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

This topic has 2 expert replies and 0 member replies

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

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

OA C

Source: Official Guide

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BTGmoderatorDC 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 r^2

OA C

Source: Official Guide
Given: On the number line, point R has coordinate r and point T has coordinate t.

Question: Is t < 0?

Let's take each statement one by one.

(1) -1 < r < 0

We do not have any information about point T. Insufficient.

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

Certainly insufficient. T can be on either side of the number line. Insufficient.

(1) and (2) together

We know that the coordinate of point R is -1 < r < 0 and the distance between R and T is equal to r^2.

Note that if -1 < r < 0, r^2 < |r|

Thus, the coordinate of point T must also be negative. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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Manhattan Review GRE Prep

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BTGmoderatorDC 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 r^2

Source: Official Guide
\[t\,\,\mathop < \limits^? \,\,0\]
\[\left( 1 \right)\,\,\, - 1 < r < 0\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {r,t} \right) = \left( { - 0.5,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,\left( {r,t} \right) = \left( { - 0.5, - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,\,\left| {r - t} \right| = {r^2}\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {r,t} \right) = \left( {0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,\left( {r,t} \right) = \left( { - 1, - 2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\,\,\left| {r - t} \right| = {r^2}\,\,\,\,\mathop \Rightarrow \limits^{{\text{squaring}}} \,\,\,\,\,{\left( {r - t} \right)^2} = {r^4}\,\,\,\,\, \Rightarrow \,\,\,\,{r^2} - 2rt + {t^2} = {r^4}\,\,\,\,\,\left( * \right)\]
\[ - 1 < r < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,{r^4} < {r^2}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\, - 2rt + {t^2} = {r^4} - {r^2} < 0\]
\[\left. \begin{gathered}
- 2rt + {t^2} < 0 \hfill \\
{t^2} \geqslant 0 \hfill \\
\end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\, - 2rt < 0\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{r\, < \,\,0} \,\,\,t < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\,\]


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

Regards,
Fabio.

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