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 ❤️
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
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
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.
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
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 🤗
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..
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'.
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....
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..😊
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
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.
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.
I appreciate d hardwork u hv put in this video to make d concept clear ..thankuuu sooo much ❤ it helps me a lot lot lot lot plz one request can u plz teach English grammar..i doo silly mistake & go wrong with that plz think about this
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.
If (x-4)=0 then x=4 Equate both the equations then substitute'x= 4' in those equations then we get 16-32+15=16-4k-1 -1=15-4k 4k=15+1 4k=16 K=16/4 'K=4'.
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'.
15) question My solution A+B = 156 HCF = 13 which means A and B both are multiples of 13, so take it common. 13(A+B) =156 i.e. (A+B) = 12 So the pairs are (1,11), (2,10),(3,9),(4,8),(5,7),(6,6) Now in these pairs choose only pairs whose HCF is 1 Therefore answer is (1,11) and (5,7).(2 pairs) Ps: I found this easier 😂...I don't like big numbers
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.
11th questioms solution doesnt make any sense its completely wrong. i appriciate the efoorts you put in making the concepts clear but i was disappointed with q3 q6 q11
@Tarun Bisht 93/13 the remainder is 2. you cant do it like 13 6za 78 and reaminder is 15.thats not possible and also reaminder can never greater than the devisor. take a pen paper and solve 93/13 you ll get what m trying to tell.
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'.
What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but which when divided by 13 leaves no remainder? a. 312 b. 962 c. 1586 d. 1562 yaravathu ans pannunga
Divisibility Inverse Geometry, Coordinate Geometry Clocks & Calendar Logarithms Algebra pls make video on above quant topics as these also come for campus placement aptitude tests
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
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?
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
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
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
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
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
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
Learned so many tricks that I am sure I wont remember when to use which in exams😅
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
There were some mistakes in the exedcution but the concepts and methods dicussed were very accurate and good. Well done
Just this one playlist and all topics are thoroughly covered with variety of questions
Thank you so much
best class i ever attended...of hcf and lcm
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
Sir,ur teaching is very useful to us.And,in the 2nd sum,I think the lcm of 135 is 27
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
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
Excellently explained!!! Commendable job 👏👏👏
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.
you : 49:40 that 19th table
my brain: we don't do that here
Us bro us😂
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
Thanqew for your wonderful teaching!😻
The methods, tricks are so wonderful!!
But the main thing is you have a catchy voice!! Very pleasant!!!😍🎶
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
Sir the way you teach
You make all Difficult sums so easy
Thanks a lot sir
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
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 :)
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 👍
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. :)
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'.
Thankyou so much sir...I understood very well👍🏼
Really
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....
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..😊
Sir in Q6 1461 doesn't leave remainder as 21 when divided by 12......kindly clear my doubt if I'm wrong
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
Sir you are amazing , you find awesome alternatives to cancilisation of any problem in just a minute..❣️
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
Q:10)114,76,152
Are divisible by 19,
Y 2 and then 19,
Wen u can do it by 19 directly.
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.
Every question is beautifully explained..thank u so much 😊
You're welcome, Bipasha! Glad that you found the content useful. Stay connected :)
Hyyyy
Excellent Work!!!❤️ there was just Calculation error in Q2 but approach was appreciable
Ya, LCM of denominator is 450 right?
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 very much sir !! this video helped me a lot
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
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
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.
Definitely ur hardwork is appreciable but there is an error in question 3 solution....Please have a look
Thank you sooo much sir.Gonna suggest this wonderful channel to all my peers.Keep making such videos:)
So nice of you. Thank you!
Sir very nice, need more new sums.....in each chapters....humble request
I appreciate d hardwork u hv put in this video to make d concept clear ..thankuuu sooo much ❤ it helps me a lot lot lot lot plz one request can u plz teach English grammar..i doo silly mistake & go wrong with that plz think about this
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
Cant express my gratitude through mere words.....thanku so much 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
Class is very use full thank u sir
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.
Ooo God.......atleast at the end someone came up with best explanatory vedio.....😊 thanks ...
Sir
In 11th question , I can't understand how k is become 4.
Will you please explain it.
Yes, even I too have the same doubt
If (x-4)=0 then x=4
Equate both the equations
then substitute'x= 4' in those equations then we get
16-32+15=16-4k-1
-1=15-4k
4k=15+1
4k=16
K=16/4
'K=4'.
Even i have the same doubt, x=4 doesn't give 0 in first equation
Excellent videos, thank u soo much for these videos. it's really helpfull🙏🙏🙏
Is anybody noticed that 4 doesn't satisfy the equations in 11th que ?
yup 4 doesnot satisfy
actually i think it should be x-5
yeah
Sir, your way of teaching is superb.
