Elementary, my dear Watson !
Sherlock Holmes and Watson were on a motorboat going downstream.
They overcame a raft at waterloo junction, an hour later they turned back* and after sometimes, passed the raft again, which was now six miles from the waterloo station.
"You know, what the velocity of the stream is?" said Sherlock Holmes.
"That is quite impossible to find Holmes" Watson exclaimed.
"Elementary, my dear Watson !" and then he explained how easily he can calculate the velocity of the stream just by knowing the fact that velocity is equal to distance divided by time.
It may be elementary for Sherlock Holmes, but can you ?
* assume the motorboat is in constant speed and no time was spend on turning the boat
They overcame a raft at waterloo junction, an hour later they turned back* and after sometimes, passed the raft again, which was now six miles from the waterloo station.
"You know, what the velocity of the stream is?" said Sherlock Holmes.
"That is quite impossible to find Holmes" Watson exclaimed.
"Elementary, my dear Watson !" and then he explained how easily he can calculate the velocity of the stream just by knowing the fact that velocity is equal to distance divided by time.
It may be elementary for Sherlock Holmes, but can you ?
* assume the motorboat is in constant speed and no time was spend on turning the boat
Labels: logic, mathemagic, puzzle
posted by Rajesh Lal at
9:20 AM
45 Comments:
this does not lok elementary to me at all :(
I think we would have to know exactly how long Holmes and Watson traveled "*back" before passing the raft the second time.
I was fooled into doing too much math, however the answer is "elementary".
By constant speed it is meant that the boat has a constant speed through the water.
Since the distance through the water is exactly the same, (the water flows with the same speed as the raft) the moment of the second meeting is exactly two hours after the first meeting. Therefore dividing the distance between the meetingpoints by two hours results in the speed of the water:
6/2=3 m/h
basic relativity !
I believe it is 6mph
I wish it were that elementary, but I think the constant speed of the boat is relative to the flow of the river (what the speedometer of the boat would read, with a constant throttle setting). Speedometers on boats work by measuring the water flow through a tube on the bottom of the boat. Otherwise, Sherlock and Holmes would have to adjust the throttle when they turned around to compensate for the flow of the river, which would be impossible if they didn't know the speed of the river when they turned around.
So if the river flows at 5 mph, and the boat travels at 10 mph, its total velocity downstream is 15 mph and its total velocity upstream is 5 mph.
I'm assuming that the raft is unpowered, and therefore it will travel downstream at the same speed as the river.
Vr = Velocity of the River/Raft
Vb = Velocity of the Boat
T = Time elapsed between the first and second passing
The total velocity of the boat while traveling downstream is Vb + Vr, and the total velocity of the boat travelling upstream is Vb - Vr.
We know that it took T hours for the raft to travel 6 miles:
T = 6mi / Vr
We also know that it too T hours for the boat to travel downstream from Waterloo for 1 hour, turn around and travel back to a point 6 miles downstream from Waterloo:
6mi = (Vb + Vr) * 1hr - (Vb - Vr) * (T - 1hr)
6mi = (Vb + Vr) * 1hr - [Vb * (T - 1hr) - Vr * (T - 1hr)]
6mi = (Vb + Vr) * 1hr - Vb * (T - 1hr) + Vr * (T - 1hr)
6mi = (Vb + Vr) * 1hr - Vb * (6mi / Vr - 1hr) + Vr * (6mi / Vr - 1hr)
6mi = (Vb + Vr) * 1hr - (Vb * 6mi / Vr) + 6mi
Which finally reduces down to:
0 = Vr^2 + Vb * Vr - Vb*6
This is a quadratic equation with a = 1, b = Vb, and c = Vb*6.
And using the quadratic formula, we can derive Vr in terms of Vb:
Vr = (-Vb + SQRT(Vb^2 + 24*Vb) -Vb )/(2)
Since Sherlock and Watson know the speed of their boat, they can simply plug it in to this equation to determine the speed of the river.
For example, if the speed of the boat is 8mph, then the speed of the river is 4mph.
