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