0 <2x + y <50, -50 <x + 2y <0, ask which of the following must be true ( you can select more than 1 that is right?
I. x> 0
II. Y> 0
III. X> y
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Jay@ManhattanReview
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From the inequality 0 < 2x + y < 50, we get that the sum of two x and one y is positive (0 < 2x + y). When one x is replaced with a y, the inequality becomes -50 < x + 2y < 0. We see that now the sum of two y and one x is negative (x + 2y < 0). This implies that x is positive and y is negative; and x > y.mihai9 wrote:0 <2x + y <50, -50 <x + 2y <0, ask which of the following must be true ( you can select more than 1 that is right?
I. x> 0
II. Y> 0
III. X> y
Hope this helps!
-Jay
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GMATGuruNY
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Inequalities can be LINKED TOGETHER.mihai9 wrote:0 <2x + y <50, -50 <x + 2y <0, ask which of the following must be true ( you can select more than 1 that is right?
I. x> 0
II. Y> 0
III. X> y
Linking together x+2y < 0 and 0 < 2x+y, we get:
x+2y < 0 < 2x+y
x+2y < 2x+y
,y < x
Thus, statement III must be true.
We can ADD TOGETHER inequalities in which the inequality symbol faces the same direction.
Adding together y<x (proven above) and 0<2x+y, we get;
y+0 < x+2x+y
0 < 3x
0 < x.
Thus, statement I must be true.
Adding together y<x (proven above) and x+2y<0, we get;
y+x+2y < x+0
3y < 0
y < 0.
Thus, statement II is NOT true.
The correct answer is I and III.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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