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mbaspire
Joined: 23 Sep 2015
Last visit: 16 Dec 2016
Posts: 12
Given Kudos: 1
Location: United States
Concentration: Healthcare, General Management
Schools: Booth '19 (D) CBS '19 (D) Stern '19 (S)
GMAT 1: 740 Q50 V40
GPA: 3.34
WE:Operations (Healthcare/Pharmaceuticals)
20 Apr 2016, 16:32
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (01:54) correct
26%
(01:56)
wrong
based on 1194
sessions
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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) \(t^2<4<u^2<v^2\)
(2) \(u<0<v\)
(1) \(t^2<4<u^2<v^2\)
(2) \(u<0<v\)
20 Apr 2016, 23:51
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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.
(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.
20 Apr 2016, 19:20
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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
Answer: C
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
Answer: C
