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

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

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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) $$p^2 + p – 1$$

OG 2017 New Question

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]

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15 Jun 2016, 06:22
2
3
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]

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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]

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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

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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =  [#permalink]

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05 Dec 2016, 17:34
3
1
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

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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =  [#permalink]

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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]

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04 Apr 2017, 21:31
1
2
$$p-\cfrac { \left( 1-{ p }^{ 2 } \right) }{ p } =\cfrac { r }{ p } \\ \left[ p\left( p-\cfrac { \left( 1-{ p }^{ 2 } \right) }{ p } =\cfrac { r }{ p } \right) \right] \\ { p }^{ 2 }-\left( 1-{ p }^{ 2 } \right) =r\\ { p }^{ 2 }-1+{ p }^{ 2 }=r\\ 2{ p }^{ 2 }-1=r$$

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]

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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]

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26 Jul 2017, 03:14
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]

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17 Sep 2017, 22:46
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]

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17 Sep 2017, 22:52
2
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: $$p^2 –1 + p^2 =r$$;

$$2p^2-1=r$$

Hope it's clear.
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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =  [#permalink]

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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

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!
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If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =  [#permalink]

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01 Mar 2018, 11:53
1
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]

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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']

Posted from my mobile device
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Re: If p ≠ 0 and p – (1 – p^2 )/p = r/p, then r =  [#permalink]

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