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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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20 Apr 2016, 19:20
11
2
St1: t^2 can take values between 0 and 4 --> -2 < t < 2 When u and v have the same sign: v > u > 2 or v < u < -2 --> t does not lie between u and v When u and v have oppositie sign: u > 2 and v < -u < -2 or u < -2 and v > -u > 2 --> t lies between u and v
Illustrating the above statements: When t = 1; u = 3; v = 4, t does not lie between u and v When t = 1; u = -3; v = 4, t lies between u and v Not Sufficient.
St2: u is -ve and v is positive --> Clearly insufficient as we have no information about t
Combining St1 and St2: -2 < t < 2; u is negative --> u < -2 v is positive --> v > -u > 2 As shown in St1, when u and v have opposite signs, t lies between u and v. Sufficient
Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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20 Apr 2016, 21:37
abhisheknandy08 wrote:
Hi Bunuel , Is this really a GMAT PREP Question .
Hi, the Q is a bit twisted but tests number properties.. and since it has been shown that the source is GMAT PREP EP2, we should be ready for such Qs
_________________
Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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20 Apr 2016, 23:51
8
15
Three points T, U and V on the number line have coordinates t, u, and v respectively. Is T between points U and V?
(1) \(t^2<4<u^2<v^2\). Since all parts of the inequality are non-negative, we can safely take the square root from it: \(|t| < 2 < |u| < |v|\). It's possible that T is between U and V, for example, if t=0, u=-3 and v=4, as well as it's possible that it's not, for example, t=0, u=3 and v=4. Not sufficient.
(2) \(u<0<v\). Clearly insufficient.
(1) Since from (2) u<0<v, then from (1) we get \(|t| < 2 < -u < v\). If break it: we'll get \(u<-2\) (from 2 < -u), \(-2<t<2\) (from |t| < 2), and \(2<v\). T is between U and V. Sufficient.
Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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01 Jun 2017, 04:24
Bunuel wrote:
Three points T, U and V on the number line have coordinates t, u, and v respectively. Is T between points U and V?
(1) \(t^2<4<u^2<v^2\). Since all parts of the inequality are non-negative, we can safely take the square root from it: \(|t| < 2 < |u| < |v|\). It's possible that T is between U and V, for example, if t=0, u=-3 and v=4, as well as it's possible that it's not, for example, t=0, u=3 and v=4. Not sufficient.
(2) u<0<v. Clearly insufficient.
(1) Since from (2) u<0<v, then from (1) we get \(|t| < 2 < -u < v\). If break it: we'll get \(u<-2\) (from 2 < -u), \(-2<t<2\) (from |t| < 2), and \(2<v\). T is between U and V. Sufficient.
Answer: C.
Hope it's clear.
When you take the sq rt of 4 in this, shouldn't the result be +/- 2?
Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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01 Jun 2017, 06:24
2
ailintan wrote:
Bunuel wrote:
Three points T, U and V on the number line have coordinates t, u, and v respectively. Is T between points U and V?
(1) \(t^2<4<u^2<v^2\). Since all parts of the inequality are non-negative, we can safely take the square root from it: \(|t| < 2 < |u| < |v|\). It's possible that T is between U and V, for example, if t=0, u=-3 and v=4, as well as it's possible that it's not, for example, t=0, u=3 and v=4. Not sufficient.
(2) u<0<v. Clearly insufficient.
(1) Since from (2) u<0<v, then from (1) we get \(|t| < 2 < -u < v\). If break it: we'll get \(u<-2\) (from 2 < -u), \(-2<t<2\) (from |t| < 2), and \(2<v\). T is between U and V. Sufficient.
Answer: C.
Hope it's clear.
When you take the sq rt of 4 in this, shouldn't the result be +/- 2?
Are you talking about t^2 < 4? What is your question?
t^2 < 4
|t| < 2 (which is the same as -2 < t < 2).
_________________
Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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10 Aug 2017, 00:14
Bunuel wrote:
ailintan wrote:
Bunuel wrote:
Three points T, U and V on the number line have coordinates t, u, and v respectively. Is T between points U and V?
(1) \(t^2<4<u^2<v^2\). Since all parts of the inequality are non-negative, we can safely take the square root from it: \(|t| < 2 < |u| < |v|\). It's possible that T is between U and V, for example, if t=0, u=-3 and v=4, as well as it's possible that it's not, for example, t=0, u=3 and v=4. Not sufficient.
(2) u<0<v. Clearly insufficient.
(1) Since from (2) u<0<v, then from (1) we get \(|t| < 2 < -u < v\). If break it: we'll get \(u<-2\) (from 2 < -u), \(-2<t<2\) (from |t| < 2), and \(2<v\). T is between U and V. Sufficient.
Answer: C.
Hope it's clear.
When you take the sq rt of 4 in this, shouldn't the result be +/- 2?
Are you talking about t^2 < 4? What is your question?
t^2 < 4
|t| < 2 (which is the same as -2 < t < 2).
Yeah, It is a medium level question. I bit of guess/application work is needed to solve this question. Mentioned question can be solved only if we were given both information A and B.
Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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18 Mar 2018, 07:17
kritikalal wrote:
How do I check the difficult level in terms of 660-650-700-750 level? It just shows easy/ medium on the timer :/
We have three grades of difficulty: easy, which corresponds to sub-600; medium, which corresponds to 600-700; and hard, which corresponds to 700+. You can check difficulty level of a question along with the stats on it in the first post. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.
_________________
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71% (00:45) correct 
