Thank you for a great session and helping us to gain knowledge. But in second question it should be 27 as a divisor for 135, so the answer obtained is 12/ 1350.
i appreciate the hardwork you have put in this video to make the concepts clear. In second question the LCM of denominator is 1350. so that the ans is 12/1350.
The best thing in your teaching is that you start from the very basic and then go for tricks so that the base become stronger.... thankyou so much for your efforts ❤️
If anyone wants to understand what just happened in 44:30 question no 9. suppose, the common remainder is k and HCF is x. 47 = ax + k 35 = bx + k 27 = cx + k we cant calculate HCF x of 47, 35, 27 because we don't know the remainder k so 47 - 35 = 12 = (a-b)x , 35 - 27 = 8 = (b-c)x , 27 - 47 = -20 = (c-a)x and HCF of all this 12, 8 and 20 has HCF x and it is free from remainder k
CareerRide is my new learning platform. Great Going .. pls don't change the way of teaching. many channels cover in just 10 to 15 min which helps nothing. loved this video.. starting from basics and covered each type of question. Thanks alot
Thank You for your efforts, Sir. In Q3, the remainder of 93/13 is not 15 as 13 * 7 = 91 and we get the remainder as 2. I think there is a problem with the question as HCF would be 13 only when the remainders are 4, 3, 2. Please correct me if I am wrong.
q11-: when we put 'x=4' in two equations , we get '-1' in both the equations -1=-1 . To find 'k' we equate both the equations that means both the equations has same value. So, i think for that question equation answers should be equal instead of equations equal to '0'.
who so ever is coming across the video don't leave it because it's of 1hr 10 min i completely assure you that it will help you a lot thank you for the knowledge 🤗
if u think that way you won't be able to solve these questions..because to leave remainder 15 a number must be greater than 15..and less than than 17...because first number is 17..so only 16 but that will leave remainder as 1 which is not asked in question....so his answer is correct....
at timestamp 36:40 isn't this question wrong, how can a number when divided by a divisor leave a remainder greater than the divisor itself? Also at in Q 11) 50:57 value of K is 15/4, and not 4, also 4 is not a factor of the equation x^2 - 8x + 15, for this eq. the factors are 3 and 5. Also this equation does not have any role in solving the question
Thank you sir these videos are very helpful and you explanation is very good !! But i have a doubt in question 11. I can be wrong so please correct me. When you put the value of x=4 in the equation they are not equal to zero. It would be correct if the equation was (x^2)-8x+16 instead of (x^2)-8x+15. And again thank you for your efforts.
But when we put 'x=4' in two equations , we get '-1' in both the equations -1=-1 . To find 'k' we equate both the equations that means both the equations has same value. So, i think for that question equation answers should be equal instead of equations equal to '0'.
@@karansiriya7796 Yes Actually ( X-4) is not HCF to those equations.. But in the question they given If (X-4) is HCF to those equations then what is K value?... So when we put X= 4 in the both equations they must get satisfied.. That means when X=4 both equations get equal value. So putting X=4 in both equations by equating we get K=4 According to question we have to find K value and they asked if X-4 is HCF then what is K value..
In question 12 the answer is 5 times because firstly they ring together at 00:00 also and after it rings 4 times so it will be 5 times. If I am wrong, please let me know.
The explaination is really wonderful. Me just want a few things. It is not a must to find HCF when greatest is mention. We have to check whether it is written "divided by" or "divides" in the question alongwith. For example, in a question like" find the greatest no of 3 digit which when divided by 6, 9, 12 leaves 3 as remainder in each case". Here we have find the LCM of the nos. and then carry on the other steps as required. Actually I derive it after practicing few problems of HCF and LCM.
@@Agl_Mansha789 Yes although Greatest was mentioned but it's asking to find the dividend as divisiors are given. GCD will find the factor of divisor itself which will be wrong so we find LCM then add the remainder
Thankyou so much, this was my weak topic from start of my school year till date when I am in engineering and I was ashamed to admit it, but because of your videos, I have understood the topic completely and can solve its questions correctly. Thankyou Sir and CarrerRide.
