Is v < 0? (1) uvu > 0 (2) v^4 < v

IMO D

Is v < 0?

(1) uvu > 0
u^2*v >0
Since, u^2 >=0 for any value, => v > 0 for uvu >0
Sufficient.

(2) v^4 < v
v (v^3-1) <0 => 0 < v < 1
So, v>0
Sufficient.

In Statement 1, we can divide both sides by u^2, because u^2 is positive, and we learn v > 0, so the answer to the question is certainly 'no'.

In Statement 2, we learn that v is bigger than v^4, which must be positive since it's an even power. If v is bigger than a positive number, v is certainly positive, and the answer to the question is again certainly 'no'. So the answer is D.

Notice that neither Statement is true if any letter equals zero, which is why there's no reason to consider that possibility.

Kindly see the attachment.

IMO D

IMO A.

u^2v is greater than 0 .
u^2 is positive here so should be v.
So Suff.

v^4 -v is less than 0
v could be fraction could be less than 0.
Cant say
Insuff.

A it is.

Posted from my mobile device

IMO D

(1) uvu > 0 ==> \(u^2\) * v >0 ; \(u^2\) will always +ve, so for expression to be +ve, v >0 SUFFICIENT
(2) \(v^4\) < v; \(v^4\) will always +ve. => \(v^4\) >0, hence v>0 SUFFICIENT

Is v < 0?

(1) uvu > 0
(2) v^4 < v

st1) For any sign and value of u (except 0), u^2 is positive. Hence v has to be +ve for the equation to work . SUFFICIENT

st2) v^4 < v , few careful number pluggins in -2 < v < 2, will tell us that the equation holds true only for positive proper fraction value of v ( 0 < v < 1). so SUFFICIENT


Hence, the answer should be D

Is v < 0?

(1) uvu > 0
(2) v^4 < v

1) u^2 v > 0, since u^2 is either 0 (which is not possible in this case) or greater than 0, v will also be greater than 0. so v < 0 is wrong. Sufficient

2) v is in the range 0 to 1. So v < 0 is not true. Sufficient.

D is the answer.

Ans -A
We need to prove v<0, so whether v is - or +

With Statement 1
Given
uvu>0, lets take possible values of u and v
+ + +>0
- + ->0
Above are only possible cases, clearly V is positive, so this statement is sufficient to prove

Statement 2
v4<v
Lets take Positive v =1/2
This proves the conditions.

But if you take Negative V=-1, condition is valid again.
So this is insufficient as it proves both + and - V

Posted from my mobile device

Is v < 0?

(1) uvu > 0
(2) v^4 < v

1 ->
uvu = (u^2)v
Square of a number is always positive or zero, hence u^2 is always positive as it can not be zero because given product in statement 1 is greater than zero
Product of two terms can only be positive if both the terms are of same sign -> 2*2>0 OR (-2)*(-2)>0
Thus, in (u^2)v>0 -> u^2 is positive; v>0
Hence this statement is sufficient. Eliminate option BCE

Statement 2 ->
v^4 < v
Even power of a number is always positive or zero; v cannot be zero as 0<0.
v^4 is always positive and v is always greater than that positive value of v^4, hence v cannot be negative
Hence v!<0
Statement 2 is also sufficient
This can be checked by plugging in some values to validate our answer. We can try putting up any values on both side of number lineand check by substituting.

Hence correct answer is D

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