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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

5

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]
22 Sep 2012, 09:28

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