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# m05 Q5 simple math help please

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17 Sep 2008, 19:00
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If $$P# = \frac{P}{P-1}$$ , what is the value of $$P##$$ ?

(A) $$\frac {P}{P-1}$$
(B) $$\frac{1}{P}$$
(C) $$P$$
(D) $$2 - P$$
(E) $$P - 1$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Can somebody walk me through this one?

Last edited by akkane on 16 Jan 2013, 17:02, edited 2 times in total.
Fixed the question stem

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17 Sep 2008, 21:34
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You really need to use for math formatting in order for us to figure out if there are any errors in the question as written. I can't determine which of these is the proper way for this question to be displayed:

#1 Option => $$\frac{p}{(p-1)-1}$$
or
#2 Option => $$\frac{p}{(p-1)}-1$$

Generally, this is a question that you can use substituting and get the correct answer, but that doesn't seem to be working. For example, if you make p = 4 such that you have 4##.

#1 gives you division by zero because you have $$\frac{4}{(4-1)-1$$ = 2, so 2# = $$\frac{2}{2-1-1} = \frac{2}{0}$$

#2 is all other the place with results. If p = 8, so you have 8##, you get the following:

$$\frac{8}{(8-1)}-1 = \frac{8}{7} - \frac{7}{7} = \frac{1}{7}$$ and you then have $$\frac{1}{7}#$$

which = $$\frac{\frac{1}{7}}{\frac{1}{7} - \frac{7}{7}}-1 = -1\frac{1}{6}$$

1/7 dvidied by -6/7 is like 1/7 * -7/6 so the 7's cancel out and you have -1/6 left. Then subtract 1 from that to give you the answer above.

If you plug in the original numbers used above for p into the answer equations, you never get the same value as you do above so it makes the question (or answers) invalid.

Someone, please show me that I'm wrong and I just didn't see the correct approach because I'm tired.

d2touge wrote:
If P# = p/(p-1) -1 , what is the value of P##?

a) p/p-1
b) 1/p
c) p
d) 2-p
e) p-1

Can somebody walk me through this one?

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17 Sep 2008, 22:09
Oops, in the question, forget the extra (-1). it's just p/(p-1)

Using substitution as you suggested, I was able to solve the answer. Thanks!

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21 Oct 2008, 11:41
As stated above the original problem was P# = p / (p-1)

Find P##

Can someone please kindly show this to me using algebra?

How do you get (p/(p-1)) / (p/(p-1) - 1) as suggested in the answer.

Thanks alot
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22 Oct 2008, 02:41
How do you get (p/(p-1)) / (p/(p-1) - 1) as suggested in the answer.

in your denominator pls see that its ( (p/(p-1) ) - 1 )

which will be (p -p + 1) / (p -1) [ This is just the denominator ]
so finally p / (p -1) / (1 / (p - 1) ) which is p.

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23 Oct 2008, 03:26
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I'll try to make this clearer by using our math syntax. See the short intro here:

Given $$P# = \frac{P}{P-1}$$, need to find $$P##$$ - original problem
We might simplify this problem by saying that this '#' sign is an operation that replaces each $$P$$ with $$\frac{P}{P-1}$$. $$P##$$ is found by replacing all the $$P$$s with $$\frac{P}{P-1}$$ in this expression:
$$\frac{P}{P-1}$$, which equals $$P#$$

This is how it looks after these replacements:

$$\frac{\frac{P}{P-1}}{\frac{P}{P-1}-1}$$

$$\frac{\frac{P}{P-1}}{\frac{P}{P-1}-\frac{P-1}{P-1}}=$$

$$\frac{\frac{P}{P-1}}{\frac{P-P+1}{P-1}}=$$

$$\frac{\frac{P}{P-1}}{\frac{1}{P-1}}=$$

$$\frac{P}{P-1}*\frac{P-1}{1}=$$

$$P$$

Hope this clears the doubts.
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10 Jan 2010, 11:52
For both picking number and algebra solution explanations are incorrect:
1. Example- picking number: If we choose a negative number (The problem didn’t state that P is a positive number), let’s say -2 then -2# = -2/-2-1 = 2/3. Then P# <> P
2. Example - Algebra: P# = P/(P-1) We agree here, but how did you come to P/(P-1)/P/(P-1) -1 for P##? Shouldn’t it be P/(P-1)/(P-1) . You are multiplying both numerator and denominator what is not right P/(P-1)* #/#?, 2/3 * 2 is not 2/3 * 2/2 for putting it in simple terms. So answer should be P/ (P-1) (P-1) Not P. Please verify.

