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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]
29 Apr 2012, 04:24

6

This post received KUDOS

Expert's post

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.

Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]
22 Sep 2012, 09:28

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.

Answer: D.

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]
05 Feb 2015, 16:10

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Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z [#permalink]
21 May 2015, 09:16

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.

gmatclubot

Re: If x^2 = y + 5, y = z - 2 and z = 2x, is x^3 + y^2 + z
[#permalink]
21 May 2015, 09:16

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