Last visit was: 19 Nov 2025, 13:43 It is currently 19 Nov 2025, 13:43
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
study
User avatar
Joined: 05 Oct 2008
Last visit: 22 Mar 2022
Posts: 159
Own Kudos:
3,933
 [83]
Given Kudos: 22
Posts: 159
Kudos: 3,933
 [83]
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 13 Jun 2010, 09:11
6
Kudos
Add Kudos
76
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

46% (02:34) correct 54% (02:48) wrong based on 1219 sessions

History

Date
Time
Result
Not Attempted Yet
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) \(y = 4\)

Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
GMAT Club Tests Check Our Products Try Free Tests
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,351
 [46]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,351
 [46]
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 13 Jun 2010, 09:33
29
Kudos
Add Kudos
17
Bookmarks
Bookmark this Post
study
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. \(y = 4\)
Show more

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.

Answer: D.

Hope it helps.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,351
 [12]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,351
 [12]
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 29 Apr 2012, 05:24
8
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
keiraria
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
Show more

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.

Answer: D.
General Discussion
User avatar
study
User avatar
Joined: 05 Oct 2008
Last visit: 22 Mar 2022
Posts: 159
Own Kudos:
3,933
Given Kudos: 22
Posts: 159
Kudos: 3,933
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 13 Jun 2010, 10:04
Kudos
Add Kudos
Bookmarks
Bookmark this Post
excellent explanation..thanks. kudos
avatar
anandnat
avatar
Joined: 10 Jun 2010
Last visit: 01 Oct 2010
Posts: 31
Own Kudos:
131
Given Kudos: 2
Posts: 31
Kudos: 131
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 21 Jul 2010, 09:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,351
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,351
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 21 Jul 2010, 09:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
anandnat
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.
Show more

Pleas read the solution above.

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 \(x\).
avatar
blink005
avatar
Joined: 20 Aug 2011
Last visit: 23 Jun 2016
Posts: 61
Own Kudos:
215
 [1]
Posts: 61
Kudos: 215
 [1]
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 07 Jan 2012, 03:37
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
shinbhu
User avatar
Joined: 29 Jul 2011
Last visit: 14 Oct 2013
Posts: 51
Own Kudos:
174
 [1]
Given Kudos: 6
Location: United States
Posts: 51
Kudos: 174
 [1]
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 07 Jan 2012, 15:39
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
User avatar
fameatop
User avatar
Joined: 24 Aug 2009
Last visit: 09 Jun 2017
Posts: 383
Own Kudos:
2,495
Given Kudos: 275
Concentration: Finance
Schools:Harvard, Columbia, Stern, Booth, LSB,
Posts: 383
Kudos: 2,495
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 22 Sep 2012, 10:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
keiraria
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.

Answer: D.
Show more

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.
User avatar
PathFinder007
User avatar
Joined: 10 Mar 2014
Last visit: 21 Oct 2018
Posts: 129
Own Kudos:
733
 [1]
Given Kudos: 13
Posts: 129
Kudos: 733
 [1]
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 27 Apr 2014, 06:17
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
study
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.

Answer: D.

Hope it helps.
Show more

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

Please clarify

Thanks.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,351
 [1]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,351
 [1]
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 28 Apr 2014, 02:41
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PathFinder007
Bunuel
study
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.

Answer: D.

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

Please clarify

Thanks.
Show more

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?
avatar
sanguine2016
avatar
Joined: 18 Mar 2015
Last visit: 13 Jul 2016
Posts: 20
Own Kudos:
26
Given Kudos: 1
Location: India
Concentration: Technology, Entrepreneurship
Schools: Anderson '18 (A) Haas '18 (D) Kellogg '18 (D) ISB '17 (S) Sloan '18 (S)
GMAT 1: 730 Q50 V38
GPA: 4
WE:Engineering (Computer Software)
Schools: Anderson '18 (A) Haas '18 (D) Kellogg '18 (D) ISB '17 (S) Sloan '18 (S)
GMAT 1: 730 Q50 V38
Posts: 20
Kudos: 26
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 21 May 2015, 10:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
stonecold
User avatar
Joined: 12 Aug 2015
Last visit: 09 Apr 2024
Posts: 2,244
Own Kudos:
3,549
 [1]
Given Kudos: 893
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Posts: 2,244
Kudos: 3,549
 [1]
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 21 Aug 2016, 07:08
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
abhimahna
User avatar
Board of Directors
Joined: 18 Jul 2015
Last visit: 06 Jul 2024
Posts: 3,514
Own Kudos:
5,728
Given Kudos: 346
Status:Emory Goizueta Alum
Products:
My Reviews
Expert
Expert reply
Posts: 3,514
Kudos: 5,728
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 21 Aug 2016, 10:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
stonecold
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
Show more

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. :-D
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 19 Nov 2025
Posts: 8,422
Own Kudos:
4,982
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,422
Kudos: 4,982
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 01 Mar 2019, 04:37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
study
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) \(y = 4\)
Show more

solve for the given eqn:

\(x^2 = y + 5\) , \(y = z - 2\) and \(z = 2x\)

we get
x=-1,+3
y= -4,4
z=6,-2

#1
\(x \gt 0\)
so x=3,y=4,z=-2
\(x^3 + y^2 + z\)
we get 49; yes
sufficeint
#2
\(y = 4\)
i.e x=3,z=-2
yes, sufficient
IMO D
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,587
Own Kudos:
1,079
Posts: 38,587
Kudos: 1,079
If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z 23 Jan 2025, 12:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Bunuel
Math Expert
105390 posts
kingbucky
496 posts