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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]  13 Jun 2010, 09:11
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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 \gt 0
2. y = 4
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Re: Divisibility by 7 [#permalink]  13 Jun 2010, 09:33
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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.

Answer: D.

Hope it helps.
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Re: Divisibility by 7 [#permalink]  13 Jun 2010, 10:04
excellent explanation..thanks. kudos
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Re: Divisibility by 7 [#permalink]  21 Jul 2010, 09:21
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.
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Re: Divisibility by 7 [#permalink]  21 Jul 2010, 09:33
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.

Pleas read the solution above.

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.
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Arithmetic [#permalink]  07 Jan 2012, 03:10
Can anyone help me in solving the below problem in 2 min
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Re: Arithmetic [#permalink]  07 Jan 2012, 03:37
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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
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Re: Arithmetic [#permalink]  07 Jan 2012, 03:52
blink005 wrote:
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

Thanks i made things complex........
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Re: Arithmetic [#permalink]  07 Jan 2012, 15:39
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
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Re: Arithmetic   [#permalink] 07 Jan 2012, 15:39
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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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