Thank you so much sir..🙏for providing content which is very helpful for competitive exams and ur explanation is awesome👏 so satisfied vth ur classes sir..😊
Hello sir the answer for clocks question is ......(4+1)=5 times they ring in the span of 8 hours..... because we have to consider the initial ringing also at 00:00 time What do you say???
This channel is free to understand concept but still ur society is not seeing the good side still sending comments on the small multiplication mistake.For those people i would like to say "SAALE AGAR ITNI HI ATA HAI TO KHUD KAR LE"
Yes I am seeing ur comment after two years but this is the fact😂 when you are doing great things noone appreciate you but if u have done a small mistake society will always criticise u
For Q3, if you consider a=bq+r, then you have 17=bq1+4, 42=bq2+3, and 93=bq3+15. Clearly, bq1=13. Since 13 is a prime number, q1 must be 1. But b is also the HFC, so the answer is 13.
I think question is wrong because if we divide x^2 - 8x + 15 by x - 4 It is not completely divisible so it is not the factor there must be mistake in equations
Your method is wrong. Remainder can never be larger than the divisor . Ques 3 , ans is 13 but the last part (93) has remainder as 15. when 93 is divided by 13 remainder is 2 . Also in Ques 6, you derived the ans as 1461 , remainder 21 isn't possible in any case as all the divisor are smaller than the remainder. 1461 gives remainder 9,5,3,1 for divisors 12,16,18,20 respectively.
A very good observation. In that sense, both the questions (3 and 6) are wrong. Also, the solution for Q15 could have been a little more elegant. The sum of 156, when divided by 13 (the HCF) gives 12. So, essentially, we need two numbers that are co-prime (so that the HCF remains 13) and that add up to 12. The only two pairs that work are 1 and 11 and 6 and 7. This is simpler than having to work with larger 3-digit numbers and checking if their HCF is 13. Still, I must add that such mistakes do creep in at times and it should not be allowed to cloud the good work done by CareerRide or other educators.
Thanks for covering different types of problems. It helped me a lot...Can you please explain general method for the last method coz I am getting answer different from the option verification. Again, Thanks a lot..... you are better than my Aptitude Sir.
But when we put 'x=4' in two equations , we get '-1' in both the equations -1=-1 . To find 'k' we equate both the equations that means both the equations has same value. So, i think for that question equation answers should be equal instead of equations equal to '0'.
Practice Questions on "HCF and LCM" available at:
www.careerride.com/mcq/problems-on-h-c-f-and-l-c-m-quantitative-aptitude-mcq-questions-13.aspx
Upload video for number system
Thank you soo much sir
Sir link is not opening
Q6 was wrong check it once
Sir 135/5=27
Thank you for a great session and helping us to gain knowledge.
But in second question it should be 27 as a divisor for 135, so the answer obtained is 12/ 1350.
12/4050
correct
Yes there is a correction in final answer.
yes i also got this answer
12/1350
Second question is wrong.
Ans is 12/1350
(135/5=27 not 29)
Ans. 24/1350
👍
@@suniln5542 24 can't divide 36 so 12 is right.
my ansr also 12/1350
Help pls. Can it still be simplified to 2/225? Or should it just be 12/1350?
This is what called a teaching from 0 to hero. I'm so lucky that youtube recommend this channel to me 😊🎉❤
Friendly way of teaching just like teaching in classroom.Thank you so much sir from bottom of heart for such lovely tutorials.
i appreciate the hardwork you have put in this video to make the concepts clear. In second question the LCM of denominator is 1350. so that the ans is 12/1350.
Yeah broo
Well said bro in a polite way
wah ehsaan kiye bata kar ata hai to khud karlo ek to free mein chanel bata rha hai upar se usko bhi rectify kar rhe hai
Indeed.. litrally I tried for 2-3 times and I got the same 25*27*2 = 1350
Yes true
The answer for question 2 is, 12/1350 because LCM = 1350 and HCF = 12.
