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# 0 < a < b < c

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Joined: 18 Apr 2015
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0 < a < b < c [#permalink]  26 Aug 2018, 02:53
Expert's post
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Question Stats:

81% (00:38) correct 18% (00:57) wrong based on 97 sessions
$$0 < a < b < c$$

 Quantity A Quantity B $$\frac{b}{a}$$ $$\frac{c}{b}$$

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.

Kudos for R.A.E
[Reveal] Spoiler: OA

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VP
Joined: 20 Apr 2016
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WE: Engineering (Energy and Utilities)
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Re: 0 < a < b < c [#permalink]  26 Aug 2018, 05:47
1
KUDOS
Carcass wrote:
$$0 < a < b < c$$

 Quantity A Quantity B $$\frac{b}{a}$$ $$\frac{c}{b}$$

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.

Kudos for R.A.E

Here no information apart from positive nos. is given for a, b and c
SO they can be integer or fraction

Taking integer value

a = 1 ; b = 2 ; c = 3

Then QTY A = 2/1 = 2

QTY B = 3/2 = 1.5

Therefore QTY A > QTY B

Taking fraction

a = 1/2 ; b = 1 ; c = 2

Then QTY A = 2

QTY B = 3/1 = 3

So QTY B > QTY A

Hence option D
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Re: 0 < a < b < c [#permalink]  07 Jun 2019, 06:30
1
KUDOS
Expert's post
Carcass wrote:
$$0 < a < b < c$$

 Quantity A Quantity B $$\frac{b}{a}$$ $$\frac{c}{b}$$

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.

Kudos for R.A.E

Pranab01's solution (above) is probably the fastest, but we can also solve this question using matching operations

Given:
Quantity A: $$\frac{b}{a}$$
Quantity B: $$\frac{c}{b}$$

Since a is POSITIVE, we can safely multiply both sides by a to get:
Quantity A: $$b$$
Quantity B: $$\frac{ca}{b}$$

Since b is POSITIVE, we can safely multiply both sides by b to get:
Quantity A: $$b^2$$
Quantity B: $$ca$$

At this point, we can see that, if a = 1, b = 2, and c = 4, then the two quantities are equal
Conversely, if a = 1, b = 2, and c = 3, then Quantity A is greater

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Re: 0 < a < b < c   [#permalink] 07 Jun 2019, 06:30
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