We can solve like this too for question no.11 8 will appear 20 times from 100 to 200 so 8 wont appear 80 times from 100 to 200 so 80*8(that means from 100 to 900)=640+8(that is from 900 to 999)=648 exclude 1000 because it is a 3 digit question. therefore (80*8+8=648)
Sir, you teach from the bottom of your heart with lots of patience so that any person irrespective of his/her IQ levels can understand the concepts. Thanks for being a great teacher ! 🙏
your each video make me feel confident about QA .heartly thanks for your hard work.thanks alot for making it available at zero cost,i really needed.thanks sir
@@bobghosh8313 but during 1 to 100 he , the 60 to 66 part he counted 66 as a + 1. According to that the digit 6 should more than 100 times between 600 to 699.
From 600-699 ,we would get 110 6's + 19(normal case six) = 129. Then add in remaining that is 171+129 = 300 Its not related to 10 th question its should be actually 6 th question.
I am preparing for upsc where csat is consider as a barrier but when i found your videos my preparation in that portion became easier and understandable.thank you.
Sir here you changed the question from many many numbers have digit 6 to how many digits 6 will appear? The both make different sense as in first me mention only how many number have 6 atleast onece and in second we have to count number of 6 that appeared it will be more than the number that have 6 in them
It can be solved. (900-252) here 252 is achieved by adding (72+72+81+8+9+9+1) where 72,72,81 is achieved for no.s having 8 only once. 8,9,9 for no.s having 8 only twice and 1 for 888.
Sir your free contents are way better than so many Paid contents. Thank you
Dai summa iru da…
anyone studying for 2024 cat
Yes bro +1
yes
Yup
Bro anyone have rodha sir notes of all concepts please share
yes bro
Sir i have completed more than half of this playlist,Is this 280 videos playlist enough to cover all the cat syllabus,please reply sir!!!🙏🏻🙏🏻🙏🏻
keep on doing. I am adding
Did you make it in iim 😊
yes IIMA @@sumitbhagat3779
By his aura I assume he didn't@@sumitbhagat3779
We can solve like this too for question no.11 8 will appear 20 times from 100 to 200 so 8 wont appear 80 times from 100 to 200 so 80*8(that means from 100 to 900)=640+8(that is from 900 to 999)=648 exclude 1000 because it is a 3 digit question. therefore (80*8+8=648)
Sir 10 th question is
How many numbers instead of how many times and answer also according to how many numbers
Same
Brilliant explanation sir!
CONGRATULATIONS FOR 1M SIR!!!
Sir for 66 you counted two sixes making it 20 times.
but in 600 to 699 you didnt consider situations likes 606, 666, 660 etc..why?
atleast once vale hi count krne h bro
@@Ankur33Goyalbut in 66 he counted it 2 times!?
legendary!
When does n(n-1) and n(n-1)÷2 formula is used.someone tell me
This is summation formula which comes in Sequence and Series
Anyone studying for CAT 25
Sir, you teach from the bottom of your heart with lots of patience so that any person irrespective of his/her IQ levels can understand the concepts.
Thanks for being a great teacher ! 🙏
iq ki hi zaroorat hai mujhe abhi
your each video make me feel confident about QA .heartly thanks for your hard work.thanks alot for making it available at zero cost,i really needed.thanks sir
sir i love the way you are teaching
In question number10, the digit 6 will appear more than 100 times in between 600 to 699...check the 3rd method
Bro 666 will be taken as 1 , then 662 will be taken as 1 , u did not understand the question
@@bobghosh8313 but during 1 to 100 he , the 60 to 66 part he counted 66 as a + 1. According to that the digit 6 should more than 100 times between 600 to 699.
@@Anurag_Sharma_Captures he wrote the Q wrong, the Q should have been how many digits contains 6 not how many time
Why are we dividing it by 2! In the last question ..if anybody could please explain
hey pls explain this question im confused@@bobghosh8313
From 600-699 ,we would get 110 6's + 19(normal case six) = 129. Then add in remaining that is 171+129 = 300
Its not related to 10 th question its should be actually 6 th question.
I am preparing for upsc where csat is consider as a barrier but when i found your videos my preparation in that portion became easier and understandable.thank you.
Apka CSAT qualify hua 23 ?
sir in q11 can we do this by subtracting from total - 3 digits having 8 atleast once
And sir here from 601-700 there are only 99 number
But from 501-600 there are -30 number instead of 19
CSAT ke kuch questions ho jynge kya is playlist se?
26:22 What I understood is:
Eliminate the bigger n-r count for numerator and choose the smaller r-value for denominator
Thank you Sir 🙏, The way you explain numbers, it becomes really easy to understand.
Sir here you changed the question from many many numbers have digit 6 to how many digits 6 will appear? The both make different sense as in first me mention only how many number have 6 atleast onece and in second we have to count number of 6 that appeared it will be more than the number that have 6 in them
600-699 ke beech m v toh 19 numbers aur aayenge na, unko kyu nhi count kiya h
600-699 m 100 numbers aayge. Sabme 1 6 hai
Great teaching sir! I have studied from 3 different paid instituitions but no one can match his level of depthness in the concepts. Free treasure.
Thankyou so much for your content, its so much better and clear
Just 2 words, Great Teacher😎
Why 11th question can't be done by subtracting no of no's in which 8 appears from all possible no.s?
It can be solved. (900-252) here 252 is achieved by adding (72+72+81+8+9+9+1) where 72,72,81 is achieved for no.s having 8 only once. 8,9,9 for no.s having 8 only twice and 1 for 888.
Why u are taking 72 72 why not 81 81 81
From 600-699 isn't there will 119 times 6 (100 times 6 at hundredth place and 19 other)? 5:00
that was for AT LEAST 1 ..so only 100 will be considered
@@yashselukar1846 wow yeah nice point, ty ty
sir, in question no 11 y in 3 digit no "0" is not possible in hundred place?
coz if u take 0 in hundredth place ,it becomes two digit number.
because we have to take from 1 to 1000, 000 is nothing but 0 so 0 is less than 1
#sndeymath1 sir in question 10 .is it how many times or how many numbers 6 will appear.i think it should be in how many numbers.thanks in advance.
same doubt
exactly
yes it is in how many nos. that question is in continuation from previous video.
Prachi yes .thats the fact.
sir aapko alphabets ki numbering bhi yaad h ....... how many more ways will you amaze us sirrr
n
C
r
Sir I can't thank you enough for giving such content to us
You are the boss !
question ask for number of times 6 appear and method is for counting total numbers in which six appears
I have same doubt
yes it is in how many nos. that question is in continuation from previous video.
hats off sir
Sir you have solved wrong the que nu 11
answer will be 728.
Sir is right.
In quedtion 3 attitude why t is take has one and before sufficient why we cant take F as 1
Thankyou sir
Sir apne konse school mein padhte the.?
Thankyou Sir.
Amazing session sir🥰
Just wow.
Sir,in qs 11can we get the and by subtracting the total no available - no which have digit 8
Yes you can do that way too
If you mean 1000-271 then you can't do that. 271 if for 1 to 1000 but q11 asks for only 3 digit no.
tysm sirrr
How many 6between 600to699
699-600+1=100
what about 666?@@ritikagaddellu1704
Confidence ke sath galat bol jate hai aap kabhi kabhi
dont judge on 1% when you are gaining 99%