Official Solution:

A steamer going upstream would cover the distance between town A and town B in 4 hours and 30 minutes. If the same steamer going downstream would cover the distance between the towns in 3 hours, how long would it take a raft moving at the speed of the current to float from town B to town A?

A. 10 hours
B. 12 hours
C. 15 hours
D. 18 hours
E. 20 hours

Let \(V\) and \(C\) denote the speed of the steamer and the speed of the current respectively. The distance between \(A\) and \(B\) \(= 3(V + C) = 4.5(V - C)\). From this equation it follows that \(V = 5*C\). The time it will take an object moving at speed \(C\) to get from \(B\) to \(A\) equals \(3*\frac{(V + C)}{C} = 18*\frac{C}{C} = 18\)hours.

Answer: D
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Let \(V\) and \(C\) denote the speed of the steamer and the speed of the current respectively.

Now while going upstream the steamer will be going against the stream, so its speed will become less = V-C..
it takes 4.5 hour to reach up.. so distance = speed * time = 4.5(c-v)..

similarly while going downstream the steamer will be going with the stream, so its speed will become more = V+C..
it takes 3 hour to reach down.. so distance = speed * time = 3(c+v)..

But the distance between \(A\) and \(B\)must be the same..
So \(= 3(V + C) = 4.5(V - C)\).
\(3V+3C=4.5V-4.5C\) ..
\(3C+4.5C=4.5V-3V\)..
\(7.5C=1.5V\)
From this equation it follows that \(V = 5*C\).

Hope it helped
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I solved in the following way,

Speed of current = (Downstream rate - upstream rate)/2

Let, d is distance.
So, Downstream rate = d/3 and Upstream rate = d/4.5

Speed of current is (d/3-d/4.5)/2 -------> d/18
So, to travel distance d at speed d/18 it would take 18 hrs.

If I am wrong, please correct me.
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Hey Bunuel,

I was wondering where the denominator of "C" came from in \(3*\frac{(V + C)}{C} = 18*\frac{C}{C} = 18\)hours.


Thanks!

Time = Distance/Rate

The distance between A and B \(= 3(V + C)\) and the rate of current is C, thus it would take Time = Distance/Rate = 3(V + C)/C hours for raft moving at the speed of the current to float from town B to town A.
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Why did you divide the speed of the current by 2?
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Its the same logic as Bunuel has explained. Speed = Distance / Time
Down Stream: V + C = d/3 (1)
Up Stream: V - C = d/4.5 (2)

(1) - (2) -> 2C = d/3 - d/4.5 = d/9
-> d/C = 18 is What we are looking for.

This typical form of task and can be quick solved by Mahmud formula.

Hi Bunuel

How did you get V=5∗CV=5∗C?

Thanks!

thanks chetan!!

This is how I arrived at the answer:

As shown in the above solutions:

Let V and C denote the speed of the steamer and the speed of the current respectively.

As shown above by Chetan2u, while going upstream the steamer will be going against the stream, so its speed will become less = V - C.

Now, since it takes 4.5 hours to reach UP, and speed is an inverse of time, the speed to reach UP = V - C = 1/4.5 = 2/9

similarly while going downstream the steamer will be going with the stream, so its speed will become more = V+C.

Now, since it takes 3 hours to reach DOWN, and speed is an inverse of time, the speed to reach DOWN = V + C = 1/3

Now we have two equations

V - C = 2/9
V + C = 1/3

Add the two equations to get 2V = 5/9 ===> V = 5/18

Substitute the value of V into any of the two equations and you have C = 1/18

Since C = 1/18 is the speed of the current, the time duration it takes the raft moving at the speed of the current will be the inverse of the speed of the current = 18

Hi Bunuel or Chetanu2, pls is anything wrong with my approach?

v= 5c - clear till this point , but could not understand what the question is trying to ask:

how long would it take a raft moving at the speed of the current to float from town B to town A?

does it mean how long will it take if streamer moves at a speed of current from B to A i.e downstream ?

if yes then pls clarify solution in detail

Yes, the question asks about the time it will take the steamer to go down from B to A with only current's speed.

To proceed: Time = Distance/Rate

The distance between A and B \(= 3(V + C)\) and the rate of current is C, thus it would take Time = Distance/Rate = 3(V + C)/C hours for raft moving at the speed of the current to float from town B to town A. We know that V = 5C, so Time = Distance/Rate = 3(5C + C)/C = 18 hours.

Hope it's clear.
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This is how I solved this question! Guess it is also a valid method by picking a smart number.

Speed of steamer - R
Speed of current = x

Since we are going against current,

(R-x) * 4.5 = D

Since we are going with the current,

(R+x) * 3 = D

Lets assume D = 90 (90 since it is divisible by both 4.5 and 3)

therefore,

(R-x) * 4.5 = 90 --> R-x = 20

(R+x) * 3 = 90 --> R+x = 30

Solving these two we get R = 25 and x = 5

The question asks for x * t = D

5 * t = 90 --> t = 18 hours

Answer is (D)

Why are we not taking total downstream speed for raft and instead considering only speed =c

Posted from my mobile device

C in the solution denotes the speed of the current, which is the downstream speed of a raft.
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Hey Bunuel,
Actually the query is that shouldn't downstream speed = c+c (speed of boat + speed of current) = 2c
But what I deduced is that since the word 'float' is mentioned it means the speed of the boat = 0 and we've to only consider the speed of the current which takes the boat with it.
Please correct me if I'm wrong. Thanks.

I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. Shouldn't there be a clarity in the question that raft is moving from B to A at the speed of the current in still water?
It can be confused that the raft is travelling downstream at the speed of current, in that case the answer would be 9 hrs.

Let the distance between A and B is 9 miles
Let the speed of boat be s miles/hr
let the speed of current be c miles/hr

so,
s-c=9/4.5=2.......(i)
s+c=9/3=3.........(ii)

solving equation (i) and (ii),
c=0.5 miles/hr
So time taken = 9/0.5 = 18 hours

Answer: D