Time Speed and Distance 3 | CAT Preparation 2024 | Arithmetic | Quantitative Aptitude
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- เผยแพร่เมื่อ 10 เม.ย. 2019
- CAT PREPARATION 2024 I QUANTITATIVE APTITUDE PREPARATION for CAT 2024
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In girl question it is easy when we take ratio
S1/S2=T2/T1
Let actual time = x min
4/6=(x-2)/(4+x)
x=14min
but how will you calculate usual distance..? from this
@@techtalkies4119 Distance = 4(14/60+4/60)
@@user-ol2pu8bk9v can u pls let me know why has it come as 4(14/60+4/60).
I didn't understand?
@@new_life- usual time is found to be 14min. For speed of 4 kmph time is 18min(14+4).
thus d = s x t. 4 is speed and (14/60 + 4/60) is time. and usual speed as explained by sir in method 2.
4 is in terms of kmph (t+4 is in mins), if u convert 4 kmph into 1/15 kmpm... u will get answer 18....
The more views he gets, the more I get scared that competition is getting tougher day by day lol
😂😂
True
The first method is more understandable for me personally. Thank you so much
2nd one also
Last question was just awesome. Discovered a new way to approach the question, Thanks a lot Sir!
one simple method is comparing SxT for both cases as D is constant, let time be T
4 x (T+4)=6 x (T-2)
we get T=14
Hello , can someone solve one doubt. Why we cannot say that the difference of 12 mins which coming. It is due to thoae 15 km so train is actually covering 15km in 12 min so . Speed = 75kph?
4 is in terms of kmph (t+4 is in mins), if u convert 4 kmph into 1/15 kmpm... u will get answer 18....
Best one👏
Method one is more easier then other ones, really helpful thankyou sir.
But how get 14 min usual time through method 1 in first ques
Sir I don’t understand why are we calculating total distance 1st. In each of the methods we could go calculating req value directly.
Thankyou so much sir ye video ka explanation bhut hi osm lga 🙏🙏
just a guess, sir was recording the lecture in the morning around 10-11 am. inference from his taking example of 10 am to explain time difference😅😅
2nd one was very understandable for me personally. Thankyou so much sir. 🙂
Bro how time taken is 18 mins
@@thepratiktalks when time is reduced by 1/3 it takes 6 minutes less.
1/3= 6 minutes
Total = 18 minutes (6*3)
@@thepratiktalks time taken at 4km/hr is 18.. Usual time is 18-4= 14 min
@@chiragshetty6140 1/3 how is equal to 3 ,can u plx explain that
@@RohitKumar-hn9dz if 3 parts of the fraction 3/3 = 18, then 1 part of fraction 1/3=6, i hope you got that.
Thank you for this beautiful concept Sir!!!
Thank you, sir.
If the change in speed is already given as 3/4th then why are we taking 1/4th please someone explain
Raju when going slower by 30 km/hr, reaches the destination late by 90 hours. If he goes faster by 20 km/hr from his original speed, he reaches early by 40 hours than the original time. Find the distance he covers.
Any short trick for this question?
Lets take speed as 'S' , time as 't' and distance as 'D'
t=D/S
[D/(S-30)] - [D/S] = (t+90) - t ------------------ eqn 1
[D/(S+20)] - [D/S] = (t-40) - t ------------------- eqn 2
(difference between time)
Solve these above equations you will get speed
then put speed in any two eqns you will get distance and then you can time also by (t = D/S)
ans. Distance D= 3900 km , Speed S= 130 km/hr , Time t = 300 hrs
verify it using this (S-30)(t+90) = (S+20)(t-40)
I hope this helps....
Sir.
In second problem they are already mentioned by the speed of 3/4 of original speed to reach 15 km is 38 min late but we are calculate it as 36 mins to travel how sir
36 min is the time taken by train without any problem to cover that 15km
if we do the first method how are we going to find speed at 4 kmph i.e 18kph?
In 1st questions solution by 2nd method (usual time taken by girl to reach the skool ) this portion is not understandable can someone explain this
Sir what's the distance? In the last question
Thanku sir
thank you sir
If in last question we were asked to find out the total distance , what would be the approach anyone ?
Hi, You can just compare the distance after 85 km as if the train would have traveled at 25kmph then time would be t, but due to reduced speed, the train is getting 38 mins late. So apply the increase decrease ratio, then you will get the time required for distance after 85 km at 25kmph, so you can get the distance = 25 * time. I solved using 2nd case (38 mins extra and cross-checked the answer in case 1(50 mins extra), you can consider any of the cases. Here's the answer I got time taken after 85 km at 25kmph = 114 mins, So total distance = 85+95/2. You can cross-check in both cases. Hope it will help.
