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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =

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AbdurRakib
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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   Updated on: 22 Jun 2016, 13:19
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If p ≠ 0 and \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\), then r =


A) p + 1

B) 2p – 1

C) \(p^2+ 1\)

D) \(2p^2 – 1\)

E) \(p^2 + p – 1\)


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D

Originally posted by AbdurRakib on 15 Jun 2016, 01:09.
Last edited by AbdurRakib on 22 Jun 2016, 13:19, edited 1 time in total.
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   05 Dec 2016, 18:34
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AbdurRakib wrote:
If p ≠ 0 and p – {\(\frac{(1 – p^2 )}{p}\)} = \(\frac{r}{p}\), then r =

A) p + 1
B) 2p – 1
C) \(p^2\)+ 1
D) 2\(p^2\) – 1
E) \(p^2\) + p – 1


We are given that p - [(1-p^2)/p] = r/p, and we must isolate r.

We can start by multiplying the entire equation by p and then simplifying:

p^2 - (1 - p^2) = r

p^2 - 1 + p^2 = r

2p^2 - 1 = r

Answer: D
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   15 Jun 2016, 07:22
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AbdurRakib wrote:
If p ≠ 0 and p – {\(\frac{(1 – p2 )}{p}\)} = \(\frac{r}{p}\), then r =

A) p + 1
B) 2p – 1
C) \(p^2\)+ 1
D) 2\(p^2\) – 1
E) \(p^2\) + p – 1

OG 2017 New Question



\(p –\frac{(1 – p^2 )}{p} = \frac{p^2 –(1 – p^2 )}{p} = \frac{p^2 –1 + p^2}{p} = \frac{2p^2 –1}{p} = \frac{r}{p}\).................
\(r= 2p^2-1\)

D
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   04 Apr 2017, 22:31
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\(p-\cfrac { \left( 1-{ p }^{ 2 } \right) }{ p } =\cfrac { r }{ p } \\ \left[ p\left( p-\cfrac { \left( 1-{ p }^{ 2 } \right) }{ p } =\cfrac { r }{ p } \right) \right] \\ { p }^{ 2 }-\left( 1-{ p }^{ 2 } \right) =r\\ { p }^{ 2 }-1+{ p }^{ 2 }=r\\ 2{ p }^{ 2 }-1=r\)

PEMDAS in the 3rd line got me thinking whether I should multiply that negative to \(\left( 1-{ p }^{ 2 } \right)\).
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   17 Sep 2017, 23:46
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Bunuel. Once we multiply both sides by p, p in the numerator becomes p^2, but what happens to the p in the denumerator? I understand that it gets cancel, but I don't see how.
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   17 Sep 2017, 23:52
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Valentina86 wrote:
If p ≠ 0 and \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\), then r =


A) p + 1

B) 2p – 1

C) \(p^2+ 1\)

D) \(2p^2 – 1\)

E) \(p^2 + p – 1\)

Bunuel. Once we multiply both sides by p, p in the numerator becomes p^2, but what happens to the p in the denumerator? I understand that it gets cancel, but I don't see how.


\(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\);

Multiply by p: \(p*(p – \frac{(1 – p^2 )}{p}) =p*\frac{r}{p}\);

Expand the left hand side and reduce by p in the right hand side: \(p^2 – p*\frac{(1 – p^2 )}{p} =r\);

Reduce by p in the second term on the left hand side: \(p^2 –(1 – p^2 ) =r\);

Open the brackets: \(p^2 –1 + p^2 =r\);

\(2p^2-1=r\)

Answer: D.

Hope it's clear.
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   31 Oct 2020, 17:48
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Video solution from Quant Reasoning:
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   12 May 2021, 00:59
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AbdurRakib wrote:
If p ≠ 0 and \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\), then r =


A) p + 1

B) 2p – 1

C) \(p^2+ 1\)

D) \(2p^2 – 1\)

E) \(p^2 + p – 1\)


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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   16 May 2021, 23:25
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Asked: If p ≠ 0 and \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\), then r =

\(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\)
\(\frac{(p^2 - 1 + p^2 )}{p} =\frac{r}{p}\)
\(\frac{(2p^2 - 1)}{p} =\frac{r}{p}\)
r = 2p^2 - 1

IMO D
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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   05 Feb 2022, 14:37
avigutman wrote:
Video solution from Quant Reasoning:


avigutman
Thank you! Would choosing a number for P to solve for R, and then testing the answer choices always work as well? Is this a good strategy? For instance, I said let P=2 and then R equaled 7 after I solved for R in the original. I then plugged in P=2 in the answer choices and only D yielded 7 for R.
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post   05 Feb 2022, 17:30
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woohoo921 wrote:
Would choosing a number for P to solve for R, and then testing the answer choices always work as well? Is this a good strategy? For instance, I said let P=2 and then R equaled 7 after I solved for R in the original. I then plugged in P=2 in the answer choices and only D yielded 7 for R.


woohoo921 thanks for your question. That's a fine strategy for test day, but you should strive to get more out of a question when practicing at home, or else you miss an opportunity to "grow" as a test taker. See here for more:

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Re: If p 0 and p (1 p^2 )/p = r/p, then r = [#permalink] New post   15 Apr 2023, 00:19
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