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

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

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19 Nov 2009, 06:54
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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 26 Mar 2012, 00:36, edited 1 time in total.
Edited the question and added the OA

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

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26 Mar 2012, 00:40
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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.

Answer: D.

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 ? [#permalink]

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19 Nov 2009, 08:33
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kairoshan wrote:
If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

answer D

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

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04 Jan 2011, 22:07
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If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

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Re: Alphabet Digits Part I [#permalink]

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04 Jan 2011, 22:51
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Here, you can manipulate inequalities.

From the givens, we know that:

r+s > 2t

therefore:
(1)
r+s > 2t
t > s

Inequalities with the same sign can be added / subtracted

=> r + s + t > 2t + s
=> r > t

SUFF

(2)
r + s > 2t
r > s

=> 2r + s > 2t + s
=> 2r > 2t
=> r > t

SUFF

Answer: D
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]

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20 Mar 2013, 23:13
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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.

Answer: D.

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]

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21 Mar 2013, 03:33
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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.

Answer: D.

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: Alphabet Digits Part I [#permalink]

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05 Jan 2011, 00:23
Wayxi wrote:
If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

r + s > 2t

(1) t > s
So, r+t > r+s > 2t
So, r+t > 2t
So, r>t
Sufficient

(2) r > s
So, r+r > s+r > 2t
So, 2r > 2t
So, r > t
Sufficient

Answer is (d)
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]

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25 Mar 2012, 17:44
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]

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

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

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

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

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30 Aug 2017, 13:48
$$r+s>2t \rightarrow \frac{r+s}{2}>t$$

So the midpoint of $$r$$ and $$s$$ is greater than $$t$$

The orange line represents possible locations on the number line for $$t$$

Statement 1)

If $$t>s$$ then only the second option is possible, and $$r$$ must be greater than $$t$$

Statement 2)

If $$r>s$$ then only the second option is possible, and again $$r$$ must be greater than $$t$$

Answer D
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r,s,t.PNG [ 7.67 KiB | Viewed 517 times ]

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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s   [#permalink] 30 Aug 2017, 13:48
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# If r + s > 2t, is r > t ? (1) t > s (2) r > s

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