Last visit was: 15 Jul 2024, 13:21 It is currently 15 Jul 2024, 13:21
Toolkit
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Manager
Joined: 05 Oct 2008
Posts: 160
Own Kudos [?]: 3612 [81]
Given Kudos: 22
Math Expert
Joined: 02 Sep 2009
Posts: 94354
Own Kudos [?]: 641104 [45]
Given Kudos: 85011
Math Expert
Joined: 02 Sep 2009
Posts: 94354
Own Kudos [?]: 641104 [12]
Given Kudos: 85011
General Discussion
Manager
Joined: 05 Oct 2008
Posts: 160
Own Kudos [?]: 3612 [0]
Given Kudos: 22
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
excellent explanation..thanks. kudos
Intern
Joined: 10 Jun 2010
Posts: 31
Own Kudos [?]: 125 [0]
Given Kudos: 2
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
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: 94354
Own Kudos [?]: 641104 [0]
Given Kudos: 85011
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
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: 66
Own Kudos [?]: 205 [1]
Given Kudos: 0
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
1
Kudos
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
Manager
Joined: 29 Jul 2011
Posts: 52
Own Kudos [?]: 166 [1]
Given Kudos: 6
Location: United States
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
1
Bookmarks
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
Senior Manager
Joined: 24 Aug 2009
Posts: 388
Own Kudos [?]: 2295 [0]
Given Kudos: 276
Concentration: Finance
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]
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.
Manager
Joined: 10 Mar 2014
Posts: 135
Own Kudos [?]: 678 [1]
Given Kudos: 13
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
1
Kudos
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: 94354
Own Kudos [?]: 641104 [1]
Given Kudos: 85011
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
1
Kudos
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?
Intern
Joined: 18 Mar 2015
Posts: 20
Own Kudos [?]: 24 [0]
Given Kudos: 1
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]
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.
Alum
Joined: 12 Aug 2015
Posts: 2271
Own Kudos [?]: 3195 [1]
Given Kudos: 893
GRE 1: Q169 V154
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
1
Kudos
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
Joined: 18 Jul 2015
Status:Emory Goizueta Alum
Posts: 3596
Own Kudos [?]: 5466 [0]
Given Kudos: 346
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
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.
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 7972
Own Kudos [?]: 4216 [0]
Given Kudos: 243
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1:
545 Q79 V79 DI73
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]
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
Non-Human User
Joined: 09 Sep 2013
Posts: 33980
Own Kudos [?]: 851 [0]
Given Kudos: 0
Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
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 [#permalink]
Moderator:
Math Expert
94354 posts