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Problem Solving Pack 4, Question 1) If -2 < t < 0...[

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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


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Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink]

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New post 19 Nov 2015, 17:55
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

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Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink]

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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.

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Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink]

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(-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.

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Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink]

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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
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OA, and explanation will be posted after the 48 hour window closes.

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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.

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Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink]

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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:
[Reveal] Spoiler:
A


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.

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Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[ [#permalink]

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Re: Problem Solving Pack 4, Question 1) If -2 < t < 0...[   [#permalink] 07 Jan 2018, 09:47
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