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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, 12:19
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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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Originally posted by AbdurRakib on 15 Jun 2016, 00:09.
Last edited by AbdurRakib on 22 Jun 2016, 12:19, 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, 17:34
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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, 06:22
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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

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

D
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  31 Oct 2020, 16:48
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  22 Jun 2016, 12:12
AbdurRakib equation is missing an indication of power for "p2".

I actually thought it was a typo for 2p. :(
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  22 Jun 2016, 12:20
arthurmatsuda wrote:
AbdurRakib equation is missing an indication of power for "p2".

I actually thought it was a typo for 2p. :(


edited

thanks for your remarks
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  19 Feb 2017, 09:28
Isolating r,
we get
r = p [ p - {(1-p^2)/p} ]

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

r= p^2 - 1 + p^2

r= 2p^2 - 1 = answer is 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, 21:31
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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  26 Jul 2017, 02:23
Bunuel

How to know if 1-p^2 is in bracket or not?

I found r=-1 because of this error.
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  26 Jul 2017, 03:14
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Shiv2016 wrote:
Bunuel

How to know if 1-p^2 is in bracket or not?

I found r=-1 because of this error.


In \(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\), it does not matter whether 1 - p^2 is in brackets.
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  17 Sep 2017, 22:46
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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, 22:52
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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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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  01 Mar 2018, 11:39
chetan2u wrote:
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

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

D



hi chetan2u,

can you please explain the last step of getting this \([m]r= 2p^2-1\) from this \(\frac{2p^2 –1}{p} = \frac{r}{p}\)

thanks!
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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  01 Mar 2018, 11:53
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dave13 wrote:
chetan2u wrote:
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



hi chetan2u,

can you please explain the last step of getting this \([m]r= 2p^2-1\) from this \(\frac{2p^2 –1}{p} = \frac{r}{p}\)

thanks!


Hey dave13

If you multiply or divide both the sides of a fraction by a variable/constant, the value remains the same

Another way to do this is as follows
\(\frac{2p^2 –1}{p} = \frac{r}{p}\) -> \(2p^2 –1= \frac{r}{p}*p\) (by cross multiplying) -> r = 2\(p^2\) - 1

Here p on the right-hand side cancel each other out(because in a fraction, common number/variables in the numerator and denominator cancel each other out)

Hope this helps you!
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  11 May 2019, 22:09
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) \(p^2 + p – 1\)


OG 2017 New Question


Simplify the expression,
P^2-1+p^2/p=r/p
=>2p^2-1=r [multiply both side by 'P']

Answer is D

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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  05 Jul 2020, 12:15
\(p – \frac{(1 – p^2 )}{p} =\frac{ r}{p}\)
\(r=p(p-\frac{(1 – p^2 )}{p} )\)
\(r=p^2-(1-p^2)\)
\(r=2p^2-1\)
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  29 Dec 2020, 06:24
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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) \(p^2 + p – 1\)


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

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

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

The answer is \(D\)
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  29 Dec 2020, 09:59
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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\(p – \frac{(1 – p^2 )}{p} =\frac{r}{p}\)

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

Or, \(2p^2 - 1 = r\), Answer is (D)
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink] New post  11 May 2021, 23:59
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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, 22:25
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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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r = [#permalink]
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