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

 It is currently 20 Sep 2018, 17:19

### 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

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

# If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 05 Oct 2008
Posts: 257
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z  [#permalink]

### Show Tags

13 Jun 2010, 09:11
2
30
00:00

Difficulty:

95% (hard)

Question Stats:

44% (01:48) correct 56% (02:09) wrong based on 913 sessions

### HideShow timer Statistics

If $$x^2 = y + 5$$ , $$y = z - 2$$ and $$z = 2x$$ , is $$x^3 + y^2 + z$$ divisible by 7?

(1) $$x \gt 0$$
(2) $$y = 4$$
Math Expert
Joined: 02 Sep 2009
Posts: 49271

### Show Tags

13 Jun 2010, 09:33
16
8
study wrote:
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

1. $$x \gt 0$$
2. $$y = 4$$

We have system of equations with three distinct equations and three unknowns, so we can solve it. As $$x$$ is squared we'll get two values for it and also two values for $$y$$ and $$z$$: two triplets. Hence we'll get two values for $$x^3 + y^2 + z$$, one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

Hope it helps.
_________________
Intern
Joined: 05 Apr 2012
Posts: 43
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z  [#permalink]

### Show Tags

Updated on: 10 Jul 2013, 02:29
1
6
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

(1) x > 0
(2) y = 4

Originally posted by keiraria on 29 Apr 2012, 02:29.
Last edited by Bunuel on 10 Jul 2013, 02:29, edited 2 times in total.
Edited the question
##### General Discussion
Senior Manager
Joined: 05 Oct 2008
Posts: 257

### Show Tags

13 Jun 2010, 10:04
excellent explanation..thanks. kudos
Intern
Joined: 10 Jun 2010
Posts: 39

### Show Tags

21 Jul 2010, 09:21
shouldn't the answer be B? when x=1 or x=2, the answer is not divisible by 7, while when x=3, it is. Hence A is insufficient.
Math Expert
Joined: 02 Sep 2009
Posts: 49271

### Show Tags

21 Jul 2010, 09:33
anandnat wrote:
shouldn't the answer be B? when x=1 or x=2, the answer is not divisible by 7, while when x=3, it is. Hence A is insufficient.

What are the values of $$z$$ and $$y$$ when $$x=1$$ or $$x=2$$? What is the value of $$x^3+y^2+z$$ when $$x=1$$ or $$x=2$$?

$$x=1$$ and $$x=2$$ do not satisfy the system of equations given in the stem, hence are not the valid solutions for $$x$$.
_________________
Manager
Joined: 20 Aug 2011
Posts: 128

### Show Tags

07 Jan 2012, 03:37
1
X^2=Y+5
Y=Z-2=2X-2
Z=2X

X^2=2X=3
X^2-2X-3=0
X=3, -1
Y=4,-4
Z=6,-2
X^3+Y^2+Z

1.
X>0
X=3, Y=4, Z=6
X^3+Y^2+Z= Some value we don't care about but can determine whether or not the value is divisible by 7.

Sufficient

2.
y=4, X=3, Z=6
X^3+Y^2+Z= Some value we don't care about but can determine whether or not the value is divisible by 7.

Sufficient

Hence D
_________________

Hit kudos if my post helps you.
You may send me a PM if you have any doubts about my solution or GMAT problems in general.

Manager
Joined: 29 Jul 2011
Posts: 94
Location: United States

### Show Tags

07 Jan 2012, 15:39
Rephrase the stem:
x^2 = y + 5
x^2 - y + 5 = 0
x^2 -(z-2) + 5 = 0
x^2 - z - 3 = 0
x^2 - 2x - 3 = 0
x = 3, -1
Substituting gives z = 6, -2 and y = 4, -4

Notice that both statements ask for same information, that is 1) x>0 -> x=3 and corresponding y=4 and z=6. 2) y=4 means x=3 and z=6

Substituting these values in the stem gives us 51, which is not div by 7. So, both suff.

D
_________________

I am the master of my fate. I am the captain of my soul.
Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution.
PS - Always look at the answers first
CR - Read the question stem first, hunt for conclusion
SC - Meaning first, Grammar second
RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Math Expert
Joined: 02 Sep 2009
Posts: 49271
Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z  [#permalink]

### Show Tags

29 Apr 2012, 05:24
8
4
keiraria wrote:
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

(1) x>0
(2) y = 4

We have system of equations with three distinct equations and three unknowns, so we can solve it. As $$x$$ is squared we'll get two values for it and also two values for $$y$$ and $$z$$: two triplets. Hence we'll get two values for $$x^3 + y^2 + z$$, one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

_________________
Senior Manager
Joined: 24 Aug 2009
Posts: 476
Schools: Harvard, Columbia, Stern, Booth, LSB,
Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z  [#permalink]

### Show Tags

22 Sep 2012, 10:28
Bunuel wrote:
keiraria wrote:
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

(1) x>0
(2) y = 4

We have system of equations with three distinct equations and three unknowns, so we can solve it. As $$x$$ is squared we'll get two values for it and also two values for $$y$$ and $$z$$: two triplets. Hence we'll get two values for $$x^3 + y^2 + z$$, one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

