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# A boat traveled upstream 90 miles at an average speed of

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A boat traveled upstream 90 miles at an average speed of [#permalink]  09 Sep 2010, 16:29
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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
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Re: Difficult PS Problem- Help! [#permalink]  09 Sep 2010, 16:54
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Expert's post
jjewkes 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?

-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.

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Re: Difficult PS Problem- Help! [#permalink]  11 Sep 2010, 20:33
90/(v-3)=90(v+3)+.5

90[v+3-v+3]/(v*v-9)=.5
90*6/(v*v-9)=.5
90*12+9=v*v
v*v=1089
v=33
v+3=36

Time required to travel downstream=90/36=2.5 hours
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Re: Boat Traveling [#permalink]  30 Sep 2010, 21:43
Michmax3 wrote:
A boat traveled upstream a distance of 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 upstream trip took half an hour longer than the downstream trip, 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

Given 90/(v-3) = 90/(v+3) + (1/2)
90[(1/(v-3)) - (1/(v+3))] = 1/2
90[6/(v^2 - 9)] = 1/2

90*6*2 = v^2 - 9 => v^2 = 1089 => v = 33.

Downstream time -- 90/(v+3) => 90/36 => 2.5 (A).
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Re: Boat Traveling [#permalink]  30 Sep 2010, 21:45
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KUDOS
Michmax3 wrote:
A boat traveled upstream a distance of 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 upstream trip took half an hour longer than the downstream trip, 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

Upstream time = downstream time + 0.5

90/(v-3) = 90/(v+3) + 1/2
90 (1/(v-3) - 1/(v+3))= 1/2
90 (6/(v^2-9)) = 1/2
v^2 = 1089
V = 33

Downstream time = 90/36 = 2.5

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Word problem need help solving [#permalink]  21 Nov 2010, 06:17
Hi I just took one gmat practice test and the pity about those test is that they don´t have explanations.

The problem i am struggling with in mathematical sense is something like this:

A boat that traveled 90 miles upstream a river at the rate average speed Y-3, on the way back he traveled the same distance (90 miles) at a average speed of Y+3, if the trip upstream took half an hour longer than the trip down stream. How long took the drip downstream ?

thanks, I shall give my teacher kudos if I find out how to do that
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Re: Difficult PS Problem- Help! [#permalink]  21 Nov 2010, 13:02
solve the below equation to reach upon speed of boat-

90/(v+3) - 90/(v-3) = 1/2
V=33

therefore time required while travelling downstream- 90/(33+3) = 90/36 = 2.5

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Re: Boat Traveling [#permalink]  22 Nov 2010, 01:19
shrouded1 wrote:
Michmax3 wrote:
v^2 = 1089
V = 33

Downstream time = 90/36 = 2.5

Hi I struggled with this one, I am wondering
Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ?

many thanks
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Re: Boat Traveling [#permalink]  22 Nov 2010, 05:57
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Expert's post
hafgola wrote:

Hi I struggled with this one, I am wondering
Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ?

many thanks

Let me take your question first:

It is a good idea to remember squares of numbers till 20. They come in handy sometimes.
To find the square root of bigger numbers (which are perfect squares), observe the following:
1. A square of a number only ends in 0/1/4/5/6/9.
-If it ends in 0, it will have even number of 0s at the end and the square root will end in 0 (I know 1340 cannot be a perfect square.)
-If it ends in 1, the square root will end in 1/9 (e.g. root 361 = 1[highlight]9[/highlight], root 441 = 2[highlight]1[/highlight])
-If it ends in 4, the square root will end in 2/8
-If it ends in 5, there will be a 2 right before it and its square root will end in 5 (e.g. root 625 = 25, root 675 is not an integer)
-If it ends in 6, the square root will end in 4/6.
-If it ends in 9, the square root will end in 3/7.
These are all derived by just observing the last digits when we square a number.. You don't need to learn this.

Since 1089 ends in a 9, its square root will end in 3/7.
Also 30^2 = 900 so root of 1089 will be greater than 30.
But 40^2 = 1600 so root of 1089 will be much less than 40. The number that can be the square root of 1089 is 33 (if 1089 is a perfect square). You can square 33 to check.

Also, an equation like \frac{90}{{v -3}} - \frac{90}{{v + 3}} = \frac{1}{2}, doesnt need to be solved.

99% chance is that v will be an integer.
The options give me values for \frac{90}{{v + 3}}

Let me say if it is 2.5, then 90/2.5 = 36 and v = 33
Then \frac{90}{{v - 3}} = 3. I get the difference of 1/2 an hour. 2.5 satisfies my condition.

