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Hi all,

I need some help please. This is the question :

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

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient

Re: DS from GMAT Club Test - Need Help ! [#permalink]

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22 Dec 2009, 16:01

I think the answer is (B).

You have x^2 = y + 5, y = z - 2, & z = 2x.

Combine y = z - 2 & z = 2x. You get y = 2x - 2. Put this new equation in x^2 = y + 5. You get x^2 = 2x - 2 + 5. Put it in quadratic form and you get x^2 - 2x -3 = 0, which is (x-3)(x+1). x = 3, or x = -1. This means that these are the only x-values that will satisfy these 3 equations simultaneously. If you use x= 3 to get the other values, you'll get z = 6 & y = 4. If you plug in these values into that equation to see if it is divisible by 7, you'll see that the result is 49 and so it is divisible. This is the only value that will both satisfy the equations and be divisible by 7. Any other x-value will either not satisfy the 3 equations and/or be divisible by 7. x = -1 does not matter since you're only looking at x > 0. So (A) does not give enough information.

For (B), y = 4. Looking at my work above, you'll see that if y = 4, then z = 6, & x = 3, which satisfies the equations and will give you a result that is divisible by 7. So I think (B) is the answer.

Re: DS from GMAT Club Test - Need Help ! [#permalink]

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23 Dec 2009, 01:20

Thank you ! Very Nice.

But can you please detail how you get the x^2-2x-3 = 0 equation. I understand that you put x in z and y, but I do not manage to reach this last equation. Otherwise, your explanations look clear.

Re: DS from GMAT Club Test - Need Help ! [#permalink]

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23 Dec 2009, 02:05

nesta wrote:

Thank you ! Very Nice.

But can you please detail how you get the x^2-2x-3 = 0 equation. I understand that you put x in z and y, but I do not manage to reach this last equation. Otherwise, your explanations look clear.

Re: DS from GMAT Club Test - Need Help ! [#permalink]

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23 Dec 2009, 02:53

I sorry but I missed something. I cannot reach the final equation. It looks easy but...

So, I agree with you :

1.x^2=2x+3 --> We need x^3, no ? How do you get it ? 2.y = 2x - 2 3.z = 2x We are looking for x^3 + y^2 + z, right ? So do you substitute these 3 equations into x^3+ y^2 + z to get x^3+(2x-2)^2+2x ?

Re: DS from GMAT Club Test - Need Help ! [#permalink]

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23 Dec 2009, 03:05

1

This post received KUDOS

nesta wrote:

Quote:

Quote:

I sorry but I missed something. I cannot reach the final equation. It looks easy but...

So, I agree with you :

1.x^2=2x+3 --> We need x^3, no ? How do you get it ? 2.y = 2x - 2 3.z = 2x We are looking for x^3 + y^2 + z, right ? So do you substitute these 3 equations into x^3+ y^2 + z to get x^3+(2x-2)^2+2x ?

Thank you xcusemeplz2009 !

not necessary it will complicate the things to prove x^3+y^2+z is div by seven we need to knw the value of x,y,and z the eqn x^2-2x-3=0 gives x=3 or -1 with x=3 we get a value for y and z using the given relations and we come to know that its div. by 7 and with x=-1 we come to knw that its not div by 7 now the task is to know wether x is 3 or -1

s1) tells x>0 , and x=3 is the only +ve root of x^2-2x-3 hence suff s2) tells x=3 same as abve hence suff...

here now we want to ensure that is it div or not and if the statements wud have shown that x=-1 even then its suff...
_________________

GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

Re: DS from GMAT Club Test - Need Help ! [#permalink]

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23 Dec 2009, 03:27

Really Thx xcusemeplz2009 !

I get it ! I tried to find x^3 + y^2 + z with substitution, that is why I do not reach your last equation. You right, we got x^2 = 2x+3 so x^2-2x-3 = 0, then factorization ect...

gmatclubot

Re: DS from GMAT Club Test - Need Help !
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23 Dec 2009, 03:27