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0 < a < 1 < b < c

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Carcass
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0 < a < 1 < b < c [#permalink] New post   17 Oct 2017, 14:05
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\(0 < a < 1 < b < c\)

Quantity A
Quantity B
\(\frac{c^2}{a}\)
\(\frac{bc}{ab}\)



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.
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A
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IlCreatore
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Re: 0 < a < 1 < b < c [#permalink] New post   18 Oct 2017, 05:04
Quantity A has at the denominator a number lower than 1 but greater than zero, this means it increases when divided by it. Moreover, it has a numerator as the highest number among the three to the power of 2.

Quantity B, instead, has a number greater than 1 at the denominator and a lower numerator, given that b < c and thus bc < c^2.

This leads to the choice of B!

Putting in some numbers as a = 1/2, b = 2 and c = 3; we would get A = 3^2/(1/2) = 18 and B = 3*2/(1/2)*2 = 6.
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0 < a < 1 < b < c [#permalink] New post   20 Jun 2019, 06:18
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Carcass wrote:



\(0 < a < 1 < b < c\)

Quantity A
Quantity B
\(\frac{c^2}{a}\)
\(\frac{bc}{ab}\)



We can solve this question using matching operations

Given:
Quantity A: \(\frac{c^2}{a}\)
Quantity B: \(\frac{bc}{ab}\)

Since we're told a is POSITIVE, we can safely multiply both quantities by a to get:
Quantity A: \(c^2\)
Quantity B: \(\frac{bc}{b}\)

Since b ≠ 0, we can cancel out the b's in Quantity B to get:
Quantity A: \(c^2\)
Quantity B: \(c\)

Since we're told c is POSITIVE, we can safely divide both quantities by c to get:
Quantity A: \(c\)
Quantity B: \(1\)

We're told that 1 < c, so the correct answer is A


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Re: 0 < a < 1 < b < c [#permalink] New post   05 Aug 2020, 10:17
now all values are above 0

let it be 0 < 0.1 < 1 < 0.2< 3
substitute

you'll get 9/0.1 and 0.6/0.02

Thus A
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Re: 0 < a < 1 < b < c [#permalink] New post   28 Dec 2021, 08:31
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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Re: 0 < a < 1 < b < c [#permalink]
28 Dec 2021, 08:31
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