The steamer going upstream would cover the distance between

18

S = Speed of Steamer
C = Speed of Current

(1) (S-C)*4.5 = d
(2) (S+C)*3 = d
Combining (1) and (2)
4.5S - 4.5C = 3S + 3C
C = S/5
We are looking for (1/C)*d
d = (S + S/5)*3 = 18S/5

(1/C)*d = 5/S * 18S/5 = 18 hour

Late, but D:

S - distance
t - time
v1 - speed upstream
v2 - speed downstream

Suppose that the distance = 9

4.5 = 9/v1
v1 = 2

3 = 9/v2
v2 = 3

v of the current = 0.5

So the time needed for a raft to travel downstream = 9/0.5 = 18

1. For a problem which requires that sum of the speed and difference of the speed respectively be taken considering the direction of an object and in which only the time taken in both the directions are given, we can find the time taken by the slower one of the two, in this case the time taken by an object traveling at the speed of the water current by,

2/ (1/t1 - 1/t2), where t1 is the time taken when sum of the speed is considered and t2 is the time taken when the difference of the speed is considered

2. So we have time time taken by the raft to travel from B to A as 2/ (1/3 - 1/4.5) = 18 hours.

Could not get the above statement.....

R=Rate of boat
S=Rate of stream

4.5(R-S) = D
3(R+S) = D
4.5(R-S) = 3(R+S)
4.5R- 4.5S = 3R + 3S
1.5R = 7.5S
R = 5S

Plug 5S in the original formula

3(5S+S) = D
15S + 3S = D
18S = D

the Answer is D, 18 hours.

c=speed of current
s=speed of steamer
1=distance one way
s+c=1/3
s-c=1/4.5
subtracting,
2c=1/9
c=1/18 (speed=distance/time)
18 hours
D

Time taken for upstream=4.5hrs
time taken for downstream=3 hours
time upstream/time downstream=4.5/3-->3/2
therefore, speed downstream/speed upstream will be in the same ratio, i.e., 3/2
plug in values and check, take 10 and 15.
Total distance =45, speed of current=(speed downstream-speed upstream)/2=2.5
so time=45/2.5=18 hours which is option (D)

——
Could u pls help me understand the formula for the speed of the current? Why is it divided by 2

Posted from my mobile device

The more intuitive, mathematical method:

to determine the time it will take for the raft to float Downstream WITH the Current, we need to have some info. about the Speed of the Current.

Let D = Distance between Town A and Town B

Let R = Speed of Steamer

Let C = Speed of Water Current


(1)Speed Downstream With Current = R + C

(2)Speed Upstream Against Current = R - C

If we SUBTRACT the 2 Speeds:

R + C - (R - C) =

R - R + C + C =

2C = 2 * (Speed of Current)


Concept: therefore, if we subtract the Upstream Speed from the Downstream Speed and then Divide the result by 2 —-> we will find the Speed of the Current

(R + C) = D m / 3 hr = D/3 = 3D / 9

(R - C) = D m / (9/2) hr = 2D / 9

——Subtracting the 2 and then Dividing by 2—-

[ (R + C) - (R - C) ] / 2 =

[ (3D/9) - (2D/9) ] / 2 =

(D/9) / 2 =

D / 18 = Speed of Current in m.p.h.

This means a raft floating with the Speed of the Current will travel the Distance between the Towns = D in 18 hours

-D- 18 hours

Posted from my mobile device

Upstream time is - 4 hours and 30 minutes = \(4 \frac{1}{2}\) hrs = \(\frac{9}{2}\) Hrs
Downstream time is 3 Hrs

Let the distance be 9 Km

Thus, Upstream spead \(( s - c ) = \frac{9*2}{2} => 2\)
And , Downstream spead \(( s + c ) = \frac{9}{3} => 3\)

\(2s = 5\)

Or, \(s = 2.5\) & \(c = 0.50\)

Hence, Time taken by raft moving at the speed of the current to float from B to A is \(\frac{9}{0.5} = 18\) hrs, Answer must be (D)

It's a good question. However, the language in the last sentence is very confusing. It says that the raft is moving at the speed of the current - which means that the speed of raft = twice the speed of the current (speed of the raft + speed of the current). In that case, it would take the raft only 9 hours to travel.

I am wondering this too! Why do we divide by 2 here? Bunuel could you provide an insight?

I tried explaining this here: https://gmatclub.com/forum/the-steamer- ... l#p1335320 Let me know if it helps.

Official Solution:

A steamer takes 4 hours and 30 minutes to travel from Town A to Town B when traveling upstream against the current. However, it only takes the steamer 3 hours to travel from Town B to Town A when traveling downstream with the current. Assuming the current flows at a constant speed, how long will it take a raft, floating downstream at the speed of the current, to travel from Town B to Town A ?

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


Assume the speed of the steamer in still water is \(s\) and the speed of the current to is \(c\).

When traveling upstream against the current, the steamer's speed is \(s - c\), and in 4.5 hours it covers a distance of \(4.5(s - c)\).

When traveling downstream with the current, the steamer's speed is \(s + c\), and in 3 hours it covers a distance of \(3(s + c)\).

Since both distances represent the distance between Towns A and B, we can equate them: \(4.5(s - c)=3(s + c)\), which gives \(s = 5c\). To express the distance in terms of only one unknown, substitute \(s = 5c\) in either of the above equations, to get \(3(s + c)=18c\).

To determine how long a raft, floating downstream at the speed of the current, will take to travel from Town B to Town A, we divide the distance between the towns by the speed of the raft: \(\frac{distance}{rate}=\frac{18c}{c}=18\) hours.


Answer: D

Hope it helps.