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

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kairoshan
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If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink] New post  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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If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink] New post  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] New post  06 Feb 2021, 19:51
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Re: If r + s > 2t, is r > t ? [#permalink] New post  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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Re: Alphabet Digits Part I [#permalink] New post  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] New post  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] New post  25 Mar 2012, 17:44
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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] New post  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 \(<\)) --> \(a-c>b-d\) (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] New post  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: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink] New post  30 Aug 2017, 13:48
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\(r+s>2t \rightarrow \frac{r+s}{2}>t\)

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] New post  24 Sep 2019, 05:29
One way to do this is to manipulate the original stem from r-s > 2t (given) into r-t > t-s (another way to see the stem info).

Qn asks if r-s is > 0 --> two ways to know if we can answer yes:

a) if (r-s) > 0 directly, or
b) if (t-s) > 0 --> therefore (r-t) MUST be > 0

Statement 1: (t-s) > 0 (sufficient)
Statement 2: (r-s) > 0 (sufficient - the question answers its own question)

Answer: D
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink] New post  05 Nov 2019, 00:53
Sir Bunuel, do you have a quick list of similar questions at hand?
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink] New post  05 Nov 2019, 00:58
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chrtpmdr wrote:
Sir Bunuel, do you have a quick list of similar questions at hand?


9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink] New post  02 May 2022, 22:30
CAN SOMEONE HELP ME BY DOING THIS BY PLUG IN VALUES?
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink] New post  04 Jun 2022, 07:05
mehtasahil56 wrote:
CAN SOMEONE HELP ME BY DOING THIS BY PLUG IN VALUES?



To be honest, I tried doing this by plugging numbers, but since there is very little info about the nature of r,s,t (integer, fraction, +ve, -ve etc) I took a lot of time.

Therefore I then chose to use the algebraic method.

Never hurts to have many times of weapon in your arsenal!

Let me know if you did manage to plug in values for this
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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink] New post  06 Feb 2023, 00:43
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.


Hi Bunuel please advice on my reasoning:

1 t>s and r+s> 2t

Let say if "s" was equal to "t" then
r+t>2t => r>t for s=t this implies any value where t>s r>t

2 r>s let's say r=s then 2r>2t => r>t

This for any value where r>s we have r>t

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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]
06 Feb 2023, 00:43
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