Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

\(sqrt(xz)\) 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.

\(sqrt(xz)\) 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.

\(sqrt(xz)\) 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?
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

\(sqrt(xz)\) 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?

\(sqrt(xz)\) 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: \(x = z^2. ==> sqrt(xz) = sqrt(z^3)\) _________________