If the speed of the boat is 10mph, then the speed of the river is 4.22 mph.
No... You have to judge the speed of the river by the speed of the raft, which is simply floating (unpowered) in the current of the river. In ~1 hour, the raft floated 6 miles down stream. The current is ~6 miles per hour. We can't evaluate "some time", and have to make the current an approximate value.
LOL, ill just pretend i know what you all are talking about... :P
the problem here is not enough information. Like Holmes said, velocity is d/t we have d=6 but currently t=>1 hour 6 / >1 = >6 mph
for the guy above
we are totally clueless and we prefer to remain anonymous
And you call yourself KingD
hahahahaha...
I approached the problem assuming the constant speed for the boat was not taking into account the speed of the river, so the actual speed of the boat was its constant speed (x) plus the rivers speed (y) while traveling down stream, and x-y while traveling upstream.
Since we know the boat turned at 1 hour, we can say the distance traveled in this time was (x+y) miles. D=S*T
Since the boat met the raft 6 miles from the first meeting, the distance traveled upstream was the distance traveled downstrem, less the 6 milles ( x+y - 6)miles.
We can hence find a result for the total time of the journey,since we know the first part took one hour, and can use T=D/S for the second part, giving
T = 1 + (x+y-6)/(x-y)
We can now forget about the motorboat altogether and focus on the raft. Since the raft is traveling at the same speed of the river (y) we have, using S=D/T and the fact that the time is the same time the motorboat took to get to the same point 6 miles away
y = 6 / T
Only follow this if like maths :p
>>>>>>>>>>
y=6 / (1+ ((x+y-6)/(x-y)))
y = 6 /(( (x-y) + (x+y-6) )/ (x-y))
y = 6(x-y)/ (x - y + x + y - 6)
y = 6(x-y)/(2x - 6)
y = 3(x-y)/(x-3)
xy - 3y = 3x - 3y
so>>>>>>>>>>
xy = 3x
Now we have 2 solutions
The first is the answer relitivity miss', if the constant speed of the boat was 0 and so it was simply drifting we will not get a constant speed for the river, as the boat is always with the rift.
So we can asume that x is not 0, and hence cancelation gives
y = 3
"
6mi = (Vb + Vr) * 1hr - (Vb - Vr) * (T - 1hr)
6mi = (Vb + Vr) * 1hr - [Vb * (T - 1hr) - Vr * (T - 1hr)]
6mi = (Vb + Vr) * 1hr - Vb * (T - 1hr) + Vr * (T - 1hr)
6mi = (Vb + Vr) * 1hr - Vb * (6mi / Vr - 1hr) + Vr * (6mi / Vr - 1hr)
6mi = (Vb + Vr) * 1hr - (Vb * 6mi / Vr) + 6mi
"
You were right until the last line here, it should read:-
6mi = (Vb + Vr) * 1hr - Vb * 6mi / Vr + Vb*1hr + Vr * 6mi / Vr - Vr * 1hr
Removing brackets and rearaging without units gives
6 = Vb + Vr -6Vb/Vr + Vb + 6Vr/Vr - Vr
6 = 2Vb - 6Vb/Vr + 6
0 = 2Vb - 6Vb/Vr
6Vb/Vr = 2Vb
6Vb=2VbVr
3Vb=VbVr
Vr = 3 (unless Vb = 0)
you guys are looking at this wrong, you do not need the speed of the boat, the only difference that would make, is nhow far away the boat would be when they find it. The velocity of the stream is d (distance traveled) divided by t (time passed) so if the boat went faster, the velocity of the stream wouldn't change, just the amount of time passed, and therefore, the distance the boat has gone. So unless we can get the time passed between when the boat was found at point A and when it was found 6 miles from point A, there will not be enough information. It is obviously, not elementary
the only other thing for now, is that if your answer is lower than 6 mph then you're wrong, because the only givens are 6 miles, and >1 hour so that answer has to be greater than 6/1, or 6 mph. I believed this was mentioned in an earlier comment
no srry, ignore that last comment i confused myself
didnt someone prove the answer was 3 already, and that the boat speed was irrelivent?
