May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 05 Oct 2008
Posts: 241

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
Question Stats:
43% (02:31) correct 57% (02:51) wrong based on 730 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) \(y = 4\)
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Divisibility by 7
[#permalink]
Show Tags
13 Jun 2010, 09:33
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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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.
_________________




Intern
Joined: 05 Apr 2012
Posts: 41

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
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




Manager
Joined: 05 Oct 2008
Posts: 241

Re: Divisibility by 7
[#permalink]
Show Tags
13 Jun 2010, 10:04
excellent explanation..thanks. kudos



Intern
Joined: 10 Jun 2010
Posts: 35

Re: Divisibility by 7
[#permalink]
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: 55271

Re: Divisibility by 7
[#permalink]
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. 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 \(x\).
_________________



Manager
Joined: 20 Aug 2011
Posts: 128

Re: Arithmetic
[#permalink]
Show Tags
07 Jan 2012, 03:37
X^2=Y+5 Y=Z2=2X2 Z=2X X^2=2X=3 X^22X3=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: 85
Location: United States

Re: Arithmetic
[#permalink]
Show Tags
07 Jan 2012, 15:39
Rephrase the stem: x^2 = y + 5 x^2  y + 5 = 0 x^2 (z2) + 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 = 34 min



Math Expert
Joined: 02 Sep 2009
Posts: 55271

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
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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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.
_________________



Senior Manager
Joined: 24 Aug 2009
Posts: 461
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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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. Your explanation is good but i want to know where am i making a mistake. z=2x y= 2x2 x^2 = 2x+3 \(x^3 + y^2 +z = x(2x+3) + (2x2)^2 + 2x = 6x^2 3x +4\) 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: 186

Re: Divisibility by 7
[#permalink]
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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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.



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Divisibility by 7
[#permalink]
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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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 > \(x^3 + y^2 + z=1+162=13\), 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. 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?
_________________



Intern
Joined: 18 Mar 2015
Posts: 21
Location: India
Concentration: Technology, Entrepreneurship
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: 2617

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
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
_________________



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3631

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 V40My MBA Journey: My 10 years long MBA DreamMy 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 placeBlog: Subscribe to Question of the Day BlogGMAT Club Inbuilt Error Log Functionality  View More. New Visa Forum  Ask all your Visa Related Questions  here. New! Best Reply Functionality on GMAT Club!Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for freeCheck our new About Us Page here.



CEO
Joined: 18 Aug 2017
Posts: 3524
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: If x^2 = y + 5 , y = z  2 and z = 2x , is x^3 + y^2 + z
[#permalink]
Show Tags
01 Mar 2019, 04:37
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) \(y = 4\) 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
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.




Re: If x^2 = y + 5 , y = z  2 and z = 2x , is x^3 + y^2 + z
[#permalink]
01 Mar 2019, 04:37






