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Retired Moderator V
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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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Question Stats: 83% (01:10) correct 17% (01:39) wrong based on 313 sessions

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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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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

$$\frac{S-P}{PR}= N$$

$$N = \frac{S-P}{PR}$$

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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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P * (1+nr) = S
1 + nr = S/P

nr = S/P - 1

nr = S-P/P

n = S-P/Pr

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
RC Moderator V
Joined: 24 Aug 2016
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GMAT 1: 540 Q49 V16 GMAT 2: 680 Q49 V33 Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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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) $$\frac{Pr}{S} - 1$$

Project PS Butler : Question #06

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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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GMAT 1: 680 Q49 V34 Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms  [#permalink]

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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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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

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Director  G
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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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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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Many of life's failures happen with people who do not realize how close they were to success when they gave up. Re: If P = S/1+nr and P, S, n, and r are positive numbers, then in terms   [#permalink] 03 Feb 2019, 05:11
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