Was there a problem with the proof?
The answer is no time at all, duh!!!!!! they traveled for an hour the turned back and passed it again so it is NO time what so ever!!!!!!
the question is vague, we don't know what they mean exactly by "constant speed". Is that relative to the river or not?
Question was: What is the velocity of the stream.
The raft apparently traveled about 6 miles in an hour, so hmmm...what do YOU think the stream's velocity is?
>:-D
srry to burst your bubble, but the boat traveled 6 miles in more than an hour, because after one hour they turned around and then later they found the boat
I honestly could not tell you the answer, all I can say is that I'm almost sure that it is not 3 MPH. If they were going at a constant speed from the junction and turned around after an hour, it would take them exactly one more hour to return to the junction, NOT to the raft, for they would not have reached it yet going the same speed after 2 hours.
Ive been testing assuming the 3 mph theroy is correct. If the river speed is 3mph making the raft get to the 6 mile point in two hours, then...
A boat traveling at one mile an hour will travel downstream at 4mph (since the river goes at 3 mph) and upstream at -2 mph. this would mean after 2 hours the boat would be at the 6 mile marker.
A boat traveling at 3 mph will travel downstream at 6mph, and upstream at 3mph, so will remain 6 miles from the start after the first hour.
A boat traveling at 5 mps will go downstream at 8mps and up at 2 mph, so would reach the 6 mile mark at the 2 hour mark.
A boat traveling 135mph will travel downstream at 138 mph and up at 132, putting it at 6 miles after 2 hours.
Seems to work out.
OK one last try to convince people...
Imagine that the event was held on dry land, and the raft never moved. If you drive away from it for an hour then turn around and drive the same speed back, how long would it take to get back?
One hour.
Next imagive the whole thing take place on a conveyer belt.
You run for an hour down the belt, then run back up it at the same speed.
how long till you reach the starting point on the conveyer?
One hour.
Although you are at the same point on the conveyer belt, you are now 6 miles from where you started on solid ground... how fast was the belt going?
it moved 6 miles in 2 hours... thats 3 miles an hour.
Ill give up now :)
we all seem to miss the comment some time.They travelled for an hour and after some time they seen old mate in his raft! define some time? we cant the answer is no i cant work it out as i do not have sufficant information.
irodov 1.1
what i've been seeing wrong in most of these explanations, especially the ones where the answer is 3mph, is that it says they are going at a constant speed. So that means the same speed always. If they go at 1mph, add the river speed (i'll humor you let's say 3mph) you get 4. So they go 4mph going downstream. Now going upstream you start with -3mph as river speed, so they'll set power so it would go 7mph to equal 4mph or same speed going downstream. This is probably confusing, so i'll write it in a simpler form.
Vb-Velocity of boat
Vr-Velocity of river
Vd-Velocity downstream(Vb+Vr)
Vu-Velocity upstream(Vb-Vr)
If Vr=3 Then Vd=Vb+3=x
If Vr=3 Then Vu=Vb-3=x
since we must have a constant speed, as is dictated in the question, x must be the same in both cases, so here is how it turns out if x=10
Vd=(7)+3=10
Vu=(13)-3=10
or
Vd=10
Vu=10
Vd=Vu
x=5
Vd=(2)+3=5
Vu=(8)-3=5
or
Vd=5
Vu=5
Vd=Vu
Yes it's complicated, and i've probably overexplained but try and understand
the quesion has oo many variables which arent accounted for ! But the 'anonymous' gus who are doing all the math shold SO add their nicknames... you rock ! Im gonna be lazy and sit back and let you do the hard work on this one ! :)
I think the people who are saying 3 mph are wrong, but have the general idea... there are 2 things keeping me from saying what the answer is, what the "sometime" is and (really doesnt have to do with the question) I just got off a 12 hour shift from work. It all seems to vivid to me right now. They never metion that the boat holmes is in is doing the same speed. If all is how I think it is.
