TSD Boat Streams 2 | CAT Preparation 2024 | Arithmetic | Quantitative Aptitude
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- CAT PREPARATION 2024 I QUANTITATIVE APTITUDE PREPARATION for CAT 2024
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At 12.10 answer should be 13.33 instead of 8m/s.
exactly
Yes off course
12:10
Ravi sir the answer is coming as 13.33 at 12.10 time of the video. Please confirm the same. Waiting for your reply
right
Su= 1kmph and distance to cover is 6 km so he requires 6 hrs
Yes correct
@@aditi9393yes
what i have found is there is at least one question in the video where there is a mistake in the solution. Sir please try to rectify such in the comment box in order to avoid unnecessary confusion. Rest all is superb and you are an inspiration.
In one of the question there is 45sec and 600 m the answer will be 13m per sec rather than 8m per sec
13.33 m/s
Hello everyone looks like many of us are having doubts in the last question. In my opinion if we use the frame of reference concept, then the speed of the river will be eliminated entirely. Relative speed of boat with respect to hat in the frame of reference of river will simply be speed of boat wrt river- speed of hat wrt river
Speed of hat wrt river will be zero as it is freely floating and speed of boat wrt to river will be 33 km/hr
Using this logic distance covered by boat wrt river =(15/60)*33
Now people noticed the hat they will return back with same speed of 33km/hr wrt river
Then the time taken will be distance bw boat and hat wrt river/ relative speed of boat and hat wrt river
The answer will come out to be 15 mins
Thank you ..hope it helps
FoR concept will itself be a new concept for non science guys
In 15 mins the hat would have hit shore considering 30kmph as the speed of the River.
@@josephbj8766 I thought the same but noticed that, it is not mentioned in question that raft was at shore initially or not, your assumption would be correct if it had been mentioned
cool concept, thanks
Sir, In first question they have asked "how much more time he need to row?", which is 6hrs. And you have given answer in speed 6kmph.
that doesnt matter when you will equate again you will get 6 hrs is left to cover the remaining 6km distance left. Guess he checked answer 6 and moved on :P
@@subham625 bhai questions me ye nahi bola hai ki remaining distance 1 hours me complete karna hai jab uska upstream me speed 1km/h hai toh 6km/h kaise hoga
Exactly, I calculated it and was puzzled as to how required speed is even relevant
u cleared CAT ??
at 12.26sec there is a mistake. the speed comes out to be 40/3 m sec^-1and not 8
Thanks For replay we will review it.
We Have started QA forum in PagulGuy to discuss QA related to Rodha youtube channel you can join to interact with faculty www.pagalguy.com/discussions/official-quant-thread-for-cat-2016-38731516
@ayush Raman can you help me to in last question why boat speed is 63 m/s if it is turms back i goes in same direction of river instead of adding, we should substract it na?
@@Abhijeetkr_2211 the speed of the boat in still water is 33km/hr, it means that if the river is still(v(river)=0) the speed with which the boat will travel in either of the direction will be 33km/hr. but here the speed of the river is 30km/hr. so when the rafting team goes downstream (means in the same direction of the river), the net speed of the boat will become 30+33-63 km/hr. Again this is the net speed of the boat and not the speed of the boat in still water. Now Let say a person is standing on the shore of the river. So the person will see that the boat and cap are floating at the speed of 63km/hr and 30km/hr respectively in the same direction.
Yes exctly... The speed is 20/3 m/sec.
@05:22 Answer should be 6 hrs I believe. Sir pls check once
yeahh bro.
I think at 12:15 the answer will be 23.33 as they asks the speed of the boat in still water. speed of the ship in upstream is 13.33m/s so in the case of speed in still water we have to add speed of the steam to speed of the ship.
No, at 10:55 you can see that relative Speed comes out "x" itself, so whatever will be the value of Relative speed would be the Speed of Boat which is 13.33 m/sec , even Sir also said that in video. "x" is the spped of boat.
