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

8

This post received KUDOS

Expert's post

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

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.
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