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CAT 2024 || Quantitative Aptitude Preparation
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CAT PREPARATION 2024
How to Prepare for Quantitative Aptitude
cat 2024 / cat 2024 syllabus / cat 2024 strategy / cat 2024 self preparation / cat 2024 classes / cat lrdi revision / cat lrdi pyq
It is easy if solved by equations i am giving my solution:
let no of hits by A=a , B=b , C=c and misses be p,q,r respectively
we know p+q+r = 35
now we know their score at end therefore :
For A: 2a+(missed by other) = 2a+(q+r) = 2a+(35-p) this equals 60
Similarly For B: 2b+(35-q) = 53
Similarly For C: 2c+(35-r) = 43
When we solve this we get:
p = 2a -25 , therefore total shot by A = 3a-25 (i)
q= 2b -18 , therefore total shot by B = 3b-18 (ii)
c= 2c -8 , therefore total shot by C = 3c - 8 (iii)
Now we know 78shots are taken (12 in each) so total is 6 rounds + 6 shots
6rounds = 24 shots
6shots = (4 , 2 , 0) these can be taken by anyone so now we hit and trial
new eqs are
3a-25 = 24+[4 or 2 or 0] (iv) for A
3b-18 = 24+[4 or 2 or 0] (v) for B
3c-8 = 24+[4 or 2 or 0] (vi) for C
If B takes 4 or 2 then from (v) b becomes non-integer so only option is 0 i.e 3rd
Similary A can take 2 shots only for a to be integer in (iv)
Hence C will take 4 shots
Hence you can calculate a,b,c and therefore p,q,r , all the best
Ok. Thanks.
cheers..nicely explained lad
!!
Nice Bro... Did you come up with it yourself..??? If yes, then you are very ahead in DILR...
Best
AT THE STARTING SIR SAID IT CAN BE SOLVED BY EQUATION, BUT EVERY WHERE EQNS DO NOT WORK SO BETTER IS WE BUILD OUR PROBLEM-SOLVING MINDSET.
Teachers like you are the ray of hope for students like me who need a tender explanation. Thank you, sir.
This was a nice set easy to solve except for the catch who goes first, if you consider in the sequence as ABC equations won't solve. So I was also frustrated and went to see the solution after that the first 10 secs of explanation we don't know who goes first, so after that I matches as per the equations which is solving giving no fraction values and the question got solved. Really nice question
Sir request you to cover all the topics of LRDI PLEASE.
IT WILL BE OF GREAT HELP🙏
I have an intresting observation -> 78 shots were made , 78/3 = 26, and we are proceeeding in steps of 4 so (26,26,26) is not possible . So if we simulate this game , first they all would have taken (24,24,24) shots and then 1st player shoots 4 , then 2nd player shoots 2 and then game over cause 60 points are reached . So the number of shots per player will be (28,26,24) but it is guaranteed that A has to be the player with 26 shots !!!! (something that was solved in a little bit more complicated way by the video)
i agree this was my approach too
Thank you for always explaining all these amazing conceptual sets so smoothly and brilliantly.
Instead of solving again from scratch for B I have considered, Total Miss- Miss by A = Miss by B and C (35-9= 26). Now for B I am considering that all the misses have been done by C and solved by same method I am getting Total hit by B + Total Miss by B (17+7)= 24. The total number of shots is same but the number of hit and miss are different so is my method also correct?
If all the misses are done by C so you also have to add the total misses of A for calculating for B because B is getting points for both A & C missing a hit!
