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A boat traveled upstream 90 miles at an average speed of

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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
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,
Brent
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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:

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

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

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New post 09 Feb 2016, 21:12
The question already told u that the upstream boat and the downstream boat has a speed difference of 1/2 hours.
that makes everything easy here.
90/(v-3) - 90/(v+3) = 1/2

this is 180/(v-3) - 180/(v+3) = 1

This means that each of the two fractions above must be consecutive integers, since there diff is one.
Quick testing of value (which logically must be multiple of 3) shows that 33 is the value of v that transforms the fractions to 6-5, which equals 1.

Remember the speed for downstream boat is
90/(v+3)
so plug in 33 and get 80/36
5/2 or 2.5

Answer is A.

You don't need any long calculation to get v.
No you don't.
You might, if it were a DS question,
when ya trying to find another possible value (quadratic comes in handy).
PS: Even in a DS question, u cud test further values)
One "truth" I've earlier learnt from some masters who've posted above is that GMAT never, yes, NEVER will give you a question that MUST demand long calculations.
If you often find urself doing long calculation, then register for a prep.

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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, 22:15
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.



Hi Lamode
GMAT is more benevolent than we think.
Look at the solutions and links provided below your comment.
Eg in my solution above i derived
180/(v-3) - 180/(v+3) = 1

the fractions above could as well be fractions instead of integers as I assumed(since a fraction minus a fraction can equal 1 as in 1/2- (-1/2) etc)
But assumed intergerhood for simplicity.
And my simplicity assumptions got rewarded by GMAT.
in fact the question was constructed to be answered based on simple math tricks that will save u calculations time.

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

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New post 10 Feb 2016, 00:10
"Quick testing of value (which logically must be multiple of 3) shows that 33 is the value of v that transforms the fractions to 6-5, which equals 1. "


Hi Nez. Could you please elaborate how you got this quoted part?

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

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New post 10 Feb 2016, 00:26
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.



Will we not consider the speed of stream?. can we assume the speed of stream as 3 m/hr.?
Also , please advise me how I can solve such quadratic equation fast. I formed the main equation quickly but got struck in the calculation which took 4 minutes. Is there any quick method.?

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

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New post 10 Feb 2016, 03:27
shreyashid wrote:
"Quick testing of value (which logically must be multiple of 3) shows that 33 is the value of v that transforms the fractions to 6-5, which equals 1. "


Hi Nez. Could you please elaborate how you got this quoted part?


Alright shreyashid,
you had this already, 180/(v-3) - 180/(v+3) = 1.

you have two fractions in the lefthandside i.e. 180/(v-3) and 180/(v+3).
you want to make each of these fraction a whole number i.e each will turn to an integer.
you do this by ensuring that the denominators in each fraction is a factor of 180 the numerator.

in the first fraction 180/(v-3), u pick a value for v that when reduced by 3 will be a factor of 180 and that
same value you picked, when increased by 3 must also still be a factor of 180 (considering the denominator of the second
fraction 180/(v+3).

This tells u that the number must be a factor of 180 and a multiple of 3!

u can test from say v=6 ---> 180/3 and 180/9 is 60 and 20, there diff(30) isn't even close to one!.
So aplying commonsense, u pick a very higher value, since u know that picking 9 next will only reduce the diff slightly.
test v=18 ------> 180/15 and 180/21 is 12 and about10. (still not close to 1!)
if v=30 -------> 180/27 and 180/33 is a little less than 7 and a little over 5, hmm seems close. now pick the next in line.
if v=33 --------> 180/30 and 180/36 is 6 and 5 respectively. Gotcha!


PS: when i picked 6 and saw a diff of 30, an instinct told me to pick 30 next to quickly smash the huge difference. U cud even have known that a larger value of 30 is likely since the numerator is 180, large enough.
Give a kudos if that helped.

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

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New post 13 Feb 2016, 03:03
kentash wrote:
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.


We are not given this information.

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

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New post 28 Mar 2016, 03:41
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!
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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]

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New post 29 Mar 2016, 06:01
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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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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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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
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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Re: A boat traveled upstream 90 miles at an average speed of [#permalink]

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New post 28 Mar 2017, 08:23
first approaching this problem I wrote down what the answer requires: 90/(v+3) or time spent going downstream
from the data let's make up an equation:
90/(v-3)-90/(v+3)=1/2
we can arrive to v^2=1089
v=33
Then be careful!
90/(33+3)=90/36=2,5
Answer is A

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

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New post 28 Mar 2017, 10:34
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


let t=downstream hours
v+3=90/t
v-3=90/(t+1/2)
subtracting, 6=90/t-[90/(t+1/2)]➡
2t^2+t-15=0
(2t-5)(t+3)=0
t=2.5 hours
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 2017, 04:57
can someone explain what is wrong with my explanation below and why I am not getting the right answer?

Upstream
D= 90 miles
S= v-3 mph
T=H

Downstream
d= 90 miles
s= v+3
t=h

H=h+1/2

distance=speed x time
Upstream

90=(v-3) (h+1/2).....(i)

Downstream

90=(v+3)h ....(ii)

with equation i and ii

(v-3)(h+1/2)=(v+3)h

After this, I am stuck. Is the above logic right at all?

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

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New post 26 Aug 2017, 08:26
pra1785 wrote:
can someone explain what is wrong with my explanation below and why I am not getting the right answer?

Upstream
D= 90 miles
S= v-3 mph
T=H

Downstream
d= 90 miles
s= v+3
t=h

H=h+1/2

distance=speed x time
Upstream

90=(v-3) (h+1/2).....(i)

Downstream

90=(v+3)h ....(ii)

with equation i and ii

(v-3)(h+1/2)=(v+3)h

After this, I am stuck. Is the above logic right at all?

pra1785, in a different kind of question, the logic would work more efficiently. Your logic is not wrong here. It just makes things a lot harder, I think.

Distances are equal, so it does seem like a natural step to set each leg's (r * t) equal, but . . .

The problem is that you have defined one unknown variable in terms of another. You have two unknown variables and one equation. The answer choices are values, not variable expressions (in which case your equation, without more, might work, see below).

You will have to use a known value at some point. IMO, it's easier to do so at the beginning.

Your equation yields \(\frac{1}{2}v - 6h = \frac{3}{2}\)

"Solve" for time, which you have called h:

h = \(\frac{(v - 3)}{12}\)

Now you must "go back" and use the known value of 90.

h\(_2\) = \(\frac{(v-3)}{12}\)

h\(_1\) = \(\frac{90}{(v+3)}\)

\(\frac{(v-3)}{12}\) = \(\frac{90}{(v+3)}\)

\(v^2 - 9 = 1080\)
\(v^2 = 1089\)
\(v = 33\)

Finally (whew!), if h = \(\frac{(v - 3)}{12}\), then

\(\frac{(33 - 3)}{12}\) = h

h = \(\frac{30}{12}\) = 2.5 hrs

If answer choices WERE variable expressions, depending on the prompt, the variable-variable solution ("h = ...") to your equation as-is (i.e., not taken further as I did, with D = 90) would be among them.

But the answer choices have actual values. So you need one variable and one known in your equation. Known D / unknown rates = known time.

When distances are equal, it often works efficiently to write \(r_1*t_1 = r_2*t_2\). I don't think it's efficient here. But if you are very quick and very accurate with algebra, it's probably fine. :-)

Does that help?

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