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Is x/y < xy ?

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Is x/y < xy ?  [#permalink]

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New post Updated on: 08 Aug 2018, 21:30
2
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

51% (02:16) correct 49% (01:56) wrong based on 73 sessions

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Is \(\frac{x}{y} < xy\) ?

1) xy > 0

2) y < -1

Originally posted by KSBGC on 08 Aug 2018, 17:20.
Last edited by amanvermagmat on 08 Aug 2018, 21:30, edited 1 time in total.
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Re: Is x/y < xy ?  [#permalink]

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New post 08 Aug 2018, 19:49
selim wrote:
Is \(\frac{x}{y} < xy\) ?

1) xy > 0

2) y < -1



\(xy-\frac{x}{y}>0...........\frac{x(y^2-1)}{y}>0.\).
So two cases
a) y>0, so X(y^2-1)>0
    # either x>0 and y^2>1 or y>1
    # x<0 and y^2<1or 0<y<1
b) y<0, so x(y^2-1)<0
    # either x>0 and y^2-1<0 or -1<y<0
    # x<0 and y^2-1>0 or y<-1

Let's see the statements
1) xy>0
Tells us both are of same sign.
From above two cases..
# x>0, y has to be greater than 1....but y could be between 0 and 1 also
# x<0, y>1 but y could be anything positive
So insufficient

2) y<-1
Insufficient, nothing about x

Combined
Only one case possible that y<0..
And when y<0, X<0 and y<-1 both rules satisfied

C
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Re: Is x/y < xy ?  [#permalink]

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New post 08 Aug 2018, 19:55
selim wrote:
Is \(\frac{x}{y} < xy\) ?

1) xy > 0

2) y < -1



St1:- xy > 0

a) x>0, y>0 so \(\frac{x}{y}\) may or mayn't be greater than xy. (1,1), (1,2), (3,2) etc.
b) x<0, y<0 so \(\frac{x}{y}\) may or mayn't be greater than xy. (-1,-1), (-1,-2) etc
Insufficient.

St2:- y < -1
No info on x.
a) say x=-2, y=-2, then \(\frac{x}{y} < xy\)
b) say x=2, y=-2, then \(\frac{x}{y} > xy\)

Insufficient.

Combining, we have x<0 and y< -1. So xy and \(\frac{x}{y}\) are always positive.
a) x=y=-2(say), then \(\frac{x}{y} < xy\)
b) x>y. x=-\(\frac{1}{2}\) and y=-2, then \(\frac{x}{y} < xy\)
c) x<y. x=-3 and y=-2, then \(\frac{x}{y} < xy\)
We have found that the question stem is consistent.
Hence, sufficient.

Ans. (C)
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Re: Is x/y < xy ?  [#permalink]

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New post 08 Aug 2018, 20:13
selim wrote:
Is \(\frac{x}{y} < xy\) ?

1) xy > 0

2) y < -1


Hi chetan2u,

I solved the stem as follows -

\(\frac{x}{y} < xy\)

On re-shuffling we get,

\(y^2 > 1\)

Now we need only 'y' for our answer -

Statement 1 does not specifically tell us anything about y, that is it could be positive or negative.

Statement 2 says y<-1, hence we can surely say that \(y^2 > 1\).

So i rounded up to option B.

Please help in explaining what went wrong for me here as clearly the OA says C.

Thanks in Advance :)
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Re: Is x/y < xy ?  [#permalink]

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New post 08 Aug 2018, 20:21
1
adstudy wrote:
selim wrote:
Is \(\frac{x}{y} < xy\) ?

1) xy > 0

2) y < -1


Hi chetan2u,

I solved the stem as follows -

\(\frac{x}{y} < xy\)

On re-shuffling we get,

\(y^2 > 1\)

Now we need only 'y' for our answer -

Statement 1 does not specifically tell us anything about y, that is it could be positive or negative.

Statement 2 says y<-1, hence we can surely say that \(y^2 > 1\).

So i rounded up to option B.

Please help in explaining what went wrong for me here as clearly the OA says C.

Thanks in Advance :)



Always remember..
you cannot cross multiply in an inequality if you are not sure about the signs.
Only cross multiply if the denominator is positive but here you cannot be sure about the sign of y..
So take all terms to one side of the inequality and then proceed
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Is x/y < xy ?  [#permalink]

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New post 09 Aug 2018, 01:03
1
adstudy wrote:
selim wrote:
Is \(\frac{x}{y} < xy\) ?

1) xy > 0

2) y < -1


Hi chetan2u,

I solved the stem as follows -

\(\frac{x}{y} < xy\)

On re-shuffling we get,

\(y^2 > 1\)

Now we need only 'y' for our answer -

Statement 1 does not specifically tell us anything about y, that is it could be positive or negative.

Statement 2 says y<-1, hence we can surely say that \(y^2 > 1\).

So i rounded up to option B.

Please help in explaining what went wrong for me here as clearly the OA says C.

Thanks in Advance :)


Hi,

The problem starts with the highlighted part.

1- You excluded an important value for x which x=0 (You can use this value to Prove answer is NO for the question)

2- If you know that x does not equal to 0, then when you cancelled x in inequality, you must consider its sign. Also you must consider the sign of y (in this example y can't be zero as x/0 is indefinite and GMAT won't deal with it. So take as fact that zero can't be in denominator)

My approach in solving this question:

Is \(\frac{x}{y}< xy?\)

I would not spend a lot of time in analyzing the question stem. Let's the statements driving how to deal with the stem.


1) xy > 0

It means that either both negative numbers or both positive numbers. Neither equals to zero. In either situation both sides are positive

The question becomes:

Is (some positive number) < (some positive numbers)? clearly no clue. Let's check quickly

Let x = 1/2 & y= 2.......................Answer Yes

Let x = 2 & y=1/2.........................Answer NO

Insufficient

Some students spend time to find different example, while reversing number could easily prove insufficiency quickly.

2) y < -1

Here y is negative.

The question becomes:

Is \(\frac{x}{y}< xy?\).........As y in negative, multiply both sides in y and flip the sign. So the question becomes:

Is \(x > xy^2?\)

Is \(x - xy^2 >0?\)

Is \(x (1 - y^2) >0?\).......As y < -1........So (1 - y^2) is always Negative........So question becomes:

Is x < 0...........No clue at all

X could equal zero or positive ...........Answer is No

OR X could equal Negative ...............Answer is Yes

Insufficient

P.S.: going in this approach for B, put 'is' to remind yourself that it is a question. Some arrive to say that x <0 and thought it is solution.

Combine 1 & 2

From 1 both have same sign & from 2, y is negative ..........So x must be negative

So From your deduction in 2........The answer is certainly yes.

Sufficient

You can go for plugging numbers after combining both

y<-1.......So let's take y =-2 and examine different x with negative numbers (NOT ONLY integers)

y =-2 & x= -10..........-10/-2<(-10)(-2).........5 < 20...............Answer is Yes

y =-2 & x= -1/2..........(-1/2)/(-2)<(-1/2)(-2).........1/4 < 1...............Answer is Yes

All cases covered

Answer: C

I hope it helps
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Is x/y < xy ?   [#permalink] 09 Aug 2018, 01:03
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