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11 Jan 2010, 08:29
This question is pretty confusing. I'll try to describe the approach that works for me. Let's see once again what this "#" operation does to $$P$$. We see from the question stem that this operation replaces $$P$$ with $$\frac{P}{P-1}$$. I've covered that more thoroughly in my above post. So, after the operation is "executed", we get $$\frac{P}{P-1}$$, but this is not the answer as we have to find $$P##$$. Now, to find $$P##$$, we need to perform this operation $$#$$ (replacing all $$P$$s with $$\frac{P}{P-1}$$) over this expression we got previously: $$\frac{P}{P-1}$$. If you replace all $$P$$s with $$\frac{P}{P-1}$$ in this expression, you'll end up with a big one (that can be simplified to $$P$$ as described in more detail in my above post).

Hope this helps.
Ikowill wrote:
For both picking number and algebra solution explanations are incorrect:
1. Example- picking number: If we choose a negative number (The problem didn’t state that P is a positive number), let’s say -2 then -2# = -2/-2-1 = 2/3. Then P# <> P
2. Example - Algebra: P# = P/(P-1) We agree here, but how did you come to P/(P-1)/P/(P-1) -1 for P##? Shouldn’t it be P/(P-1)/(P-1) . You are multiplying both numerator and denominator what is not right P/(P-1)* #/#?, 2/3 * 2 is not 2/3 * 2/2 for putting it in simple terms. So answer should be P/ (P-1) (P-1) Not P. Please verify.

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12 Jan 2010, 06:53
d2touge wrote:
If $$P# = \frac{P}{P-1}$$ , what is the value of $$P##$$ ?

(A) $$\frac {P}{P-1}$$
(B) $$\frac{1}{P}$$
(C) $$P$$
(D) $$2 - P$$
(E) $$P - 1$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Can somebody walk me through this one?

Let X=P#=P/(P-1)
therefore, X#=X/(X-1) put the value for X.
ans P
hence option C
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12 Jan 2010, 06:57
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given p#=p/(p-1)
then p##= p/(p-1)# ==>

numerator => p/(p-1)
denominator => p/(p-1)-1 ==> 1/(p-1)

so p##==> p/(p-1)/1/(p-1) ==> p

so ans is C

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12 Jan 2010, 07:15
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p# = p/(p-1)

(p#)# = p/(p-1)] / [ {p/(p-1)}-1]
= p/(p-1)] / [ {( p - (p-1)}/(p-1)]
= p/(p-1)] / [ 1/(p-1)]
= p ( p-1) / (p-1)
= p

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17 Jan 2011, 07:46
I tried with

# = 1, Solvng we get P = 2

P## = 2

Ony option 'C' satisfies this ....

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17 Jan 2011, 11:47
C

p# = p/(p-1)
p##=( p/(p-1))/(p/(p-1) -1)
= p/(p-1) * (p-1)/(p-p+1)
= p

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17 Jan 2011, 12:11
thanks guys. can anyone tell me whats the lvl of this question?

thanks.
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18 Jan 2011, 03:02
I thought this was a very simple problem.

I dint think of substituting with number - But just went with (p#) #

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11 Jun 2011, 18:50
P## = (P/(P-1)) #

= $$P/(P-1) / ((P /(P-1))-1)$$

= P

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19 Jan 2012, 08:05
Why do you subtract a 1 for the second part of the equation?

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19 Jan 2012, 09:39
C. multiplying buy the reciprocal is a very useful thing

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22 Jan 2013, 10:46
atojha wrote:
I tried with

# = 1, Solvng we get P = 2

P## = 2

Ony option 'C' satisfies this ....

But,
option (1) also satisfies it.

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28 Jul 2013, 21:46
If P@=P1/P−1, what is the value of P@@?

(A) P/P−1

(B) 1/P

(C) P

(D) 2−P

(E) P−1

[Reveal] Spoiler:
C

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# m05 Q5 simple math help please

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