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Is xy > x2y2?

Expert replies
by Sanket05 » Fri Oct 09, 2009 2:57 am
Is xy > x2y2?
(1) 14x2 = 3
(2) y2 = 1

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Source: — Data Sufficiency |

by NikolayZ » Sat Oct 10, 2009 11:16 am
Hello ! i think it is E.

Simplify the expression first. xy(1-xy)>0.
so , must be that xy>0 and 1-xy>0 (1>xy)
OR xy<0, 1-xy<0 (1>xy)

(1) 14x^2=3, x=+sqrt(3/14) and x=-sqrt(3/14). We have nothing about y. so Insufficient.

(2) y^2=1, y=+1,-1. Again nothing about x.

(1)+(2)
e.g. consider x=sqrt(3/14), y=1.
==> xy>0, and xy<1. so, yes, expression is >0.
and x=-sqrt(3/14), y=1.
==> xy<0, and (1-xy)>0, so the expression is <0.

I think it is E.
Am i right ?
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by vaibhav.iit2002 » Sun Oct 11, 2009 9:26 am
IMO E

same explanation as above
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by sumitha_ambati » Tue Oct 13, 2009 8:10 am
ya its E
bcoz v dont know whether x,y are +ve or -ve
n also x2 n y2 values doest determine tat
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by mmslf75 » Thu Jan 14, 2010 9:21 pm
NikolayZ wrote:Hello ! i think it is E.

Simplify the expression first. xy(1-xy)>0.
so , must be that xy>0 and 1-xy>0 (1>xy)
OR xy<0, 1-xy<0 (1>xy)

(1) 14x^2=3, x=+sqrt(3/14) and x=-sqrt(3/14). We have nothing about y. so Insufficient.

(2) y^2=1, y=+1,-1. Again nothing about x.

(1)+(2)
e.g. consider x=sqrt(3/14), y=1.
==> xy>0, and xy<1. so, yes, expression is >0.
and x=-sqrt(3/14), y=1.
==> xy<0, and (1-xy)>0, so the expression is <0.

I think it is E.
Am i right ?
In such a sum, do we take all possibilites
I mean
x=+sqrt(3/14) y = 1
x=+sqrt(3/14) y = -1

x=-sqrt(3/14 y=1
x=-sqrt(3/14 y=-1

????????
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by bhumika.k.shah » Wed Jan 20, 2010 7:32 am
But the OA is E

Though i aint sure whether it is correct
Last edited by bhumika.k.shah on Wed Jan 20, 2010 7:54 pm, edited 1 time in total.
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by sallespadua » Wed Jan 20, 2010 12:36 pm
I found the answer as C

Considering both statements, every combination of values provides the negative afirmation, therefore sufficient to answer whether xy>x^2y^2!
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by rohan_vus » Wed Jan 20, 2010 12:50 pm
I disagree ..combining 2 stmnts you get x^2*y^2 = 4/13---> xy could be (4/13)^.5 or -(4/13)^.5...

If xy = (4/13)^.5 then xy > x^2*y^2 ...
if xy = -(4/13)^.5... then xy < x^2*y^2...

So combining you get 'yes' and 'no'.. So C cant be the ans
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by bedazzled » Mon Jan 25, 2010 7:53 am
I agree with rohan. There will b ambiguity with C therefore its E.
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by bhumika.k.shah » Mon Jan 25, 2010 8:58 am
well i guess then the OA is correct!
thanks guys :D
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by shashank.ism » Fri Feb 19, 2010 6:16 am
Sanket05 wrote:Is xy > x2y2?
(1) 14x2 = 3
(2) y2 = 1

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
for, xy>x2y2 --> 0<xy<1
st.1: 14x2 =3 x= sqrt (3/14) <1 not sufficient we don't know about y
st.2 : y2=1 --> y=+/- 1 not sufficient we don't know about x
combined we get 2 values of xy =+/-sqrt (3/14) certainly 1 value does not lead to xy > x2y2 ...not sufficient.
Ans E
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