Yess
Yeah that's what made me doubt the answer
Exactly ☺️☺️
I cross checked myself thinking that I gone wrong while simplifying for 10 minutes 🤣🤣
He made mistake in calculating 5*27 =135 but he said 5*29=135
this channel was recommended by our institute really awesome videos i feel instead paying fees in my institute would have watched your videos before
The best thing in your teaching is that you start from the very basic and then go for tricks so that the base become stronger.... thankyou so much for your efforts ❤️
Yes
In 2nd question of finding hcf of 36/75, 48/150, 72/135 .
LCM of (75, 150,135) = 1350 because all those are divisible by 15. So the answer is 12/1350.
12/1350
@@rambabupalla959 please ask mee 7 question why lcm is 480
@@bhuvidhana Bro 480 is given pls read que carefully
If anyone wants to understand what just happened in 44:30 question no 9.
suppose, the common remainder is k and HCF is x.
47 = ax + k
35 = bx + k
27 = cx + k
we cant calculate HCF x of 47, 35, 27 because we don't know the remainder k
so 47 - 35 = 12 = (a-b)x , 35 - 27 = 8 = (b-c)x , 27 - 47 = -20 = (c-a)x
and HCF of all this 12, 8 and 20 has HCF x and it is free from remainder k
best class i ever attended...of hcf and lcm
There were some mistakes in the exedcution but the concepts and methods dicussed were very accurate and good. Well done
51:47 When you put x= 4 to X^2-8X+15 it will never be 0.
yeahh.. same doubt
@@misahhere9224 the right answer is k= 15/4
Explaination:
(4)^2 - 4k - 1 = 0
16 - 4k - 1 =0
15 - 4k =0
4k = 15
Therefore k = 15/4
Exactly his answer is wrong
it should be 16 instead of the 15 it is question or typing mistake
x-4 cannot be the HCF of these equations
CareerRide is my new learning platform. Great Going .. pls don't change the way of teaching. many channels cover in just 10 to 15 min which helps nothing.
loved this video.. starting from basics and covered each type of question. Thanks alot
Just this one playlist and all topics are thoroughly covered with variety of questions
Thank you so much
Learned so many tricks that I am sure I wont remember when to use which in exams😅
Sir the way you teach
You make all Difficult sums so easy
Thanks a lot sir
It is very helping during lockdown..
you : 49:40 that 19th table
my brain: we don't do that here
Us bro us😂
Sir,ur teaching is very useful to us.And,in the 2nd sum,I think the lcm of 135 is 27
Excellently explained!!! Commendable job 👏👏👏
Thank You for your efforts, Sir. In Q3, the remainder of 93/13 is not 15 as 13 * 7 = 91 and we get the remainder as 2. I think there is a problem with the question as HCF would be 13 only when the remainders are 4, 3, 2. Please correct me if I am wrong.
You are right. I think question is misprint. Remainder would be 2 not 15.
Yes. Bro.. I just checked now
Yes question is wrongly printed
Correct 💯😅
it may be some tricky question made to confuse us
The Selection Set of Questions in this vid are DAMN GOOD!!
It was a nice video but Question 2: LCM is 1350 (75,150,135).
Yes, Bcoz here by mistake he took 29 in place of 27. i.e. 135/5= 27.
Thiyagu yes u r correct
Yes sir it's not 29
Yes
Yess ua crct
The way you teach sir has developed a interest in me in studying it ....thanks a lot
hey can i continue with this channel for apptitude
I am really thankful for your efforts of helping students.I learnt a great deal from you,as I was struggling before finding your channel
q11-:
when we put 'x=4' in two equations ,
we get '-1' in both the equations
-1=-1 .
To find 'k' we equate both the equations that means both the equations has same value.
