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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Problem Solving Pack 4, Question 1) If -2 < t < 0...[  [#permalink]

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QUANT 4-PACK SERIES Problem Solving Pack 4 Question 1 If -2 < t < 0...

If -2 < t < 0 and 0 < v < 2, then which of the following must be true?

I. tv - v < 0

II. $$t^{2}$$ – $$v^{2}$$ > 0

III. $$(t – v)^{2}$$< 4

A) I only
B) II only
C) III only
D) I and III
E) I, II and III

48 Hour Window Answer & Explanation Window
Earn KUDOS! Post your answer and explanation.
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

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# Rich Cohen

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Senior Manager  B Joined: 28 Feb 2014 Posts: 294 Location: United States Concentration: Strategy, General Management Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink] ### Show Tags If -2 < t < 0 and 0 < v < 2, then which of the following must be true? I. tv - v < 0 v(t-1)<0 this is always true as t-1 is always negative and v is always positive II. $$t^{2}$$ – $$v^{2}$$ > 0 if t=-1 and v=1 this statement is false III. $$(t – v)^{2}$$< 4 the square turns this equation positive. since t and v will always be less than 2 the statement will always be less than 4. Answer: D) I and III Manager  Joined: 17 Oct 2013 Posts: 50 Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink] ### Show Tags 2 peachfuzz wrote: If -2 < t < 0 and 0 < v < 2, then which of the following must be true? I. tv - v < 0 v(t-1)<0 this is always true as t-1 is always negative and v is always positive II. $$t^{2}$$ – $$v^{2}$$ > 0 if t=-1 and v=1 this statement is false III. $$(t – v)^{2}$$< 4 the square turns this equation positive. since t and v will always be less than 2 the statement will always be less than 4. Answer: D) I and III I think answer is A for III, what if t = -1.9 and v =1.9 we have -1.9 -1.9 = -3.8, now square of this > 4. Intern  Joined: 21 May 2015 Posts: 11 Concentration: Marketing, Nonprofit GMAT 1: 720 Q48 V41 WE: Analyst (Non-Profit and Government) Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink] ### Show Tags 1 (-2)-----(t)--------(0)----(1)-(v)------(2) I. v(t-1)<0: Always true. Because t<1, v>0. II. (t-v)(t+v)>0: * t-v < 0 always * t+v < 0 only when |t|>|v| because t<0. As we see in the number line this is not always the case. III. |t-v|<2: t could be further to -2 and v could be further to 2. Their maximum distance is 4, so nothing guarantees they would be at these respective positions for the inequality to hold. So only I is correct. CEO  S Joined: 20 Mar 2014 Posts: 2626 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink] ### Show Tags 3 EMPOWERgmatRichC wrote: QUANT 4-PACK SERIES Problem Solving Pack 4 Question 1 If -2 < t < 0... If -2 < t < 0 and 0 < v < 2, then which of the following must be true? I. tv - v < 0 II. $$t^{2}$$ – $$v^{2}$$ > 0 III. $$(t – v)^{2}$$< 4 A) I only B) II only C) III only D) I and III E) I, II and III 48 Hour Window Answer & Explanation Window Earn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4-Pack series Scroll Down For Official Explanation This is a type of question that can be solved either by taking certain values for t and v. As t<0 and v>0 --> v>t --> v-t>0 Now for a must be true question, ALL cases must satisfy for a TRUE!! i) $$tv-v^2<0$$ ---> v(t-1) < 0 and as v>0 , the only case possible for t is for t<1 and we have indeed been given that t<0 --> t<1 . Thus this is a must be true statement. Eliminate B and C. ii) $$t^2-v^2 > 0$$ ---> $$(t+v)(t-v) > 0$$ ---> t-v < 0 as v> 0 and t<0 (given). But is t+v <0 ? This may or may not be true. You can check it by taking the following values: t=-1.5 and v=0.5, t+v <0 but t=-1 and v=1.5, t+v >0 This makes this statement not always true. Eliminate E. iii) $$(t-v)^2>4$$ this statement can be negated by taking t=-1.5, v=1.5 --> $$(t-v)^2 = 3^2 = 9 > 4$$ making this statement false. Eliminate D A is the correct answer. EMPOWERgmat Instructor V Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14209 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink] ### Show Tags 2 1 EMPOWERgmatRichC wrote: QUANT 4-PACK SERIES Problem Solving Pack 4 Question 1 If -2 < t < 0... If -2 < t < 0 and 0 < v < 2, then which of the following must be true? I. tv - v < 0 II. $$t^{2}$$ – $$v^{2}$$ > 0 III. $$(t – v)^{2}$$< 4 A) I only B) II only C) III only D) I and III E) I, II and III Hi All, In Roman Numeral questions, the answer choices can often be used to approach the prompt in the most efficient way possible (and sometimes avoid doing unnecessary work). Given the ranges for t and v (-2 < t < 0 and 0 < v < 2), we're asked 'which of the following MUST be true?' This essentially means "which of the following is ALWAYS TRUE no matter how many different examples we can come up with. This prompt can be beaten by TESTing VALUES and/or by using Number Properties. From the Roman Numeral frequencies in the answer choices, it looks like 1 or 3 should be dealt with first. Roman Numeral 1 looks relatively easy, so let's start there. I. tv - v < 0 Since t is NEGATIVE and v is POSITIVE, tv = (negative)(positive) = NEGATIVE. Thus, we have... tv - v = ? negative - positive = negative The end result of this calculation will ALWAYS be NEGATIVE, so Roman Numeral 1 is TRUE. The correct answer MUST include Roman Numeral 1. Eliminate Answers B and C. From the remaining 3 answers, Roman Numeral 3 should be dealt with next (if we can eliminate it, then we'll have the final answer...). III. $$(t – v)^{2}$$< 4 For this Roman Numeral, we should look to DISPROVE the idea that the calculation will always be less than 4. From the given ranges, t can get 'really close' to -2 (think -1.99999) and v can get 'really close' to +2 (think +1.999999). $$(-1.9999999 - 1.999999)^{2}$$= about $$(-2 -2)^{2}$$= about $$(-4)^{2}$$ = about 16. This is clearly NOT less than 4, so Roman Numeral 3 is NOT always true. Eliminate Answers D and E. With only one answer remaining, there's no need to deal with Roman Numeral 2. Final Answer: For the sake of working through Roman Numeral 2 though, we should take the same general approach that we used with Roman Numeral 3: try to prove that it's NOT always true... II. $$t^{2}$$ – $$v^{2}$$ > 0 IF.... t = -1 v = +1 $$(-1)^{2}$$ - $$(1)^{2}$$ = 1 - 1= 0 This result is NOT greater than 0, so Roman Numeral 2 is NOT always true. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** # Rich Cohen Co-Founder & GMAT Assassin Follow Special Offer: Save$75 + GMAT Club Tests Free
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Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[  [#permalink]

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1
I. $$tv - v < 0$$
$$v(t-1)$$ < 0
as v is always positive,
$$(t-1) < 0$$
$$t < 1$$ ? yes always true

2. $$t^2 - v^2 > 0$$ ?
$$t^2 > v^2$$ ?
$$|t| > |v|$$ ?
need not be, if t = -0.5 and v = 1

3. $$(t-v)^2 < 4$$ ?
$$|t-v| < 2$$ ?
need not be if t = -1.8, v = 1.8

Answer (A) Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[   [#permalink] 26 Jan 2018, 04:42
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