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If P = S/1+nr and P, S, n, and r are positive numbers, then in terms

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If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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New post 08 Nov 2018, 04:42
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If \(P = \frac{S}{1+nr}\) and \(P, S, n,\) and \(r\) are positive numbers, then in terms of \(P, S\) and \(r\) what does \(n\) equal?


(A) \(\frac{S-P}{Pr}\)

(B) \(\frac{S}{rP} - 1\)

(C) \(\frac{S-P}{r}\)

(D) \(\frac{S}{P} - r\)

(E) \(\frac{Pr}{S} - 1\)


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Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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New post 08 Nov 2018, 05:29
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HKD1710 wrote:
If \(P = \frac{S}{1+nr}\) and \(P, S, n,\) and \(r\) are positive numbers, then in terms of \(P, S\) and \(r\) what does \(n\) equal?


(A) \(\frac{S-P}{Pr}\)

(B) \(\frac{S}{rP} - 1\)

(C) \(\frac{S-P}{r}\)

(D) \(\frac{S}{P} - r\)

(E) \(\frac{Pr}{S} - 1\)



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We need to find N in terms of P, R , S :)

\(P = \frac{S}{1+nr}\) divide both sides by 1+nr

\(\frac{P}{(1+NR)}= S\)

\(S= P+PNR\)

\(S-P = PNR\) divide by PR

\(\frac{S-P}{PR}= N\)

\(N = \frac{S-P}{PR}\)

IMO: A :)
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Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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New post 08 Nov 2018, 05:57
P * (1+nr) = S
1 + nr = S/P

nr = S/P - 1

nr = S-P/P

n = S-P/Pr

Answer choice A

Try n = 1, r = 2 and s = 6

P = 6/(1+2) =6/3 = 2

If we rearrange the first formula it would be 6 - 2 = 4/(2*2) = 4/4 = 1
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Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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New post 08 Nov 2018, 09:39
1
HKD1710 wrote:
If \(P = \frac{S}{1+nr}\) and \(P, S, n,\) and \(r\) are positive numbers, then in terms of \(P, S\) and \(r\) what does \(n\) equal?


(A) \(\frac{S-P}{Pr}\)

(B) \(\frac{S}{rP} - 1\)

(C) \(\frac{S-P}{r}\)

(D) \(\frac{S}{P} - r\)

(E) \(\frac{Pr}{S} - 1\)


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\(P = \frac{S}{1+nr}\)
or, \(\frac{P}{S} = \frac{1}{1+nr}\)
or, \(\frac{S}{P} = 1+nr\)
or, \(\frac{S}{P} -1= nr\)
or, \(\frac{S-P}{Pr} =n\).... Ans A.
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Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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New post 09 Nov 2018, 08:38
u1983 wrote:
HKD1710 wrote:
If \(P = \frac{S}{1+nr}\) and \(P, S, n,\) and \(r\) are positive numbers, then in terms of \(P, S\) and \(r\) what does \(n\) equal?


(A) \(\frac{S-P}{Pr}\)

(B) \(\frac{S}{rP} - 1\)

(C) \(\frac{S-P}{r}\)

(D) \(\frac{S}{P} - r\)

(E) \(\frac{Pr}{S} - 1\)


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\(P = \frac{S}{1+nr}\)
=> \(\frac{S}{P} = 1+nr\)
=> \(\frac{S}{P} -1= nr\)
=> \(\frac{S-P}{Pr} =n\)

A is the correct Answer
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Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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New post 11 Nov 2018, 19:23
1
HKD1710 wrote:
If \(P = \frac{S}{1+nr}\) and \(P, S, n,\) and \(r\) are positive numbers, then in terms of \(P, S\) and \(r\) what does \(n\) equal?


(A) \(\frac{S-P}{Pr}\)

(B) \(\frac{S}{rP} - 1\)

(C) \(\frac{S-P}{r}\)

(D) \(\frac{S}{P} - r\)

(E) \(\frac{Pr}{S} - 1\)


Simplifying, we have:

P(1 + nr) = S

P + Pnr = S

Pnr = S - P

n = (S - P)/Pr

Answer: A
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Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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New post 03 Feb 2019, 04:11
HKD1710 wrote:
If \(P = \frac{S}{1+nr}\) and \(P, S, n,\) and \(r\) are positive numbers, then in terms of \(P, S\) and \(r\) what does \(n\) equal?


(A) \(\frac{S-P}{Pr}\)

(B) \(\frac{S}{rP} - 1\)

(C) \(\frac{S-P}{r}\)

(D) \(\frac{S}{P} - r\)

(E) \(\frac{Pr}{S} - 1\)


Substitute the values carefully



P =S / (1+nr)

S = 4 n = 3 r =1, P = 1

We need to find an expression which will give n=3

A does that
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Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms   [#permalink] 03 Feb 2019, 04:11
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