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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 25 Aug 2012, 04:59
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A
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4)>0?

(1) xy > z^4
(2) x > z
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 25 Aug 2012, 05:19
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Given that X, Y, Z are non zero integers. Is (X^3)(Y^5)(Z^4)>0?

Notice that since z is a non-zero integer, then z^4>0, so we can reduce the given inequality by it and the question becomes: is x^3*y^5>0? or: is xy>0?

(1) XY > Z^4 --> since z^4>0, then we have that xy>z^4>0. Sufficient.
(2) X > Z. Not sufficient as we don't know anything about y.

Answer: A,
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 25 Aug 2012, 05:23
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Given that X, Y, Z are non zero integers. Is (X^3)(Y^5)(Z^4)>0?

(1) XY > Z^4
(2) X > Z

The sign of the given expression depends on the sign of the product \(XY\) because \(X^3Y^5Z^4=XY*X^2Y^4Z^4\) and \(X^2Y^4Z^4\) is always non-negative. The given product can be 0 as well, if at least one of the variables is 0.

(1) Not sufficient, because Z can be 0.
(2) Not sufficient, because Z can be 0.

(1) and (2): Although from (1) we can deduce that \(XY>0,\) (because \(Z^4\geq0\)) again not sufficient, the same fast reasoning, Z can still be 0.
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 25 Aug 2012, 05:34
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EvaJager
vinay911
Given that X, Y, Z are non zero integers. Is (X^3)(Y^5)(Z^4)>0?

(1) XY > Z^4
(2) X > Z

The sign of the given expression depends on the sign of the product \(XY\) because \(X^3Y^5Z^4=XY*X^2Y^4Z^4\) and \(X^2Y^4Z^4\) is always non-negative. The given product can be 0 as well, if at least one of the variables is 0.

(1) Not sufficient, because Z can be 0.
(2) Not sufficient, because Z can be 0.

(1) and (2): Although from (1) we can deduce that \(XY>0,\) (because \(Z^4\geq0\)) again not sufficient, the same fast reasoning, Z can still be 0.

Eva, notice that we are told that X, Y, Z are non zero integers.
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 25 Aug 2012, 05:47
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vinay911
Given that X, Y, Z are non zero integers. Is (X^3)(Y^5)(Z^4)>0?

(1) XY > Z^4
(2) X > Z

The sign of the given expression depends on the sign of the product \(XY\) because \(X^3Y^5Z^4=XY*X^2Y^4Z^4\) and \(X^2Y^4Z^4\) is always non-negative. The given product can be 0 as well, if at least one of the variables is 0.

(1) Not sufficient, because Z can be 0.
(2) Not sufficient, because Z can be 0.

(1) and (2): Although from (1) we can deduce that \(XY>0,\) (because \(Z^4\geq0\)) again not sufficient, the same fast reasoning, Z can still be 0.

Eva, notice that we are told that X, Y, Z are non zero integers.


Oooooops! Thanks.

So, forget about any of the variables being 0. We need to check whether \(XY>0.\)
Then the above solution changes:

(1) Sufficient, since \(XY>0,\) (because \(Z^4>0\)).
(2) Not sufficient, as we don't know anything about Y.

Answer A
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 26 Aug 2012, 10:34
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Given that X, Y, Z are non zero integers. Is (X^3)(Y^5)(Z^4)>0?

(1) XY > Z^4
(2) X > Z


My approach :-

(i) XY>Z^4
RHS is always +ve , so XY both can either be +ve or -ve
a)if XY both +ve then (X^3)(Y^5)(Z^4)>0 (because Z^4 is +ve
b)if XY both -ve then (X^3)(Y^5)(Z^4)>0 (because - - + is +ve

Suffficient

(ii) no clue about Y -insufficient

(A) wins
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 26 Aug 2012, 22:07
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it is given that x, y and z are no zero integer, if so z^4 is always positive what ever the z is.

to be positive the given expression we have to determine whether x and y have the same sign or not, keeping in mind that x and y have odd power.

statement 1

xy > z^4
XY > positive that means X and Y have the same sign

so the given expression must be positive

statement 1 is sufficient

statement 2 does not tell anything about y so insufficient

please correct me if i am wrong
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 21 Dec 2018, 11:12
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Question, whether \(X^3Y^5Z^4 > 0\)

From statement 1:

\(XY > Z^4\)
Since Z^4 is always positive.
Either X and Y are both negative or X and Y are both positive.
Hence, \(X^3Y^5Z^4\) will always be greater than 0.
Sufficient.

From statement 2:

\(X > Z\)
If X = -2 and Z = -3.
-2 > -3
But, \(X^3Y^5Z^4 < 0\)
If x = 2 and Z = 1.
Then \(X^3Y^5Z^4 > 0\)
Hence, Insufficient.

A is the answer.
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 08 Jan 2019, 19:33
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We need to see whether both x and y are positive or not for the inequality to be true.

1. product of xy is greater than z^4. We already know that z^4 will be positive so xy is positive. Hence Sufficient

2. we cannot conclude the value of y (+ve or -ve). Insufficient.

Hence A is the answer
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 09 Jan 2019, 07:29
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To prove: X^3*Y^5*Z^4>0
For this statement to be true, following cases must be satisfied:
Case 1: All terms: x^(3), y^(5) and z^(4) are positive
so possible values :X,Y should be positive, Z can be positive or negative
Case 2: z^(4) is positive, y^(5)*x^(3) is negative
Possible values:Z is positive/negative, y is positive, x is negative...xy is negative
Z is positive/negative, x is positive, y is negative.....xy is negative

Statement 1: XY>Z^4
z^(4) is always positive. so XY is also positive. so from above, case 1 is satisfied. Statement is sufficient

Statement 2: X>Z
Insufficient

Answer A
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 09 Jan 2019, 10:54
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chand567
Given that X, Y and Z are non zero integers. Is X^3*Y^5*Z^4>0 ?

(1) XY>Z^4

(2) X > Z


(1) XY> Z^4. Since X>4 must >0, this means XY>0. This is good because we can factor out XY without changing the sign of our expression

XY(x^2Y^4Z^4) >0 Now, since xy>0, x^2,y^2, z^4 must >0 then the entire expression must >0 Suff

(2) X> Z Suppose x=-2 z=-1 y=1 our expression becomes (-2)^3(1)^5(1)^4<0 giving us a NO. Now suppose x=1, y=1 Z=1, our expression becomes 1>0 giving us a yes. Since we get a Yes and a No by testing different appropriate values (2) is NS

Answer is A
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Given that x, y, z are non zero integers. Is (x^3)(y^5)(z^4) 10 Jan 2024, 21:14
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