@@deepakvijay5809 isnt that speed wrt to the boat
I think final answer will be 13.33 m/s. Rel. speed 600/45 = 13.33 m/s, so subtracting 10 m/s we'll get the speed of the ship in upstream, then to get the speed of the ship in still water again we have to add 10 m/s.
u can easily check your ans,
if 23.33m/s then in 45s boat will cover 1045m but it it covering 600m in 45s i.e. S = 13.33m/s
@@bestshorts2762 Yeah bro seriously, Had totally forgotten.
thankyou so much sir, you're literally jeetu bhaiya for us :)
In the first question , the answer would be 6H time would be required.
Even the 3rd question with 600 m boat, speed comes out to be 40/3 m/s.
sir , the answer is 13.33 m/s for the problem solved at 12.12
I want to say everybody silly mistakes can be made by teachers we ourselves should correct them
if anyone have doubt in last question you can understand by this exam one man going in East direction and another is going in west direction the have speed 5 m/s and 10m/s respectively, what will the distance b/w both after 10 sec, 10x5= 50, and 10x10 = 100, TD = 150, or we can use formula that sir use in last question 10x (S1+S2) = 10x15 = 150,
Now if the man who is going in the west direction turns and try to Cath the man who is going in the east direction in how much time he will Cath him, RD = 150 and Relative speed = S2-S1 = 10-5= 5, so time 150/5 = 30 sec
alternative apprch to q1=>
Dds=3*Sb + 3*1 & Dus = 3*Sb -3*1=> Dus=Dds-6
4:53 Required time is 6 hrs
In every video you get 2-3 questions wrong and i get confused.
Disliking only because of this
In first question it's asking for time and not speed . It should be 6 hours and not 6km/h.
Please rectify these because these are a repeated mistakes coming from past lectures .
In the last question, as after getting the distance variation between both of the things after 15min , you have not added some extra distance that hat will travel further with the flow of river..
Can you clarify my doubt
there is no need to add extra distance as it is relative distance between boat and hat
relative distance at the time of noticing , but once they start moving , even hat will travel some distance na ?
@@kartikchodvadiya8735
speed of ship in still water in Q3) is 23.33 m/s.
In second ques ans would be 18 m/s because speed is asked in still water..since Sr is 10 km but ans which sir gave is in Su
But max distance that hat can travel is only upto shore in last qstn. Sir, you haven't considered this.
@7:30 averaged speed of the Journey is (SD+SU)/2=which equals to SB. therefore, the answer should be 100m/s, no calculation required
When time is not constant, speed average is always 2xy/(x+y), therefore the way he did is right.
the best part of every videos of ravi sir is......hii folks my name is ravi prakash......and lets solve this
Last question is so good..!!!
sir, in the last question before the boat take a return the hat reaches the shore if it is freely floating. so the whole calculation is wrong.
Its not mentioned that they started from the shore (even I got confused)
In the last question, till the time the boat will reach the hat, it will also move ahead because the speed of river is 30km/hr. We haven't considered that??
You are right, but when we consider relative distance that comes into picture because we are also considering relative speed. You can also verify other way round. In 15 minutes the boat travelling at 63kph will travel 15.75km whereas the hat travelling at 30kph will travel 7.5km. So the boat has travelled 8.25km more than the hat. This was the gap between them initially.
Same doubt
i think sir should have taken that distance along with 33/4
@@swapneelrao8276 ohhh okay thanks bro
This is the reason exactly why in this question relative speed concept is used to find the time.
In last question in 10 minutes boat can travel 0.5 km. At same time hat falls.
How can hat travel 7.5 km ? Though it's irrelevant still it didn't make sense and some one can mark cannot be determined as option.
