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

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If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]  19 Nov 2009, 05:54
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If r + s > 2t, is r > t ?

(1) t > s

(2) r > s
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Mar 2012, 23:36, edited 1 time in total.
Edited the question and added the OA
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Re: If r + s > 2t, is r > t ? [#permalink]  19 Nov 2009, 07:33
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kairoshan wrote:
If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

1.
r + s > 2t
s<t

subtract inequalities and you get r>t so sufficient

2. r>s or r-s>0
r+s>2t

add equations and you get 2r>2t or r>t
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]  25 Mar 2012, 16:44
thx lagomez. just forgot that we could add inequalities and equations to help simplify an equation.
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]  25 Mar 2012, 23:40
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If r + s > 2t, is r > t ?

(1) t > s --> since the signs of two equations (t > s and r + s > 2t) are the same direction we can sum them: $$t+(r+s)>s+2t$$ --> $$r>t$$. Sufficient.

(2) r > s --> the same here: since the signs of two equations (r > s and r + s > 2t) are the same direction we can sum them: $$r+(r+s)>s+2t$$ --> $$2r>2t$$ --> $$r>t$$. Sufficient.

THEORY:
You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]  20 Mar 2013, 22:13
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Hi Bunnel, since we don't know the signs of t and s, how can we subtract s on both sides to simplify the inequality R+s+t>S+2T? Am I missing something here?

Bunuel wrote:
If r + s > 2t, is r > t ?

(1) t > s --> since the signs of two equations (t > s and r + s > 2t) are the same direction we can sum them: $$t+(r+s)>s+2t$$ --> $$r>t$$. Sufficient.

(2) r > s --> the same here: since the signs of two equations (r > s and r + s > 2t) are the same direction we can sum them: $$r+(r+s)>s+2t$$ --> $$2r>2t$$ --> $$r>t$$. Sufficient.

THEORY:
You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]  21 Mar 2013, 02:33
Expert's post
AnnT wrote:
Hi Bunnel, since we don't know the signs of t and s, how can we subtract s on both sides to simplify the inequality R+s+t>S+2T? Am I missing something here?

Bunuel wrote:
If r + s > 2t, is r > t ?

(1) t > s --> since the signs of two equations (t > s and r + s > 2t) are the same direction we can sum them: $$t+(r+s)>s+2t$$ --> $$r>t$$. Sufficient.

(2) r > s --> the same here: since the signs of two equations (r > s and r + s > 2t) are the same direction we can sum them: $$r+(r+s)>s+2t$$ --> $$2r>2t$$ --> $$r>t$$. Sufficient.

THEORY:
You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

You are mixing subtraction/addition with multiplication/division. We are only concerned with sign when we multiply/divide an inequality.
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]  15 Jul 2014, 05:56
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s   [#permalink] 15 Jul 2014, 05:56
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