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# If r + s > 2t, is r > t ? (1) t > s (2) r > s

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Senior Manager
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If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]

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12 Dec 2008, 05:12
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If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

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12 Dec 2008, 11:32
study wrote:
If r + s > 2t, is r > t ?

(1) t > s
(2) r > s

If r + s>2t, either r>t or s>t or each is >t..
1: if t>s, r has to be > t otherwise, r + s > 2t is not true.
2: if r>s, r must be >t.

D.
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12 Dec 2008, 11:39
study wrote:
If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

r+s>2t is r>t?

r+s>2t
+
t>s
gives us
r+s+t>2t+s => r>t
sufficient

r+s>2t
+
r>s
2r+s>2t+s
2r>2t or r>t

sufficient

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Manager
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12 Dec 2008, 13:56
fresinha12 wrote:
study wrote:
If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

r+s>2t is r>t?

r+s>2t
+
t>s
gives us
r+s+t>2t+s => r>t
sufficient

r+s>2t
+
r>s
2r+s>2t+s
2r>2t or r>t

sufficient

can you explain how u did the addition of two inequalities.

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12 Dec 2008, 20:17
r+s>2t= t + t
1. t>s => t+t > t+s
then r+s>2t> t+s => r>t suff
2.r>s => r+r>r+s
then 2r= r+r > r+s >2t => r>t suff

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12 Dec 2008, 23:27
For stmt1:

r+s > 2t
and t - s > 0

r + t > 2t or, r > t.

For stmt2:

r+s > 2t
r-s > 0

2r > 2t
or, r > t.

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Senior Manager
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13 Dec 2008, 01:16
scthakur - thanks for the solution.

question for you:

how do we decide that 2a > 2b will lead to a > b?
If a or b - one of them is negative, won't the inequality sign change to a < b?

how do we decide to change signs?

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13 Dec 2008, 11:46
study wrote:
scthakur - thanks for the solution.

question for you:

how do we decide that 2a > 2b will lead to a > b?
If a or b - one of them is negative, won't the inequality sign change to a < b?

how do we decide to change signs?

In inequality, dividing by a positive number does not alter the inequality.

Hence, if 2a > 2b then by dividing both sides by 2, a > b.
However, if -2a > -2b, then dividing both sides by -2 will change inequality sign and a < b.

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Senior Manager
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14 Dec 2008, 01:18
got it. thanks.

i was confused...

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14 Dec 2008, 07:00
Both statements independelntly proves r > t .

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Re: Inequalities   [#permalink] 14 Dec 2008, 07:00
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