Abhishek Kumar after hat has fallen boat moves for 15min. in those 15mins distance travelled by hat will be 7.5km. this 7.5km must be added to 33/4km to calculate the total time taken at 63-30 i.e. 33kmph
@@shishir2243
The relative distance between boat & hat when they notice it=distance travelled by raft(upstream)+distance travelled by hat(downstream)
Total relative distance=15/60 hr * 3 km/hr + 15/60 hr * 30 km/hr
=15/60 hr * (3+30) km/hr
=15/60 hr * 33 km/hr
=33/4 km
(no need to add 7.5km to 33/4km)
Sir at 12.23 45s=600m/x so how come x is 8 m/s .. any one!! plzz explain and iam getting 13.33ms as answer
Last question was very simple if we solve the problem with respect to river instead of solving with respect to shore
Here it is
With respect to river (consider you are in another boat still in water)
Hat won't move
And so boat move far for 15 minutes
And with respect to river boat speed don't change whether it is downstream or upstream
So speed is also contact
Both distance and speed are same
So time is also same as 15 minutes
So answer 15 minutes
Without solving
And also we don't need any values also just need time
Think once
Sir in 12:10 , boat has covered a distance of 450m in 45 sec and the ship has covered only 150m , so you can't say that the boat has done neligible work , the answer will be 7.66 m/s
How did u get 7.66 m/s
In the last question, when hat has fallen then distance between that point and shore is 0.5km , and when they came to know about the hat then hat has already been reached at the shore because of river's speed .. that means hat's speed is zero when they came to know about the hat ... Can we not thik like that... And if not then why... Can anyone explain?
Bro...it’s not given that water is coming towards the shore in downstream....Aisa hoga to flood aa jaaega😂....movement is along the shore
Anyone can tel that .
Why we are taking R.s only
@Rodha sir, in the 600 m ship question, I think the answer should be 23.33 m/s instead of 13.33m/s. Because 600/45 will give the speed of the upstream. To get the speed of the ship we will have to add 10m/s in the (600/45). Please clarify sir
Yeah bro you are right I think so
yes, the speed of boat should be 23.33m/s
No he has assumed x as ship speed , upstream speed is x-10 that’s why effective speed is x-10 +10 = x
@7:27 is it safe to do the question this way instead (as the speeds are in AP not HP):
Su = 80m/s
Sb = 100 m/s
Sd = 120 m/s
Speeds are in AP, therefore Time is in HP.
Therefore time in stillwater = (2.80.120)/200 i.e. 96s per side, therefore distance per side = 9600 m.
Avg speed = Total Distance/Total time
=) Avg Sp = 2.9600/200 = 96m/s
Hope this is not flawed in any way, please let me know. I know this takes slightly longer but im more comfortable here as theres no change in approach wrt AP HP treatment.
When distances covered are same, avg speed is HM of speeds, concept was covered in previous videos
In 600m ship ques how can the speed of ship i.e 8m/s is less than the speed of stream? If speed of the ship is less than the stream than the ship will flow downstream with the stream with 2 m/s speed
Only calculation mistake, 600/45= 13.33 m/s not 8
at 7.37 how come speed is in HP. when d is constant then if speed is in ap then time will be in HP not speed or am i missing some thing ????
Same doubt here
@@mrsahilrohilla Total distance/total time = 2d/ (d/80 + d/120)
do it like these same answer
For constant distances, speed will be in HP. Whereas for constant time gaps, speed will be in AP.
Sir, at 5:23, time is asked and not speed. Thus answer should be 6 more hours.
Yes. 6 hours must be the answer
3 hours is the answer . 6/2
@@madhuramvs20 6 hrs is the answer.... (Remaining time = 6km / 1km/hr = 6hrs)
Still confused for the last question
sir, please suggest some revision strategies, to remember all these concept!!!!! facing A tough time here
Done 👍🏻
23 September 2024
hat didn't flow?😨
Right answer at 12:22 is 23.33 m/s 13.33 + 10m/s (because it was going upstream)
Guys in 12:12 the question asked for spped of ship in still water. Then we just go for (Sb+Sr). Then the answer will be (13.33+10) = 23.33 .
Correct me if im worng
Because the water is still so there is no need of Sr and the answer is 13.33 m/sec
12:16 Let's think logically, A 600m huge ship is sailing upstream without the help of river's speed. And a small boat is set freely (obviously downstream) on the river with speed of 10m/sec. The question is "what is the speed of ship in still water?" That means ship has 0 speed and completely dependent on the river's speed, if a small boat moves 10m/sec then how come a huge ship move 13.33m/sec without it's own speed? Length of a ship maybe huge but huge objects doesn't move faster in water but a small boat can be sailed easily without it's own speed. If you think logically 8 is the correct answer but according to the information given in the question is 13.33 at the end a huge ship cannot move faster than a small boat in a still water.
