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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 (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
[Reveal] Spoiler: OA
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Re: Difficult PS Problem- Help! [#permalink]

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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.
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Re: Boat Traveling [#permalink]

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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 ? :roll:

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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Re: Boat Traveling [#permalink]

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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

Answer is (a)
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Re: A boat traveled upstream a distance of 90miles at an average [#permalink]

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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

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shrouded1 wrote:
Michmax3 wrote:
v^2 = 1089
V = 33

Downstream time = 90/36 = 2.5

Answer is (a)


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 ? :roll:

many thanks
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problem is time consuming rather difficult for memorizing square root.
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Re: Difficult PS Problem- Help! [#permalink]

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

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lamode wrote:
This is the hardest problem I have seen in my GMAT studies so far. Not because of the maths involved, but the time required. Even when I know the answer, I can't go through the calculations in under 5 mins (I just spent an hour trying over and over). Ouch.



Check out these posts lamode:

http://www.veritasprep.com/blog/2013/03 ... culations/
http://www.veritasprep.com/blog/2013/03 ... s-part-ii/

You will learn how to do such questions much faster.
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]

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RussianDude wrote:
it took me more than 5 minutes to solve this question. Is there a way to solve this type of questions faster? Is it even worth it to spend time on such question on the real exam, or is it better to dump it and move on? The chances of making a silly mistake are very high in my opinion. Looking forward to your answers!


Your questions present a little puzzle for the gmat taker.
But I will answer the first question.
YES there is a way of solving this question faster on the exam day. Learn to solve it faster now. How do you learn to solve it faster now? First let your mind enter into it. I guess you've done that or you wouldnt be slugging it out for 5 minutes. Secondly learn the shortcut and practice on it. Now what is the shortcut? Check out the two posts above from Nez and the link by
VeritasPrepKarishma. I bet you if you do that and praactice on what you learnt, then you could solve that question in about 2mins plus with a very low margin of error. Now that's a plus for you considering that such questions in the GMAT are likely to be 700 level and you are rewarded much more for that.
Everythng depends on the score YOU want on the gmat. If you want 550, please jettison my advice and forget this problem cos you wouldnt even see similar question on gmat. But if you want anything around 680 and up, then heed.
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]

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ameyaprabhu wrote:
I am a little confused as to how we arrived at the highlighted equation, from the previous equation. How did we get 90 * 6. Can someone please explain?

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.


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

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

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

\(\frac{90*6}{v^2-9}=\frac{1}{2}\).

Hope it's clear.
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New post 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]

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New post 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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New post 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 ?

right answer is 2,5

please help me this is so annoying :)

thanks, I shall give my teacher kudos if I find out how to do that
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New post 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

Answer:- A
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Re: Difficult PS Problem- Help! [#permalink]

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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.
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Re: Downstream Rate Question [#permalink]

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New post 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..
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Re: Downstream Rate Question [#permalink]

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New post 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].
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Re: Boat Traveling [#permalink]

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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 ? :roll:

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??
Re: Boat Traveling   [#permalink] 14 Mar 2013, 01:11

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