That make sense
@@mahimanigam2845exactly, I have done the same mistake
It took 40 min for me to solve
Your approaches are impeccable sir, be it quant or LRDI, you have the best solutions!! Thank you so much for making this free of cost! 🤩
but what if we fix A as 17 shot and 9 missed, and continue solving for B and C, where we left Total hits remained as 43-17 = 26 and total miss remains as 35-9 = 26, we get B as 17H and 7M and C as 9H and 19M which satisfy the given conditions.. Can anyone care to explain?
if you calculate the total for C which is 9*2=18+9+7 is not equal to 43
also maths equation for solving this let x= no of hits , y= no of miss now -for a - 2x +(35-y)=60 x+y= 24/26/28 whichever returns an integer will be a value of x and y. in this case x+y=26 will return an integer.
simillarly for b - 2x + (35-y)= 53 , x+y=24/26/28, it will return integer value with x+y=24 , hece you will get values of x and y
Take 3hours to understand:,but finally understood
Where ia the lecture one of games and tournament
Amazing explanation! Thank you so much!
hello sir, firstly I would like to thank you for providing such amazing content to us. I have a small request to you to kindly make an introduction video on games and tournaments with tricks and tips. It would be really helpful.
Many of you have asked for the equation solution
Here you go !!
Assume Hits of A,B,C are a, b and c
Assume Misses of A,B,C are a1,b1,c1
a+ b + c= 43 - Equation 1
a1 + b1+ c1 = 35 - Equation 2
2 (a) + 1 ( b1 + c1 ) = 60 - Equation 3
2 (b) + 1 ( a1 + c1 ) = 53 -Equation 4
2 (c) + 1 ( a1 + b1 ) = 43 - Equation 5
Put 2 in 3
2a-a1= 35 - Equation 6
So a+a1 - Equation 7 can be either of 24,26,28
When you add Equation 6 and Equation 7 , you get a multiple of 3 so choose 26 as 26 +35 is 51 which is the only multiple of 3 possible for the case of a.
==> a+ a1= 26 - Equation 8
Solve 6 and 7 , a=17, a1=9
Similarly do it for others
Courtesy : This many numbering of equations may seem scary. Have just taken for better clarity.
This method of solving should be possible in 4-5 mins max.
Cheers !!
thnx a lot😄🥳
Sir your level of explanation is unmatched!
Sir your are doing best but It looks very difficult to use the thinking to solve the question plz tell us how to develop these thinking what should we do 🙏🙏
Wait! If there is given that out of total 78 shots there are 43 shots and rest 35 misses soo how can A get 26 points and b get 25 points it is going more than total no of misses according to what i thought is that out of 35 miss it should be divided between all three of them!
Thank you sir! Amazing approach 🙌🙌🙌
Thank you. If you like the channel please share among the friends who are preparing for CAT exam
Very well explained....worth 21 minutes 👌...keep doing this sir 👏
Where can we find more questions like these?? Please answer Sir!!
if all the miss was done by(b and c)..how is a's min number of hit points be 25..it has to be the max number of hit points ryt...am i missing somethng here
Total hits are 43 , if A's hit are 25 so rest 18 hits are shared by B nd C ..so A has max hits among 3
Sir is there free test series or topic wise questions or mocks you are providing with or without solutions apart from video ones?
Super duper question ,
And super duper approach,
By super duper Ravi sir. 😀
Beat teacher best explanation best lectures ❤️😍
Hloo sir I need a help.... kindly reply me
Thank uh sir.. You are simply amazing
Thanks
Done 👍🏻
30 September 2024
AMAZING SET EXPLAINED VERY EASILY SIR THANKS A LOT
It really opens up my mind and level of thinking
Thank u sir 🙏
Game finishes when someone scores 60 points. A was the winner that means 77th and 78th shots were by A so clearly A had 26 shots. Thats how I arrived at it.
Nope not necessarily
The only channel which directly comes onto the cat level questions with covering the basics.
Great approach sir
Why the sound in every video is so low, sir if you can improve upon the sound it would be of great help!!
You boost our thinking capacity!
Thanks
amazing explaination thnku sir for providing this free videos for all
Nice explanation boss.
Keep up the good work. :)
Doubt: How we concluded A's hits and misses = 28 or 26 or 24??
