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EMPOWERgmatRichC
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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 19 Nov 2015, 17:32
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A
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61% (02:04) correct 39% (02:02) wrong based on 745 sessions

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

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A
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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 19 Nov 2015, 18:55
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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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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 19 Nov 2015, 19:07
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peachfuzz
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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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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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 19 Nov 2015, 21:58
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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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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 23 Nov 2015, 09:17
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EMPOWERgmatRichC
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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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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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 24 Nov 2015, 00:46
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EMPOWERgmatRichC
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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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:
Show 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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Rich
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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 26 Jan 2018, 04:42
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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)
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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 04 Apr 2020, 18:15
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-2<t<0<v<2 --> t always negative, v always postitive
(a) tv-v < 0 --> v(t-1)<0 and since t is negative, t-1 will be negative, so -v<0 is always true
(b) t^2-v^2 > 0 --> if t=-1 and v=1, then this is 0, not true
(c) (t-v)^2<4 ---> if t=-1 and v=1, then 2^2=4, so not true

Answer is A, I only
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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 03 Feb 2025, 17:17
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T could be -1.5 and V could be 1.5,

both will be added in the equation 3,
as a result we will get (-1.5-1.5)^2=(3)^2=9, which is greater than 4

so the answer is actually A

peachfuzz
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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If -2 < t < 0 and 0 < v < 2, then which of the following must be true? 06 Feb 2025, 10:02
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Why is everyone on this thread taking such complex values to explain option III. Why cant we just consider:

t= -1 which falls under the condition: -2 < t < 0

v = 1 which falls under the condition: 0 < v < 2

now, (t-v)^2 => (-1-1)^2 => (-2)^2 => 4

Since, the condition says <4 and does NOT say <=4, it any which way will be false.

Am i missing something here?
EMPOWERgmatRichC
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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QUANT 4-PACK SERIES Problem Solving Pack 4 Question 1 If -2 < t < 0...

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