@rodha team please check the answer given at 12:05 time frame. I think sir have done some calculation error
yess sir ans for ship and boat question is 13.33 not 8
In 15 mins the hat would have hit shore!
Why 600 m the length of boat is considered the distance
at 6:58, why it is 2*a*b/a+b and not just a+b/2, it is just the avg speed right? @rodha
That happens when avg speed is found at constant intervals of time, here its the case of constant distance
Can anyone please explain the 2nd question, that if speeds are in AP, why the harmonic mean is used to find the average speed?
Sir would have derived this , abg speed = total distance / total tme = [d+d]/(d/80 + d/120) this is implified as given by sir In this video
When distances covered are same, avg speed is HM of speeds, concept was covered in previous videos
Can somebody please explain why we have taken the relative speed of boat as 3 but later as (33+30)?
Why have we not taken speed as 33 earlier?
boat speed in standstill=33 , 1st case boat in upstream=33-30=3 , 2nd case boat in downstream=33+30=63
@@gunasaivikranth4396 thank you
Is there any doubt solving group ?
last question answer is wrong according to me
we have not assumed hat in a moving state while calculating time plzz look in to this..!!!
in the second ques why didnt we take avg speed as 1/2(speed of boat upstream + speed of boat downstream) = 1/2(80 + 120) ? Harmonic Mean method is used to find the speed of boat in still water, right? not he avg speed. Can anyone explain?
Exact same doubt! here the speed is in AP, so time should be in HP, How he is taking Harmonic mean of speeds, instead of AM.
If it were given in the question, that time intervals are equal for each part of the total distance covered @ different speeds, then Avg. speed would've been the AM of all speeds.
@@saptarshidas488 but they have asked avg of 2 speeds. We have 2 speeds. So it should be AM .
@@AtulyaPaul the distance is same for both cases that means speeds are in A.P and Time is in H.P as simple as that and hence the answer should be 100 km/hr rather than 96 for the second question
Indeed downstream speed, boat and upstream speed are in AP, so time are in HP.
lets take distance =480 km.
time taken = 480/120=4 hours (down)
time taken = 480/80=6 hours (up)
total distance travelled =480*2=960
total time=10 hrs
avg speed= 96 km/hr
12:26 can anyone tell me why he took 600m as distance but its actually a length of the ship
In the question its mentioned that the ship took 45sec to cross the boat. so 45 sec is the time taken to cross the distance of 600m which is the length of the boat.
@@Nerdypadosi but shouldn't the length of the boat be also considered? 600 m is length of ship only
@@saptarshidas488Length of boat is negligible
Someone please tell me how to calculate relative distance in last question.
If somebody having problem in last question, then to clear it up, the hat will still be floating and wont stop, thus relative speed between hat and boat is still 33 kmph only !
but the distance between them will be 8.5 ?? not 8.25 ?
@@saarthakbadani7863 No, it will be 8.25!
Last question will be 16.5
Hi sir, In the last question, I think we should consider the 10 mins after that the Hat falls. Because in 10 minutes the hat is 500m (with an upstream speed of 3km/hr) away from the shore and when they noticed it after 15 mins, the hat traveled a distance of 7500m (with a downstream speed of river that is 30 km/h) and while they came back and got it in 15 mins, the hat must travel a distance of 15km according to your solution. But as we saw, the hat is only 500m away from shore. So after 500m, the hat stops floating. So to solve this question we must use distance = 500m + 15/60*3(0.750KM=750m). and speed =63km/hr.
i can confirm that sir's solution is correct, because i applied Vector (Class 11th Physics) to solve the question before looking at sir's solution. You are considering their motion perpendicular to the shore, however it is parallel to the shore. Relative to the shore means the observer is at the shore.
thanks @@vaibhavraj6512 i was considering the motion of boat as perpendicular to shore.
