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Knewton test problem

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Knewton test problem

by punitkaur » Sun Oct 11, 2009 12:20 pm
If y>0 and 0<1+(x/y)<1 which of the following must be true?

I)x<0
II)x/y < 0
III)x^2 + y^2 > 1


(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III

I know how to check for I & II algebraically. But I dont know how to check for III.

Please give a solution which does not pick numbers to test the inequality. I am looking for an algebraic proof

Thanks

OA is C
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by John@Knewton » Mon Oct 12, 2009 2:28 pm
Unfortunately, since it is not the case that III must be correct (since C, rather than E, is the correct answer), it is extraordinarily difficult (and arguably impossible, depending on what you are looking for in an algebraic proof) to use pure algebra to establish that III need not be true. Testing the two pairs of numbers (-1/4, 1/2) and (-2, 4) is certainly the best way to approach this question.

You could reason that since the given inequalities relate x and y, they represent regions in the xy-plane (0 < 1 - x/y < 1 represents the region bounded on the right by the y-axis, and on the bottom/left by the line y = -x, while x^2 + y^2 > 1 represents the region outside the unit circle), and then see that, since neither region is a proper subset of the other, we can draw no conclusions about whether being in one region implies being in the other. However, this is clearly an indirect way to the solution.

The fundamental difficulty here is that on this question, as on many difficult inequality/number properties questions, we are asked to prove a relationship between two essentially unrelated facts: 0 < 1 + x/y < 1, and x^2 + y^2 > 1. Attempting an algebraic proof that there is NO relationship is an enormous waste of time, and will certainly cost you points in the end on rushed questions. By far the better approach here is to find numbers to plug in (and, since you're studying for the test now, this is a great time to get more comfortable with plugging in numbers!).
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Re: Knewton test problem

by Stuart@KaplanGMAT » Mon Oct 12, 2009 3:21 pm
punitkaur wrote:If y>0 and 0<1+(x/y)<1 which of the following must be true?

I)x<0
II)x/y < 0
III)x^2 + y^2 > 1


(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III

I know how to check for I & II algebraically. But I dont know how to check for III.

Please give a solution which does not pick numbers to test the inequality. I am looking for an algebraic proof

Thanks

OA is C
As John notes, there's really no way to prove algebraically that something need not be true. However, we can often reason things through much more quickly than we can do algebra.

Let's start by simplifying the original inequality:

0<1+(x/y)<1

if we subtract 1 from all 3 parts, we get:

-1 < x/y < 0

and now it's much easier to see that x/y must be a negative fraction.

Further, since y is positive, we know that x must be negative. Moreover, since x/y is a fraction, |x| < |y|.

So, what does this tell us about the 3 roman numerals:

I) we've already shown this must be true.

II) we've already shown this must be true.

III) we know that x/y is a fraction, but does that mean than x and y have to be integers? Of course not... x and y could be fractions, as long as y is "bigger".

For example, we could pick x=-1/3 and y=1/2 and end up with x^2 + y^2 < 1.
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by TM » Mon Oct 12, 2009 4:35 pm
john@knewton wrote:Unfortunately, since it is not the case that III must be correct (since C, rather than E, is the correct answer), it is extraordinarily difficult (and arguably impossible, depending on what you are looking for in an algebraic proof) to use pure algebra to establish that III need not be true. Testing the two pairs of numbers (-1/4, 1/2) and (-2, 4) is certainly the best way to approach this question.

You could reason that since the given inequalities relate x and y, they represent regions in the xy-plane (0 < 1 - x/y < 1 represents the region bounded on the right by the y-axis, and on the bottom/left by the line y = -x, while x^2 + y^2 > 1 represents the region outside the unit circle), and then see that, since neither region is a proper subset of the other, we can draw no conclusions about whether being in one region implies being in the other. However, this is clearly an indirect way to the solution.

The fundamental difficulty here is that on this question, as on many difficult inequality/number properties questions, we are asked to prove a relationship between two essentially unrelated facts: 0 < 1 + x/y < 1, and x^2 + y^2 > 1. Attempting an algebraic proof that there is NO relationship is an enormous waste of time, and will certainly cost you points in the end on rushed questions. By far the better approach here is to find numbers to plug in (and, since you're studying for the test now, this is a great time to get more comfortable with plugging in numbers!).
I think we can solve this problem with graphic approach. Don't know if it's counts for algebra solution.

y>0, 0<1+(x/y)<1 means:
1. y>0, 2. x/y>-1 and 3. 1+x/y<1 or x/y<0 (which already on this stage shows that I and II are always true)

If we draw it we'll have the green area for x and y which is the half of II quadrant (between the diagonal and Y-axis). From this it's obvious that:

I. x<0 and II. x/y<0 are always true
as for
III. x^2 + y^2 > 1, as x^2+y^2=1 is a formula of a circle centered at the origin, we have points out of the circle. But from our given statements (y>0, 0<1+(x/y)<1) we have the point which are in the circle (close to the origin) so this not must be true.

It can be easily seen if one draws it.

Please correct me if I'm wrong.
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