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
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) \(y = 4\)
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 49271

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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




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
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




Senior Manager
Joined: 05 Oct 2008
Posts: 257

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



Intern
Joined: 10 Jun 2010
Posts: 39

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: 49271

Re: Divisibility by 7
[#permalink]
Show Tags
21 Jul 2010, 09:33



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: 94
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: 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
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



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: \(x^2 = y + 5\) \(y = z  2\) > \(y=2x2\). \(z = 2x\) \(x^2 =2x2+ 5\) > \(x^22x3=0\) > \(x=3\) 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: 205

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: 49271

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?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Current Student
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: 2648

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
_________________
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 GMATQuant(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 Blog: Subscribe to Question of the Day Blog
GMAT 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 free



NonHuman 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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




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






