GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 May 2019, 22:08

### GMAT Club Daily Prep

#### 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

# M26-30

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55276
M26-30  [#permalink]

### Show Tags

16 Sep 2014, 01:26
20
00:00

Difficulty:

95% (hard)

Question Stats:

25% (01:19) correct 75% (01:16) wrong based on 286 sessions

### HideShow timer Statistics

If $$x$$, $$y$$, and $$z$$ are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ an integer?

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.

(2) $$\sqrt{z}$$ is not an integer.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 55276
Re M26-30  [#permalink]

### Show Tags

16 Sep 2014, 01:26
1
5
Official Solution:

Note: a perfect square, is an integer that can be written as the square of some other integer. For example $$16=4^2$$, is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, $$27=3^3$$ is a perfect cube.

Make prime factorization of 2,700: $$xyz=2,700=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube. If $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3= \text{odd perfect square}$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3= \text{odd perfect cube}$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z} \ne integer$$ then $$z \ne 1$$, therefore it must be $$3^3$$ (from 1), so $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

Answer: C
_________________
Intern
Joined: 11 Aug 2014
Posts: 5
Re: M26-30  [#permalink]

### Show Tags

16 Nov 2014, 00:21
1
1
Bunuel wrote:
Official Solution:

Note: a perfect square, is an integer that can be written as the square of some other integer. For example $$16=4^2$$, is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, $$27=3^3$$ is a perfect cube.

Make prime factorization of 2,700: $$xyz=2,700=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube. If $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3= \text{odd perfect square}$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3= \text{odd perfect cube}$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z} \ne integer$$ then $$z \ne 1$$, therefore it must be $$3^3$$ (from 1), so $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

Answer: C

can you please explain 2) \sqrt{z} is not an integer. Clearly insufficient. and sqrt{z} ne integer then z ne 1
Math Expert
Joined: 02 Sep 2009
Posts: 55276
Re: M26-30  [#permalink]

### Show Tags

16 Nov 2014, 06:06
parameswaranprasad wrote:
Bunuel wrote:
Official Solution:

Note: a perfect square, is an integer that can be written as the square of some other integer. For example $$16=4^2$$, is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, $$27=3^3$$ is a perfect cube.

Make prime factorization of 2,700: $$xyz=2,700=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube. If $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3= \text{odd perfect square}$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3= \text{odd perfect cube}$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z} \ne integer$$ then $$z \ne 1$$, therefore it must be $$3^3$$ (from 1), so $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

Answer: C

can you please explain 2) \sqrt{z} is not an integer. Clearly insufficient. and sqrt{z} ne integer then z ne 1

$$z \ne 1$$ because if it were 1, then $$\sqrt{z}$$ would be an integer and that would violate the second statement.
_________________
Intern
Joined: 30 Jun 2012
Posts: 12
M26-30  [#permalink]

### Show Tags

07 Dec 2014, 16:21
Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.
Board of Directors
Joined: 17 Jul 2014
Posts: 2552
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
M26-30  [#permalink]

### Show Tags

07 Dec 2014, 16:57
rsamant wrote:
Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

would like as well answer to this question.
Math Expert
Joined: 02 Sep 2009
Posts: 55276
Re: M26-30  [#permalink]

### Show Tags

08 Dec 2014, 05:10
mvictor wrote:
rsamant wrote:
Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

would like as well answer to this question.

There are other cases possible. For example, z=3, y=2*3*5^2 and x=2.
_________________
Intern
Joined: 23 Apr 2016
Posts: 21
Location: Finland
Concentration: General Management, International Business
GPA: 3.65
Re: M26-30  [#permalink]

### Show Tags

07 Nov 2016, 10:24
Really nice question Bunuel !
Intern
Joined: 04 Jan 2017
Posts: 5
Re: M26-30  [#permalink]

### Show Tags

10 Apr 2017, 09:19
Awesome question!
Intern
Joined: 24 Jun 2017
Posts: 14
GMAT 1: 750 Q49 V44
GPA: 3.1
Re: M26-30  [#permalink]

### Show Tags

30 Sep 2017, 09:21
Wow what a question! Very difficult without blatantly trying to trick you.

This one took me a while, but I'm very proud to say I got it right!
Intern
Joined: 01 May 2017
Posts: 18
Location: United States (IL)
Concentration: Technology, Strategy
GPA: 3.95
Re M26-30  [#permalink]

### Show Tags

29 Jul 2018, 03:43
I think this is a high-quality question and I agree with explanation. Awesome!!!
Manager
Joined: 06 Nov 2016
Posts: 61
Location: Viet Nam
Concentration: Strategy, International Business
GPA: 3.54
Re: M26-30  [#permalink]

### Show Tags

29 Jul 2018, 10:33
I think this is a high-quality question and I agree with explanation.
Didn't think abt the case z=1^3 in statement 1 urgggg... Need to be more careful.
_________________
Intern
Joined: 28 Aug 2018
Posts: 1
Location: India
GMAT 1: 650 Q49 V30
Re: M26-30  [#permalink]

### Show Tags

28 Aug 2018, 07:56
Great Question, forgot to check the case for z= 1^3.
Manager
Joined: 20 Jul 2018
Posts: 88
GPA: 2.87
Re: M26-30  [#permalink]

### Show Tags

30 Aug 2018, 08:26
1
by prime factorization of 2700 = $$2^2 * 3^3 * 5^2$$

1) if y is perfect square and z is perfect cube then y could be $$2^2 * 5^2$$ or $$2^2$$ or $$5^2$$ and z could be $$3^3$$ or $$1^3$$. so if z is $$3^3$$ then $$\sqrt{x}$$ is an integer but if z is $$1^3$$ then $$\sqrt{x}$$ is not an integer. INSUFFICIENT

2) if $$\sqrt{z}$$ is not an integer, its clearly INSUFFICIENT

by combining both we know than z could only take value of $$3^3$$, so $$\sqrt{x}$$ would always be an integer. SUFFICIENT

Answer C
_________________
Hasnain Afzal

"When you wanna succeed as bad as you wanna breathe, then you will be successful." -Eric Thomas
Intern
Joined: 19 Nov 2012
Posts: 28
M26-30  [#permalink]

### Show Tags

29 Sep 2018, 09:10
Bunuel wrote:
mvictor wrote:
rsamant wrote:
Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

would like as well answer to this question.

There are other cases possible. For example, z=3, y=2*3*5^2 and x=2.

Hi Bunuel: shouldn't one additional 3 be in included in the factorization of y? I guess that's a typo.
Needless to say Brilliant question. even understanding the given solution takes lot of efforts; forget solving it correctly.
Manager
Joined: 01 Nov 2018
Posts: 74
GMAT 1: 690 Q48 V35
GPA: 3.88
Re: M26-30  [#permalink]

### Show Tags

23 Dec 2018, 13:26
Great Question. 50-51 level. Concepts isn't difficult but LOTS to think about, especially during a time crunch. Picked D because short on time but well though out question.
Re: M26-30   [#permalink] 23 Dec 2018, 13:26
Display posts from previous: Sort by

# M26-30

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.