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Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]
25 Mar 2012, 23:40

11

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

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 \(5>1\) --> \(3-5<4-1\).

Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]
20 Mar 2013, 22:13

1

This post received KUDOS

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 \(5>1\) --> \(3-5<4-1\).

Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]
21 Mar 2013, 02:33

Expert's post

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. _________________

Re: If r + s > 2t, is r > t ? (1) t > s (2) r > s [#permalink]
15 Jul 2014, 05:56

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It has been a fairly long time since I have posted here, but I definitely did not want to sign off without giving readers a quick update on my personal...