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# For all non zero integers n, n*=(n+2)/n. What is the value

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Manager
Joined: 21 May 2011
Posts: 171
For all non zero integers n, n* = (n+2)/n . What is the  [#permalink]

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24 Jul 2011, 12:41
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For all non – zero integers n, n* = (n+2)/n . What is the value of x ?

(1) x* = x

(2) x* = – 2 – x
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Intern
Joined: 22 Jul 2013
Posts: 12
GMAT 1: 650 Q48 V31
For all non zero integers n, n*=(n+2)/n. What is the value  [#permalink]

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Updated on: 27 Oct 2013, 06:40
1
5
For all non zero integers n, n*=(n+2)/n. What is the value of x ?

(1) x* = x.
(2) x* = -2-x.

Answer = C.

For statement 2 , the solution says
=>(x+2)/x= -2-x.
=> Multiply both sides by x.
=> (x+2) = -x(x+2)
=> x+2= -$$x^2$$ - 2x
=>$$x^2$$ + 3x + 2 = 0.
(x+1)(x+2) = 0
x = -1 . x = -2

My question is why cant we cancel out the (x+2) on LHS and RHS instead of multiplying it by x in step.
We will get x = -1.
So answer should be B instead of C.

Why cant we reduce the equation ? Shall we never do it in GMAT ?
Could anyone please explain where equations should be reduced and where they shouldn't be

Originally posted by NeetiGupta on 26 Oct 2013, 18:20.
Last edited by Bunuel on 27 Oct 2013, 06:40, edited 2 times in total.
Edited the question.
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Joined: 20 Dec 2010
Posts: 1588
Re: DS - 700 level - n*  [#permalink]

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24 Jul 2011, 16:46
1
bschool83 wrote:
For all non – zero integers n , n* = (n+2)/n . What is the value of x ?

(1) x * = x

(2) x * = – 2 – x

1)
(x+2)/x=x
x+2=x^2
x^2-x-2=0
x^2-2x+x-2=0
(x+1)(x-2)=0
x=-1, x=2
Not Sufficient.

2)
(x+2)/x = -2 -x
x+2=-2x-x^2
x^2+3x+2=0
(x+1)(x+2)=0
x=-1, x=-2
Not Sufficient.

Together;
x=-1
Sufficient.

Ans: "C"
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Joined: 08 Jul 2011
Posts: 4
Re: DS - 700 level - n*  [#permalink]

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26 Jul 2011, 13:17
I don't have any issues with the mechanics of solving this question. I can factor and derive the roots for each equation pretty handily. However, I don't understand conceptually what the roots {2, -2} are. Are they correct values for x in statements (1) and (2) respectively but not for the entire system of equations. And when a question asks for a value, must there always be only a single value?

Thanks, and I'm happy to attempt to clarify my question if it's confusing.
Intern
Status: If I play my cards right, I can work this to my advantage
Joined: 18 Jul 2011
Posts: 13
Location: India
Concentration: Operations, Social Entrepreneurship
GMAT Date: 11-12-2011
WE: Information Technology (Telecommunications)
Re: DS - 700 level - n*  [#permalink]

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26 Jul 2011, 23:37
1
elementbrdr wrote:
I don't have any issues with the mechanics of solving this question. I can factor and derive the roots for each equation pretty handily. However, I don't understand conceptually what the roots {2, -2} are. Are they correct values for x in statements (1) and (2) respectively but not for the entire system of equations. And when a question asks for a value, must there always be only a single value?

Thanks, and I'm happy to attempt to clarify my question if it's confusing.

yes DS questions always ask for a definite (one) value only from what i've solved till now from OG / Kaplan ...
any solution in this case quadratic, having 2 roots; is not sufficient

that is the reason the definate solution is by combining the two solutions of (1) and (2) option
Intern
Joined: 05 Oct 2013
Posts: 20
Re: For all nonzero integers n,n*=(n+2)/n.What is the value of x  [#permalink]

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26 Oct 2013, 18:51
1
NeetiGupta wrote:
Q. For all non zero integers n,n*=(n+2)/n. What is the value of x ?
1. x* = x.
2. x* = -2-x.

Answer = C.

