|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 03 Aug 2005
Posts: 137
Followers: 1
Kudos [?]:
2
[0], given: 0
|
If xy is different from 0, is x < y? (1) x^4 < y^4 (2) [#permalink]
 04 Nov 2005, 01:02
If xy is different from 0, is x < y?
(1) x^4 < y^4
(2) x^(-3) < y^(-3)
|
|
|
|
|
|
|
Manager
Joined: 04 Oct 2005
Posts: 248
Followers: 1
Kudos [?]:
3
[0], given: 0
|
man, I'm quite weak at number properties as well I just found out, trying this one... well lets give it a shot:
xy != 0 means x and y can be +ve or -ve but none is zero
(1) the ^4 neutralizes the sign and says that the absolute value of y > x
The true value cannot be know.
- both could be +ve then y>x
- both could be -ve then y<x
---> INSUFF
(2) rewrite to 1/(x^3) < 1/(x^3), and signs play a role
For this statement to be true, the true values could be:
- both could be +ve then x>y
- both could be -ve then x<y
---> INSUFF
(1)&(2)
On the premise, that the absolute value of y > x from (1),
in order for (2) to be true, x must be -ve.
Hence, y>x
SUFF.
|
|
|
|
|
|
Manager
Joined: 21 Sep 2005
Posts: 231
Followers: 1
Kudos [?]:
1
[0], given: 0
|
Boy..I would mark it as E.
|
|
|
|
|
|
Manager
Joined: 28 Jun 2005
Posts: 219
Followers: 1
Kudos [?]:
2
[0], given: 0
|
B:
x^3>y^3; cube root on both side x>y
|
|
|
|
|
|
Intern
Joined: 22 Mar 2005
Posts: 39
Location: Houston
Followers: 0
Kudos [?]:
0
[0], given: 0
|
I like B as x will always be greater than y.
Positive whole #
x = 3
y = 2
1/27 < 1/4
Negative whole #
x = -3
y = -4
-1/27 < -1/64
Fractions
x = 1/2
y = 1/4
8 < 64
x=-1/4
y=-1/2
-64 < -8
|
|
|
|
|
|
Manager
Joined: 04 Oct 2005
Posts: 248
Followers: 1
Kudos [?]:
3
[0], given: 0
|
what if x is -1 and y is 1?
then x < y
plugging this into (2) yields:
1/(-1)^3 < 1/1^3
-1 < 1.... still holds but x < y
|
|
|
|
|
|
Director
Joined: 27 Jun 2005
Posts: 518
Location: MS
Followers: 1
Kudos [?]:
3
[0], given: 0
|
I am with B too
cause
A) X^4 < y^4 => X^2<Y^2=> X<Y or X>Y Not Suff
B) X^-3<Y-3 = Y^3 - X^3< 0
Y^3<X^3 so Y<X Sufficient
B
what is OA ?
gotoknow3 wrote: B:
x^3>y^3; cube root on both side x>y
|
|
|
|
|
|
SVP
Joined: 24 Sep 2005
Posts: 1913
Followers: 7
Kudos [?]:
55
[0], given: 0
|
Re: number properties 3 [#permalink]
04 Nov 2005, 09:31
jdtomatito wrote: If xy is different from 0, is x < y?
(1) x^4 < y^4
(2) x^(-3) < y^(-3)
(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0
since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0
We have to cases:
case1 : x-y<0 and x+y>0 ---> -y<x<y
case2: x-y>0 and x+y<0 ---> y<x<-y
---> insuff
(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0
--> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0
1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0
---> 1/x-1/y < 0 --> (y-x)/xy<0
we have two cases:
case1 : y-x<0 and xy>0 ---> y<x and xy>0
for example: x=3, y=2 we have 1/3-1/2<0
case2: y-x>0 and xy<0 ---> y>0>x
--->insuff
Last edited by laxieqv on 04 Nov 2005, 09:42, edited 1 time in total.
|
|
|
|
|
|
SVP
Joined: 24 Sep 2005
Posts: 1913
Followers: 7
Kudos [?]:
55
[0], given: 0
|
(1) and (2),
refer to nero's explanation.
yes, C it is.
|
|
|
|
|
|
Manager
Joined: 03 Aug 2005
Posts: 137
Followers: 1
Kudos [?]:
2
[0], given: 0
|
I also get C. Very nice work nero44.
OA is B, but it is obviously wrong.
|
|
|
|
|
|
Manager
Joined: 04 Oct 2005
Posts: 248
Followers: 1
Kudos [?]:
3
[0], given: 0
|
jdtomatito wrote: I also get C. Very nice work nero44.
OA is B, but it is obviously wrong.
thanks but,
whats the source of your questions? there seem to be two questions from you now, which have a possibly wrong OA... are there any OEs?
|
|
|
|
|
|
VP
Joined: 22 Aug 2005
Posts: 1133
Location: CA
Followers: 1
Kudos [?]:
9
[0], given: 0
|
Re: number properties 3 [#permalink]
04 Nov 2005, 10:25
laxieqv wrote: jdtomatito wrote: If xy is different from 0, is x < y?
(1) x^4 < y^4
(2) x^(-3) < y^(-3) (1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff (2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff you nailed it lexieqv.
|
|
|
|
|
|
Current Student
Joined: 28 Dec 2004
Posts: 3439
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
134
[0], given: 2
|
I get C
1/x^3 < 1/Y^3
so pick numbers..
say x=1, y=0.8
x>y
works out just fine
say x=-2 y=-3
well x >y
but if x=-1 y=1
the 1/x^3 < 1/y^3; x<y Insuff
(2) tells us that the |x| <>|y| so last condition above doesnt exist...
|
|
|
|
|
|
Manager
Joined: 28 Jun 2005
Posts: 219
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Re: number properties 3 [#permalink]
04 Nov 2005, 11:00
laxieqv wrote: jdtomatito wrote: If xy is different from 0, is x < y?
(1) x^4 < y^4
(2) x^(-3) < y^(-3) (1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff (2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff good explanation; but the bold part does not satisfy statement 2. pick some numbers and try
Answer is good: B is suff
|
|
|
|
|
|
SVP
Joined: 24 Sep 2005
Posts: 1913
Followers: 7
Kudos [?]:
55
[0], given: 0
|
Re: number properties 3 [#permalink]
04 Nov 2005, 11:05
gotoknow3 wrote: laxieqv wrote: jdtomatito wrote: If xy is different from 0, is x < y?
(1) x^4 < y^4
(2) x^(-3) < y^(-3) (1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff (2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff good explanation; but the bold part does not satisfy statement 2. pick some numbers and try Answer is good: B is suff x=-2, y=2
x^-3 = -1/8
y^-3= 1/8
-1/8<1/8
Hik,still wrong ? 
|
|
|
|
|
|
Manager
Joined: 03 Aug 2005
Posts: 137
Followers: 1
Kudos [?]:
2
[0], given: 0
|
Actually there are three questions I think have a wrong OA. Questions are from Princeton +48 Gmat math.
OE: 1 is insuf because negative values for the variables give you the opposite value than negative values.
In statement 2, negative numbers retain their sign but we need to look closely at how fractions react. If you plug in x=2 and y=1 that gives you a "no". If you plug in x=1/2 and y=1/4 you also get a "no". Therefore statement 2 is sufficient and answer is B.
But if you plug in x=-1 y=1 you get a yes with statement 2, thus OA/OE is wrong.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
