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# M05- Question 28

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Senior Manager
Joined: 16 Apr 2009
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01 Mar 2010, 19:01
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. $$y = 4$$

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Senior Manager
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Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

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01 Mar 2010, 21:16

Stmt 1) simplifying the given equation in terms of x,
x^3 + 4*x^2 - 14*x + 16.

for any value of x > 0, say x = 1, 2, it may or may not be divisible by 7. Insufficient.

Stmt 2) simplifying the above equation in terms of y,
(sqrt(y + 5))[y - 9] + [y + 21].

Substituting y = 4, (+/-3)(-5) + 25, no matter 3 is positive or negative, it is still not divisible by 7. Standard answer, confirmed not divisible - sufficient.
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Senior Manager
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02 Mar 2010, 18:36
OA is mentioned as D
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Math Expert
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04 Mar 2010, 01:30
1
KUDOS
Expert's post
ichha148 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 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.
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04 Mar 2010, 10:01
d

x=-1 or x=3
x>0
x=3

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Senior Manager
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06 Mar 2010, 14:44
Thanks a lot Bunuel , as always your explanation is great and lucid +1 from my side
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Re: M05- Question 28   [#permalink] 06 Mar 2010, 14:44
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# M05- Question 28

Moderator: Bunuel

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