Had 2.5 not been the first option, you might say that it would take me a long time to check each option.
No. 90/2.4 = 900/24 which is not an integer since 900 is not divisible by 8 so it will not be divisible by 24
90/2.3. No ways 900 is divisible by 23
90/2.2. No ways 900 is divisible by 11
90/2.1. No ways 900 is divisible by 7
In fact, when I look at the options, the first one I try is 2.5. With practice, you will gain the instinct.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Director Status: GMAT Learner Joined: 14 Jul 2010 Posts: 655 Followers: 29 Kudos [?]: 152 [0], given: 32 Re: Difficult PS Problem- Help! [#permalink] 25 Mar 2011, 10:22 problem is time consuming rather difficult for memorizing square root. _________________ I am student of everyone-baten Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html SVP Joined: 16 Nov 2010 Posts: 1698 Location: United States (IN) Concentration: Strategy, Technology Followers: 28 Kudos [?]: 261 [0], given: 36 Re: Difficult PS Problem- Help! [#permalink] 27 Mar 2011, 00:18 90/(v-3) = 90/(v+3) + 1/2 => 90( V + 3 - v + 3)/(v-3)(v+3) = 1/2 => 90 * 6 * 2 = v^2 - 9 => v^2 = 1080 + 9 => v^2 = 9 (120 + 1) => v^2 = 3^2 * 11^2 => v = 3*11 = 33 So Downstream time = 90/36 = 30/12 = 2.5 hrs Answer - A _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) Director Joined: 01 Feb 2011 Posts: 774 Followers: 12 Kudos [?]: 75 [0], given: 42 Re: Difficult PS Problem- Help! [#permalink] 25 Aug 2011, 16:19 upstream v-3 t+1/2 downstream v+3 t (v+3)t = (v-3)(t+1/2) vt + 3t = vt +v/2 - 3t - 3/2 => t = (v-3)/12 = 90/v+3 => v = 33 => t = 30/12 = 2.5 hours Answer is A. Current Student Joined: 14 May 2011 Posts: 277 Location: Costa Rica Concentration: Healthcare, International Business GMAT 1: 640 Q V0 GMAT 2: 710 Q42 V45 GMAT 3: 740 Q48 V42 GMAT 4: Q V0 GMAT 5: Q V0 GPA: 3.3 WE: Research (Consulting) Followers: 10 Kudos [?]: 52 [0], given: 134 Re: Downstream Rate Question [#permalink] 17 Aug 2012, 14:13 Can someone pls help with the above question? What is the most efficient way to do this. Here's what i've got.. If one takes the time for the trip downstream to to be T, then one gets UPSTREAM: V-3 = 90 / (T + 1/2) DOWNSTREAM V+3 = 90 / T Now no matter how much I simplify I end up with a complicated equation, and then it will be plugging in off answer choices but that takes >3 minutes at least. Any suggestions for a sohrtcut, I'm sure there's an easy way.. _________________ How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing) Intern Joined: 01 Jan 2011 Posts: 22 Location: Kansas, USA Schools: INSEAD, Wharton Followers: 2 Kudos [?]: 5 [0], given: 9 Re: Downstream Rate Question [#permalink] 17 Aug 2012, 16:40 I think the answer to the question posted here should be 2.5. Basically, frame the equation in terms of the difference in time (= distance/ speed): [90][/v-3]- [90][/v+3]= 1/2. Once you get this you could solve for V. Remember that the question is asking for the value of [90][/v+3]. Moderator Joined: 02 Jul 2012 Posts: 1122 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Followers: 48 Kudos [?]: 499 [1] , given: 92 Re: A boat traveled upstream a distance of 90miles at an average [#permalink] 19 Nov 2012, 00:10 1 This post received KUDOS Amateur wrote: A boat traveled upstream a distance of 90miles at an average speed of (v-3) miles per hour and then traveled down stream at an average of (v+3) miles per hour. If upstream look 30 min more than downstream time, how much time did the boat travel downstream? a) 2.5 b) 2.4 c) 2.3 d) 2.2 e) 2.1 how to solve these kind of questions in 2 min? any short way... \frac{90}{(v-3)}=\frac{90}{(v+3)} +1/2 solving for v, value of \frac{90}{(v+3)} is tedious... and answer choices are in too close range to make an estimation.... how should I be tackling these? The equation itself is simple enough.. \frac{90}{v-3}- \frac{90}{v+3} = \frac{1}{2} \frac{540}{v^2-9} = \frac{1}{2} v^2 - 9 = 1080 v^2 = 1089 = 3^2 * 11^2 v = 3*11 = 33 Answer = 90/36 = 2.5 Kudos Please... If my post helped. _________________ Did you find this post helpful?... Please let me know through the Kudos button. Thanks To The Almighty - My GMAT Debrief GMAT Reading Comprehension: 7 Most Common Passage Types Intern Joined: 27 Dec 2012 Posts: 11 Followers: 0 Kudos [?]: 0 [0], given: 10 Re: Boat Traveling [#permalink] 14 Mar 2013, 01:11 VeritasPrepKarishma wrote: hafgola wrote: Hi I struggled with this one, I am wondering Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ? many thanks Let me take your question first: It is a good idea to remember squares of numbers till 20. They come in handy sometimes. To find the square root of bigger numbers (which are perfect squares), observe the following: 1. A square of a number only ends in 0/1/4/5/6/9. -If it ends in 0, it will have even number of 0s at the