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Time and Distance please help!

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by DavidG@VeritasPrep » Thu Jan 29, 2015 11:14 am
This is a tough one. What you'll find, if you set this one up algebraically, is that it's going go be unpleasant to solve this conventionally. Here's how I think about it. If the speed is v + 3 going with the current and v - 3 against, I know that there is a difference of 6 between up and down stream. Now I'll backsolve and see what answer gives me that difference of 6 in the speeds.

Looking at the answers, I want to start with the number that will give me the least unpleasant arithmetic. In this case, 2.5 is the most likely to give me integer values, so I'll try that one. If it took 2.5 hours to do downstream, and it took .5 hours more to go upstream, then it will have taken 3 hours to go upstream.

The rate for down stream is 90/2.5, which is 36. (I'll simplify like this: 900/25 = 9 *100/25 = 9*4 = 36)

The rate for upstream is 90/3 = 30. Difference of 6, I'm done.

Takeaway: backsolving isn't just valuable, it's sometimes necessary. Remember that the GMAT isn't testing your math skills per se, but your ability to make good decisions under pressure, and the good decision here is to work with the numbers that will make your life easiest. (It's not a coincidence that you don't have to test any of the really painful answer choices on this one.)
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by [email protected] » Thu Jan 29, 2015 11:15 am
Hi Mallika,

This is a layered story-problem and takes a lot of effort to solve using a traditional "math approach." 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 Brent@GMATPrepNow » Thu Jan 29, 2015 11:16 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,
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by MartyMurray » Thu Jan 29, 2015 12:28 pm
A boat traveled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then traveled 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
If you want to do it algebraically, here's one way.

Let v be the velocity of the boat on it's own. v - 3 is the speed upstream. v + 3 is the speed downstream.

.5 + (90/(v + 3)) = 90/(v - 3) This is the description of the situation.

.5(v^2 - 9) + 90(v - 3) = 90(v + 3) Multiply to make all denominators 1.

.5v^2 - 4.5 + 90v - 270 = 90v + 270 Multiply.

.5v^2 - 274.5 = 270 Add and subtract like terms.

v^2 - 549 = 540 Multiply by 2.

v^2 - 1089 = 0 Put everything on one side.

(v + 33)(v - 33) = 0 Factor. This may seem a little difficult, but it's the GMAT and some integer is probably going to work as the square root of 1089. In this case it's a number between 30 and 40 that ends in 3.

v = 33 or -33

Only 33 can be a velocity.

V + 3 = 36. 90/36 = 2.5 hours to go downstream.

Choose A.

Might be better off plugging in. Depends I guess.
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by GMATGuruNY » Fri Jan 30, 2015 8:15 am
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 VivianKerr » Sat Jan 31, 2015 3:46 pm
Agreed, backsolving is definitely the way to go here!

Plugging In (also known as Backsolving) is a MAJOR asset in many Problem Solving questions, especially helpful in difficult Word Problems. When we "plug in," we essentially avoid doing algebra by testing out the 5 answer choices AS IF THEY ARE CORRECT, and seeing if we arrive at the original numbers given in the question-stem. If we do, then we must have landed on the correct answer choice! If an answer choice does NOT make sense with the givens from question-stem, it is obviously incorrect and can be eliminated.

Here's two great pages to see it explained:

https://www.800score.com/content/gre/guidec5b3.html

https://magoosh.com/gmat/2013/backsolving-on-gmat-math/

Backsolving teaches us to BE FLEXIBLE and USE THE ANSWER CHOICES. Here's a tough question we can solve by playing with the answer choices:

If j and k are positive integers, j - 2 is divisible by 4 and k - 5 is divisible by 4, all of the following could be the value of j - k EXCEPT:

A) 43
B) 33
C) 21
D) 13
E) 5

--

(j - 2) / 4

(k - 5) / 4

Let's say j = 6 and k = 9. The difference will be a multiple of 3. ELIMINATE B and C.

Now let's keep k = 9 and just keep adding 4 to j, seeing how the difference changes.

If j = 14 and k = 9, the difference is 5. ELIMINATE E.

If j = 22 and k = 9, the difference is 13. ELIMINATE D.

The answer is (A).

I also attached some extra Work/Rate problems if you need more practice with these concepts!
Attachments
30 Rates and Work questions.doc
(82.5 KiB) Downloaded 87 times
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