Can someone please explain how to calculate 33/4/30 at 20:45
bro that is 33/4/33 =33/4x33 =1/4
63-30=33 not 30
At 7:25 sir uses harmonic mean formula, why ?? ... because speeds are in AP
Bcz when distance is equal
Avg speed will be harmonic mean of the speeds
Distances are equal so speeds are in HP, time is in AP. Hence avg speed is HM.
thank you so much
your earlier videos are good,but there are too many mistakes in 2 videos of tsd boat streams.
In the last question when we are calculating time required to reach the hat at distance 33/4 then why are we using relative speed as 63-30=33 and not 33+30=63 as whole moving downstream the speed of boat i.e. 33 and speed of stream i.e. 30 must be added.
Please someone clear this doubt.
as hat and raft both are now moving in same direction so when two objects are moving in same direction we take their speed and subtract it (relative speed)
ok we will take raft speed and hat speed as 63 and 30. What if the water is still? The hat speed will be zero and the raft will be 63-30 which is 33.
the relative speed is 30+33=63 but hat is also moving, so 63-30=33
thanks for asking same que
@@bitcoinguruji1741 thanks for telling doubt
How to know whether we should take speeds in AP or HP ? In the previous video we took in AP here in HP ?
Yeah
Speed will always be in AP whereas time will always be in HP
As ravi sir taught earlier
When distance is equal
Avg speed will be harmonic mean of the speeds
When time taken is equal
Avg speed will be arithmetic mean of the speeds
According to relative speed question at 12:25. If speed is 8m/s of ship then how it's moving forward as S(downstream) = -2
Answer is 13.33 . He has done a calculation mistake
@@soundsmeller5586 23.33
600/45 is not 8. He has done calculation mistake.
can someone tell me why at 7:00 harmonic mean is used and not arithmetic mean ??
Bcoz time is inversely proportional to speed and relation between AM and HM is same as that of speed and time
why the boat will travel 33/4 km upstream! while going upstream, upstream speed of boat should be taken which is 3kmph and not 33kmph.
also in 15mins, hat would have travelled 7.5km downstream. so relative distance should be distance moved by hat in 15mins + distance travelled by boat in 15mins.
anyone?
@shishir 33/4km is not the distance travelled by boat in upstream. 33/4km is the gap between the boat and the hat after 15min when they notice it.
as you correctly said,
relative distnace=distance moved by hat in 15mins(downstream) + distance travelled by boat in 15mins(upstream)
=15/60*30 + 15/60*3
=30/4 + 3/4
=7.5 +0.75
=8.25km
=33/4 km
GR8
Doubt in last ques ....m getting 13.6 as answer
No its correct...it will be 15mins...do u need any help?
@@sonty6589 Yaa it's correct, while doing it at first I did a calculation mistake so it went wrong ......bt later on I got it right....thanks btw!!
how can someone be so wrong ,almost every answer was wrong in the video🥵
in the ship and boat sum i did another method i don't know whether it's right or wrong but do hear out my logic
since boat is freely floating speed of boat = 10 m/s (downstream)
since the distance isn't mentioned but the focus of the problem is Speed,Therefore,Time covered by the boat while crossing the ship=Time required by the ship while crossing boat=45 seconds (Since, S=D/T therefore T=D/S)
T of Boat should be proportional to T of Ship,so
{Assume A as the starting point of Distance of (Ship+Boat) and B as the end point,since we are calculating the Distance when both Ship and Boat cross each other,therefore Distance AB is equal to the length of Ship)
Since T=D/S
T(boat)=600/10
T(Ship)=600/(Speed of Ship-10) ~[upstream]
Since T is same for both crossing each other i.e. 45,
(Speed of Ship -10) = 600/60
Speed of Ship= 10+10 = 20 m/s
Answer is different,but i think logic is correct,do let me know if there's a mistake
I think it's wrong
for boat u cant assume d to be 600 i guess, for ship it travels its own length thats why the boat's length is considered negligible.
21:00
Is there a part 3 to this video? Can somebody share the link if there is, pls.
no
5:30
thanks every one those who leave there commentes for sec. q.
Thankyou sir
Can anyone please explain me how in first question Sr is 1km/h?
Ok
hahahah
@@DSA_Coding y u r laughing
Be happy to live stress free ,😎 find happiness in little things
@@DSA_Coding ok thank you for your advice
☠☠