end and the square root will end in 0 (I know 1340 cannot be a perfect square.) -If it ends in 1, the square root will end in 1/9 (e.g. root 361 = 19, root 441 = 21) -If it ends in 4, the square root will end in 2/8 -If it ends in 5, there will be a 2 right before it and its square root will end in 5 (e.g. root 625 = 25, root 675 is not an integer) -If it ends in 6, the square root will end in 4/6. -If it ends in 9, the square root will end in 3/7. These are all derived by just observing the last digits when we square a number.. You don't need to learn this. Since 1089 ends in a 9, its square root will end in 3/7. Also 30^2 = 900 so root of 1089 will be greater than 30. But 40^2 = 1600 so root of 1089 will be much less than 40. The number that can be the square root of 1089 is 33 (if 1089 is a perfect square). You can square 33 to check. Also, an equation like \frac{90}{{v -3}} - \frac{90}{{v + 3}} = \frac{1}{2}, doesnt need to be solved. 99% chance is that v will be an integer. The options give me values for \frac{90}{{v + 3}} Let me say if it is 2.5, then 90/2.5 = 36 and v = 33 Then \frac{90}{{v - 3}} = 3. I get the difference of 1/2 an hour. 2.5 satisfies my condition. Had 2.5 not been the first option, you might say that it would take me a long time to check each option. No. 90/2.4 = 900/24 which is not an integer since 900 is not divisible by 8 so it will not be divisible by 24 90/2.3. No ways 900 is divisible by 23 90/2.2. No ways 900 is divisible by 11 90/2.1. No ways 900 is divisible by 7 In fact, when I look at the options, the first one I try is 2.5. With practice, you will gain the instinct. Hi karishma, I have been able to solve most of the questions from work, speeed distane & mixtures. However i always solve them using the algebraic approach which eats up a lot of time. I have seen you solving questions by a logical approach that saves a considerable bit of time. I have tried solving that way but I just cant get going. Could you please advice me on how to develop this skill?? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4038 Location: Pune, India Followers: 862 Kudos [?]: 3625 [0], given: 144 Re: Boat Traveling [#permalink] 14 Mar 2013, 19:36 Expert's post sauravleo123 wrote: Hi karishma, I have been able to solve most of the questions from work, speeed distane & mixtures. However i always solve them using the algebraic approach which eats up a lot of time. I have seen you solving questions by a logical approach that saves a considerable bit of time. I have tried solving that way but I just cant get going. Could you please advice me on how to develop this skill?? It needs practice. You need to believe that you can solve the question very quickly for you to be able to think up of logical approaches. In GMAT, you can solve almost every question within two minutes so if you get a big complicated equation, you can be sure that you are missing something. If the question looks really complicated, we know there has to be a trick (In fact, one tends to overlook the so called short-cuts when questions are simple and don't take much time anyway). I would suggest you to read up and practice the logical solutions - once you understand the logic, try using it in other questions. Don't worry about the time factor. I discuss some logic based ideas regularly on my blog. Check it out: http://www.veritasprep.com/blog/categor ... om/page/3/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]  08 May 2013, 03:27
I would prefer substituting answers rather than solving the equation to save time. In actual test, i was way past time, took more than 3 mins to solve the Q. Later offline, i used substitution after arriving at the initial set of below equations for Time. It was so easy and less troublesome.

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

DOWNSTREAM
V+3 = 90 / T
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A boat traveled upstream a distance of 90 miles at an averag [#permalink]  26 Jan 2014, 07:15
A boat traveled upstream a distance of 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 half an hour longer than the trip downstream, 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
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Re: A boat traveled upstream a distance of 90 miles at an averag [#permalink]  26 Jan 2014, 08:23
Expert's post
TheZezet wrote:
A boat traveled upstream a distance of 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 half an hour longer than the trip downstream, 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

Merging similar topics. Please refer to the solutions above.
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Re: A boat traveled upstream a distance of 90 miles at an averag   [#permalink] 26 Jan 2014, 08:23
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