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| If r + s > 2t, is r > t ? (1) t > s (2) r > s |
| My solution |
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Borcho
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If r + s > 2t, is r > t ?
(1) t > s (2) r > s 1) t>s <=> t-s>0 r+s > 2t => r > t+(t-s) where t-s>0 => r>t SUFFICIENT 2) r>s => r+r>r+s 2r>r+s>2t r>t Sufficient Answer D. |
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Harish Dorai
Guest
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We should not add/subtract/multiply/divide variables to the inequalities without knowing whether the integers presented are non-zero or non-negative. I feel the answer is still (D).
Given r + s > 2t Statement (1) says t > s which means s < t So the given inequality r + s > 2t can be re-written as r + (<t) > 2t ==> This implies r has to be greater than t to satisfy the inequality r + s > 2t. SUFFICIENT. Statement (2) says that r > s which can be re-written as s < r So the given inequality can be re-written as r + (<r) > 2t ==> this also implies that r has to be greater than t. SUFFICIENT. So the answer is (D). |
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Stacey Koprince
MGMAT STAFF
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You can, indeed, add inequalities AS LONG AS the signs are identical and pointing in the same direction. And, in fact, when we can do this, it is often the easiest way to handle the math.
Trying numbers is a great fall-back method, and on some problems it should be the main method, but the one drawback is that we might not try all the right numbers or find the circumstances we need to prove or disprove something. So just have to be careful there. Separately, for inequalities, you do not want to multiply or divide by variables if you do not know whether those variables are positive or negative. |
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Last edited by Stacey Koprince on Thu Sep 04, 2008 8:32 am; edited 1 time in total |
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| If r + s > 2t, is r > t ? (1) t > s (2) r > s |
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Guest
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Stacey,
In the above post you have mentioned that we can add or substract ineqalities, I have read other of your posts and you have mentioned that we can add but not substract them. So I was just making sure that this adding is allowed and substracting not allowed and that this was just by mistake |
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Stacey Koprince
MGMAT STAFF
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Oops - yes, it should've just said "add" not "add or subtract." Sorry about that! I've edited the post.
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| If r + s > 2t, is r > t ? (1) t > s (2) r > s |
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