if X^2=y+5 , y=z-2 , z=2x Is x^3 + y^2 +z divisible by 7 a) : DS Archive
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# if X^2=y+5 , y=z-2 , z=2x Is x^3 + y^2 +z divisible by 7 a)

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if X^2=y+5 , y=z-2 , z=2x Is x^3 + y^2 +z divisible by 7 a) [#permalink]

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13 Sep 2007, 13:36
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if X^2=y+5 , y=z-2 , z=2x
Is x^3 + y^2 +z divisible by 7

a) x>0
b) y=4
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13 Sep 2007, 14:19
Is x^3 + y^2 +z divisible by 7 ?

x^2=y+5 , y=z-2 , z=2x

Statement 1

y = 2x-2

x^2 = 2x-2+5

x^2 = 2x+3

since x > 0 the only solution for this equation is x=3

3^2 = 2*3 + 3

x = 3, z = 6, y = 4

sufficient

Statement 2

Edited - see b14kumar post !

y = 4

we can solve for z = 6

x has to be x = 3

3^3 + 4^2 + 6 = 49/7 ---> true

sufficient

Last edited by KillerSquirrel on 13 Sep 2007, 22:49, edited 1 time in total.
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13 Sep 2007, 21:00
fresinha12 wrote:
if X^2=y+5 , y=z-2 , z=2x
Is x^3 + y^2 +z divisible by 7

a) x>0
b) y=4

I think, it's D.

Each of the options is sufficient.

If we solve the given equation then we will have two sets of the values of x,y and z.

Set1 = {x=-1, y= -4 , z= -2}
Set2 = {x=3, y= 4 , z= 6}

Stm1: x>0

Then x = 3 and hence, y=4 and z=6.
=> x^3 + y^2 +z = 49 which is divisible by 7

Stm2: y=4

Then from set2, x=3 and z=6
=> x^3 + y^2 +z = 49 which is divisible by 7

Hence D.

- Brajesh
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13 Sep 2007, 21:05
b14kumar wrote:
fresinha12 wrote:
if X^2=y+5 , y=z-2 , z=2x
Is x^3 + y^2 +z divisible by 7

a) x>0
b) y=4

I think, it's D.

Each of the options is sufficient.

If we solve the given equation then we will have two sets of the values of x,y and z.

Set1 = {x=-1, y= -4 , z= -2}
Set2 = {x=3, y= 4 , z= 6}

Stm1: x>0

Then x = 3 and hence, y=4 and z=6.
=> x^3 + y^2 +z = 49 which is divisible by 7

Stm2: y=4

Then from set2, x=3 and z=6
=> x^3 + y^2 +z = 49 which is divisible by 7

Hence D.

- Brajesh

I agree - good catch !

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14 Sep 2007, 12:49
fresinha12 wrote:
OA is D

Thanks

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15 Sep 2007, 07:45
It is D,

If everything solved before going into 1 and 2, it will be clear that both of them sufficient.

Ans: D
15 Sep 2007, 07:45
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