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0 < t < 1

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0 < t < 1 [#permalink] New post 07 Nov 2017, 03:08
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A
B
C
D
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0 < t < 1

Quantity A
Quantity B
1 - t
\frac{1}{2} + t




A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Last edited by Carcass on 09 Nov 2017, 12:16, edited 1 time in total.
Edited by Carcass
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Re: 0 < t < 1 [#permalink] New post 07 Nov 2017, 05:40
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Quantity B is equal to 3/2 that is greater than 1. Quantity A is 1 minus a number greater than 0, so it must be less than 1. Thus, quantity B is greater.

Answer B
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Re: 0 < t < 1 [#permalink] New post 09 Nov 2017, 03:49
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OA Added.

Thank you
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Re: 0 < t < 1 [#permalink] New post 09 Nov 2017, 08:24
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Carcass wrote:
OA Added.

Thank you



Are you sure there are no typos in the text of the question? I mean the answer must be B if the quantities are the ones reported. If t is between 0 and 1, it must be positive. Then, if we subtract 1 from both quantities we get a comparison between -t and 1/2. Now, -t is -(something positive) so it is negative, while 1/2 is positive. Answer is still B
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Re: 0 < t < 1 [#permalink] New post 09 Nov 2017, 12:15
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Yes, you are truly right. The second quantity is not 1 but t. Sorry for the inconvenience :) but the OA is still D

Edited
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Re: 0 < t < 1 [#permalink] New post 12 Nov 2017, 10:17
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Written in this way we can sum t, subtract 1/2 and divide by 2 on both sides in order to get a comparison between 1/4 and t. Then, given that t is between 0 and 1 it may be greater than 1/4 (e.g. if equal to 1/2) but also smaller (if equal to 1/6). Answer D
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Re: 0 < t < 1 [#permalink] New post 13 Nov 2017, 11:40
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0 < t < 1

Suppose t = 0.9 then,
Quantity A : 1 - t = 1 - 0.9 = 0.1, and
Quantity B : 1/2 + t = 0.5 + 0.9 = 1.4
In this case: B) Quantity B is greater.

Suppose t = 0.1 then,
Quantity A : 1 - t = 1 - 0.1 = 0.9, and
Quantity B : 1/2 + t = 0.5 + 0.1 = 0.6
In this case: A) Quantity A is greater.

So the correct answer is : D) The relationship cannot be determined from the information given.
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Re: 0 < t < 1 [#permalink] New post 13 Nov 2017, 11:54
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A great way to approach this question is by subtracting one quantity from another. That way, you can compare the value 0 to an expression, and it'll be easier to see which quantity is greater. So let's start by subtracting Quantity A from both quantities. That would give us:

Quantity A: 1-t - (1-t) = 0
Quantity B: 1/2+t - (1-t) = 1/2 + t - 1 + t = -1/2 + 2t

Now we can plug in the extreme values for t. Since t can go all the way down to just above 0, Quantity B in this case could go all the way down to just above -1/2. This is less than Quantity A, so we can eliminate B and C as the correct answers. Next, since t can go all the way up to just below 1, Quantity B in this case could go all the way up to just below 3/2. Since this is greater than Quantity A, we can eliminate A as the correct answer. Thus, the only answer remaining is D.
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Re: 0 < t < 1 [#permalink] New post 14 Nov 2017, 07:54
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Carcass wrote:


0 < t < 1

Quantity A
Quantity B
1 - t
\frac{1}{2} + t






We can solve this question using matching operations

Given:
Quantity A: 1 - t
Quantity B: 1/2 + t

Add t to both quantities to get:
Quantity A: 1
Quantity B: 1/2 + 2t

Subtract 1/2 from both quantities to get:
Quantity A: 1/2
Quantity B: 2t

Divide both quantities by 2 to get:
Quantity A: 1/4
Quantity B: t

We're told that 0 < t < 1

So, for example, t COULD equal 1/4, in which case the quantities are EQUAL
t COULD also equal 1/2, in which case the Quantity B is greater
t COULD also equal 1/10, in which case the Quantity A is greater

Answer: D

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Re: 0 < t < 1   [#permalink] 14 Nov 2017, 07:54
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