Your explanation is good but i want to know where am i making a mistake.
z=2x
y= 2x-2
x^2 = 2x+3
$$x^3 + y^2 +z = x(2x+3) + (2x-2)^2 + 2x = 6x^2 -3x +4$$
Why cant we use this equation to solve the question.
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS.
Kudos always maximizes GMATCLUB worth
-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Manager
Joined: 10 Mar 2014
Posts: 205

### Show Tags

27 Apr 2014, 06:17
Bunuel wrote:
study wrote:
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

1. $$x \gt 0$$
2. $$y = 4$$

We have system of equations with three distinct equations and three unknowns, so we can solve it. As $$x$$ is squared we'll get two values for it and also two values for $$y$$ and $$z$$: two triplets. Hence we'll get two values for $$x^3 + y^2 + z$$, one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

Hope it helps.

Hi bunnel,

In second statement y=4 if we replace we found x^2 = 4+5 = 9

so can we write x= +3 and -3. if we replace x= -3 then sum will be -24+16+6 = -1. then how it can be divisible by 7

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 49271

### Show Tags

28 Apr 2014, 02:41
PathFinder007 wrote:
Bunuel wrote:
study wrote:
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

1. $$x \gt 0$$
2. $$y = 4$$

We have system of equations with three distinct equations and three unknowns, so we can solve it. As $$x$$ is squared we'll get two values for it and also two values for $$y$$ and $$z$$: two triplets. Hence we'll get two values for $$x^3 + y^2 + z$$, one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$

$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$

$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.

(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.

Hope it helps.

Hi bunnel,

In second statement y=4 if we replace we found x^2 = 4+5 = 9

so can we write x= +3 and -3. if we replace x= -3 then sum will be -24+16+6 = -1. then how it can be divisible by 7

Thanks.

Given: $$x^2 = y + 5$$, $$y = z - 2$$ and $$z = 2x$$.

(2) $$y = 4$$:

$$y = z - 2$$ --> $$z=6$$;
$$z = 2x$$ --> $$x=3$$.

So, x cannot be -3.

Does this make sense?
_________________
Current Student
Joined: 18 Mar 2015
Posts: 21
Location: India
Concentration: Technology, Entrepreneurship
GMAT 1: 730 Q50 V38
GPA: 4
WE: Engineering (Computer Software)
Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z  [#permalink]

### Show Tags

21 May 2015, 10:16
If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z divisible by 7?

What is provided to us: z = 2x and y = z - 2 => y = 2x -2.
and x^2 = y + 5 => x^2 = 2x -2 + 5
=> x^2 - 2x - 3 = 0.
Solving this Quad. Eqn. we get x = -1 or x = 3.
So, following sets for possible values of x,y and z :: (x=-1,y=-4,z=-2) OR (x=3,y=4,z=6)

Now,
(1) x > 0
if x > 0, then we will have x =3, y = 4 and z = 6. And we can easily evaluate the expression: x^3 + y^2 + z = 49. So Yes divisible by 7.
So, Sufficient.
(2) y = 4
Again we have, x =3, y = 4 and z = 6. And x^3 + y^2 + z = 49. So Yes divisible by 7.
So, Sufficient.
Hence,
Ans:: D.
Current Student
Joined: 12 Aug 2015
Posts: 2648
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z  [#permalink]

### Show Tags

21 Aug 2016, 07:08
1
Remember => we dont need to solve this scary looking question.
we need the value of x,y,z
now solving the system of three equations its clear that xcan be 3 and -1
now statement 1=> x>0 => sufficient as x=3 hence we can get the value of y and z
statement 2 => again if y=4 => x=3 and z=6 so we have the values of x,y,z

Note => divisiblity by 7 can be replaced by 13 or 5 or 17 or 20 even 138483461349273794739939738972 that would still be immaterial here. we just need the number a.k.a the values of x,y,z

Smash that D

Stone Cold
_________________

MBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE!

Getting into HOLLYWOOD with an MBA!

The MOST AFFORDABLE MBA programs!

STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)

AVERAGE GRE Scores At The Top Business Schools!

Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3674
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z  [#permalink]

### Show Tags

21 Aug 2016, 10:34
stonecold wrote:
Remember => we dont need to solve this scary looking question.
we need the value of x,y,z
now solving the system of three equations its clear that xcan be 3 and -1
now statement 1=> x>0 => sufficient as x=3 hence we can get the value of y and z
statement 2 => again if y=4 => x=3 and z=6 so we have the values of x,y,z

Note => divisiblity by 7 can be replaced by 13 or 5 or 17 or 20 even 138483461349273794739939738972 that would still be immaterial here. we just need the number a.k.a the values of x,y,z

Smash that D

Stone Cold

I also followed the same approach and got x =3 or -1. I never looked for the scary equation and both the options gave me x=3 and hence will always give a unique value of the scary equation.
_________________

My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place

GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.

Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free

Non-Human User
Joined: 09 Sep 2013
Posts: 8105
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z  [#permalink]

### Show Tags

04 Jan 2018, 08:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z &nbs [#permalink] 04 Jan 2018, 08:09
Display posts from previous: Sort by

# Events & Promotions

 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®.