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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]

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New post 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]

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New post 09 Sep 2010, 16:54
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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: A boat traveled upstream a distance of 90miles at an average  [#permalink]

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New post 19 Nov 2012, 00:10
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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

Kudos Please... If my post helped.
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Re: Difficult PS Problem- Help!  [#permalink]

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New post 11 Sep 2010, 20:33
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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
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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


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]

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New post 30 Sep 2010, 21:45
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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: Boat Traveling  [#permalink]

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New post 22 Nov 2010, 01:19
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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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Re: Boat Traveling  [#permalink]

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New post 22 Nov 2010, 05:57
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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: Difficult PS Problem- Help!  [#permalink]

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New post 27 Mar 2011, 00:18
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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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Re: Difficult PS Problem- Help!  [#permalink]

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New post 25 Aug 2011, 16:19
2
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: Boat Traveling  [#permalink]

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New post 14 Mar 2013, 01:11
1
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??
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Re: Boat Traveling  [#permalink]

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New post 14 Mar 2013, 19:36
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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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A boat traveled upstream 90 miles at an average speed of  [#permalink]

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New post 10 Feb 2014, 21:12
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CfaMiami 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


Here is a post discussing how to solve such equations easily: http://www.veritasprep.com/blog/2013/03 ... culations/
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Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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New post 20 Mar 2014, 12:19
1
2.5 hrs..A set of best possible option spares u the horror of solving the equations..

90 miles is the distance traveled downstream..2.5 is the only only option that divides 90..I am pretty sure the speeds are never (25.12+3) & (25.12-3)..So we can safely assume an integer..2.5 & 3 are the set of speeds that aptly fit in.

took me just 10 secs..
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Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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New post 08 Feb 2016, 15:03
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.
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Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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New post 08 Feb 2016, 16:18
2
Quote:
A boat travelled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then travelled 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


I like to begin with a "word equation."
We can write:
travel time upstream = travel time downstream + 1/2

Time = distance/rate
So, we can replace elements in our word equation to get:
90/(v-3) = 90/(v+3) + 1/2

Now solve for v (lots of work here)
.
.
.
v = 33

So, travel time downstream = 90/(v+3)
= 90/(33+3)
= 90/36
= 5/2
= 2 1/2 hours
Answer: A

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

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New post 09 Feb 2016, 02:55
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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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New post 25 Aug 2016, 05:04
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.
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Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

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New post 25 Aug 2016, 07:15
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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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A boat traveled upstream 90 miles at an average speed of  [#permalink]

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New post 26 Aug 2016, 08:55
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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


Since time = distance/rate, therefore the time going upstream = 90/(v – 3) and the time going downstream = 90/(v + 3). Since the time going upstream is ½ hour more than the time going downstream, we can set up an equation as follows:

90/(v – 3) = 90/(v + 3) + 1/2

Let’s multiply the equation by 2(v – 3)(v + 3) to get rid of the denominators:

2(90)(v + 3) = 2(90)(v – 3) + (v – 3)(v + 3)

180v + 540 = 180v – 540 + v^2 – 9

540 = v^2 – 549

v^2 = 1089

v = \(\sqrt{1089}\)

v = 33

Since the time going downstream = 90/(v + 3) and we’ve found that v = 33, so the time going downstream = 90/(33 + 3) = 90/36 = 5/2 = 2.5 hours.

Answer: A
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