a number is 256 the we can write it as 250 + 6 in the same way 25P can be written as 250 + P

[@AnIndian142]

@@iasraviranjan4082 बहुत बहुत धन्यवाद भाई 🙏🙏🙏💞

[@sanafarooq308]

Thank you very much

[@PriyaSingh-qn9pj]

This session is better than other paid courses. Thanku so much please keep continue this initiative.

[@naveenthayyil9164]

Answer Option B - 6,
Explanation
3798125P369
037 981 25P 369 (As per 3 pair)
981 - 037 + 369 - 250 P (25P to 250 P for easy simplification)
944 + 119 P
1063 P
P = 1063/7 (as divisible by 7)
Remainder= 6

[@ishasingh8841]

Why have you taken 25p as 250p here? Could you please explain?

[@preetisangwan363]

Too good sir…it seems csat is easy enough if your classes are followed and practiced well👍👍👍

[@adityasah3137]

The class is good sir but please provide the solution for the last homework question also. Thank you!!

[@SachinKumar-ol7nb]

❤❤❤
With each leacture i am getting the confidence to crack CSAT

[@humerakhan5463]

Amazing video sir... You are indeed a legend teacher 👏

[@tusharsrivastava1879]

Ans Option b- 6
Thankyou for the lecture Sir.

[@vishnudutt7889]

Your answer is correct.

[@aqengaming2556]

my answer is also b

[@PujaKumari-pc2nc]

6 is answer of hw..thank you so much sir for this wonderful class.

[@soniyalashkan8637]

Option b is correct....I go through the long procedure...that sir has explained.....but in that way answer will always same.......put the value that given in option at the place of p....and check it

[@d.Boi-ii]

yeah i did the same, hectic but works.

[@ananyajaiswal6952]

You are best teacher

[@anganaborah9828]

Thanks sir your explanation help me so much❤❤❤

[@sugyanisahu2752]

23:38 I think 'n' should be ODD. Becoz 10^5-1=9999 divisible by 11

[@Elakkiyasjoy]

But it is not 9999 sir it will be 99999 then it is not divisible by 11 sir

[@Monika_051]

Sir jo 11 ka divisibility rule ka first question h usme to n = 0 bhi le sakte h to fir vo even wala rule follow nhi kar rha

[@kirtikohli4636]

Amazing videos sir may Mahadev bless you with infinite wealth n health

[@chandrashekharrajput1024]

Very good teaching skill and explanation

[@muskii1807]

Thank you so much😊❤ sir you 🌹❤🙏are very talented teacher👩‍🏫‍ and best teacher👩‍🏫‍ in this world because you gave me a wonderful✨😍 sessions in free🆓 of cost.

[@NishaMaurya-o9h]

P=7 so may be option C is right answer

[@shalinikumari-p3b]

Wonderful explantion 😊 please complete this csat math session 🙏🙏

[@Hindu.NATIONALIST]

Why guide is not coming

[@rishabhgaming2.082]

Thankyou so much sir❤❤

[@nidhiaggarwal-cx6uz]

Your lectures are very useful for us sir.. thank you ♥️...for such a great explanation

[@happychhabra4228]

Thanku so much sir

[@nitheeshe4051]

May god bless you sir, thank you

[@NehaKumari-th3yh]

H.w.- P= 6 , thanku sir.

[@bootedakpe6698]

Sir your teaching techniques is brilliant 👍

[@user-roshni]

P=1

[@mirachavan3809]

Ans C (7) Thank u sir🌼

[@mohammadtasaleem8339]

Your teaching skill I awesome

[@Podcastrise]

Amazing vedio you are doing well

[@vishalnaik3423]

Have you taught taking -,+ alternatively in group of 3 pairs under 7 divisibility rule ?

[@nikitapatel8728]

No

[@MohanKumarYuvaTeja]

SIr please also include other previous year upsc questions too

[@upscmotivation769]

Thank you sir for giving me csat preparation full guidance and helping my ❤

[@vishnudutt7889]

Thank you dear.
Keep watching keep learning.

[@lovelyboy746]

Respected sir ,
I need some suggestion that how to crack upsc prelims and mains
Because the main reason behind me is I am in working period 9:30 am to 6 pm. So its difficult to get time for study.
So please suggest me how i manage timings for study ?

[@nandansingh6175]

Nice Class Sir.

[@SaritaYadav-xu6su]

Option b (6)

[@SahilKhan-ri8ex]

How

[@yadavdeepika]

​@@SahilKhan-ri8exhave you got answer?

[@yerriswamygm5328]

Unbelievable teaching sir...

[@gaganyadavgaganyadav8712]

Thank you sir for this valuable series

[@nikhilyadav9991]

Thanku so much sir ☺️ It is very helpful for upsc aspirants

[@sukhjinderrocky8993]

Finally found some valueable lectures ..thankyou sir .

