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

Question

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

rishi02 · Answer

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

FightToSurvive · Answer

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.

LightYagami · 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

Kurtosis · Answer

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

ScottTargetTestPrep · Answer

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

cuhmoon · Answer

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

Bunuel · Answer

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

BrentGMATPrepNow · Answer

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

fskilnik · Answer

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

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

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

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

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

\left. \begin{gathered}
 - 2rt + {t^2} < 0 \hfill \
 {t^2} \geqslant 0 \hfill \ 
\end{gathered} ight\}\,\,\,\,\,\, \Rightarrow \,\,\,\, - 2rt < 0\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{r\, < \,\,0} \,\,\,t < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{	ext{YES}}} ightangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{	ext{SUFF}}.\,

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

Regards,
Fabio.

GMATBusters · Answer

(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

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=lxJ5ERVp5xg

GMATinsight · Answer

Answer: Option C

Video solution by  GMATinsight

https://www.youtube.com/watch?v=EaUrTpBJVVc

Nrj0601 · Answer

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.

ChandlerBong · Answer

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.

KarishmaB · Answer

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

ChandlerBong · Answer

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

SirSanguine · Answer

Here's an alternate solution without maths.

Its clear that s-1 is not enough (doesnt even mentions t)

Combing the two : -1 <r <0...Understand that :r^2 will be lesser in magnitude than r. now no matter how small or big r^2 is on adding with r (to get t) we wont be able to cross 0..ie the smaller is r even smaller is r^2 and the larger is r still r^2 is not larger enough (<1) to help you cross 0.

Great Question!!

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

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