GMAT Prep Exam Math Questions

[ddohnggo]

Hi,

I have a few (2) math questions from the GMAT Prep Exam, mainly figuring out faster/efficient ways to solve them:

1. 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

Q: I have seen an explanation to the problem that would take more than 2 minutes to complete. My question is what is an alternative way to figure out the answer within a reasonable amount of time or a method that could immediately narrow down the answer choices (I think it's hard since the answer choices do not deviate much from one another).
The answer choice I viewed:

Distance = Rate * Time, so...

90 = (v-3) * (t+1/2) - since it took half an hour longer to go upstream, and 90 = (v+3) * t.

Expand these out:

90 = vt - 3t + v/2 - 3/2
90 = vt + 3t

Subtract the first from the second:

0 = 6t - v/2 + 3/2

Solve for t in terms of v:

6t = v/2 - 3/2
t = v/12 - 1/4

Substitute into the second equation so we can solve for t:

90 = vt+3t = v(v/12-1/4)+3(v/12-1/4)
= v^2/12-v/4+v/4-3/4
= v^2/12 - 3/4

v = sqrt((90+3/4)*12)
= sqrt(363/4*12)
= sqrt(1089) = 33

And since t = v/12-1/4:
t = 33/12-1/4 = 33/12-3/12 = 30/12 = 5/2. It took 2 1/2 hours downstream and 3 hours upstream.

2. The number 75 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers?
A: 17
B: 16
C: 15
D: 14
E: 13

Q: Is there a quick and easy way to solve this problem? My first intuition was to break 75 into its primes, but realized that did not help. After reviewing the question I wrote out all numbers from 1-8 and their squares and eye balled to see what squared values would sum to 75.

[Guest]

For the first one i think thats the only method i know off

For the second one 1^2 + 5^2 + 7^2 = 75 Hence 13

[StaceyKoprince]

As a general rule, please split different problems into separate threads. I'll go ahead and answer both of these now, but please do this in future. Thanks!

I'll give the intuitive way first and the algebraic way second. Totally up to you which you prefer.

#1
The distance is the same for both trips. The difference in speed for the two trips is 6mph (that is, the boat goes 6mph faster downstream than upstream). So, for every hour that the boat travels downstream, it travels 6 miles more in that time than it does during an hour spent traveling upstream. Imagine two boats doing the downstream and upstream portions simultaneously. At the one hour mark, the downstream boat is 6 miles ahead. At the two hour mark, the downstream boat is 12 miles ahead. And so on.

A 6mph difference over 90 miles translates to a half hour difference in travel time (given by the problem). So the upstream boat needs half an hour extra to make up the distance it loses by going 6mph slower.

Your choices are all between 2.1 and 2.5 hours, so let's start at 2 hours to make this a little easier. Assume that the downstream boat takes 2 hours, which means it is 12 miles ahead of the upstream boat. Also, calculate the speed: to go 90 miles in 2 hours, the boat must be traveling 45mph. If the upstream boat is 6mph slower, it must be traveling 39mph. In half an hour, that boat would travel 19.5 miles. So that upstream boat must be going significantly slower, because it only has 12 miles to make up in that half hour, not 19.5. 2 hours is obviously not the right answer, because it isn't there, but neither are the numbers very close to 2, because 12 and 19.5 are pretty far apart.

So swing to the other end of the spectrum: try 2.5. If the downstream boat takes 2.5 hours, it will be 15 miles ahead of the upstream boat and its speed will be 90/2.5 = 180/5 = 36 mph. The upstream boat, then, must be traveling 30mph and, in half an hour, would travel 15 miles. Hey! That's exactly the distance we said it would be behind! That's the right answer.

Or, if you want to do something more like the above algebraic solution, there are a lot of things you can do to make it easier. Start by replacing (v-3) and (v+3) with r = rate for upstream trip and r+6 = rate for downstream trip. That gives you:
90 = r(t+0.5)
90 = (r+6)t
and these are much nicer to work with. Make sure to set the plain variable t for the downstream portion of the trip, since that's what the problem asks you to find.

