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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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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

Originally posted by rishi02 on 16 Jun 2016, 06:09.
Last edited by rishi02 on 17 Jun 2016, 21:57, edited 2 times in total.
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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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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

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.
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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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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





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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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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
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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
Thanks. can you please explain this in context of the problem?

ScottTargetTestPrep wrote:
Bunuel wrote:
On the number line, point R has coordinate r and point T has coordinate t. Is t < 0?

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.
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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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gzimmer wrote:
Bunuel wrote:
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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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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Bunuel 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


\(t\,\,\mathop < \limits^? \,\,0\)

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

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

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

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

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


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

Regards,
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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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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

Bunuel 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
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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
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Bunuel 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


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Answer: Option C

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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
Let's assume r and t on below line,
------t---r---t------
is t<0?

1. -1<r<0, no info on t. (Insufficient)
2. distant b/w R and T is \(r^2\). This doesn't tell where t is. it could +ve or -ve. (Insufficient)

Combining 1 & 2, Since r ranges b/w -1 and 0, not inclusive, making square of any values b/w it would less than the original value. TF t can never be >0. (Sufficient)
Ans C.
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On the number line, point R has coordinate r and point T has coordinat [#permalink]
Hi KarishmaB,

Can you please share your approach to this question? I figured that both statements are individually insufficient, but want to be clear on how we are getting sufficient using both statements.

Thanks in Advance. :)
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On the number line, point R has coordinate r and point T has coordinat [#permalink]
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Bunuel 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


This is what I am thinking: "Is t < 0?" means we want to find out whether point T is to the left of 0

(1) -1 < r < 0
No clue where T is located. Could be to the left or to the right of 0. Insufficient.

(2) The distance between R and T is equal to r^2
Don't know where R is and where T is. Both could be to the left of 0 or both could be to the right of 0 and hence insufficient.

Both together, I imagine a number line. r lies between 0 and -1, say r = -1/2. Then r^2 = 1/4


______________ -1 ______________ -1/2______________ 0 _____________________________

Point T is at a distance 1/4 away from -1/2. Whether it is to the left or to the right of -1/2, it will be to the left of 0 only because magnitude of 1/4 is less than magnitude of -1/2. Then the number property comes to mind -
For all numbers between 0 and -1, the magnitude of the square obtained is smaller. So -1/3's square will be 1/9 which has magnitude less than 1/3 and so on. Then the placement of R and T will look something like this:

______________ -1 _______(T)_______ R_______(T)_______ 0 ______________________________
or
______________ -1 ___________________(T)___ R___(T)___ 0 ______________________________
or
_______(T)____ -1 ____ R_____________(T)______________ 0 ______________________________

Hope you see that in every case, T will be to the left of 0. Sufficient.

Answer (C)

Check this video on using Number Line to solve questions: https://youtu.be/3gxVx3Y9xJA
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Re: On the number line, point R has coordinate r and point T has coordinat [#permalink]
KarishmaB wrote:
Bunuel 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


This is what I am thinking: "Is t < 0?" means we want to find out whether point T is to the left of 0

(1) -1 < r < 0
No clue where T is located. Could be to the left or to the right of 0. Insufficient.

(2) The distance between R and T is equal to r^2
Don't know where R is and where T is. Both could be to the left of 0 or both could be to the right of 0 and hence insufficient.

Both together, I imagine a number line. r lies between 0 and -1, say r = -1/2. Then r^2 = 1/4


______________ -1 ______________ -1/2______________ 0 _____________________________

Point T is at a distance 1/4 away from -1/2. Whether it is to the left or to the right of -1/2, it will be to the left of 0 only because magnitude of 1/4 is less than magnitude of -1/2. Then the number property comes to mind -
For all numbers between 0 and -1, the magnitude of the square obtained is smaller. So -1/3's square will be 1/9 which has magnitude less than 1/3 and so on. Then the placement of R and T will look something like this:

______________ -1 _______(T)_______ R_______(T)_______ 0 ______________________________
or
______________ -1 ___________________(T)___ R___(T)___ 0 ______________________________
or
_______(T)____ -1 ____ R_____________(T)______________ 0 ______________________________

Hope you see that in every case, T will be to the left of 0. Sufficient.

Answer (C)

Check this video on using Number Line to solve questions: https://youtu.be/3gxVx3Y9xJA




Thankyou so much KarishmaB for such a clear explanation, understood the concept! :D
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