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Re: 0 < t < 1 [#permalink]
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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]
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OA Added.

Thank you
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Re: 0 < t < 1 [#permalink]
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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]
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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

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Re: 0 < t < 1 [#permalink]
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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]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



\(0 < t < 1\)

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]
Such questions must be solved while putting extreme values, (i.e maximum or minimum, which can be imagined).
Max: t = 0.99999999
Min: t = 0.00000001

There's how, you can easily and quickly manipulate that in first case, Quantity B is greater, while on second case, Quantity A will be greater.

So it's just process of thinking, rather than just putting values to find answer.
So Answer D.
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0 < t < 1 [#permalink]
Carcass wrote:

\(0 < t < 1\)

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.


\(0 < t < 1\)
This means \(t\) is a positive fraction

Col. A: \(1 - t\)
Col. B: \(\frac{1}{2} + t\)

Adding \(t\) to both sides;

Col. A: 1
Col. B: \(\frac{1}{2} + 2t\)

Subtracting \(\frac{1}{2}\) from both sides;

Col. A: \(\frac{1}{2}\)
Col. B: \(2t\)

Dividing by \(2\) both sides;

Col. A: \(\frac{1}{4}\)
Col. B: \(t\)

If \(t = \frac{1}{2}\), Col. A < Col. B

If \(t = \frac{1}{4}\), Col. A = Col. B

If \(t = \frac{1}{8}\), Col. A > Col. B

Hence, option D
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Re: 0 < t < 1 [#permalink]
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Re: 0 < t < 1 [#permalink]
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