So, i think for that question equation
answers should be equal instead of equations equal to '0'.
who so ever is coming across the video don't leave it because it's of 1hr 10 min i completely assure you that it will help you a lot thank you for the knowledge 🤗
Thanks for assuring other viewers, Hritika. Appreciate it :)
Thanqew for your wonderful teaching!😻
The methods, tricks are so wonderful!!
But the main thing is you have a catchy voice!! Very pleasant!!!😍🎶
At 36:35
least number to find...
The differance is same value can be subtracted
The differance is different what should be do
Ooo God.......atleast at the end someone came up with best explanatory vedio.....😊 thanks ...
Thank you sir ! Thank you so much for teaching. Now I placed CTS company with the help of your videos and your teaching. Great job sir.
That's a fantastic news, Arun. Congratulations. Thank you for coming back and letting us know. All the best. :)
Cant express my gratitude through mere words.....thanku so much sir
Class is very use full thank u sir
Sir, your way of teaching is superb.
Thankyou so much sir...I understood very well👍🏼
Really
U r the best Teacher ... Thanks for making this video
39:33 how came answer would be 1461...
When 1461 is divided by 12 the answer will be quotient 121 and reminder be 9
If u take one less ,i.e add 12 to 9 then it will be 21... How else a no. Like 12 can leave remainder 21??
Thank you so much for helping me to clear TCS nqt ,got selected for TCS ninja role🎉
That's a great news. Congrats.
And, thank you for coming back to share the news with us.
hi bro can you guide me.
Thank you sooo much sir.Gonna suggest this wonderful channel to all my peers.Keep making such videos:)
So nice of you. Thank you!
27:28 if we divide 93/13 then we get remainder 2 not 15, we get remainder 15 when we have 13*6= 78.
Pls explain 93 divided by 13 leaves reminder 15.🤔
Check out at minute 29:51.
Yeah when 93 divided by 13 we get reaminder as 2 not 15
93/13 reminder 15
if u think that way you won't be able to solve these questions..because to leave remainder 15 a number must be greater than 15..and less than than 17...because first number is 17..so only 16 but that will leave remainder as 1 which is not asked in question....so his answer is correct....
Sir bht achaa smjhaya aapne
Doubts the jo bhi saare clear ho gye ..😀
Sir you are amazing , you find awesome alternatives to cancilisation of any problem in just a minute..❣️
at timestamp 36:40 isn't this question wrong, how can a number when divided by a divisor leave a remainder greater than the divisor itself?
Also at in Q 11) 50:57 value of K is 15/4, and not 4, also 4 is not a factor of the equation x^2 - 8x + 15, for this eq. the factors are 3 and 5. Also this equation does not have any role in solving the question
x-4 is not a factor of x^2 - 8x + 15.
@monuxyz yes but how is it relevant
thank you very much sir !! this video helped me a lot
See its so easyyyy ....you made our life easier bro...thanks
Most welcome 😊
Thank you sir these videos are very helpful and you explanation is very good !!
But i have a doubt in question 11. I can be wrong so please correct me. When you put the value of x=4 in the equation they are not equal to zero. It would be correct if the equation was (x^2)-8x+16 instead of (x^2)-8x+15. And again thank you for your efforts.
K= 15/4
No need to equate.. simply put x=4 in x^2-kx-1
But when we put 'x=4' in two equations ,
we get '-1' in both the equations
-1=-1 .
To find 'k' we equate both the equations that means both the equations has same value.
So, i think for that question equation
answers should be equal instead of equations equal to '0'.
But hcf is also a factor.. so it should divide perfectly.. i.e. remainder = 0 ,, so why equate
@@karansiriya7796
Yes
Actually ( X-4) is not HCF to those equations..
But in the question they given
If (X-4) is HCF to those equations then what is K value?...
So when we put X= 4 in the both
equations they must get satisfied..
That means when X=4 both equations get equal value.
So putting X=4 in both equations
by equating we get K=4
According to question we have to find K value and they asked if X-4 is HCF then what is K value..
Ohh.. ok.. I got it.. just a little mistake in video.. "equations = 0" ,, instead directly equate..