Holmes bout is doing it's powered speed(assumed at 100% power) + the speed of the river, going down river for 1 hour. Then they turn around and go up river at the powered boats speed - the rivers speed, this is where it get's weird to me.. The time Holmes goes up stream isnt defined.
At this point Im at a stand still. If this helps anyone with a possible answer cool. Im going to stop rambling for now and sleep on it.
Sir,
I also came up with 3 Mph solution in a very similar way.
The conveyor belt analogy is very concise and clear. Well done.
I also got 3 mph. I used three variables and algebra. Only had two equations but after substitution, the variable for the speed of the boat got divided out from both sides (safe to assume the boat speed was not zero) leaving just 2r = 6.
Sir is exactly right, the speed of the boat is irrelevent. If you can't understand his conveyer belt explaination, then I dont think anything will convince you. Good game!
i think you all missed the point i made earlier the boat speed is not part of the equation, the equation, as Holmes said, is distance divided by time. distance of course is 6 miles, and time is undefined, other than >1 hour. What we need here, is some way of figuring out how much time passed between Holmes and Watson leaving point A, and Holmes and Watson reaching the point 6 miles downstream of point A. The conveyer belt theory, while useful in other places, does not quite work here. If the boat speed is to stay the same, then the item representing Watson's and Holmes' boat would have to change speed when it turned around, in order to keep a cponstant speed.
In one hour the raft moved exactly 6 miles, with v = s/t (where v is velocity, s is distance and t is timee) therefore the raft moved with the speed of 0.1 miles per minute or 6 miles per hour... what ever the unit of time is desired :o)
The speed of the river IS the speed of the raft.
Forget the speed of the motor boat, it is a 'Red Herring'!
Concentrate on the raft:
We know the raft travels 6 miles and that the elapsed time for the raft to do this is 1hr. So the speed of the raft, and therefore the river, is 6 miles/hour!
The question has been very cleverly posed to fool us into forming complicated quadratic equations etc. I, too, nearly went down that route.
So, concentrate on the distance travelled by the raft in the known time - and you have the answer.
It's a good one though!
what about the return time , you did not considered that ?
like the guy above said, you're missing the return time curiosity kid, it says "an hour later they turned back* and after sometimes, passed the raft again", the problem in the end is we need a more precise amount than "sometimes", otherwise, you're just re-iterating what i've said, and i congratulate you on noticing the irrelevance of the speed of the motor boat, to the equation (unlike most people). So what I believe is the closest to correct answer we have, as i said earlier, is V = 6 / >1, or V = <6
I noticed the last comment came a couple weeks after the question was posted, and there is no comment from the owner.
COME ON MR. LAL
YOU CAN DO BETTER THAN THAT!!!!
Patrick, you have to wait for a month i think because this is a Friday Special and accordingto the FAQ's page
http://trickofmind.com/2006/10/answers-to-questions.html
this will be answered after a month.
I agree with 3 MPH.
Let the speed of :
Motorboat= Y MPH
Stream or Raft = X MPH
Travelling downstream, the Boat speed = (Y+X) MPH
1 Hour later:
<--------------------(Y+X) Mile------------------------->Boat
<--X Mile-->Raft
Note: distance between Raft & Boat = Y Mile
Travelling upstream, the Boat speed = (Y-X) MPH
So, 1 Hour later the Boat will passed the Raft.
<--------------------(Y+X) Mile------------------------->Boat
<--X Mile-->--X Mile--><--------(Y-X) Mile------------
2X Mile = 6 Mile in 2 Hours
the Speed = 3 MPH.
i figured 2mph for river flow
ie
boat meets raft
1 hour later turns back
1 hour later reaches 1st meeting place (this assumes constant speed of boat up and down stream)
to make it easy assuming boat speed 6 mph
therefore 1 hour later meets raft again 6 miles down river
this means 3 hours have elapsed since 1st meeting and the raft was
6 miles from 1st meeting so
6/3=2
you can check this with boat speed
6 miles then turn back 6 miles to
1st meet catches raft 6 miles later = 18 miles / 3 hours = 6mph
the boat speed is important hence it was noted as being constant.
the speed is three mph when they preceded past the raft at waterloo they continued for another hour BEFORE turning around thus its an hour trip bak v= dist/time 6mile/2hrs=3mph
"No... You have to judge the speed of the river by the speed of the raft, which is simply floating (unpowered) in the current of the river. In ~1 hour, the raft floated 6 miles down stream. The current is ~6 miles per hour. We can't evaluate "some time", and have to make the current an approximate value."
