On the number line, point R has coordinate r and point T has coordinat

We are given that on the number line, point R has coordinate r and point T has coordinate t, and we need to determine whether t < 0.

Statement One Alone:

-1 < r < 0

Statement one tells us that r is a negative proper fraction. However, without knowing anything about t, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

The distance between R and T is equal to r^2.

The distance between two values on the number line is the absolute value of the difference between the two values. Thus statement two gives us the equation |r - t| = r^2. However, without knowing anything about r and t, we can’t determine whether t is less than zero. For instance, r could be 2 and t could be -2; or r could be -2 and t could be 2. In each of the cases, |r - t| = 4 = r^2; but in one case t > 0 and in the other t < 0. Statement two is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using statements one and two, we know that r is a negative proper fraction and |r - t| = r^2. Thus, r - t = r^2 OR r - t = -r^2. Solving each of these for t, we get: t = r - r^2 OR t = r + r^2.

Since r is a negative proper fraction, no matter what the value of r is, t will always be a negative number. For instance, if r = -1/2, then r^2 = 1/4 and t will either be -1/2 - 1/4 = -3/4 or -1/2 + 1/4 = -1/4.

The reason why t cannot be positive is that when we square r (a negative proper fraction), the value of r^2 (though positive) will be less than the absolute value of r. Recall that t = r - r^2 or t = r + r^2. When a positive proper fraction with a smaller absolute value is added to (or subtracted from) a negative proper fraction with a larger absolute value, the sum (or difference) will always be less than zero.

Answer: C
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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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On the number line, point R has coordinate r and point T has coordinate t. Is t < 0?

(1) -1 < r < 0: Tells us r is a (negative) proper fraction
Does not tell us anything about the coordinates of t. Hence INSUFFICIENT

(2) The distance between R and T is equal to r^2:

If r is +ve and >1 t can be positive or negative
If r is +ve and <1 t will be positive since square of a proper fraction is less than the fraction itself
If r is -ve and <-1 t can be positive or negative
If r is -ve and >-1 t will be negative since square of a proper fraction is less than the fraction itself

Multiple cases hence INSUFFICIENT

Combining, it conforms to case 4.

Hence Ans = C

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 r^2

r can take values from -0.99 to -.01 approx
if r = -0.99 then distance between r an t is r^2 = .98 the absolute value is the distance which can be both right side and left side. But in whichever case it will less than 0. Hence C is the answer.

1) -1 < r < 0

This statement by itself is insufficient as it does not establish any relationship with t.

2) the distance between r and the t is = r^2

| t-r | = r^2

When t - r >=0 --> t >= r

t - r = r^
t = r^2 + r
t = r (r+1) ---------- eq 1

When t - r < 0 --> t < r

- t + r = r^2
t = r (1-r) ------------ eq 2

Insufficient

Taking 1 and 2 together

eq1 --> t = negative for the given value of r.

And as per the condition of the case t >= r

eq2 --> t = negative for the given value of r

And as per the condition of the case t < r

Either ways t < 0 in negative.
Sufficient

C is the correct answer





Sent from my SM-G935F using GMAT Club Forum mobile app

Very nicely done. Thanks!

Question: Is t negative?

St1: -1 < r < 0 --> Clearly insufficient

St2: \(r - t \geq {0}\)
Possible solutions: 1) +ve - (+ve)
2) -ve - (-ve)
3) +ve - (-ve)
Not Sufficient

Combining St1 and St2: We know that r is negative --> Only possible solution is -ve - (-ve) --> t < 0.
Sufficient.

Answer: C

Thanks. can you please explain this in context of the problem?

[quote="Bunuel"]On the number line, point R has coordinate r and point T has coordinate t. Is t < 0?

Can someone help? I keep on seeing this explained in questions, but I am not getting it ("Point R has coordinate r"). Can someone give me a break down of what this is saying. EX: on an xy plane point Q could be at (1,0). It's x coordinate is 1, and it's Y coordinate is 0....

what is ("Point R has coordinate r") saying? can you give me an example if it were in number format?

It means that point R is at r on the number line. For example if r = -3, then we'd have the below case:

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so the key is that squaring a fraction always results in a smaller number... right ?
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\(t\,\,\mathop < \limits^? \,\,0\)

\(\left( 1 \right)\,\,\, - 1 < r < 0\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {r,t} \right) = \left( { - 0.5,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,\left( {r,t} \right) = \left( { - 0.5, - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,\left| {r - t} \right| = {r^2}\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {r,t} \right) = \left( {0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,\left( {r,t} \right) = \left( { - 1, - 2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,\,\left| {r - t} \right| = {r^2}\,\,\,\,\mathop \Rightarrow \limits^{{\text{squaring}}} \,\,\,\,\,{\left( {r - t} \right)^2} = {r^4}\,\,\,\,\, \Rightarrow \,\,\,\,{r^2} - 2rt + {t^2} = {r^4}\,\,\,\,\,\left( * \right)\)

\(- 1 < r < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,{r^4} < {r^2}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\, - 2rt + {t^2} = {r^4} - {r^2} < 0\)

\(\left. \begin{gathered}\\
- 2rt + {t^2} < 0 \hfill \\\\
{t^2} \geqslant 0 \hfill \\ \\
\end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\, - 2rt < 0\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{r\, < \,\,0} \,\,\,t < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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(1) -1 < r < 0
no info about t.
NOT SUFFICIENT

(2) The distance between R and T is equal to r^2
No info about the r .
NOT SUFFICIENT

Combining Statement 1 & 2
the coordinate of T : t
is either \(r+r^2\) or \(r-r^2\)

Now, when -1 < r < 0, both \(r+r^2\) or \(r-r^2\) will be <0
Hence t <0
SUFFICIENT

Answer C


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This is sincerely one of the most detailed explanations, for any question covered so far, by anyone.

Thank you!

Video solution from Quant Reasoning:
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_________________

Answer: Option C

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ScottTargetTestPrep

This is very helpful, thank you I have a follow-up question. I am a bit thrown off by the coordinates given in the problem (coordinate r for point R and coordinate t for point T). For number lines is there always only one coordinate for each point (e.g. r for point R) versus two coordinates for a point on a plane (e.g. x,y)? Thank you in advance for any help that you can provide. You rock!

So in that case, doesn't it mean that is -1<r<0, r=0?

Because -1<R<0

Hi,
Not always.
Be careful about the 0 and 1 and also the 0 and -1 regions.

Concept:
> from 0 to 1 if a number is multiplied to itself any number of times (e.g. squared, cubed ) the number decreases
> from -1 to 0 if the number is multiplied even number of times(squared) to itself the value increases as compared to the original number(which is -ve) because the -ve * -ve cancels out and results in a +ve. Though observe that the absolute value of the original number is greater than the squared result. In this question this fact gets exploited.

Say r=-0.5
1---------r---------0
The distance between r and 0 = 0.5 (the absolute value)
T is \(r^2\) away from r

-1---t---------r---------t------0
.......|--0.25-||--0.25--|
...................|-------0.5------|
\(r^2\) = 0.25
so T cannot lie beyond 0 as the distance between R and 0 = 0.5 and R and T = 0.25 (<0.5)
Hence T is also -ve

If \(r^2 > |r| \)then T could be beyond 0 and be either positive or negative. One can be tempted to think that in the -1 to 0 region\( r^2 > r\) but here we take the distance so must only see the absolute value .