Thanks 👍
In question 12 the answer is 5 times because firstly they ring together at 00:00 also and after it rings 4 times so it will be 5 times. If I am wrong, please let me know.
Q:10)114,76,152
Are divisible by 19,
Y 2 and then 19,
Wen u can do it by 19 directly.
Ye mera favorite channel hai I love this channel bcz mujhe yahan se bahut kuchh sikhne mil rah hai thank you sir
I studied this video for more than 2 weeks sir. You are so awesome sir. Thank you sir and CARRER RIDE for updating this. Thankyou...
Most welcome :).
Do take the advantage of our interview videos as well.
@@CareerRideOfficial Surely
Awesome , just awesone . Best and simplest video you're gonna find on HCF/LCM.
finally, I choose the best platform to learn aptitude...Thank you for the best content
You're welcome :)
Literally this is a nyc video and full of information.... nyc playlist (most wanted)
The explaination is really wonderful. Me just want a few things. It is not a must to find HCF when greatest is mention. We have to check whether it is written "divided by" or "divides" in the question alongwith. For example, in a question like" find the greatest no of 3 digit which when divided by 6, 9, 12 leaves 3 as remainder in each case". Here we have find the LCM of the nos. and then carry on the other steps as required. Actually I derive it after practicing few problems of HCF and LCM.
But in all the qs discussed, where greatest no was mentioned wether along with divided or divides - HCF WAS USED....CAN U EXPLAIN...
When will we use lcm when greatest no is mentioned...i don't understand
@CareerRide sir can u plz explain
@@Agl_Mansha789 Yes although Greatest was mentioned but it's asking to find the dividend as divisiors are given.
GCD will find the factor of divisor itself which will be wrong so we find LCM then add the remainder
@@Agl_Mansha789 So say simply if question says to find:-
Divisor (Divides) - GCD/HCF
Dividend (Divide by) - LCM
Dhanyavad Sir Ji 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
Every question is beautifully explained..thank u so much 😊
You're welcome, Bipasha! Glad that you found the content useful. Stay connected :)
Hyyyy
Thankyou so much, this was my weak topic from start of my school year till date when I am in engineering and I was ashamed to admit it, but because of your videos, I have understood the topic completely and can solve its questions correctly. Thankyou Sir and CarrerRide.
Thank you so much sir..🙏for providing content which is very helpful for competitive exams and ur explanation is awesome👏 so satisfied vth ur classes sir..😊
Fantastic session🎉worth watching ❤jst loved it ...very helpful 😊
For what you are preparing?
your communication skill is too good.....
Hello sir the answer for clocks question is ......(4+1)=5 times they ring in the span of 8 hours..... because we have to consider the initial ringing also at 00:00 time
What do you say???
Definitely ur hardwork is appreciable but there is an error in question 3 solution....Please have a look
Bro you literally cleared my concepts thnks alot
This channel is free to understand concept but still ur society is not seeing the good side still sending comments on the small multiplication mistake.For those people i would like to say "SAALE AGAR ITNI HI ATA HAI TO KHUD KAR LE"
Yes I am seeing ur comment after two years but this is the fact😂 when you are doing great things noone appreciate you but if u have done a small mistake society will always criticise u
Best channel.....this video solve my all doubts👍
Sir Q(11) is wrong . (x-4 ) is not the factor of (x2-8x+15) . so, (x-4) can't be HCF of both equations . Hope you correct it Sir.
Exactly
Is here the question wrong only or the method and concept are also wrong?
@@OvishaSanyal The methodology is correct, but the question is wrong.
Excellent videos, thank u soo much for these videos. it's really helpfull🙏🙏🙏
For Q3, if you consider a=bq+r, then you have 17=bq1+4, 42=bq2+3, and 93=bq3+15. Clearly, bq1=13. Since 13 is a prime number, q1 must be 1. But b is also the HFC, so the answer is 13.
But 93%13=2.