At 36:35
least number to find...
The differance is same value can be subtracted
The differance is different what should be do
Ye mera favorite channel hai I love this channel bcz mujhe yahan se bahut kuchh sikhne mil rah hai thank you sir
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'.
15) question
My solution
A+B = 156
HCF = 13 which means A and B both are multiples of 13, so take it common.
13(A+B) =156
i.e. (A+B) = 12
So the pairs are (1,11), (2,10),(3,9),(4,8),(5,7),(6,6)
Now in these pairs choose only pairs whose HCF is 1
Therefore answer is (1,11) and (5,7).(2 pairs)
Ps: I found this easier 😂...I don't like big numbers
how your pairs satisfy a+b=156
13(5+7) = 156
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.
You are excellent.May God bless you ❤️
I absolutely enjoyed this session 😄❤️
Dhanyavad Sir Ji 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
Bro you literally cleared my concepts thnks alot
Please make classes on Data interpretation and sufficiency
+1
The Selection Set of Questions in this vid are DAMN GOOD!!
LCM of 75,150,135 is =1350
5x27=135...what you did was 5x29!
Yes I too feel the same
S i feel same
Enjoyed a lot by learning...thanks for this video..☺️
your communication skill is too good.....
In Q.3, why would 93/13 leave 15 as remainder? How could remainder be greater than the divisor?
11th questioms solution doesnt make any sense its completely wrong. i appriciate the efoorts you put in making the concepts clear but i was disappointed with q3 q6 q11
@Tarun Bisht 93/13 the remainder is 2. you cant do it like 13 6za 78 and reaminder is 15.thats not possible and also reaminder can never greater than the devisor. take a pen paper and solve 93/13 you ll get what m trying to tell.
@@shreedharbadiger7594 then I think the question was wrong..
Thank You Sir, For the great explanation!!
I understood every question but in 11th question value of equations is not 0 will any one explain plz!
Same doubt here
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'.
@@surekhagiri8023 correct
What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but which when divided by 13 leaves no remainder?
a. 312
b. 962
c. 1586
d. 1562
yaravathu ans pannunga
finally, I choose the best platform to learn aptitude...Thank you for the best content
You're welcome :)
Awesome , just awesone . Best and simplest video you're gonna find on HCF/LCM.
Such an amazing explanation by u sir.....got it like that I'll never forget it ....thanks for ur efforts for us....👌👌
Best channel.....this video solve my all doubts👍
In 2nd question answer is wrong...
U write 29 instead of 27
Yup
Yes
Yes ans will be 12/1350
Literally this is a nyc video and full of information.... nyc playlist (most wanted)
It is very helping during lockdown..
Sir 3 problem 93 further divided by 13 7 time remaining 2 alone remainder
that what i got as well. it cannot be remainder 15 because 13 can go in 15
It is a bit clumsy. But very excellent
explanation and very important problems selected
fraction second ques wrong ..its 12 / 1350
Second question that you have cleared the way is wrong.. (135/5=27) not 29..so that we get the answer as 12/1350 is correct answer.
Correct
answer is 12/4050
I solemnly appreciate your efforts sir. Very helpful honestly!
Divisibility
Inverse
Geometry, Coordinate Geometry
Clocks & Calendar
Logarithms
Algebra
pls make video on above quant topics as these also come for campus placement aptitude tests
i.) Calendar Video: ua-cam.com/video/oxc4G14nyUY/v-deo.html
ii.) Clocks: ua-cam.com/video/edEvlh0tqzk/v-deo.html
iii.) Logarithms: ua-cam.com/video/w-7mbazOx6o/v-deo.html
Hi in our material there are some questions like they will product of unknown numbers and give LCM or HCF and say to find numbers or find HCF or LCM
Thank you sirrr
Please sir will you make number on system videos also it's my request
U r the best Teacher ... Thanks for making this video
Clarrified my doubt how to decide whether LCM or HCF
Thank you sir
It supports me alot to prepare fir campus, tq so much sir making it..keep on doing like this
Yo're welcome, Harshita! Stay connected :)
lcm of 75,150 & 135 is 1350
Right...I also got confused that what i am doing wrong..i have tries almost 5 times..
Yes..I thought I was worng
But its 1350 I was right
Yeah he did wrong multiplication, see 24:42
@@dheeru0025 sir third questions also wrong plz an upload
Yes the lcm was 1350.
Sir, Please check Question numberr 11 ... Here x^2 -8x +15 is not equal to 0, Where x=4.