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29 Dec 2011, 06:24
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (02:57) correct
67%
(02:28)
wrong
based on 33
sessions
History
Date
Time
Result
Not Attempted Yet
For any two numbers X and Y , define X#Y=X+Y -XY . If X,Y>0.5, then X#X < Y#Y if
i. 1>X>Y
ii. 1>Y>X
iii. Y>X>1
A) i only
B) ii only
C) i and ii only
D) ii and iii only
E) i and ii only
i. 1>X>Y
ii. 1>Y>X
iii. Y>X>1
A) i only
B) ii only
C) i and ii only
D) ii and iii only
E) i and ii only
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29 Dec 2011, 06:42
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Anasthaesium
i. 1>X>Y
Plug numbers
Y = 0.8
X = 0.9
Y#Y = 0.8 + 0.8 - 0.8*0.8 = 1.6 - 0.64 = 0.96
X#X = 0.9 + 0.9 - 0.9*0.9 = 1.8 - 0.81 = 0.99
Y#Y < X#X
(i) is wrong
Eliminate Option A, C and E
THe remaining options are B and D
Both include (ii)
so it must be true
Check (iii)
iii.Y>X>1
X = 2
Y = 3
X#X = 2 + 2 - 2*2 = 0
Y#Y = 3 + 3 - 3*3 = -3
Y#Y < X#X
not true
Oprion D is eliminated
Option
29 Dec 2011, 07:00
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X#X = 2X-X^2 and Y#Y=2Y-Y^2
X#X <Y#X
=> 2X - X^2 < 2Y - Y^2
=> 2(X-Y) < (X-Y)(X+Y)
=> X+Y>2 and X-Y>0 or X+Y<2 and X-Y<0
=> X+Y>2 and X>Y or X+Y<2 and X<Y
Using (i), if 1>X>Y, then the first case applies but X+Y>2 is not satisfied. False.
Using (ii), if 1>Y>X, then the second case applies and X+Y<2 is satisfied. True.
Using (iii), if Y>X>1, then the second case applies but X+Y<2 is not satisfied. False.
Therefore only (ii) is true, and choice (B) is correct.
X#X <Y#X
=> 2X - X^2 < 2Y - Y^2
=> 2(X-Y) < (X-Y)(X+Y)
=> X+Y>2 and X-Y>0 or X+Y<2 and X-Y<0
=> X+Y>2 and X>Y or X+Y<2 and X<Y
Using (i), if 1>X>Y, then the first case applies but X+Y>2 is not satisfied. False.
Using (ii), if 1>Y>X, then the second case applies and X+Y<2 is satisfied. True.
Using (iii), if Y>X>1, then the second case applies but X+Y<2 is not satisfied. False.
Therefore only (ii) is true, and choice (B) is correct.
