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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]

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New post Updated on: 25 Mar 2012, 23:36
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If r + s > 2t, is r > t ?

(1) t > s

(2) r > s

Originally posted by kairoshan on 19 Nov 2009, 05:54.
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 ? (1) t > s (2) r > s  [#permalink]

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New post 25 Mar 2012, 23:40
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17
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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New post 19 Nov 2009, 07:33
1
2
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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Re: Alphabet Digits Part I  [#permalink]

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New post 04 Jan 2011, 21:51
1
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: Alphabet Digits Part I  [#permalink]

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New post 04 Jan 2011, 23:23
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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New post 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]

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New post 20 Mar 2013, 22:13
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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New post 21 Mar 2013, 02:33
1
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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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 30 Aug 2017, 12:48
\(r+s>2t \rightarrow \frac{r+s}{2}>t\)

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

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