Practice Exam #2 DS Number Properties

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Practice Exam #2 DS Number Properties

by fangtray » Sun Apr 22, 2012 4:38 am
is root(x-5)^2 = 5-x?

1. -x|x| >0
2. 5-x >0

please help![/img]

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by simone88 » Sun Apr 22, 2012 8:47 am
for a real number a, |a|=a if and only if a>=0. Moreover for every number a, |a|=|-a|. In our case root(x-5)^2=|x-5|=|5-x| so we are asked if |5-x|=5-x and this correspond to 5-x>=0 or x<=5.
(P.S: when you arrive to such a conclusione just give your self a little check and try some numbers! So the statement should be true for x=4 and root(4-5)^2=root(1)=1 while 5-4=1...ok! for x=6 it should be false so root(6-5)^2=root(1)=1 while 5-6=-1!)
So now the very question is: is x<=5?
1)|x|>=0 for every x so, simplifying, this assumption is really saying that -x>0 or x<0. (Also here, just check with some value of x, as x=-2, or x=1...). If x<0 is it true that x<=5? Of course!
2) this assumption is saying that x<5. If x<5 is it true that x<=5? Of course!
So the right answer is: every statement alone is sufficient.

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by klmehta03 » Sun Apr 22, 2012 8:56 am
Agree IMO D OA pls?

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by fangtray » Sun Apr 22, 2012 2:55 pm
simone88 wrote:for a real number a, |a|=a if and only if a>=0. Moreover for every number a, |a|=|-a|. In our case root(x-5)^2=|x-5|=|5-x| so we are asked if |5-x|=5-x and this correspond to 5-x>=0 or x<=5.
(P.S: when you arrive to such a conclusione just give your self a little check and try some numbers! So the statement should be true for x=4 and root(4-5)^2=root(1)=1 while 5-4=1...ok! for x=6 it should be false so root(6-5)^2=root(1)=1 while 5-6=-1!)
So now the very question is: is x<=5?
1)|x|>=0 for every x so, simplifying, this assumption is really saying that -x>0 or x<0. (Also here, just check with some value of x, as x=-2, or x=1...). If x<0 is it true that x<=5? Of course!
2) this assumption is saying that x<5. If x<5 is it true that x<=5? Of course!
So the right answer is: every statement alone is sufficient.
how did you simplify the question for is x <=5? How did you spot that? You just plug in numbers ? Algebraically, how do you solve |5-x| = 5-x? and I guess i kinda feel dumb on this one, i wasn't sure that it simplified to |5-x|... I mean the square gets rid of any negative value regardless of what X is right? so why do we have the |x-5|?

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by simone88 » Sun Apr 22, 2012 10:11 pm
fangtray wrote: how did you simplify the question for is x <=5? How did you spot that? You just plug in numbers ? Algebraically, how do you solve |5-x| = 5-x?
First of all we have sqrt((x-5)^2)... do you agree that this is equal to |x-5|? lesson for the future:
"for every number a sqrt(a^2)=|a|"(so in this case sobstitute a=x-5): let's give some exempla:
sqrt((-3)^2)=sqrt(9)=3 but 3=|-3|; sqrt(7^2)=sqrt(49)=7 but 7=|7|; sqrt((-3)^2)=3 but 3=|-3|...did you get it?
Ok so we have simplified the main statement this way: Is |x-5|=5-x?
Now I have just noticed that |x-5|=|5-x|...just for convenience! The reason why I can do this is that 5-x=-(x-5) and that for every number a |a|=|-a| (so in this case sobstitute a=5-x). Let's give some exampla: |-3|=3; |2|=2 |-sqrt(2)|=sqrt(2)...and so on for every a!
This is why I was able to do that. The reason why I did that is that in this way I have similar things on the left and right hand side! (|5-x|, 5-x contain both 5-x)
So now we can rearrange our main statement like that: is |5-x|=5-x?
Now you say: "but Algebraically, how do you solve |5-x| = 5-x?"
simone88 wrote: for a real number a, |a|=a if and only if a>=0
so we just need to apply this principle for a=5-x: sobstituting in the last principle a=5-x we have
"for a real number x, |5-x|=5-x if and only if 5-x>=0 (if and only if x<=5)".
So we got the answer guys! we have simplified the question "is sqrt(x-5)^2=5-x" with the more easy question "is x<=5?"

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by GMATGuruNY » Mon Apr 23, 2012 4:18 am
fangtray wrote:is root(x-5)^2 = 5-x?

1. -x|x| >0
2. 5-x >0

please help![/img]
It helps to know the following:
√ means the POSITIVE ROOT ONLY.
Thus, √(x²) = |x|.
|x-y| = the DISTANCE between x and y.

In the problem at hand:
√(x-5)² = |x-5| = the DISTANCE between x and 5.
A distance must be greater than or equal to 0.

5-x = the DIFFERENCE between 5 and x.
A difference can be negative, 0, or positive.

The DIFFERENCE between two values is equal to the DISTANCE between the two values whenever the DIFFERENCE is greater than or equal to 0.

Thus, |x-5| = 5-x whenever 5-x≥0.
Question rephrased: Is x≤5?

Statement 1: -x|x| > 0
Since |x| cannot be negative, both factors (-x and |x|) must be positive.
Thus:
-x > 0
x<0.
Since x<0, we know that x≤5.
SUFFICIENT.

Statement 2: 5-x > 0
Thus, x<5.
SUFFICIENT.

The correct answer is D.
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