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M05#28 [#permalink]

06 Apr 2012, 20:20

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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

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.

[Reveal] Spoiler: OA

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Re: M05#28 [#permalink]

06 Apr 2012, 21:58

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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.

Answer: D
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Re: M05#28 [#permalink]

07 Apr 2012, 03:00

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vix wrote:

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.

Answer: D.

Hope it helps.
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Manager

Status: Back to (GMAT) Times Square!!!

Joined: 15 Aug 2011

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Re: M05#28 [#permalink]

07 Apr 2012, 17:09

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, 17:09

M05#28

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