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For all counting numbers X,

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For all counting numbers X, [#permalink] New post 04 Feb 2018, 15:02
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Question Stats:

64% (00:37) correct 35% (01:09) wrong based on 14 sessions


For all counting number x, \(x^* = \frac{2}{x}\)

Quantity A
Quantity B
\(x\)
\((x^*)^*\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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[Reveal] Spoiler: OA

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Re: For all counting numbers X, [#permalink] New post 04 Feb 2018, 20:03
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This problem is a disguised function problem. The testmakers want you to see that asterisk and freak out because you've never studied anything with an asterisk in it before and clearly can't be expected to do this one. But here's the trick: don't freak out. Nobody else has studied this asterisk thing either, because the testmakers just made it up! But they will always give you instructions. You just need to follow the instructions. Let's try this one.

What they're telling us is that when you see the asterisk, take the thing under it and divide 2 by it. Fair enough. If you look under Quantity B, we see (x*)*. It's nested! Just follow PEMDAS and do the function inside the parenthesis first. So for x*, we've been shown that it's 2/x. Great. But now we have (2/x)*. This just means do it again. Take the 2/x and divide 2 by it. What we get is 2/2/x, which doesn't look great when typed that way, but you know what I mean.

How to simplify a three-level fraction? Back up. We have 2 divided by 2/x. We can pretend the first 2 is also a fraction and call it 2/1. So now we have 2/1|2/x. Now we have a four-level fraction! We made it worse! Not really though.

If you have a four-level fraction, just flip the bottom fraction and then multiply it by the top fraction, so 2/1|2/x should become 2/1 times x/2. This is easier to write on a whiteboard than in a text editor, but you should see that the 2s cancel out and we're left with just x. So the answer is C.
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Re: For all counting numbers X, [#permalink] New post 05 Feb 2018, 03:21
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Carcass wrote:


For all counting number x, \(x^* = \frac{2}{x}\)

Quantity A
Quantity B
\(x\)
\((x^*)^*\)





Here QTY A = x and we can leave it as it is.

Now for QTY B = \((x^*)^*\) = \((\frac{2}{x})^*\) (since it is given, \(x^* = \frac{2}{x}\))

= \((\frac{2}{2}) * x\)
= x

Hence QTY A = QTY B. i.e option C
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Re: For all counting numbers X, [#permalink] New post 05 Feb 2018, 20:57
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I am a bit confused on this.

Let us take \(x = 1\)
For \(X*\) we get, \(\frac{2}{1}\)\(= 2\)
Again we repeat the operation for 2 and end up with 1

However, Let us take a larger number than 2 for eg. 10
we have \(x = 10\)
For\(X*\) we get, \(\frac{2}{10}=\)\(\frac{1}{5}\)
At this point I think we cannot perform the 2nd operation because the function is defined for counting numbers only and we now have a fraction
I think the ans should be (D)
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Re: For all counting numbers X, [#permalink] New post 07 Feb 2018, 02:00
Answer: C
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Re: For all counting numbers X, [#permalink] New post 17 May 2018, 23:50
When ever I encounter such a question where there is some unknown operator then I really can not comprehend what to do..

as Here I am flabbergasted to discern the function of * and what does it mean in mathematical term (x^*)^*.??
Re: For all counting numbers X,   [#permalink] 17 May 2018, 23:50
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