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1 of toughest TSD question ever
by 50 cent - Sunday, 15 July 2012, 12:08 AM
  a swimmer jumps from abridge over a canal and swims 1 km upstream. after that first km he passes a floating cork. He continues swimming for half an hour and then'turns around and swims back to the bridge. The swimmer and the cork arrive at the same time. What is the speed of the water if the speed of the swimmer has been'constant? (if know the answer then describe the full solution as well)
option: a- 1kmph
b- 2 kmph
c- 3 kmph
d- 4 kmph

Re: 1 of toughest TSD question ever
by Mohit Sharma - Sunday, 15 July 2012, 02:14 AM
  hi siddharth,
ans for this ques is 1kmph. Suppose the swimmer is swimming in still water, in that case swimmer swims for 1/2hr upwards and 1/2hr downwards to get back to the cork. Now consider the river is flowing, in this case swimmer swims for 1/2hr upstream and 1/2hr downstream to get back to the cork. So total time taken is 1hr, and the cork crosses 1km in 1hr, hence, 1kmph.
Hope u hv understood.
Re: 1 of toughest TSD question ever
by TG Team - Sunday, 15 July 2012, 04:31 PM
  Good work Mohit smile
Re: 1 of toughest TSD question ever
by 50 cent - Sunday, 15 July 2012, 08:24 PM
 

hi mohit,

i got what you aer tryin to say but we cannot take the time of going upward and downward as same as the speed would vary when the man would go upstream and downstream. So as the speed would vary the resp. and time would also change. Please solve it as taking the speed of Man as X and speed of the cork as Z.

Re: 1 of toughest TSD question ever
by Mohit Sharma - Monday, 16 July 2012, 01:40 AM
  Hi siddharth,
You need to grasp this logic. Otherwise you can verify it by assuming some values. Time taken will be same.
Re: 1 of toughest TSD question ever
by TG Team - Monday, 16 July 2012, 07:28 AM
 

Hi Siddharth smile

The man is going upstream for 1/2 hr, right? Now whether I see normally or relative to cork, the man travells for 1/2 hour upstream.

For simplicity, tell me what is the relative speed of man with respect to cork in upstream motion and downstream one.

Re: 1 of toughest TSD question ever
by dhwani bhatt - Friday, 27 July 2012, 09:36 PM
  Sir , I am not getting it.
Please explain !
Re: 1 of toughest TSD question ever
by Fareed Un - Thursday, 12 September 2013, 11:34 PM
  I think this question has some technical error.

If the man travels 1 hour (0.5 hour upstream and 0.5 hour downstream), he will get to the position where he previously met the cork, and we know that this position is 1 km away from the bridge. Now, if the speed of the water is exactly 1 kmph than the cork should reach the bridge when the man is 1 km away from the bridge.

But the speed of the water is even lesser than 1 kmph, because the cork reaches the bridge at the same time when the man reaches there.

So the real answer is lesser than 1 kmph
Re: 1 of toughest TSD question ever
by Fareed Un - Thursday, 12 September 2013, 11:37 PM
 
Re: 1 of toughest TSD question ever
by 50 cent - Sunday, 15 July 2012, 08:24 PM
 

hi mohit,

i got what you aer tryin to say but we cannot take the time of going upward and downward as same as the speed would vary when the man would go upstream and downstream. So as the speed would vary the resp. and time would also change. Please solve it as taking the speed of Man as X and speed of the cork as Z.

The question states that the man is traveling at constant speed. It means that we have to assume that the speed is constant and does not change.


If you still want to clear your confusion about the upstream and downstream issue, than just assume that the man applies more force while going upstream and comparatively lesser force while coming downstream, but don't assume on your own that the speed is not constant...

hope it clears the scenario a bit
Re: 1 of toughest TSD question ever
by Prateek Pathak - Tuesday, 6 May 2014, 07:37 PM
  just equate the total time of man and cork...
1/vr=1/2+(1+1/2(vm-vr))/(vr+vm)
after solving u'll get vr=1
vr=speed of cork (or river)
vm-speed of man
Re: 1 of toughest TSD question ever
by MUHAMMED SAVAD M K - Thursday, 8 May 2014, 12:08 PM
  1 KM/hour



let 'x' be the start position,
'y' be the meeting position ( 1 KM from x)
and 'z' be the position until which the man swims

Let speed of stream= 'a' kmph and speed of man(constant) ='b' kmph
xy= 1 km

yz = (a-b)*1/2

the time taken by the cork to reach the bridge after meeting with the man = 1/b

so from the passage we can say that,
1/b= ½ + (time taken by the man to travel ‘zy’) + (time taken by the man to travel ‘yx’)
 1/b= ½+ (a-b)/(2*(a+b)) + 1/(a+b)
 1/b= (a+b)/ 2*(a+b) + (a-b)/(2*(a+b)) + 2/ 2*(a+b)
 1/b = (a+1)/(a+b)
 b=1 and a can hold any value.
Re: 1 of toughest TSD question ever
by 2013277 tg - Tuesday, 29 July 2014, 12:59 AM
  Consider the speed of the stream to be zero. In that scenario, the swimmer would have reached the hat in 1 hour (half hour to go forward +half hour to move back to the position where he first saw the cork). But, because the hat was 1 km behind that point (reverting to the original case), it means that the hat covered 1km in 1 hour i.e. the speed of water= 1km/hr
Re: 1 of toughest TSD question ever
by Anupam Mazumdar - Sunday, 14 September 2014, 09:12 PM
  let speed of swimmer = x, water = y
time taken by the swimmer to meet the cork after turning back = t

so yt = 1
if you check the distance, (1/2)(x-y) + y((1/2)+t) = (x+y)t
solve and get y = 1

First question of IE Irodov..Brings back memories