90 = rt + 0.5r upstream
90 = rt + 6t downstream
subtract
0 = 0.5r - 6t

next, solve for r to substitute, not t, since we want to solve for t eventually
6t = 0.5r
12t = r

substitute into 2nd equation (downstream) b/c that one's easier
90 = 12t(t) + 6t
90 = 12t^2 + 6t
0 = 12t^2 + 6t - 90 simplify: divide by 6
0 = 2t^2 + t - 15
= (2t-5)(t+3)
t = either 2.5 or -3. And it can't equal a negative, so it must be 2.5.

[pratik.munjal]

Very surprised to see an "upstream/downstream" question in GMAT. Such questions are common in India's b school entrance exams.

Anyway, the most "viable" method would be the obvious one:

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

Now comes the time taking part:

Solve for v.

V comes to 33.

And since downstream's time has been asked,

Distance/downstream's speed

90/(33+3)=2.5

But yes, this is a time consuming problem. Perhaps, one could compensate by spending lesser time on a following DS question.

[jnelson0612]

Thanks all!

[angierch]

Hi Stacey,

How did you solve this?

= (2t-5)(t+3)
t = either 2.5 or -3

Thank you much,
Angelica

[RonPurewal]

Hi Stacey,

How did you solve this?

= (2t-5)(t+3)
t = either 2.5 or -3

Thank you much,
Angelica

The left-hand side of that equation is zero. Does that help?

[angierch]

yes. sorry, I got a little confused. dahhh for me...

[RonPurewal]

yes. sorry, I got a little confused. dahhh for me...

No worries.

If the left-hand side of the equation is "missing" (i.e., a new line just starts with an equals sign, rather than with a mathematical expression), the implication is that the left-hand side is the same as it was on the previous line.
(I don't think it's possible to do the same with the right-hand side; the right-hand side must be written every time.)

When I write out derivations like that, I tend to write both sides of the equation every time, even if one side isn't changing. Especially if the left side is as short as "0". But, yeah.

[angierch]

Hi Ron,

so sorry, I meant to ask the following:

How do you get from here:
0 = 2t^2 + t - 15

to here 0= (2t-5)(t+3)?

What I am confused about is the 2t^2. I am used to solve this type of equation without the constant next to the t^2 (two numbers that summed equal 1 and that multiplied equal -15)
Angelica

[georgepa]

To find the roots of an equation of the form ax^2 + bx + c = 0, find two factors for the product (a . c ) that sum to the coefficient of the middle term b. Note that you are basically doing this when the coefficient of the x^2 term is 1 (i.e a = 1).

Code: Select All Code

2t^2 + t -15 = 0
2(-15) = -30.
Find factors of 30 that sum to +1
+6 and -5 are the solutions since
+6 -5 = +1
+6 x -5 = -30
Now, rewrite the equation with the two factors as coefficients for the split middle term
2t^2 +6t -5t -15 =0
Factoring:
2t(t+3) - 5(t+3) = 0
Factor out the  (t+3)
(t+3)(2t-5) = 0


[RonPurewal]

Note that, on the upstream/downstream problem, you can avoid all this algebraic silliness by just backsolving"”i.e., starting from an answer choice and working through the given information.

2.5 hours is the clear choice to try first, since that's the only choice that makes "half an hour longer" into anything halfway decent (= 3 hours).

The given speeds are supposed to be "v + 3" and "v - 3" miles per hour. So, just divide 90 by the two times, and see whether you get two speeds that differ by 6 miles per hour.

* Downstream:
90 mi/2.5 hrs
= 180/5
= 36 miles per hour

* Upstream:
90 mi/3 hrs = 30 miles per hour

The difference is 6 miles per hour, so we're done. This is the correct answer.

[RonPurewal]

I'm going to close the thread at this point, since each thread is only supposed to contain one problem. If you have further questions about either of the problems in this thread, please search the forum for other threads on the same problems.

Thanks.