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GMAC Number Properties Problem

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Source: — Data Sufficiency |
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by mehravikas » Thu Oct 15, 2009 4:47 pm
answer should be 'C'

Statement 1 - r and m could be anything
Statement 2 - possible values can be

A. 6 - 12 - 18
B. 0 - 12 - 24
C. -6 - 12 - 30

Now if we use statement 1, i.e. distance between r and 0 is 3 times the distance between m and 0.

Therefore, only A is able to satisfy statement 1.
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by life is a test » Thu Oct 15, 2009 10:02 pm
IMO c
--------------0------m-----------r
stmt1: <- x -><---2x----->
stmt 2: x=12 hence r =36

we know m and r on the positive side of the line from statement 2 which says midpoint is a +ve number.

is E the OA?
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by DCJ » Sun Oct 18, 2009 8:44 am
OA is E. I think it has to do with distance being an absolute value which doesn't indicate the signs of the two numbers.
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by mehravikas » Sun Oct 18, 2009 11:33 am
I doubt...
DCJ wrote:OA is E. I think it has to do with distance being an absolute value which doesn't indicate the signs of the two numbers.
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by DCJ » Sun Oct 18, 2009 8:43 pm
The OA is E. This is from an official GMAT Practice test. Anyway I figured out that the OA is E because combining the two statements allows you to determine a pair of values that make the statements true but you still don't know which value belongs to which variable. If there was additional info that one variable was less than the other then C would be the correct answer but we don't know which is the lesser and which is the greater value and we cannot assume that we are giving the variables in ascending order.
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by Talkativetree » Sun Oct 18, 2009 9:15 pm
If m and r are two numbers on a number line what is the value of r?

1. The distance between r and 0 is 3 times the distance between m and 0
2. 12 is halfway between m and r


The answer is E because r could be negative.

(1) |0-r|=|3(0-m)|

R could be -3 and m could be 1. OR r=9 and m=3

(2) Now, if 12 is halfway between m and r.

now, I would draw out two instances, but also pay attention to how many -'s we have.

123 4
r---0-m
|--3x--|-x-|

or
1 23
0-m--r
|-x-|-2x-|
|---3x---|

this shows that (1) is insufficient.

now (2) is insufficient on sight (or should be), but let's look at the problem together

123 4
r---0-m
|-3x-|x|

or
1 23
0-m--r
|x|2x|
|-3x-|

so what I did was count each dash as a unit, seeing that "x=2" theoretically (this is much easier to explain drawn out and in person). x is not actually a number, more like apart of a proportion.

so we know that 12 is halfway in between m and r, so lets place it into the dashes.

0-m--r
|x|2x|
0-m-12-r

So here we have 3 -'s. From this we see that m=1/3r but also, that 12= 2/3r, so r=18, and m=6

BUT!

r------0-12-m
|--3x--|-x-|

in this instance, we have 8 -'s (it was harder to "draw" with 4-'s so I made it 8) so x=2(12)=24.
r=3x
r=-72

So we see that r=18, or -72

together they are insufficient so the answer is E
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Re: GMAC Number Properties Problem

by Stuart@KaplanGMAT » Sun Oct 18, 2009 10:01 pm
DCJ wrote:If m and r are two numbers on a number line what is the value of r?

1. The distance between r and 0 is 3 times the distance between m and 0
2. 12 is halfway between m and r

Ans: E
Picking numbers is really the way to go on this question.

Assuming we've eliminated (1) and (2) separately, let's look at them in combination.

We know that r has to be further from 0 than does m and that 12 is in the middle of r and m.

Let's try making them both positive. We know that one number will be bigger than 12, 1 smaller. Trying all the integers smaller than 12:

m=1, r=23
m=2, r=22
m=3, r=21
m=4, r=20
m=5, r=19
m=6, r=18... bingo! r is 3 times as far from 0 as is m, satisfying statement (1)

(Of course in practice we're probably not going to actually start with m=1; we'll pick values that seems likely to work, maybe trying m=3 first and then m=6 after.)

Now let's try to make r negative and m positive. Well, with a bit of common sense, we can see that's impossible; if r and m must be the same distance from 12 and r is negative, m must be bigger than 12. With 12 as our "dividing line" between m and r, m will always be further from 0 than will be r, making it impossible to satisfy statement (1).

(As an aside, Talkativetree's solution of r = -72 doesn't work - we'd end up with m = 84 and -72 isn't 3 times as far from 0 as is 84.)

So, let's try r positive and m negative. Since our dividing line is still +12, this will give us an r that's further from 0 than is m, making it likely that we can satisfy statement (1).

The first value of r that generates a negative value of m is 25; but m=-1 and we haven't come close to satisfying statement (1). Let's try multiples of 12, since that's an important number in this question.

r = 36, m = -12

kaching! r and m are equidistant from 0 and |36| = 3|12|, making this pair legal.

So, r=18 and r=36 are both possible: choose (E).
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