In the question, it's 15 (which is again 13+2). So I am a bit confused if the question is correct or not.
You are excellent.May God bless you ❤️
Excellent Work!!!❤️ there was just Calculation error in Q2 but approach was appreciable
Ya, LCM of denominator is 450 right?
I solemnly appreciate your efforts sir. Very helpful honestly!
I absolutely enjoyed this session 😄❤️
Enjoyed a lot by learning...thanks for this video..☺️
11th question:if both the equations are equal to 0,why can't we substitute X=4 directly in eq 2
The equations are not equal to 0. Both the equations will be equal to each other if 4 is substituted in them.
@@kowshikatmakuri4193 why??
I think question is wrong because if we divide x^2 - 8x + 15 by x - 4
It is not completely divisible so it is not the factor there must be mistake in equations
Absolute gem of lectures ❤
Sir very nice, need more new sums.....in each chapters....humble request
Thank You Sir, For the great explanation!!
Very helpful. Respect from Pakistan
India always helps u...
It was an amazing lecture. thank you very much.
Such an amazing explanation by u sir.....got it like that I'll never forget it ....thanks for ur efforts for us....👌👌
Super explanation sir I completely understand
In question 12 we will add 1 to the answer so it will become 5 cuz we also count the bell which rang just before .
Question says....from the moment they start...not the moment they rang together
Well this is correct
Very useful videos.. tq careerride
Your method is wrong. Remainder can never be larger than the divisor . Ques 3 , ans is 13 but the last part (93) has remainder as 15. when 93 is divided by 13 remainder is 2 .
Also in Ques 6, you derived the ans as 1461 , remainder 21 isn't possible in any case as all the divisor are smaller than the remainder. 1461 gives remainder 9,5,3,1 for divisors 12,16,18,20 respectively.
A very good observation. In that sense, both the questions (3 and 6) are wrong. Also, the solution for Q15 could have been a little more elegant. The sum of 156, when divided by 13 (the HCF) gives 12. So, essentially, we need two numbers that are co-prime (so that the HCF remains 13) and that add up to 12. The only two pairs that work are 1 and 11 and 6 and 7. This is simpler than having to work with larger 3-digit numbers and checking if their HCF is 13.
Still, I must add that such mistakes do creep in at times and it should not be allowed to cloud the good work done by CareerRide or other educators.
Nicely explain.... Thanks 🙏 so much , it help me to clear all my doubts..
Thanks for covering different types of problems. It helped me a lot...Can you please explain general method for the last method coz I am getting answer different from the option verification. Again, Thanks a lot..... you are better than my Aptitude Sir.
It is a bit clumsy. But very excellent
explanation and very important problems selected
Thank u sir your all tricks are very easy to understanding...
6th question:how can 1461 leaves reminder 21 when divided by 12 while it actually leaves a reminder of 9 with quotient 121
@@gourvi6339 yes it is correct
yes. your correct it leaves the remainder 9. not 21
Yes
For 16,18 only it leaves remainder 21
In 6th question method was right but question was wrong.
@@gourvi6339 it is wrong
in 52:18 the question and solution are wrong.
How can be the remainder of 13 be greater than 13...in the 3rd question the remainder should be 2 i.e. 13 * 7 = 91
93 = 13 * 7 + 2
yes
Tq u so much sir for such a Meaningful video making for the students
Clarrified my doubt how to decide whether LCM or HCF
Thank you sir
Please make classes on Data interpretation and sufficiency
+1
Thank you soo much sir 😊 . Yours videos helping me a lot . 🎉
review Q11 : how come when putting x=4 gives zero for x^2-8x+15
Does it come -1?
@@jeswaniamit553 yes
But when we put 'x=4' in two equations ,
we get '-1' in both the equations
-1=-1 .
To find 'k' we equate both the equations that means both the equations has same value.
So, i think for that question equation
answers should be equal instead of equations equal to '0'.
It's very useful and understand very easy ...
mistake occurred in hcf and lcm at 2 problem