[@anganaborah9828]

Right

[@shiprasri6707]

Lectures are free and very very very informative thanks a lot

[@aryaankitbaisla3004]

Sir tussi great teacher ho 🙏🙏🙏

[@shivani28282]

P=6
2nd technique used

[@MohanasaiLakshmidurga]

Tq sir good teaching

[@unicornakash]

Mazaa aa gya sirr gazab!!

[@NandiniKammar-u8s]

Answer option c sir and very nice explanation sir

[@SunitaDevi-mo8my]

You are awesome 😎 sir .thanks for your efforts.

[@gowthamarumugam29]

Thank you sir. CSAT has become interesting with your teaching

[@shivanksingh6054]

Thank you so much sir for this amazing session

[@gajuFadke]

If we take x=2 and Y =0 in this question, 65320/80 = 8169 then complete divisible...
Option no.( a) ......kiv nahi ho sakta. 13:59

[@divyakharey]

you have divided wrong... how could you get 8169... you'll get decimals

[@santrosharma9074]

When we are taking last two pairs 369-25p according to this 7 is correct because 369-257 is devisible by 7.
So lease confirm if we take only last two pairs or all pairs we need to take???

[@nancyraj1308]

B 6

[@AdityaKumar-tm9ty]

Sir m confused which playlist of you should i follow old one or this new one...?? i have not watched your old series yet nd m preparing for 2024 attempt...!! Is this new series would be sufficient for csat of my 2024 attempt

[@vishnudutt7889]

Follow the updated one also . Old +new both.

[@VanshuThakurr]

Sir, I am understanding everything you said ❤

[@shreyasikarapu397]

Such beneficial series!

[@AnIndian142]

Sir please H.W. ka answer bhi bta diya kro 🙏🙏🙏

[@praveendhere756]

Best teacher❤

[@mr_honey__97]

very helpful, wish we could reach there earlier.
thanks sir.

[@Mysterious-o5s]

Q6) (c)7 by 3 digit rule

[@manishaboora9117]

Explain

[@gunjankhatter5796]

Sir, thank you so much for your efforts and time that you are paying to teach us best in free. Your learning and teaching are worth 👌 ✨️. I am blessed to find this series. God bless you too for your kind work 😊

[@darshan27charan]

these classes are very helpful, thank you sir

[@sanamsingh3016]

Thnq sir your teaching style is awsm ❤

[@Iamjassxkaur]

Option : C is the answer 👍🏻👍🏻

[@shankarkumargupta2946]

Questions no 2 m
8 se last ke 3digit ko divide karna h n to x ki jgh 2 bhi to put kar sakte h 320/8=40

[@IASachin]

Correct

[@shameenasaiyad1376]

Thank you so much sir

[@tmr.mohsin8020]

Option 1

[@RamzanAli-s1e]

god bless u sir

[@devchauhan8901]

Thnx sir

[@amishadubey663]

Best series

[@vishnudutt7889]

Thanks

[@SonaliGaikwad746]

Thankyousomuch sir🙏🏻🙏🏻

[@ghjiklfk123]

And is b
Thankyou so much sir

[@sakshidiwakar256]

Ans is B (6)

[@RohitKumar-ut2yh]

yes

[@surajprajapati7761]

thanks for your efforts ☺️☺️

[@Alleey2311]

very nice explanation sir

[@girlishtalk7561]

Homework question 6 answer option b 6 28:30

[@abhisheksamantaray9078]

28:29 Ans. (c) 6

[@arunkumars3902]

option _ b

[@HindiMeTayari]

Thankyou sir jee ur thr best educator....love u sir

[@vijaybharati5687]

Thank You Sir

[@ankushsaroj4037]

HW ans 6
Going by option

[@anjaliawasthi7953]

Sir is question ko easily kar sakte hai 7 se divide kare aur diye hue option ki value rakhe ke check kar le 6 aa gya

[@iasraviranjan4082]

sir i have watched your previous lecture i.e available on this channel for 2021 and my first attempt is in 2024 should I watch the current ongoing series for 2024

[@vishnudutt7889]

Yes, this is the updated series.

[@mamtakaushik8707]

Thankuu sir👍😊

[@ayusdas6040]

These classes sufficient for 2026 & 27 aspirants??

[@amartripathi6985]

Thank u so much dear sir......

[@nitikasamra3338]

Home work answer (6)
Option (b) : 6 is correct

[@vishnudutt7889]

Yes your answer is correct.

[@pvartiya]

How?

[@ImVed-k9k]

Thanks for Classes 🎉

[@sirishabudumuru7185]

P=6 sir

[@shradhanjalidas8267]

Question 6
Answ- 1 i.e option A

[@Mumbai1880]

Thank you sir please continue this series

[@Hmmmmmm172]

Thank u so much sir for these lectures

[@nandikishorchhatria9713]

Thanks for the awesome series Sir.....

[@saketcr7416]

best youtube course on csat

[@danishraza5345]

Best teacher of csat

[@Tanu_auraa]

Thanku sir❤

[@priyaaggarwal9035]

very relavant classes