IMO A
xyz>0 gives us 4 case
1... x(+)y(+)z(+)...all positive or 2 negatives in remaining three cases below:
2....x(+)y(-)z(-)
3....x(-)y(-)z(+)
4....x(-)y(+)z(-)
1. y<0 so case 2 &3 above apply
(x)(y^2)(z^3).... case 2 give<0 & case 3 gives<0 again...sufficient
2. x>0 now case 1 & 2 apply
for (x)(y^2)(z^3)... case 1 is >0 while case 2 is <0 ...insufficient
ans:A
Variables signs problem
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earth@work
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Very good explanation.earth@work wrote:IMO A
xyz>0 gives us 4 case
1... x(+)y(+)z(+)...all positive or 2 negatives in remaining three cases below:
2....x(+)y(-)z(-)
3....x(-)y(-)z(+)
4....x(-)y(+)z(-)
1. y<0 so case 2 &3 above apply
(x)(y^2)(z^3).... case 2 give<0 & case 3 gives<0 again...sufficient
2. x>0 now case 1 & 2 apply
for (x)(y^2)(z^3)... case 1 is >0 while case 2 is <0 ...insufficient
ans:A
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kanha81
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q: xyz>0, is x(y^2)(z^3) < 0? Y/N
(i) y<0
=> x>0 & z<0 or x<0 & z>0
x>0 & z<0: y^2>0, z^3<0 => xy^2z^3<0. Yes
x<0 & z>0: y^2>0, z^3>0=> xy^2z^3<0. Yes
Suff.
[A] or [D]
(ii) x>0
=> y>0 & z<0 or y<0 & z>0
y>0 & z<0: => y^2>0, z^3<0 => xy^2z^3<0. Yes
y<0 & z>0: => y^2>0, z^3>0 => xy^2z^3>0. No
Insuff.
hence, [A]
(i) y<0
=> x>0 & z<0 or x<0 & z>0
x>0 & z<0: y^2>0, z^3<0 => xy^2z^3<0. Yes
x<0 & z>0: y^2>0, z^3>0=> xy^2z^3<0. Yes
Suff.
[A] or [D]
(ii) x>0
=> y>0 & z<0 or y<0 & z>0
y>0 & z<0: => y^2>0, z^3<0 => xy^2z^3<0. Yes
y<0 & z>0: => y^2>0, z^3>0 => xy^2z^3>0. No
Insuff.
hence, [A]
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