Thankyou sir
Sir how we can find the distance in between the station
In Q1, LCM(Speeds) gives us distance units? Why can't it betime units? Anyone knows on what basis we are able to decide if its time/distance?
i think we assume a distance which has to be an intiger if devided by speed thats why we take lCM of speed as a assumed distance
hope it helps
i solved the first question really quickly but i doubted if it was correct, i really feel like ive gotten really better at quant thanks to you
God's work 🙏
please take the cat level questions in time speed and distance and time and work no video on time and work please make a video on that
NIKHIL SANJU there are 3 videos on time and work. Please have a look at arithmetic playlist
Sir, please make a video for this : n/d =n/d+n.
Percentages lecture me hai
Confused with a question actually.
If speed increases by 87.5% (7/8) then will the time decrease by 7/15 times? (as per n/n+d formula)
yes
Hello , can someone solve one doubt. Why we cannot say that the difference of 12 mins which coming. It is due to thoae 15 km so train is actually covering 15km in 12 min so . Speed = 75kph?
M having the same doubt
Even I have same doubt
No this is all reference game, you cant determined with surety that with 3/4 speed it will travel 15 km in 12 minuted difference
since the distance to school from home is constant,
let distance be D. let x be the time she normally takes.
D = S X T
Case 1: D = 4. (x+4) -- when she is late 4 mins
Case 2: D = 6. (x - 2) -- when she is 2 mins early
equate 1,2 cases for x. you will get x = 14 mins.
Substitute value of x in either of the equation above, you will get D = 72 kms.
you have D = 72 and t = 14 mins,
S = 72/14 ==> 5 1/7 ans.
Very easy technique . Pls can u tell from where else u r learning?
@@hussainpunawala6113 self study.
@@hussainpunawala6113it’s class 9 maths actually
14/60 krna chahiye tha km/h mein kr rhe ho na isley
Cat 2023 prep anyone?
Yes
How much did you scored?
@@priyanshiagarwal7827what about you?
easier eq for ques. 1 will be...=> (2x+8) = (3x-6)
How?
@@newuser1282 direct apply s=dt
@@visuder7412d= st😅😅😅
can someone explain why t=36 mins for the 15 kms in the train problem?
1/3t = 12
Hence t = 36
❤❤❤❤
In the first question, only the distance was calculated by first method and not the rest of the required sub-questions. And the whole question was solved using the second method. Can i get the solution using the first method???
Yes, just divide 1.2 by 4 and you'll get 18 mins.
Very helpful!!
Sir your teaching and explainations are awesome but plz make the audio a bit louder
Hndfre lgakr dkh
@@aadishjain7454 phir bhi sound quality achcha nhi ho rha tha.... ab to CAT ho gaya h mera
@@arpankhan8485 acha🤣🤣kya rha score bhai?
@@aadishjain7454 bass 78.92%ile....mere jaise idiot se itna hua yahi kaafi h...lol..😁
Koi bat nh bhai zruri thdi har vykti hi ek kam me acha ho dsre kam bh hte h jeevn me cat jeevan thdi na h koi pura
Isn’t 25 kph the speed after reduction? At 23:52
Exactly my thought too. 25 kph should be 3/4 of the original speed
That is due to relative change. Here we are considering the change from second condition(original speed in 15km track) -> first condition (train broke down and speed is decreased), so we will get the original time as we specified in the first condition.
Sir why we are calculating distance in the first question. Just the usual time has been asked
Cause without calculating distance you can not calculate speed
And usual speed of the girl is also asked
Can anyone explain me the last question
In the last question : I have done speed = 15/(12/60) = sp = 75 which is is 3/4,
SO,orignal speed is 100 km/hr .Can someone explain why Am i wrong ?
12/60 hr is the time difference between the two scenarios and not the actual time taken in covering this 15km.
Sir through ur video
My speed for conceptual clarity is increasing at rate of 100%, study time has decreased 50% 😂❤❤
10:42 how does time taken at 4km/h become 18 minutes
We're considering the speed changes from 4 to 6kmph... The time decreases by 1/3rd of the initial case(when speed is 4kmph) when speed becomes 6kmph,which is equal to 6 minutes. So 1/3rd of the time taken when she travels at 4kmph is 6 minutes. Therefore total time taken is 6*3=18 minutes
It's 4 mins late, i.e., 14min + 4min = 18min
@abdulkareem , Imagine total time taken @4km/hr ='x' .It is given that time difference is 6mins. As the time got reduced by 1/3, so 1/3*x=6. x=18(totsl time taken @4km/hr)
Train's original speed should be 100km/hr.
ho sakta hai passenger train ho
Sir how did you get the time take at 4km/hr =18
6*3=18
I shall make u understand that
The speeds are 4 and 6 respectively
So the ratio is of Speed 1: speed 2
Is 4:6
Now time is inversely proportional
So it would be 6:4
Now if you concentrate and understand the question , between the higher speed and lower speed the diff the girl is able to make is of 6 minutes
So the in ratio 6:4 the 6-4= 2
Hence 2 parts make up 6 minutes
Henc 6 parts of time while make 6*3
18 minutes
Now distance is constant
So multiply 4km/60min * 18 min
You will get usual distance
10:55 😅😂
21:50 sir ,how 36 has been calculated here
same doubt
12x3
@@dishabhati4061 thanku 🥺
Outstanding beauty content on utube.
