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marcodonzelli
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inequalities 30 Dec 2007, 07:32
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Please explain it very carefully:

If x/y>, is 3x+2y<18

1.x-y<2
2.y-x<2



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eschn3am
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inequalities 30 Dec 2007, 08:02
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Is the problem missing something? x/y is greater than what?
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mtgmat
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inequalities 30 Dec 2007, 22:41
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Ans is A.

I have done it with the help of graph.

St 1
x/y>2 & x-y<2

Region comprises area between y=0, y=x/2 when x is -ve and between y=0, y=x/2 & y =x -2 when x is +ve.

Intersection of line y = x/2 & y = x -2 i.e. (4,2) & (0,0) lie on the same side of line 3x + 2y =18.

Thus the region in question always lies below the line 3x + 2y =18

hence 3x + 2y < 18 always SUFF
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neelesh
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inequalities 23 Feb 2008, 07:27
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Is there any other generic way to solve these problems ?

Any alternate solutions please....
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tamg08
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inequalities 27 Mar 2008, 08:01
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My take:

x/y > 2

If Y>0: X > 2Y
If Y0
given X > 2Y
2 + Y > X

Hence 2Y 0) 0 highest value
16 > highest value possible (since highest value must be less than for Y =2 and X = 4)
Answer if Y >0, also true.

A sufficient


B: Y < X + 2
If X and Y both - value, then true

If X = 10 and Y =9 (where x < 2Y and y < X + 2), False
Insufficient

A



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