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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
how am i getting this wrong?
time = t for downstream and 1.5t for upstream
Distance upstream = distance downstream (90miles = 90miles)
therefore,
(v-3) 1.5t = (v+3)t
1.5vt-4.5t = vt + 3t
therefore, v = 15mph
Downstream speed = v+3 = 15+3 = 18mph
therefore, time t= d/s = 90/18 = 5hrs
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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SohamS wrote:
how am i getting this wrong?
time = t for downstream and 1.5t for upstream
Distance upstream = distance downstream (90miles = 90miles)
therefore,
(v-3) 1.5t = (v+3)t
1.5vt-4.5t = vt + 3t
therefore, v = 15mph
Downstream speed = v+3 = 15+3 = 18mph
therefore, time t= d/s = 90/18 = 5hrs


The question says "the trip upstream took a half hour longer than the trip downstream"
So, if t = the travel time downstream (in hours), then t + 0.5 = the travel time upstream (in hours)

I have a feeling that you accidentally wrote: t + 0.5t = the travel time upstream (which simplifies to be 1.5t)
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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


We can PLUG IN THE ANSWERS, which represent the number of hours to travel downstream.
The time upstream is 0.5 hour longer than the time downstream.
The correct answer choice is almost CERTAIN to be A -- the only option that implies two times that divide evenly into 90 (2.5 hours and 3 hours).
When the correct answer is plugged in, the difference between the rate downstream (v+3) and the rate downstream (v-3) will be 6 mph:
(v+3) - (v-3) = 6

Answer choice A: 2.5 hours to travel downstream, implying 3 hours to travel upstream
Rate downstream \(= \frac{distance}{time} = \frac{90}{2.5} = 36\) mph
Rate upstream \(= \frac{distance}{time}= \frac{90}{3} = 30\) mph
Rate difference \(= 36-30 = 6\) mph
Success!

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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
Bunuel wrote:
jjewkes wrote:
Please help!

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?

-2.5
-2.4
-2.3
-2.2
-2.1


Trip upstream took \(\frac{90}{v-3}\) hours and trip downstream took \(\frac{90}{v+3}\) hours. Also given that the difference in times was \(\frac{1}{2}\) hours --> \(\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}\);

\(\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}\) --> \(\frac{90(v+3)-90(v-3)}{v^2-9}=\frac{1}{2}\) --> \(\frac{90*6}{v^2-9}=\frac{1}{2}\) --> \(v^2=90*6*2+9\) --> \(v^2=9*(10*6*2+1)\) --> \(v^2=9*121\) --> \(v=3*11=33\);

Trip downstream took \(\frac{90}{v+3}=\frac{90}{33+3}=2.5\) hours.

Answer: A.

—-

Hi Bunuel,

Thanks for explaining this so well. Could u pls post a list of similar boat stream Qs

Posted from my mobile device
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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kajaldaryani46 wrote:
Bunuel wrote:
jjewkes wrote:
Please help!

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?

-2.5
-2.4
-2.3
-2.2
-2.1


Trip upstream took \(\frac{90}{v-3}\) hours and trip downstream took \(\frac{90}{v+3}\) hours. Also given that the difference in times was \(\frac{1}{2}\) hours --> \(\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}\);

\(\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}\) --> \(\frac{90(v+3)-90(v-3)}{v^2-9}=\frac{1}{2}\) --> \(\frac{90*6}{v^2-9}=\frac{1}{2}\) --> \(v^2=90*6*2+9\) --> \(v^2=9*(10*6*2+1)\) --> \(v^2=9*121\) --> \(v=3*11=33\);

Trip downstream took \(\frac{90}{v+3}=\frac{90}{33+3}=2.5\) hours.

Answer: A.

—-

Hi Bunuel,

Thanks for explaining this so well. Could u pls post a list of similar boat stream Qs

Posted from my mobile device


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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
Dear experts, can you help me in this approach. I am getting it wrong.

(i) Upstream
\(v-3 = \frac{90}{t+\frac{1}{2}}\) .... eqn 1

(i) Downstream
\(v+3 = \frac{90}{t}\)

\(v = \frac{90}{t} - 3\) , substituting this in eqn 1

\(\frac{90-6t}{t} = \frac{180}{2t+1}\)

(90-6t)(2t+1)= 180t

90=18t

t=5.

where have i gone wrong?
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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GRS wrote:
Dear experts, can you help me in this approach. I am getting it wrong.

(i) Upstream
\(v-3 = \frac{90}{t+\frac{1}{2}}\) .... eqn 1

(i) Downstream
\(v+3 = \frac{90}{t}\)

\(v = \frac{90}{t} - 3\) , substituting this in eqn 1

\(\frac{90-6t}{t} = \frac{180}{2t+1}\)

(90-6t)(2t+1)= 180t

90=18t

t=5.

where have i gone wrong?


Hi GRS,

This equation does NOT 'simplify' in the way that you wrote.