Please, let me know.
same doubt
amazing explanation Mr Prakash.......
thank you
Thank you very much sir.
78/12 =6 ..how it comes
How to go further the que with this calculation.. anyone pls reply
In ques it is mentioned that A,B,C shot 4 shots each and ends when total 78 shots are shot . So by dividing 78/12 ( total shots/total shots by A,B,C in one round),so you'll get 72 + 6 (6 as a rem)..now remaining 6 will be divided as 4 shots to the first shooter and remaining 2 shots to the second shooter(as it is mentioned in the question that everybody gets to shot 4 shots)
Hope it helps😅
Sir please sir put IPMAT or NPAT entrance exam coaching plz plz olz sir......
At 18:30 u said b is the 2nd shooter but actually b is the 3rd shooter
I think u meant the same but said it incorrectly
C can also take 26:
12 wins: 24
14 self lost= 21 lose gains
Therefore 24+21= 43
how to solve this ambiguity
how come you know you have to chk for A first and not B or C.
If you solve C takes 22 (10 hits), 25 (11 hits )and 28 (12 hits) and no possibility of 28
You made a calculation mistake. 24+21 = 45. Not 43.
C has total 43 points.
If C has 12 hits, it gets 24 points.
Then 43-24 = 19 points have to be got from misses of A and B.
19 points means 19 misses are from A and B.
Total number of misses is 35.
Therefore C has 35-19 = 16 misses.
Therefore, 12 hits + 16 misses = 28 shots.
As established earlier, the first shooter gets 28 shots. Hence C is the first shooter.
Also sir first checked for A, then showed how it would be if one first checked for B. You can check in any order, answer remains same.
I got different answer ...
C=. 9hits 10miss
A= 17hits 9miss
C= 17hits 16miss
I think u didn't follow the rounds concept (1st round ,2nd round,3rd round ,4thround and so on....)
But your shooters have taken 19 26 and 33 shots so your approach is wrong
Great,really cant believe it.you have made so simple
very important .Gotta rewatch!
Mere ko is qustion ka answer mila hi nhi
sir i'm trying to visit your website but unable to open it.
Thank you guruji ❤
thankyou sir for this brilliant set
Amazing concepts
Thank you Sir🙏
Great sir ❤
thank you sir
Thank you sir
thanku sir , hats off to u
really an interesting set
Sir C is not being satisfied under the given conditions.. Thats why your solution is wrong .
Conditions are satisfying for C too. C's no of hits should be >=4
C hits 12 shots and scores 24 pts. Rest 19 points from shots missed by A & B. So his no of misses is 35-19=16
You can also arrive at the conclusion like this:
Out of 78 shots, 43 were hit. So shots hit by C= 43-17-14=12
2
I got the sequence as 1st shooter-B (HIT-14+MISS 14=28 TOTAL), 2nd shooter-C ( HIT-10 +MISS 16= TOTAL 26, 3rd shooter A( HIT-15+MISS 9= TOTAL 24)
Points
B-> 14*2+16+9=53
C-> 10*2+14+9=43
A-> 15*2+14+16=60
Can anyone explain what mistake i have done?
Its wrong...your total number of hits is 39 but in question condition it is given as 43. So there lies your mistake.
If you take the hits of B as 14, his points via hits would be 28 , his total points is 53, this means he scores 53- 28 = 25 points through the misses of others i.e. A & B . Now total no. Of shots missed are 35. If B is getting 25 points through missed shots then obviously 35-25 =10 shots are done by him. Hence the total no. Of shots become 14+10 = 24.
Similarly apply the same for others...
Your answer is not satisfying the conditions mentioned.
@@maaztermaaz 12x6=72,remaining is 6. 72+6=78. that 6 is divided into all the shooters 1st,2nd,3rd as 4 shots ,2 shots ,0 shots orderwise
Sir , even 43 points can be achieved by 24 shots...(14+ 20)>>>>(28+15)
Plz plz plz sir.....you teach very gooood