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Algebra - JSQ91

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Senior Manager
Joined: 25 Nov 2006
Posts: 333
Schools: St Gallen, Cambridge, HEC Montreal

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07 Feb 2009, 16:59

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Senior Manager
Joined: 30 Nov 2008
Posts: 482
Schools: Fuqua

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07 Feb 2009, 17:36
IMO E.

from stmt 1,

$$sqrt(xz)$$ is an integer ==> product of x and z is a square number like 4,9,16 etc

let $$x = 2, z = 8 ==> sqrt(xz) = sqrt(16)$$= which is an integer. But $$sqrt(z)$$ which is $$sqrt(8)$$ is not a integer.
let $$x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =$$which is an integer. And $$sqrt(z)$$ which is $$sqrt(9)$$ is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

$$x = z^2. ==> sqrt(xz) = sqrt(z^3)$$

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then $$sqrt(z) = 1$$which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.
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Joined: 07 Nov 2007
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07 Feb 2009, 18:12
agree with E.

same explanation.
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Joined: 04 Jan 2009
Posts: 235

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07 Feb 2009, 19:09
mrsmarthi wrote:
IMO E.

from stmt 1,

$$sqrt(xz)$$ is an integer ==> product of x and z is a square number like 4,9,16 etc

let $$x = 2, z = 8 ==> sqrt(xz) = sqrt(16)$$= which is an integer. But $$sqrt(z)$$ which is $$sqrt(8)$$ is not a integer.
let $$x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =$$which is an integer. And $$sqrt(z)$$ which is $$sqrt(9)$$ is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

$$x = z^2. ==> sqrt(xz) = sqrt(z^3)$$

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then $$sqrt(z) = 1$$which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.

I agree with you that both by themselves not sufficient.
Now, if taken together we have the following:
z*sqrt(z) is integer. Z is integer known already.
if z=1, sqrt z is integer.
if z=2, sqrtz is noninteger. But B is not satisfied.
If z=4, sqrtz is integer. and this satisfies both a and B.
Both the statements are satisfied for a and b to obe true.

Hence, C.
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tusharvk

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07 Feb 2009, 19:32
tusharvk wrote:
mrsmarthi wrote:
IMO E.

from stmt 1,

$$sqrt(xz)$$ is an integer ==> product of x and z is a square number like 4,9,16 etc

let $$x = 2, z = 8 ==> sqrt(xz) = sqrt(16)$$= which is an integer. But $$sqrt(z)$$ which is $$sqrt(8)$$ is not a integer.
let $$x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =$$which is an integer. And $$sqrt(z)$$ which is $$sqrt(9)$$ is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

$$x = z^2. ==> sqrt(xz) = sqrt(z^3)$$

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then $$sqrt(z) = 1$$which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.

I agree with you that both by themselves not sufficient.
Now, if taken together we have the following:
z*sqrt(z) is integer. Z is integer known already.
if z=1, sqrt z is integer.
if z=2, sqrtz is noninteger. But B is not satisfied.
If z=4, sqrtz is integer. and this satisfies both a and B.
Both the statements are satisfied for a and b to obe true.

Hence, C.

good catch..

sqrt(z^3) = z*sqrt(z) = integer.. --> sqrt(z) must be integer

how did I miss that?
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Joined: 30 Nov 2008
Posts: 482
Schools: Fuqua

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07 Feb 2009, 19:35
tusharvk wrote:
mrsmarthi wrote:
IMO E.

from stmt 1,

$$sqrt(xz)$$ is an integer ==> product of x and z is a square number like 4,9,16 etc

let $$x = 2, z = 8 ==> sqrt(xz) = sqrt(16)$$= which is an integer. But $$sqrt(z)$$ which is $$sqrt(8)$$ is not a integer.
let $$x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =$$which is an integer. And $$sqrt(z)$$ which is $$sqrt(9)$$ is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

$$x = z^2. ==> sqrt(xz) = sqrt(z^3)$$

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then $$sqrt(z) = 1$$which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.

I agree with you that both by themselves not sufficient.
Now, if taken together we have the following:
z*sqrt(z) is integer. Z is integer known already.
if z=1, sqrt z is integer.
if z=2, sqrtz is noninteger. But B is not satisfied.
If z=4, sqrtz is integer. and this satisfies both a and B.
Both the statements are satisfied for a and b to obe true.

Hence, C.

Tusharvk, I will have to agree with you. By using both the eq, we definitely know that $$sqrt(z)$$is an integer. Hence together is sufficient. here is my approach.

From stmt 2, we know that $$x = z^2 ==> sqrt(xz) = sqrt(z^3) ==> z sqrt(z).$$

from stmt 1, we know that $$sqrt(xz)$$ is an integer ==> $$z sqrt(z)$$ is an integer. And this possible only when $$sqrt(z)$$ is an integer.

I strongly believe that skill of reasoning is what is needed to master the DS questions in addition to the basic maths. But is there any how can help me know, whether we can cross verify the approach to confirm the correctness?
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Posts: 2453

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07 Feb 2009, 20:41
mrsmarthi wrote:
IMO E.

from stmt 1,

$$sqrt(xz)$$ is an integer ==> product of x and z is a square number like 4,9,16 etc

let $$x = 2, z = 8 ==> sqrt(xz) = sqrt(16)$$= which is an integer. But $$sqrt(z)$$ which is $$sqrt(8)$$ is not a integer.
let $$x= 4, z = 9 ==> sqrt(xz) = sqrt(36) =$$which is an integer. And $$sqrt(z)$$ which is $$sqrt(9)$$ is an integer.

Hence stmt 1 alone is not sufficient to say if z is an integer or not.

From stmt 2,

$$x = z^2. ==> sqrt(xz) = sqrt(z^3)$$

If z is anything other than 1 then sqrt(z) is not an integer
But if z = 1 then $$sqrt(z) = 1$$which is an integer.

Hence stmt 2 alone is not sufficient to say if z is an integer or not.

Togerther also, does not provide enough information.

You mixed st 1 in st 2: $$x = z^2. ==> sqrt(xz) = sqrt(z^3)$$
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Senior Manager
Joined: 30 Nov 2008
Posts: 482
Schools: Fuqua

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07 Feb 2009, 20:50
GMAT TIGER wrote:
You mixed st 1 in st 2:

Hum....ok now I got the mistake. well Let me see if I can do it correctly in similar example. Thank you.

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Re: Algebra - JSQ91   [#permalink] 07 Feb 2009, 20:50
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Algebra - JSQ91

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