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Re: Data sufficiency +- exponents question [#permalink]

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10 Jun 2010, 14:22

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1

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TheGmatTutor wrote:

Bunuel, I agree about S1. But on S2, that quantity y^2 will always be positive, correct?

No, not correct. \(y^2\) is not always positive, it's never negative: \(y^2\geq{0}\).

Inequality \(x^7*y^2*z^3>0\) to be true \(x\) and \(z\) must be either both positive or both negative (note that both positive or both negative excludes the possibility of either of them to be zero) AND \(y\) must not be zero. Because if \(y=0\), then \(x^7*y^2*z^3=0\). _________________

Response: x^7*y^2*z^3 is same as: (xz)*(x^6)*(y^2)*(z^2) xz is positive based on assumption #2. x^6 is (x^2)^3, since x^2 is positive for all real numbers, x^6 is also positive. z^2 is positive for all real numbers. y^2 is positive for all real numbers.

So product of 4 positive real numbers is also positive. SUFFICIENT.

Hence, B - Statement 2 Alone is Sufficient.

Per the gmatclub math25 test, the response is C - both statements together are sufficient. What incorrect assumptions am I making?

Thanks!

The red parts are not correct. Square of a number is nonnegative and not positive as you've written. So for (2) if y=0 then x^7*y^2*z^3=0.

Complete solution: Is x^7*y^2*z^3 > 0 ?

Inequality \(x^7*y^2*z^3>0\) to be true \(x\) and \(z\) must be either both positive or both negative (in order \(x^7*z^3\) to be positive) AND \(y\) must not be zero (in order \(x^7*y^2*z^3\) not to equal to zero).

(1) \(yz<0\) --> \(y\neq{0}\). Don't know about \(x\) and \(z\). Not sufficient.

(2) \(xz>0\) --> \(x\) and \(z\) are either both positive or both negative. Don't know about \(y\). Not sufficient.

Re: Is x^7*y^2*z^3 > 0 ? (1) yz < 0 (2) xz > 0 [#permalink]

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23 Mar 2012, 06:03

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kraizada84 wrote:

x^7*y^2*z^3>0 y is positive, we need to know regarding x and z 1) no information about x insufficient

2) xz>0

either of them is negative hence

the answer to our question is no

sufficient

hence B

OA for this question is C, not B.

(2) \(xz>0\) means that\(x\) and \(z\) are either both positive or both negative. So, \(x^7*z^3 > 0\) but y can still be zero and in this case \(x^7*y^2*z^3=0\), hence this statement is not sufficient.

Answer is C. However, for reaching to the conclusion of YES/NO, we have to be certain that: (1) x, y, z are not ZERO, (2) [highlight]x and y are not NEGATIVE[/highlight]

Combined from (1) and (2) we can say that x, y, z are not ZERO. However, I think they are not helpful in deciding the certainty that x and y are not NEGATIVE.

Can some one please explain the answer?

First of all, a trick in the question is 'x, y, z are not ZERO' so good that you figured it. Next, we don't need to know that x and z are not negative. We need to know [highlight]whether they have the same sign or opposite signs[/highlight] because question asks you whether (x^7)(z^3) is positive. (Ignoring y for now) For the product to be positive, either both should be positive or both negative. Then, answer will be 'YES' For the product to be negative only one of them should be negative. Then answer will be 'NO' In either case, if we get a definite YES/NO, the statements will be sufficient. If xz> 0, then either x and z both are positive or both are negative. They have the same sign. So (x^7)(z^3) is positive. y, we know is not 0, so YES, (x^7)(y^2)(z^3) is greater than 0. Sufficient. _________________

Hi, When I solved this i get an answer as B. i.e. only 2nd statement is sufficient. But GMAT Club test says its C. Can anyone please help.

According to me, if xz > 0 then the inequality can be written as [(xz)^3 (x^4) (y^2)] Now since x^4 and y^2 are going to be positive always then the statement in itself gives us the answer.

Re: Is (x^7)(y^2)(z^3) > 0 ? (1) yz < 0 (2) xz > 0 [#permalink]

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17 May 2015, 20:35

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octenisept wrote:

Guys, is there any difference between (y^2) and (y)^2. We can be sure that (y)^2 is positive but can we also be sure that (y^2) is positive? (forget about y=0 for a moment)

thank you

(y^2) and (y)^2 are the same. In both cases, y is squared. y could be anything - the whole of it is squared in both cases.

