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Algebra

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Manager
Joined: 02 Sep 2008
Posts: 103
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Kudos [?]: 31 [0], given: 1

Algebra [#permalink]  27 Apr 2009, 18:27
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Manager
Joined: 08 Feb 2009
Posts: 147
Schools: Anderson
Followers: 3

Kudos [?]: 33 [0], given: 3

Re: Algebra [#permalink]  27 Apr 2009, 19:32
Hi milind,

Given the inequalities, you should get 0 < c < d < 1. I will prove it below. This should explain ONLY I and II are TRUE.

let c/d = x (just to simplify in the next step).
Given 0 < 1 - x < 1 ---> it implies that x is less than 1 and less than 0. There is no other scenario possible. If x > 1, then (1-x)<0.
Therefore, 0 < x < 1.
which, is 0 < c/d < 1, which implies c/d is a fraction where c < d.

Thus, 0 < c < d < 1.

Hope it helps.
Manager
Joined: 02 Mar 2009
Posts: 138
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Re: Algebra [#permalink]  27 Apr 2009, 22:30
Hey guys,

I dont think d and c Need to be a fraction.

1 - c/d > 0
c/d < 1
Condition 2 has to be true

1 - c/d < 1
c/d > 0
Since d is positive, we can multiply both sides by d without changing the sign.
Therefore c>0
Condition 1 fulfilled.

Now For condition 3:
If c=5 and d=6
The two conditions above are fulfilled as so is this condition
BUT
If c=1/3 and d=1/2
Again the two conditions are fulfilled but not this one.

Thus only conditions 1 and 2 MUST be true.
Re: Algebra   [#permalink] 27 Apr 2009, 22:30
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