absolute value in an equation

This topic has expert replies
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Thu Mar 18, 2010 3:23 pm

absolute value in an equation

by Kogan » Wed Apr 21, 2010 10:28 am
Hello, the question is this:

t,q,s are different numbers. is t<s<q ?

1) q-t = |q-s|+|s-t|
2) q-t>0

obviously it's not B, and also not C (because you can see from statement 1 that q>t, no need for statement 2).
I don't know how to approach the equality given in statement 1. Any ideas?

(from all examples I took for statement 1, the answer seems to be "yes", but I couldn't find an algebraic way.)
Last edited by Kogan on Thu Apr 22, 2010 5:27 am, edited 1 time in total.

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Wed Apr 21, 2010 12:35 pm
Kogan wrote:Hello, the question is this:

t,q,s are different numbers. is t<s<q ?

1) q-t = |q-s|+|s-t|
2) q-t>0
As you noted from (1), we can quickly determine that q > t, since the sum of the terms on the right is certainly going to be positive.

Since q, s and t are different numbers, we know that neither term on the right is 0; so, we can think about (1) as:

"The distance between q and t is equal to the sum of the distances between q and s and between s and t."

Since the term on the left side is equal to the sum of the two terms on the right side, we can say:

"The distance between q and t is greater than either of the other two distances."

Well, the only way the distance between q and t could be the greatest is if q and t are furthest apart on the number line. If q and t have to be the furthest apart, then s must be in the middle.

Accordingly, from (1) we can deduce that q>s>t... sufficient.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

Junior | Next Rank: 30 Posts
Posts: 26
Joined: Mon Nov 17, 2008 9:07 pm
Thanked: 3 times

by el_torero » Wed Apr 21, 2010 12:49 pm
Wow this is a frickin' hard one. Let me roll up my sleeves...

Here's how I attacked the problem:

Let's start with (2) first because it's the easiest path to take:
(2) q - t > 0 means of course, q > t. Great. But how does this relate to s? Who knows. INSUFFICIENT.


Let's look at (1):
q - t = |q - s| + |s - t|

First thing to notice is that q - t MUST be > 0. Why? Notice we know q, s, t are DIFFERENT numbers, so q - t can't be 0 (FACT 1). This is key! Furthermore, since the right hand side has two absolute values, in conjunction with FACT 1, it must be > 0. Therefore, q > t.

But where does s lie in relation to q and t? We want to show if s lies in between q and t, which will satisfy the answer. I drew it on a number line to visualize:

(My really bad improvisational number line)
+ <-------------------------------------------------------------------------------------- > -
q ------------------- s1? ---------------------- t ------------------------ s2?

I drew two examples of s: s1 that is between q and t, and s2, which is < t. (Drawing an s3 bigger than q is redundant and you'll see why).

Notice that the distance between q and s1 is precisely |q - s|. Great. Furthermore, the distance between s1 and t is precisely |s - t|.

Add those two distances together and you'll get q - t! Awesome! According to 1, then s MUST like between q and t and furthermore we've proved that q > s > t.

But what about s2? Well, if s2, lies where it is, then |q - s2| + |s2 - t| would be much greater than q - t, and therefore, contradicts what (1) says. You can visualize this on the bad number line. Notice that the same argument is true if we created an s3 that is > q, i.e. s3 is also impossible.

[spoiler]ANSWER: A[/spoiler]

Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Thu Mar 18, 2010 3:23 pm

by Kogan » Thu Apr 22, 2010 5:30 am
Thank guys, both explanations are quite similar, and both are very good :) !