Time taken at 4 km/h=18minute kaise hua nhi samaj aaya
⭐️⭐️⭐️⭐️⭐️⭐️⭐️
how to solve 15/(3/4)s-15/s?
take 1/s common in left side and solve left hand side part which comes 5 i.e. now you have 5/s = 1/5, now after cross multiplication it is s =25.
But if you take 2.5km/hr 6mins late and 3.5 km/hr 6 mins early can you explain in method 2 sir
sir don't do so many methods we will learn everything half and get wrong answer.
U r fool than
Only watch one then sir is trying to build concept
@@caster4824yes
U should gain a overall perception of all methods which will help u solve different sums as per u convenience
In 2nd qts.. we have to find original speed...from qts we understand that diff. of of 85km-70km= 15km & required time to travel 15km is diff. of time 50mins - 38mins= 12mins. its means train travels 15km in 12mins so..it should be as below
S=D/T
S=15/(12*60)
S=(15*60)/12
S=75km/hr should be the ans.
Rechecking Distance covered in 1hr - 15km/12mins=1.25km/hr....so 1.25km/hr x 60mins=75km.
Ravi sir thank you very much. Thank you @Rodha
I was having the same doubt. Any clarity ?
Its wrong . I tried by equations. Sir is right
It's not 12 min time taken for 15 min is 36 min for 85 km ... 12 min is the difference between them
See we can't tell in 12 km it will reach 15 km because Speed Is changing right! That 12 minute is time difference you cant say that Tarin with 3/4 sped can travel the same distance in 12 mintue distance
10:51 in this question...time taken at 4km/h is 18 mins....how???
Got my answer @ 17:03
In train question you have taken speed s to 3/4s and then calculate everything
If we do opposite means
3/4s to s ---> 1/4 increase in speed
Then 1/5 decrease in time. Now this decrease time is 12 mins
1/5T = 12 mins = 60 mins = 1 hr
3/4 s = 15/1 ==> s =20km/hr
I'm not able to understand why it is wrong. Anyone help me
12 mins = 60 mins? Wtf dude?
@@arunadityarajjain1129 1/5T = 12
T = 60 min (5*12)
Calculation dude WTF you don't know this
Bro , its 1/4 decrease , so time will increase by 1/3 . Therefore time taken to cover that 15km will be 3x12 =36 min.
@@rishikeshverma9249 that I understood....we are calculating every thing in terms of s
What I have done is in terms of 3/4s
3/4 s to s -->1/4 increase in speed so
So time will be decrease by 1/5
@@AbhishekKumar-vw3mo from 3/4S to S the increase is not 1/4. ( By your ans - 1/4 inc in 3/4s would mean (1/4of 3/4 + 3/4) , Which will never come 1 ) Its a silly mistake people do .
For Example -
If from 1 to 1.5 there is a inc of 1/2 doesnt means the same in vice versa , that from 1.5 to 1 there would be a decrease of 1/2 , this can never happen.
Hope i solved your problem.
15/36/60=25KM/H ??????
15/(36/60) = (15*60)/36 = 25
Sir I wanted to know that after we have calculated the speed of the train which is 25km/hr, how will we calculate the total distance and total time taken for the same question?
It's not possible to calculate total distance or total time unless one more parameter that is.. Total dis or total time is given...
th-cam.com/video/TuSz8777COo/w-d-xo.html
Train wale ques mein 3/4 kse aaya? Smjh ni aaya
Read the question properly ....the speed of the train is becoming 3/4th of the original speed
@@rahulshrivastava8720 fir uska 1/4th kaise hua fir 15km?
In girl questions we can even use equation where distance is equal which means
4(x+4)=6(x-2)
x=14 minutes
So here usual time is 14 minutes and we can do further from here.
Yes, we can use that too. I generally use this method on the T&D ques. with standard time difference b/w two countries.
Please guide me
sir I am appearing for CAT 2024, Sir I wanted to ask you If I complete all your videos will it be sufficient for me to crack cat?
Do you need gego sir varc?
@@Aaditya2k03yes I need it.
@@Aaditya2k03yes I need it
@@shreyoshiroy6955 connect to me
@@shreyoshiroy6955 profile
regarding girl & school question
4(t+4) = 6(t-2)
t=14 min
those who are able to understand the solution Like👍
rest Dislike👎
Please sir reply please🙏
Do you need gejo sir varc?
Yes@@Aaditya2k03
bohot ganda topic h yar jara bhi smjh ni ata
Starting me lagta haii bhaii
50 questions kar loo uske baad maza aane lagega
In 1st questions solution by 2nd method (usual time taken by girl to reach the skool ) this portion is not understandable can someone explain this
thank you sir