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A boat traveled upstream 90 miles at an average speed of [#permalink]
09 Sep 2010, 16:29

4

This post received KUDOS

10

This post was BOOKMARKED

00:00

A

B

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D

E

Difficulty:

55% (hard)

Question Stats:

75% (03:56) correct
25% (04:43) wrong based on 333 sessions

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 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};

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 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?

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 ?

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

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

1

This post received KUDOS

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 ?

Re: Boat Traveling [#permalink]
22 Nov 2010, 05:57

5

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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. _________________

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)

Re: A boat traveled upstream a distance of 90miles at an average [#permalink]
19 Nov 2012, 00:10

2

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.

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??

Re: Boat Traveling [#permalink]
14 Mar 2013, 19:36

1

This post received KUDOS

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:

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.

Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
20 Mar 2014, 12:19

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.. _________________

Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

Re: A boat traveled upstream 90 miles at an average speed of [#permalink]
25 Aug 2014, 05:38

Let's solve this problem by using ratios. As the trip upstream (U) took a half hour longer than the trip downstream (D), the ratios of time - U:D = 3:2. So the ratios of speed equals to 2:3.

Given speed: U = s-3, D = s+3.

(s-3) / (s+3) = 2/3. When you solve you will find that s = 15. Then 90 / (15+3) = 5. But as 5 isn't an aswer choice, we divide it by 2 and get 2.5. The answer is 2.5.

gmatclubot

Re: A boat traveled upstream 90 miles at an average speed of
[#permalink]
25 Aug 2014, 05:38

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