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D
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Difficulty:
65%
(hard)
Question Stats:
54% (01:42) correct
46%
(01:54)
wrong
based on 1178
sessions
History
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Not Attempted Yet
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:
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:
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 \(>\)) --> \(a+c>b+d\).
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).
Hope it helps.
(1) t > s:
Since the signs of two equations (t > s and r + s > 2t) are the same direction we can sum them:
\(t+(r+s)>s+2t\);
\(r>t\).
\(r>t\).
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:
\(r+(r+s)>s+2t\);
\(2r>2t\);
\(r>t\).
\(2r>2t\);
\(r>t\).
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 \(2<5\);
\(3+2<4+5\).
\(3+2<4+5\).
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 \(5>1\);
\(3-5<4-1\).
\(3-5<4-1\).
Hope it helps.
General Discussion
19 Nov 2009, 08:33
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kairoshan
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
