A boat traveled upstream 90 miles at an average speed of

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

The trip upstream took \(\frac{90}{(v-3)}\) hours, while the trip downstream took \(\frac{90}{(v+3)}\) hours. The difference in times was given as \(\frac{1}{2}\) hours. Therefore, \(\frac{90}{(v-3)}-\frac{90}{(v+3)}=\frac{1}{2}\).

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

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

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

\(v^2=90*6*2+9\);

\(v^2=9*(10*6*2+1)\);

\(v^2=9*121\);

\(v=3*11=33\).

The trip downstream took \(\frac{90}{(v+3)}=\frac{90}{36}=2.5\) hours.

Answer: A.
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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.

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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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

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).

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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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

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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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.

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.
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Check out the blog posts here: https://anaprep.com/blogs/
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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..

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.

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,
Brent

Check out these posts lamode:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/03 ... culations/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/03 ... s-part-ii/

You will learn how to do such questions much faster.
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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?

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

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

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

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

Hope it's clear.
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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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