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Is root(5-x)^2=5-x?

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Is root(5-x)^2=5-x? [#permalink] New post 02 May 2010, 14:51
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Question Stats:

50% (01:24) correct 50% (00:52) wrong based on 24 sessions
Is \sqrt{(x-5)^2}=5-x?

(1) -x|x| > 0
(2) 5 - x > 0
[Reveal] Spoiler: OA

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Re: DS GMAT PREP 2 [#permalink] New post 03 May 2010, 02:07
My few cents

1. Considering the set formula

U - union
^ - Intersection
AUBUC = A + B + C - [A^C] - [A^B] - [B^C] + A^B^C

We have all the values given in the brackets.
We need A^B^C

1st clause (if we draw a diagram its easier)
It gives A^B = 16 out of which 9 are in C. i.e. indirectly its giving (A^B)^C and thats what we want
A^B^C = 9 (** imp)

2nd clause gives A , B and C values but as we do not know AUBUC (according to formula)
we cannot calculate A^B^C.

Hence the answer is A

3. Formulating using the given formula

20 S.P. - 20 C.P = Gross Profit.

1st clause says that Gross Profit = 2400 if S.P = 2 C.P ( ** imp step)
i.e (20*2 *C.P) - 20*C.P = 2400
Hence C.P = 120 and S.P = 240 to get a profit of 2400. However we cannot come to any
consensus about the Gross Profit really earned as we cannot derive anything about current
C.P or S.P.

2nd Clause

Suppose actual S.P = C.P + x
According to conditions
20 * ( C.P. + x + 2) - 20 * C.P = 440
Therefore, 20x + 40 = 440
x = 20
Therefore original S.P = C.P. + 20
And Gross Profit = 20 * (C.P+ 20) - 20 * C.P
G.P = 400 $

Hence the Answer B.
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Re: DS GMAT PREP 2 [#permalink] New post 03 May 2010, 19:18
The official answers are on page 2 of the word document.
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Re: DS GMAT PREP 2 [#permalink] New post 03 May 2010, 22:15
For problem 3, on solution 2,

Is there another way to solve it instead of stating that SP = CP+X. I would have never thought to define it that way so 20CP cancels out.
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Re: DS GMAT PREP 2 [#permalink] New post 03 May 2010, 23:28
You can directly take
20 (S.P. + 2) - 20 C.P = 440
that gives (S.P. - C.P) = 20
That is the profit for single item
We have 20 items hence the Gross Profit = 20 * 20 = 400 $.

In fact more suitably 20 S.P - 20 .C.P = 400 $ and here LHS is what we have assumed as our formula.
I hope this is a better alternative to understand.
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Re: DS GMAT PREP 2 [#permalink] New post 04 May 2010, 12:35
Reply for inequality question number 2:

Simplify question: taking the square root of (x-5)² is equal to consider solutions for the absolute value of x-5

|x-5| could equal either x-5 or 5-x, in other words we need to determine whether x is positive or negative to answer the question

Statement 1: sufficient as we are told that -x multiplied by the absolue value of x is negative. that means that on of -x or |x| is negative and the other is positive. As |x| HAS to be positive then -x is negative.
We therefore know that x is positive and can answer the question with NO.

Statement 2: I cannot understand here why this one is sufficient. If we simplify this statement we get x < 5 which i believe is not sufficient alone.

Can someone help???

Thx!
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Re: DS GMAT PREP 2 [#permalink] New post 04 May 2010, 14:12
For 2, i believe X is negative since -(X)>0, so x would have to be a negative number.
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Re: DS GMAT PREP 2 [#permalink] New post 04 May 2010, 14:14
Expert's post
nifoui wrote:
Reply for inequality question number 2:

Simplify question: taking the square root of (x-5)² is equal to consider solutions for the absolute value of x-5

|x-5| could equal either x-5 or 5-x, in other words we need to determine whether x is positive or negative to answer the question

Statement 1: sufficient as we are told that -x multiplied by the absolue value of x is negative. that means that on of -x or |x| is negative and the other is positive. As |x| HAS to be positive then -x is negative.
We therefore know that x is positive and can answer the question with NO.

Statement 2: I cannot understand here why this one is sufficient. If we simplify this statement we get x < 5 which i believe is not sufficient alone.

Can someone help???

Thx!


Is \sqrt{(x-5)^2}=5-x?

Remember: \sqrt{x^2}=|x|.

So "is \sqrt{(x-5)^2}=5-x?" becomes: is |x-5|=5-x?

Now LHS=sqrt=absolute value>=0, (LHS\geq{0}, as absolute value is never negative), hence RHS also must be more than zero --> RHS=5-x\geq{0} --> x\leq{5}.

For x\leq{5} --> \{RHS=|x-5|=5-x\}=\{LHS=5-x\}. So we have that if x\leq{5}, then |x-5|=5-x is true.

Basically question asks is x\leq{5}?

(1) -x|x| > 0 --> |x| is never negative (positive or zero), so in order to have -x|x| > 0, -x must be positive -x>0 --> x<0. Sufficient.

(2) 5-x>0 --> x<5. Sufficient.

Answer: D.

Hope it helps.
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Re: DS GMAT PREP 2 [#permalink] New post 04 May 2010, 22:04
Very helpful indeed! Thanks Bunuel
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Re: DS GMAT PREP 2 [#permalink] New post 05 May 2010, 06:30
Thank you for all the help folks.
Re: DS GMAT PREP 2   [#permalink] 05 May 2010, 06:30
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