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If r + s > 2t, is r > t ? (1) t > s (2) r > s
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Updated on: 26 Mar 2012, 00:36
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If r + s > 2t, is r > t ? (1) t > s (2) r > s
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Originally posted by kairoshan on 19 Nov 2009, 06:54.
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
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26 Mar 2012, 00:40
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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Hope it helps.
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Re: If r + s > 2t, is r > t ?
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19 Nov 2009, 08:33
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 rs>0 r+s>2t add equations and you get 2r>2t or r>t



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Re: Alphabet Digits Part I
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04 Jan 2011, 22:51
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
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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
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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
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20 Mar 2013, 23:13
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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\).
Hope it helps.



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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s
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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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\).
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
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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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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s
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