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31 May 2008, 09:49
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If r + s > 2t, is r > t ?
(1) t > s
(2) r > s
(1) t > s
(2) r > s
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31 May 2008, 10:06
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Statement 1:
Tells us t > s => t = s + x => s = t - x (where x is some +ve number).
=> r + s = r + t - x > 2t => r > t + x. As x is +ve so r is greater than t. So answer can only be A or D.
Statement 2:
Tells us r > s => r = s + y => s = r - y (where y is some +ve number).
=> r + s = r + r - y > 2t => 2r > 2t + y => r > t + y/2. As y is +ve so r is greater than t. So answer is D.
Tells us t > s => t = s + x => s = t - x (where x is some +ve number).
=> r + s = r + t - x > 2t => r > t + x. As x is +ve so r is greater than t. So answer can only be A or D.
Statement 2:
Tells us r > s => r = s + y => s = r - y (where y is some +ve number).
=> r + s = r + r - y > 2t => 2r > 2t + y => r > t + y/2. As y is +ve so r is greater than t. So answer is D.
31 May 2008, 10:44
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mrwaxy
D.
Use the addition of inequalities...
(1)
r + s > 2t
t > s
get: r + s + t > 2t + s
=> r + t > 2t
=> r > t
SUFFICIENT
(2)
r + s > 2t
r > s
get 2r + s > 2t + s
=> 2r > 2t
=> r > t
SUFFICIENT