(90-6t)(2t+1)= 180t

FOIL it out, combine like terms and then divide by 3... You will end up with 2 solutions (because it's a Quadratic) and the one positive solution is the answer to the original prompt.

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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
Top Contributor
Upstream speed = V - 3
Downstream Speed = V + 3

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

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

\(\frac{90(v+3-v+3)}{(v^2 -9)} = \frac{1}{2}\)

\(90*6 *2 = v^2 -9\)
\(v^2 = 1080 + 9 = 1089\)
v= 33

So the time taken by boat to travel downstream = 90/(v+3) = 90/36 = 2.5 hrs

Option A is the correct answer.

Another approach I would suggest is to try logical elimination of the options.
Most of the cases, V will be an integer.
so the Downstream speed = V + 3 will also be an integer.
\(\frac{90}{Downstream speed }= time \)
=> 90/time taken = Downstream Speed
That means, when you divide 90 by the options choices ,you should get an integer as well.

Option A will give an integer .You can also crosscheck by finding Upstream time and see if it is 3 hrs i.e. 2.5 + .5 = 3 hrs
Opt C, D,E can be eliminated as it give a recurring decimal.

Thanks,
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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Normally with this type of question the answer choices will be easy to work backwards from, so it's surprising to see decimals here.
Here are links to two other official problems of the same type:
https://gmatclub.com/forum/a-store-currently-charges-the-same-price-for-each-towel-that-144780.html
https://gmatclub.com/forum/at-his-regular-hourly-rate-don-had-estimated-the-labor-cost-136642.html

In this case, we're told that the 90 miles take half an hour longer if the speed is 6mph slower. So I would just try answer choice A, because adding half an hour to it gets us an integer.

(A) 90 miles in 3 hours means 30mph, and I must check whether 2.5 * 36mph is also 90 miles. Yes, it is (I check quickly by "borrowing a factor of 4 from the 36 and "giving" it to the 2.5, thereby getting 10*9). So the answer is A.

*it is inappropriate, in my opinion, to solve this problem, or the other two that I linked to, using algebra.
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
Hey guys, this took me just under 3.5 mins to solve on the Mock Exam GMAT 2 exam, the part that really held me up was the square root of 1089 step.

I have just been reviewing the paper now, and I was wondering how I could've solved this faster, I stupidly just tried numbers and by trial and error I found 33 works (when squared yields 1089).


In hindsight, do you guys think that prime factorisation would have been the most optimal way to root such a large number?

My logic (had I instead used prime factorisation) is that 1089 is divisible by 11 via inspection (1-0+8-9 is divisible by 11, therefore 1089 is), therefore:

1089 = 11*99 = 11*33*3 = 33*33, so the root is 33.

Is this the best way to handle rooting larger numbers on the GMAT in terms of speed / effectiveness? Or is there better?
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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PVOG wrote:
Hey guys, this took me just under 3.5 mins to solve on the Mock Exam GMAT 2 exam, the part that really held me up was the square root of 1089 step.

I have just been reviewing the paper now, and I was wondering how I could've solved this faster, I stupidly just tried numbers and by trial and error I found 33 works (when squared yields 1089).


In hindsight, do you guys think that prime factorisation would have been the most optimal way to root such a large number?

My logic (had I instead used prime factorisation) is that 1089 is divisible by 11 via inspection (1-0+8-9 is divisible by 11, therefore 1089 is), therefore:

1089 = 11*99 = 11*33*3 = 33*33, so the root is 33.

Is this the best way to handle rooting larger numbers on the GMAT in terms of speed / effectiveness? Or is there better?



30^2 is 900. 40^2 is 1600. We need something between 30 and 40 that is closer to 30 than to 40 and that, when squared, will have a units digit of 9. 33 sure sounds good to me!
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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


As always, I'd approach this problem by making sure to look at the answers in addition to the question right out of the gate. There are only five possible answers and the test writers give them to us before we do any work. Only on of the answer choices is going to give an integer value when we divide it into 90 and only one other is going to give something aside from a repeating decimal. The odds are VERY good that it's going to be A, and in the wonky scenario it's not A, it would be VERY likely to be B. Let's just try A.

The boat travelled 90 miles downstream in 2.5 hours, so 36mph. That means it travelled upstream at 30mph, so it took 3 hours. Is that a half hour longer than downstream? Yep.
Answer choice A.

Remember that business schools and GMAC are trying to test how you think rather than how well you can tackle middle/high school math. The test has TONS of opportunities to use logic in lieu of the math. Getting really good at incorporating the answer choices into your thinking is a HUGE opportunity.
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
Why cant you set the distances equal to each other and solve for T?
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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srinigaj wrote:
Why cant you set the distances equal to each other and solve for T?


Then you would have two variables, v and t and you will not be using the data that distance is 90. You will not be able to get the value of t.
Instead you don't want to introduce a new variable and want to use the given data.
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
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