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# M05#28

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Manager
Status: Back to (GMAT) Times Square!!!
Joined: 15 Aug 2011
Posts: 189
Location: United States (IL)
GMAT 1: 650 Q49 V30
WE: Information Technology (Computer Software)
Followers: 3

Kudos [?]: 71 [0], given: 25

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06 Apr 2012, 19:20
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

Please explain me how the equation got reduced to x^2 - 2*x - 3 = 0. Hence the question.

Please gimme the link, I could not find it. From 2, y=4, so I got z=6 and x = +-3, so if x>0 from 1, I had C.
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06 Apr 2012, 20:58
1
KUDOS
Let me try,

The given equations:
$$x^2=y+5$$
$$y=z-2$$
$$z=2x$$

substitute $$z=2x$$ into $$y=z-2$$
$$y=2x-2$$

substitute $$y=2x-2$$ into $$x^2=y+5$$
$$x^2=2x-2+5$$
$$x^2=2x+3$$
$$x^2-2x-3=0$$
$$(x-3)(x+1)=0$$
$$x=3$$ or $$x=-1$$

Statement 1: x>0
So x must be equal to 3.
$$y=2(3)-2=4$$
$$z=6$$
Substitute all these figures into $$x^3+y^2+z$$
$$3^3+4^2+6=27+16+6=49$$
49 is divisible by 7 so statement 1 is sufficient.

Statement 2: y=4 (remember to look at this in isolation from statement 1)
$$y=z-2=4$$
$$z=6$$

$$6=2x$$
$$x=3$$
All these figures are the same as we get in statement 1 so we don't have to perform the calculation again. Statement 2 is sufficient.

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07 Apr 2012, 02:00
1
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Expert's post
vix 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

Please explain me how the equation got reduced to x^2 - 2*x - 3 = 0. Hence the question.

Please gimme the link, I could not find it. From 2, y=4, so I got z=6 and x = +-3, so if x>0 from 1, I had C.

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.
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Manager
Status: Back to (GMAT) Times Square!!!
Joined: 15 Aug 2011
Posts: 189
Location: United States (IL)
GMAT 1: 650 Q49 V30
WE: Information Technology (Computer Software)
Followers: 3

Kudos [?]: 71 [0], given: 25

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07 Apr 2012, 16:09
Thanks, that was really helpful. Methinks I thought a lot in trying this one.
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Re: M05#28   [#permalink] 07 Apr 2012, 16:09
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# M05#28

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