For statement 2 , the solution says
=>(x+2)/x= -2-x.
=> Multiply both sides by x.
=> (x+2) = -x(x+2)
=> x+2= -$$x^2$$ - 2x
=>$$x^2$$ + 3x + 2 = 0.
(x+1)(x+2) = 0
x = -1 . x = -2

My question is why cant we cancel out the (x+2) on LHS and RHS instead of multiplying it by x in step.
We will get x = -1.
So answer should be B instead of C.

Why cant we reduce the equation ? Shall we never do it in GMAT ?
Could anyone please explain where equations should be reduced and where they shouldn't be

you can only cancel a factor if it is nonzero. In this case, if you cancel (x+2), you also skip (loose) the root (x+2=0).
Intern
Joined: 22 Jul 2013
Posts: 12
GMAT 1: 650 Q48 V31
Re: For all nonzero integers n,n*=(n+2)/n.What is the value of x  [#permalink]

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26 Oct 2013, 19:19
tuanle wrote:
NeetiGupta wrote:
Q. For all non zero integers n,n*=(n+2)/n. What is the value of x ?
1. x* = x.
2. x* = -2-x.

Answer = C.

For statement 2 , the solution says
=>(x+2)/x= -2-x.
=> Multiply both sides by x.
=> (x+2) = -x(x+2)
=> x+2= -$$x^2$$ - 2x
=>$$x^2$$ + 3x + 2 = 0.
(x+1)(x+2) = 0
x = -1 . x = -2

My question is why cant we cancel out the (x+2) on LHS and RHS instead of multiplying it by x in step.
We will get x = -1.
So answer should be B instead of C.

Why cant we reduce the equation ? Shall we never do it in GMAT ?
Could anyone please explain where equations should be reduced and where they shouldn't be

you can only cancel a factor if it is nonzero. In this case, if you cancel (x+2), you also skip (loose) the root (x+2=0).

Can you please elaborate "you can only cancel a factor if it is nonzero"
Do you mean when we take x=-2. x=2 becomes 0 and hence we cannot cancel it?
Math Expert
Joined: 02 Sep 2009
Posts: 58335
Re: For all non zero integers n, n*=(n+2)/n. What is the value  [#permalink]

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27 Oct 2013, 06:50
2
1
NeetiGupta wrote:
For all non zero integers n, n*=(n+2)/n. What is the value of x ?

(1) x* = x.
(2) x* = -2-x.

Answer = C.

For statement 2 , the solution says
=>(x+2)/x= -2-x.
=> Multiply both sides by x.
=> (x+2) = -x(x+2)
=> x+2= -$$x^2$$ - 2x
=>$$x^2$$ + 3x + 2 = 0.
(x+1)(x+2) = 0
x = -1 . x = -2

My question is why cant we cancel out the (x+2) on LHS and RHS instead of multiplying it by x in step.
We will get x = -1.
So answer should be B instead of C.

Why cant we reduce the equation ? Shall we never do it in GMAT ?
Could anyone please explain where equations should be reduced and where they shouldn't be

For all non zero integers n, n*=(n+2)/n. What is the value of x ?

(1) x* = x --> $$\frac{x+2}{x}=x$$ --> $$x^2-x-2=0$$ --> $$x=-1$$ or $$x=2.$$ Not sufficient.

(2) x* = -2-x --> $$\frac{x+2}{x}=-2-x$$ --> $$x+2=-x(x+2)$$ --> $$(x+2)(1+x)=0$$ --> $$x=-1$$ or $$x=-2.$$ Not sufficient.

If you divide (reduce) $$x+2=-x(x+2)$$ by x+2 you assume, with no ground for it, that x+2 does not equal to zero thus exclude a possible solution (notice that both x=-1 AND x=-2 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

(1)+(2) Common value of x from (1) and (2) is $$x=-1$$. Sufficient.

Answer: C.

Similar question to practice: for-all-integers-n-n-n-n-1-what-is-the-value-of-x-155982.html

Hope this helps.
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Re: For all non zero integers n, n*=(n+2)/n. What is the value   [#permalink] 27 Oct 2013, 06:50
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# For all non zero integers n, n*=(n+2)/n. What is the value

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