If y = 2a + b, both representations give (2a + b)^2 = 4a^2 + b^2 + 4ab _________________

Re: Data sufficiency +- exponents question [#permalink]

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11 Jan 2011, 06:47

Bunuel wrote:

TheGmatTutor wrote:

Bunuel, I agree about S1. But on S2, that quantity y^2 will always be positive, correct?

No, not correct. \(y^2\) is not always positive, it's never negative: \(y^2\geq{0}\).

Inequality \(x^7*y^2*z^3>0\) to be true \(x\) and \(z\) must be either both positive or both negative (note that both positive or both negative excludes the possibility of either of them to be zero) AND \(y\) must not be zero. Because if \(y=0\), then \(x^7*y^2*z^3=0\).

Thanks a ton Bunuel , this example is the perfect for learning that while considering signs we should consider +ve , -ve and zero as well . Superb collection .

Re: Data sufficiency +- exponents question [#permalink]

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13 Jun 2012, 12:24

Hi,

Simplifying the expression, \(x^7y^2z^3\), it can be written as, \((xz)^3x^4y^2\) or \((yz)^2x^6(xz)\)

Clearly, in both the expressions \(x^4y^2\) as well as \((yz)^2x^6\) are positive. Thus the sign of expression depends on sign of xz, but since value of y can be 0, Using (1), we can say y is not equal to 0.

Response: x^7*y^2*z^3 is same as: (xz)*(x^6)*(y^2)*(z^2) xz is positive based on assumption #2. x^6 is (x^2)^3, since x^2 is positive for all real numbers, x^6 is also positive. z^2 is positive for all real numbers. y^2 is positive for all real numbers.

So product of 4 positive real numbers is also positive. SUFFICIENT.

Hence, B - Statement 2 Alone is Sufficient.

Per the gmatclub math25 test, the response is C - both statements together are sufficient. What incorrect assumptions am I making?

Thanks!

The red parts are not correct. Square of a number is nonnegative and not positive as you've written. So for (2) if y=0 then x^7*y^2*z^3=0.

Complete solution: Is x^7*y^2*z^3 > 0 ?

Inequality \(x^7*y^2*z^3>0\) to be true \(x\) and \(z\) must be either both positive or both negative (in order \(x^7*z^3\) to be positive) AND \(y\) must not be zero (in order \(x^7*y^2*z^3\) not to equal to zero).

(1) \(yz<0\) --> \(y\neq{0}\). Don't know about \(x\) and \(z\). Not sufficient.

(2) \(xz>0\) --> \(x\) and \(z\) are either both positive or both negative. Don't know about \(y\). Not sufficient.

Response: x^7*y^2*z^3 is same as: (xz)*(x^6)*(y^2)*(z^2) xz is positive based on assumption #2. x^6 is (x^2)^3, since x^2 is positive for all real numbers, x^6 is also positive. z^2 is positive for all real numbers. y^2 is positive for all real numbers.

So product of 4 positive real numbers is also positive. SUFFICIENT.

Hence, B - Statement 2 Alone is Sufficient.

Per the gmatclub math25 test, the response is C - both statements together are sufficient. What incorrect assumptions am I making?

Thanks!

The red parts are not correct. Square of a number is nonnegative and not positive as you've written. So for (2) if y=0 then x^7*y^2*z^3=0.

Complete solution: Is x^7*y^2*z^3 > 0 ?

Inequality \(x^7*y^2*z^3>0\) to be true \(x\) and \(z\) must be either both positive or both negative (in order \(x^7*z^3\) to be positive) AND \(y\) must not be zero (in order \(x^7*y^2*z^3\) not to equal to zero).

(1) \(yz<0\) --> \(y\neq{0}\). Don't know about \(x\) and \(z\). Not sufficient.

(2) \(xz>0\) --> \(x\) and \(z\) are either both positive or both negative. Don't know about \(y\). Not sufficient.

Hi, When I solved this i get an answer as B. i.e. only 2nd statement is sufficient. But GMAT Club test says its C. Can anyone please help.

According to me, if xz > 0 then the inequality can be written as [(xz)^3 (x^4) (y^2)] Now since x^4 and y^2 are going to be positive always then the statement in itself gives us the answer.

Hi, When I solved this i get an answer as B. i.e. only 2nd statement is sufficient. But GMAT Club test says its C. Can anyone please help.

According to me, if xz > 0 then the inequality can be written as [(xz)^3 (x^4) (y^2)] Now since x^4 and y^2 are going to be positive always then the statement in itself gives us the answer.

Got it. Forgot to consider y = 0. As always. Thanks anyways.

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