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M05- Question 28

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M05- Question 28 [#permalink] New post 01 Mar 2010, 19:01
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

Please provide explanation as well
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Re: M05- Question 28 [#permalink] New post 01 Mar 2010, 21:16
Answer must be B.

Stmt 1) simplifying the given equation in terms of x,
x^3 + 4*x^2 - 14*x + 16.

for any value of x > 0, say x = 1, 2, it may or may not be divisible by 7. Insufficient.

Stmt 2) simplifying the above equation in terms of y,
(sqrt(y + 5))[y - 9] + [y + 21].

Substituting y = 4, (+/-3)(-5) + 25, no matter 3 is positive or negative, it is still not divisible by 7. Standard answer, confirmed not divisible - sufficient.
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Re: M05- Question 28 [#permalink] New post 02 Mar 2010, 18:36
OA is mentioned as D
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Re: M05- Question 28 [#permalink] New post 04 Mar 2010, 01:30
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ichha148 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

Please provide explanation as well


We have system of equations with three 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: M05- Question 28 [#permalink] New post 04 Mar 2010, 10:01
d

x=-1 or x=3
x>0
x=3
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Re: M05- Question 28 [#permalink] New post 06 Mar 2010, 14:44
Thanks a lot Bunuel , as always your explanation is great and lucid :) +1 from my side
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Re: M05- Question 28   [#permalink] 06 Mar 2010, 14:44
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