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by Brent@GMATPrepNow » Sun May 11, 2014 11:53 am
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

Cheers,
Brent
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by [email protected] » Sun May 11, 2014 1:15 pm
Hi mukherjee.tanuj3,

This is a layered story-problem and takes a lot of effort to solve using a traditional "math approach" (Brent notes it in his explanation: "lots of work here"). Here's how you can solve it with a bit of logic and TESTing THE ANSWERS:

From the prompt, we can create 2 equations:

D = R x T

90 = (V-3)(T + 1/2)
90 = (V+3)(T)

We're asked for the value of T.

From the prompt, I find it interesting that the distance is a nice, round number (90).... because when looking at the answer choices, most of them are NOT nice decimals. When multiplying two values together (as we do in BOTH equations), if you end up with a round number, chances are that either....

1) both numbers are round numbers
2) one of the numbers includess a nice fraction (e.g. 1/2) which can be multiplied and the end result will be a round number.

This gets me thinking that 2.5 is probably the answer, but I still have to prove it....I'm going to plug in THAT value for T and see what happens to the 2 equations....

90 = (V-3)(3)
90 = (V+3)(2.5)

30 = (V-3)
36 = (V+3)

33 = V
33 = V

Notice how both values of V are THE SAME? That means that we have the solution. V=33 and T=2.5

Final Answer:A

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by GMATGuruNY » Sun May 11, 2014 4:35 pm
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 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
We can PLUG IN THE ANSWERS, which represent the number of hours that the boat took to travel downstream.
The most likely answer choice is A -- the only option that divides evenly into 90.

Answer choice A: 2.5 hours to travel downstream.
Rate downstream = d/t = 90/(2.5) = 36 miles per hour.
Thus, v+3 = 36, implying that v=33.
Rate upstream = v-3 = 33-3 = 30 miles per hour.
Time upstream = d/r = 90/30 = 3 hours.
Time upstream - time downstream = 3-2.5 = .5.
Success!

The correct answer is A.
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by Amrabdelnaby » Sun Nov 29, 2015 6:11 pm
Hi Brent,

I noticed while solving this question that it takes - if you know what you are doing - between 3-4 minutes to solve this question for V and do all the calculations.

I was wondering if i encounter such a question in the assessment day if it is worthy to spend such time on it and i also wanted to ask you what is the level of this question?

i think its 700+ right?
Brent@GMATPrepNow wrote:
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

Cheers,
Brent
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by [email protected] » Sun Nov 29, 2015 10:48 pm
Hi Amrabdelnaby,

When you say that it took 3-4 minutes of work, what were you doing exactly during those 3-4 minutes? While this question is 'thick' and takes longer than average for most Test Takers to solve, it might be that 'your way' of tackling this question is why it took so long to solve. If you TEST THE ANSWERS and think about the Number Properties involved, then you likely would have answered it much faster.

If you're facing overall pacing problems during the Quant section of your CATs, and you can't answer these types of questions faster than you currently are, then you might need to dump them so that you can spend more time on the rest of the section.

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by prada » Sat Feb 13, 2016 4:30 pm
Brent@GMATPrepNow wrote:[

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

Cheers,
Brent
Hey Brent,
When you say lots of work here, am I do it wrong?

90/(v-3) = 90/(v+3)+1/2

Find a common denom for the right side of equality so

90/(v-3)={180/2v+6}+{v+3/2v+6]

combine terms under same denom

90/(v-3) = (183+v)/(2v+6)

cross multiply

183v-549+v^2-3v=180v+540

v^2-9=0

(v-3) (v+3) = 0

v=3 or -3

I didn't get v=33
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by GMATGuruNY » Sat Feb 13, 2016 5:23 pm
prada wrote: When you say lots of work here, am I do it wrong?

90/(v-3) = 90/(v+3)+1/2

Find a common denom for the right side of equality so

90/(v-3)={180/2v+6}+{v+3/2v+6]

combine terms under same denom

90/(v-3) = (183+v)/(2v+6)

cross multiply

183v-549+v^2-3v=180v+540
Through the equation in green, your work is correct.

Combining like terms, we get:
v² + 183v - 180v - 3v - 549 - 540 = 0

v² + 0 - 1089 = 0

v² - 1089 = 0

(v + 33)(v - 33) = 0.

v = ±33.

Since the rate cannot be negative, the only valid solution is v=33.
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by Brent@GMATPrepNow » Sat Feb 13, 2016 6:03 pm
Here's another upstream/downstream question to practice with - https://www.beatthegmat.com/speedboat-t40512.html

Cheers,
Brent
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by Matt@VeritasPrep » Sun Feb 14, 2016 9:26 pm
prada wrote:90/(v-3) = 90/(v+3)+1/2
Multiply both sides by (v-3)(v+3)*2:

180(v+3) = 180(v-3) + (v-3)(v+3)

Subtract 180*(v+3) from both sides:

0 = 180(v-3) - 180(v+3) + (v-3)(v+3)

Factor:

0 = 180((v - 3) - (v + 3)) + (v - 3)(v + 3)
0 = 180(-6) + (v-3)(v+3)
0 = v² - 9 -1080
1089 = v²

v = ±33
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by Gurpreet singh » Sun May 22, 2016 11:04 pm
90/(v-3)-90/(v+3)=30/60(convert min into hrs) (Difference of time going up & down)
90(v+3)-90(v-3)/(v-3)(v+3)=1/2

simplify

90v+270-90v+270/v^2-9=1/2

540*2=v^2-9

1080+9=V^2

0r V=1089^1/2=33

time taken to come down=90/33+3= 90/36= 5/2=2.5hrs
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