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

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

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29 Apr 2012, 02:29
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Jul 2013, 02:29, edited 2 times in total.
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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]

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29 Apr 2012, 05:24
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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.

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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]

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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.
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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]

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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]

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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.
Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z   [#permalink] 21 May 2015, 10:16
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