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Manager  S
Joined: 03 Jan 2017
Posts: 131
Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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first approaching this problem I wrote down what the answer requires: 90/(v+3) or time spent going downstream
from the data let's make up an equation:
90/(v-3)-90/(v+3)=1/2
we can arrive to v^2=1089
v=33
Then be careful!
90/(33+3)=90/36=2,5
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Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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can someone explain what is wrong with my explanation below and why I am not getting the right answer?

Upstream
D= 90 miles
S= v-3 mph
T=H

Downstream
d= 90 miles
s= v+3
t=h

H=h+1/2

distance=speed x time
Upstream

90=(v-3) (h+1/2).....(i)

Downstream

90=(v+3)h ....(ii)

with equation i and ii

(v-3)(h+1/2)=(v+3)h

After this, I am stuck. Is the above logic right at all?
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A boat traveled upstream 90 miles at an average speed of  [#permalink]

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pra1785 wrote:
can someone explain what is wrong with my explanation below and why I am not getting the right answer?

Upstream
D= 90 miles
S= v-3 mph
T=H

Downstream
d= 90 miles
s= v+3
t=h

H=h+1/2

distance=speed x time
Upstream

90=(v-3) (h+1/2).....(i)

Downstream

90=(v+3)h ....(ii)

with equation i and ii

(v-3)(h+1/2)=(v+3)h

After this, I am stuck. Is the above logic right at all?

pra1785, in a different kind of question, the logic would work more efficiently. Your logic is not wrong here. It just makes things a lot harder, I think.

Distances are equal, so it does seem like a natural step to set each leg's (r * t) equal, but . . .

The problem is that you have defined one unknown variable in terms of another. You have two unknown variables and one equation. The answer choices are values, not variable expressions (in which case your equation, without more, might work, see below).

You will have to use a known value at some point. IMO, it's easier to do so at the beginning.

Your equation yields $$\frac{1}{2}v - 6h = \frac{3}{2}$$

"Solve" for time, which you have called h:

h = $$\frac{(v - 3)}{12}$$

Now you must "go back" and use the known value of 90.

h$$_2$$ = $$\frac{(v-3)}{12}$$

h$$_1$$ = $$\frac{90}{(v+3)}$$

$$\frac{(v-3)}{12}$$ = $$\frac{90}{(v+3)}$$

$$v^2 - 9 = 1080$$
$$v^2 = 1089$$
$$v = 33$$

Finally (whew!), if h = $$\frac{(v - 3)}{12}$$, then

$$\frac{(33 - 3)}{12}$$ = h

h = $$\frac{30}{12}$$ = 2.5 hrs

If answer choices WERE variable expressions, depending on the prompt, the variable-variable solution ("h = ...") to your equation as-is (i.e., not taken further as I did, with D = 90) would be among them.

But the answer choices have actual values. So you need one variable and one known in your equation. Known D / unknown rates = known time.

When distances are equal, it often works efficiently to write $$r_1*t_1 = r_2*t_2$$. I don't think it's efficient here. But if you are very quick and very accurate with algebra, it's probably fine. Does that help?
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Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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1
Hi All,

This is a layered story-problem and takes a lot of effort to solve using a traditional "math approach". Here's how you can solve it with a bit of logic and TESTing THE ANSWERS:

From the prompt, we can create 2 equations:

D = R x T

90 = (V-3)(T + 1/2)
90 = (V+3)(T)

We're asked for the value of T.

From the prompt, I find it interesting that the distance is a nice, round number (90)…. because when looking at the answer choices, most of them are NOT nice decimals. When multiplying two values together (as we do in BOTH equations), if you end up with a round number, chances are that either….

1) both numbers are round numbers
2) one of the numbers includess a nice fraction (e.g. 1/2) which can be multiplied and the end result will be a round number.

This gets me thinking that 2.5 is probably the answer, but I still have to prove it….I'm going to plug in THAT value for T and see what happens to the 2 equations….

90 = (V-3)(3)
90 = (V+3)(2.5)

30 = (V-3)
36 = (V+3)

33 = V
33 = V

Notice how both values of V are THE SAME? That means that we have the solution. V=33 and T=2.5

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GMAT 1: 640 Q48 V28 Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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1
Took me about 2.5hrs to do this question.
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Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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Cellchat wrote:
Took me about 2.5hrs to do this question.

Quote:
A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

Let's try this way

B = Speed of Boat
S = Speed of Stream

Upstream Speed = Net speed against the flow of current = B - S
Downstream Speed = Net speed with the flow of current = B + S

Time upstream = (1/2)Hour + Downstream Time

$$\frac{90}{v-3} = (\frac{1}{2}) + \frac{90}{v+3}$$

90/30 = 3
90/36 = 2.5
Diffference = 1/2

i.e. v-3 = 30 and v+3 = 36 i.e. v = 33

Downstream time = 90/36 = 2.5

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Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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DSGB wrote:
A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

compare both times of upstream and down stream
90/v-3= 30/60+ 90/v+3
or say
90v+270-90v+270 = v^2-9 *1/2
540*2 = v^2-9
1089 = v^2
v= 33
so downstream speed ; 33+3 ; 36
time taken 90/36 ; 2.5
OPTION A Re: A boat traveled upstream 90 miles at an average speed of   [#permalink] 24 May 2020, 07:18

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