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A power boat and a raft both left dock A on a river and headed downstr

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A power boat and a raft both left dock A on a river and headed downstr  [#permalink]

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New post 31 Mar 2019, 21:53
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C
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E

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Question Stats:

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A power boat and a raft both left dock A on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock B downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock A. How many hours did it take the power boat to go from A to B?

A. 3
B. 3.5
C. 4
D. 4.5
E. 5

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Re: A power boat and a raft both left dock A on a river and headed downstr  [#permalink]

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New post 03 Apr 2019, 17:34
My solution:

From the question, I understood that "The raft drifted at the speed of the river current."
Let's assume that the speed of the river current is q kmph.

"The powerboat maintained a constant speed with respect to the river."
Let's assume that the speed of the boat (only) is p kmph.

"The powerboat reached dock B downriver, then immediately turned and travelled back upriver. It eventually met the raft on the river 9 hours after leaving dock A."
Let's assume the distance covered by Raft is X and the total distance is D.

We know that \(Speed = \frac{Distance}{Time}.\)

Let's calculate the time for both of them.
For Raft, \(Time =\frac{X}{q} hrs.\)

As Raft and Powerboat met after 9hrs => Time travelled by Raft is 9hrs.
Now, we get that X = 9q.

For Powerboat, \(Time = \frac{D}{(p+q)} + \frac{(D-X)}{(p-q)} = 9\)

Now, we can plug in values and check.

Let's try C - 4

Time taken by power boat to go from A to B is 4 hrs.
=> \(\frac{D}{(p+q)} = 4\) => D = 4(p+q)

=> \(\frac{(D-X)}{(p-q)} = 5\)

As we know X = 9q and D = 4(p+q). Let's put that in this then the equation is, 4p + 4q - 9q = 4(p-q) => 4p - 5q = 4(p-q)
LHS and RHS donot match.

We need larger value for \(\frac{D}{(p+q)}\) and when subtracted from 9 it also should give back the value we choose. What we need is a 9/2 = 4.5
Let's try D 4.5hrs

Time taken by power boat to go from A to B is 4.5 hrs.
=> \(\frac{D}{(p+q)} = 4\) => D = 4.5(p+q)

=> \(\frac{(D-X)}{(p-q)} = 4.5\)

As we know X = 9q and D = 4(p+q). Let's put that in this then the equation is, 4.5p + 4.5q -9q = 4.5(p-q) => 4.5p - 4.5 q = 4.5(p-q)
LHS = RHS.

Answer: D
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A power boat and a raft both left dock A on a river and headed downstr  [#permalink]

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New post 03 Apr 2019, 23:39
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Bunuel wrote:
A power boat and a raft both left dock A on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock B downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock A. How many hours did it take the power boat to go from A to B?

A. 3
B. 3.5
C. 4
D. 4.5
E. 5



Let the time taken to cover the distance from A to B by power boat = T hrs. We need the value of T.

Speed of boat = B
Speed of river = S

Distance covered by boat downstream - Distance covered by boat upstream = Distance covered by raft downstream

(B + S)*T - (B - S)*(9 - T) = 9S
BT + ST - 9B + BT + 9S - ST = 9S
2BT - 9B = 0
B(2T - 9) = 0

Either B = 0 or T = 9/2

Answer (D)

And yes, it is not a GMAT level question.
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Re: A power boat and a raft both left dock A on a river and headed downstr  [#permalink]

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New post 07 Apr 2019, 18:19
Bunuel wrote:
A power boat and a raft both left dock A on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock B downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock A. How many hours did it take the power boat to go from A to B?

A. 3
B. 3.5
C. 4
D. 4.5
E. 5


We can let r = the speed of the current and p = the speed of the power boat in still water. Also, we can let d = the distance between A and B.

Let’s analyze the answer choices. Let’s start with choice E and go backward.

E. 5

If it takes 5 hours for the power boat to go downstream from A to B, then we have (notice that the power boat will then spend 4 hours upstream) :

Downstream: (p + r) x 5 = d

Upstream: (p - r) x 4 + r x 9 = d

Simplifying the two equations, we have 5p + 5r = d for downstream and 4p + 5r = d for upstream. However, the two equations don’t agree (notice that if we subtract the two equations, we will have p = 0, which is impossible).

D. 4.5

If it takes 4.5 hours for the power boat to go downstream from A to B, then we have (notice that the power boat will then spend 4.5 hours upstream then) :

Downstream: (p + r) x 4.5 = d

Upstream: (p - r) x 4.5 + r x 9 = d

Simplifying the two equations, we have 4.5p + 4.5r = d for downstream and 4.5p +4.5r = d for upstream. We see that the two equations agree with each other. So D is the correct answer.

Alternate Solution:

We can let r = the speed of the current and p = the speed of the power boat in still water, d = the distance between A and B and t = the time (in hours) it takes for the powerboat to reach B from A.

Since it takes t hours for the boat to travel from A to B, the distance between A and B is t*(r + p). Therefore, we have d = t*(r + p) or, equivalently, d = tr + tp.

Since the rate of the raft is also r, the raft travels a distance of 9r in 9 hours. In t hours, the boat travels t*(r + p) and reaches B and then, the boat turns back and meets the raft, after traveling a distance of (9 - t)*(p - r) at a rate of p - r (since it is now traveling upstream). In total, they have traveled a distance of 9r + t*(r + p) + (9 - t)*(p - r) = 9r + tr + tp +9p - 9r - tp + tr = 2tr + 9p. Notice that the total distance they traveled is twice the distance between A and B, thus we must have 2tr + 9p = 2d. Dividing each side by 2, we obtain tr + 4.5p = d.

Since we have tr + 4.5p = d and we obtained tr + tp = d earlier, let’s equate the two expressions we have for d:

tr + 4.5p = tr + tp

4.5p = tp

t = 4.5

Since t was the time it took for the boat to get to B from A, the answer is 4.5.

Answer: D
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Re: A power boat and a raft both left dock A on a river and headed downstr  [#permalink]

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New post 10 Apr 2019, 07:56
A__________C___M_____B

Let the speed of the river current be 'r' kph (which is also the speed of the raft(R)), 'p' kph be the original speed of the power boat (PB) and 't' hrs the time it takes PB to go downriver from A to B.
In the above diagram, C is the point RT reaches in 't' hrs and M the point at which the two crafts meet after PB reaches B, turns around, and heads back to A.
So, 't' hrs after starting from A, PB reaches B having traveled t(p+r) kms (Since PB's original speed is increased by the speed of the current while travelling downriver). So, AB = t(p+r)
Meantime, RT has traveled 'tr' kms in 't' hrs and reached Point C.
So, 't' hrs after starting, the distance separating the two crafts (CB) is 'tp' kms [t(p+r) - tr].
PB meets RT (9 - t)hrs after it heads back from B since we know that the total time taken by PB to travel from A to B and back to the meeting point M is 9 hrs of which he has already expended 't' hrs traveling from A to B.
Since RT and PB start simultaneously from Points C and B respectively, the time taken RT to reach M is also (9 - t)hrs. Also, PB's speed reduces to (p - r) kph during the return trip since it has to travel against the current.Thus:

(Distance traveled by RT from C to M) + (Distance traveled by PB from B to M) = CB
r(9 - t) + (9 - t)(p - r) = tp
p(9 - t)= tp-->t = 4.5 hrs.

My explanation might seem overly long but I tried to be as detailed as possible to avoid any confusion. But once the approach is known the calculation is pretty simple so I think it could be done within the GMAT time frame.
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Re: A power boat and a raft both left dock A on a river and headed downstr   [#permalink] 10 Apr 2019, 07:56
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