You are correct in assuming that we cannot calculate "some time", but that does not imply that we have to approximate so roughly. Because they traveled at a constant speed DOWNRIVER, obviously the speedometer is judging how fast the motor is propelling the boat, not factoring in the help that the current is providing. Assuming it took no time to turn around, the boat would travel at the same constant speed but now against the current. The time it took to return to the raft would have to be at LEAST an hour. Therefore the raft was travelling approx. 3mph.
like omg. this is just wayyy 2 much math 4 my blond little pretty head. like hello... cudnt yall just post yalls aim if yall r cute guys. like w/e... =]
Another way to solve this would be available if you knew the length of your boat. On the return trip you could measure how long it took for the raft to pass your boat and thus calculate it speed in relation to you. Then subtract the speed of you boat from that and you would get the river speed.
a=speed of boat
b=speed of raft/current
c = 1st meet(waterloo jct/station)
d = turnback
e = c
f = 2nd meet (catching raft up)
c to d = 1hr
d to e = 1hr
e to f = xhr or 6 miles
m=miles h=hours
ergo: b = 6m/(2h+ (6 m/ a))
if a =12 mph then
b = 6m/(2h+(6/12))
b = 6 miles/(2 hrs + 1/2)
so: 6m/2.5 = 2.4 mph
let a = 6 mph
b = 6 / (2 + 1) = 2 mph
its a good trick - now spin it
lets find a if b is know
b = 2 mph (current)
b = 6m /(2h+(6m/a))
2 = 6 /2+(6/a)
2 *( 2 + (6 / a)) = 6
2+(6/a) = 6/2
6/a = 6/2 -2
a = (6/2-2)*6
a = (3-2)* 6
a = 1 * 6
a=6
may i had that sir is wrong
u keep coming back to the place u started from
u forget the raft is moving which is the true indicator to current speed
time to raft
not 6 miles / by time to where u started from
what your not seeing is the raft is`t 6 miles away when u arrive
back to where u started.
ANSWER: 3 Miles/Hour
--------------------------
Let the velocity of the motorboat be V and that of the stream be v.
Considering the Waterloo station as the origin we can say that the motorboat has travelled a distance of (V + v) miles in 1 hour.
The raft has floated to a distance of v miles downstream in 1 hour.
Say, the motorboat has passed the raft after a time t (after 1 hour) while moving upstream.
Distance travelled by the raft (downstream) = v*t
Distance travelled by the motorboat (upstream) = (V - v)*t
v*t + (V - v)*t = V
==> t = V/V = 1 hour
v + v*t = 6
==> v + v*1 = 6
==> v = 3 miles per hour .... is the velocity of the stream.
GOT RIGHT
--------------------------
3rd Guy (Anonymous), Sir, joseph v., JediTony, Ydrameos,
great explaination by Sir
"Imagine that the event was held on dry land, and the raft never moved. If you drive away from it for an hour then turn around and drive the same speed back, how long would it take to get back?
One hour.
Next imagive the whole thing take place on a conveyer belt.
You run for an hour down the belt, then run back up it at the same speed.
how long till you reach the starting point on the conveyer?
One hour.
Although you are at the same point on the conveyer belt, you are now 6 miles from where you started on solid ground... how fast was the belt going?
it moved 6 miles in 2 hours... thats 3 miles an hour."
um the Anonymous guy who wrote like an essay about how to do it even though u got it wrong U REALLY NEED TO GET A LIfE AND DO SOMETHING ELSE INSTEAD OF ANSWERING A PROBLEM FOR 10 MINUTES
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