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x < y < 1 and c > d
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23 Feb 2020, 06:14
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Question Stats:
70% (01:16) correct
30% (01:31) wrong based on 30 sessions
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\(x < y < 1\) and \(c > d\)
Quantity A
Quantity B
\( x^{- (c - d)}\)
\(y^{(d - c)}\)
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.
Since \(-(c - d) = -1(c - d) = -c + d = d - c\), we can the rewrite Quantity A as follows: Quantity A: \( x^{(d - c)}\) Quantity B: \(y^{(d - c)}\)
Let's test some possible values of c, d, x and y
case i: Since \(x < y < 1\) and \(c > d\), one possible set of values is x = -1, y = 0.5, c = 2 and d = 1 We get: Quantity A: \( x^{(d - c)}=(-1)^{(1 - 2)}=(-1)^{(-1)} = \frac{1}{(-1)^1} = \frac{1}{-1}=-1\)
Quantity B: \( y^{(d - c)}=(0.5)^{(1 - 2)}=(0.5)^{(-1)} = \frac{1}{(0.5)^1} = \frac{1}{0.5}=2\) In this case, Quantity B is greater
case ii: Since \(x < y < 1\) and \(c > d\), one possible set of values is x = 0.25, y = 0.5, c = 2 and d = 1 We get: Quantity A: \( x^{(d - c)}=(0.25)^{(1 - 2)}=(0.25)^{(-1)} = \frac{1}{(0.25)^1} = \frac{1}{0.25}=4\)
Quantity B: \( y^{(d - c)}=(0.5)^{(1 - 2)}=(0.5)^{(-1)} = \frac{1}{(0.5)^1} = \frac{1}{0.5}=2\) In this case, Quantity A is greater
Re: x < y < 1 and c > d
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05 Feb 2022, 05:58
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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