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Is yz > wz?

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Tuck School Moderator
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Is yz > wz?  [#permalink]

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New post 22 Oct 2018, 08:22
2
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

56% (01:17) correct 44% (00:57) wrong based on 60 sessions

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Is \(yz > wz?\)

1. \(y > w\)
2. \(z^3 < z\)

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Re: Is yz > wz?  [#permalink]

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New post 22 Oct 2018, 08:46
Is \(yz > wz?\)
z(y-w)>0....
Two cases..
I. y>w, and z>0
II. y<w and z<0
So we are looking for either of the two cases...

1. \(y > w\)
If z>0, yes
If z<0, or z=0, no
Insufficient

2. \(z^3 < z\)
This happens when 0<z<1 or when z<-1
Insufficient

Combi Ned..
y>w..
If z<-1...yz<wz.....no
If 0<z<1....yz>wz....yes
Insufficient

E
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Re: Is yz > wz?  [#permalink]

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New post 22 Oct 2018, 08:56
starting with statement 2, we know that z lies between 0 and 1 (positive) or z is less than -1 (negative)

using these 2 possibilities in the ques stem, we get either
1) is y>w , the answer to which is yes using (1)
or
2) is w>y, the answer to which is no using (2)

IMO the ans is e
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Re: Is yz > wz?  [#permalink]

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New post 22 Oct 2018, 09:46
PeepalTree wrote:
Is \(yz > wz?\)

1. \(y > w\)
2. \(z^3 < z\)


chetan2u Bunuel

By stmt 1 we get to know that y>w

But since we do not know anything about z we cannot say with certainty that y>z

Insuff

By stmt 2, we get the eqn of the form

z^3 -z < 0
(z-1) * z * (z+1) < 0

This is possible when z is negative

So by combining we get a definitive answer.


Where did i go wrong?

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Re: Is yz > wz?  [#permalink]

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New post 22 Oct 2018, 11:50
1
saurabh9gupta wrote:
PeepalTree wrote:
Is \(yz > wz?\)

1. \(y > w\)
2. \(z^3 < z\)


chetan2u Bunuel

By stmt 1 we get to know that y>w

But since we do not know anything about z we cannot say with certainty that y>z

Insuff

By stmt 2, we get the eqn of the form

z^3 -z < 0
(z-1) * z * (z+1) < 0

This is possible when z is negative

So by combining we get a definitive answer.


Where did i go wrong?

Posted from my mobile device


z need not be an integer. It could be 1/2 then (1/2-1)*1/2*(1/2+1)=(-1/2)1/2*3/2<0
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Re: Is yz > wz?  [#permalink]

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New post 22 Oct 2018, 12:25
PeepalTree wrote:
Is \(yz > wz?\)

1. \(y > w\)
2. \(z^3 < z\)


First inequality rearranged to yz - wz > 0 then z (y - w) > 0

Statement 1) tells us y > w and that can be rearranged to y - w > 0

z ( y - w) > 0 we know one part is > 0 but don't know about Z so insufficient.

For Statement 2)

\(z^3 < z\) can be rearranged to \(z > z^3\) then \(z - z^3 > 0\) then \(z (1 - z^2) > 0\)

we know \(z > 0\) and \(1 - z^2\) can be rearranged to square of a difference \((1 - z) (1 + z) > 0\)

so 1 > z and 1 > - z so -1 < z

0 < z < 1 and z > - 1

since we have no information on (y-w) we cannot tell.

Insufficient.

Now combine both.

we know from first statement (y-w) is positive.

We also know from second statement that z can be positive or negative for example z can be 1/2 or -1/2

Insufficient.

Answer choice E
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Re: Is yz > wz?   [#permalink] 22 Oct 2018, 12:25
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