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On the number line, point R has coordinate r

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jjjinapinch Senior | Next Rank: 100 Posts Default Avatar
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On the number line, point R has coordinate r

Post Mon Aug 07, 2017 9:12 am
On the number line, point R has coordinate r and point T has coordinate t. Is t < 0?

(1) -1 < r < 0
(2) The distance between R and T is equal to r²

Official Guide question
Answer: C

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Post Sat Nov 04, 2017 9:17 am
jjjinapinch wrote:
On the number line, point R has coordinate r and point T has coordinate t. Is t < 0?

(1) -1 < r < 0
(2) The distance between R and T is equal to r²

Official Guide question
Answer: C
Target question: Is t NEGATIVE?

Statement 1: -1 < r < 0
No information about t
So, statement 1 is NOT SUFFICIENT

Statement 2: The distance between R and T is equal to r²
There are several values of r and t that satisfy statement 2. Here are two:
Case a: r = -1 and t = -2. The distance between r and t is 1 (aka r²). So, these values of r and t satisfy statement 2. In this case, t IS negative
Case b: r = -1 and t = 0. The distance between r and t is 1 (aka r²). So, these values of r and t satisfy statement 2. In this case, t is NOT negative
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that -1 < r < 0
ASIDE: If j and k are on the number line, then |j - k| = the distance between j and k
So, from statement 2, we can write: |t - r| = r²

-----------------ASIDE-------------------------------------
There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots
--------BACK TO THE QUESTION---------------------------

Since |t - r| = r², we'll examine two possible cases:
t - r = r² and t - r = -(r²)

case a: t - r = r²
Rearrange to get: t = r + r²
Factor: t = r(1 + r)
Since -1 < r < 0, we can conclude that (1 + r) is POSITIVE
So, t = r(1 + r) = (NEGATIVE)(POSITIVE) = NEGATIVE
So, t is negative

case b: t - r = -(r²)
Rearrange to get: t = r - r²
Factor: t = r(1 - r)
Since -1 < r < 0, we can conclude that (1 - r) is POSITIVE
So, t = r(1 - r) = (NEGATIVE)(POSITIVE) = NEGATIVE
So, t is negative

In both of the two possible cases, t is negative
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

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Top Reply
Post Mon Aug 07, 2017 3:58 pm
Hi jjjinapinch,

We're told that point R and point T are on a Number Line (as r and t, respectively). We're asked if t < 0. This is a YES/NO question. This question can be solved by TESTing VALUES.

1) -1 < r < 0

Fact 1 gives us a range of values for r, but tells us NOTHING about t.
Fact 1 is INSUFFICIENT

2) The distance between R and T is equal to r²

IF.... r=1 and t=2, then the distance between them is 1^2 and the answer to the question is NO.
IF.... r=2 and t = -2, then the distance between them is 2^2 and the answer to the question is YES.
Fact 2 is INSUFFICIENT

Combined, we know that -1 < r < 0 ... so.... 0 < r^2 < 1.
Since we know that r is a NEGATIVE FRACTION, we can deduce that |r| > (r)^2. Since r is negative, this means that the distance between r and t will ALWAYS limit t to the 'negative side' of the Number Line (in basic terms, t can never get 'far enough away' from r to equal 0 or a positive). Thus, the answer to the question is ALWAYS YES.

For example, if r = -1/2, then r^2 = +1/4.... so t would have to be either -3/4 or -1/4... and BOTH of those results yields a YES answer.
Combined, SUFFICIENT

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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