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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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08 Nov 2018, 04:42
1
3
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Difficulty:

15% (low)

Question Stats:

77% (01:16) correct 23% (01:32) wrong based on 179 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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08 Nov 2018, 05:29
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) $$\frac{Pr}{S} - 1$$

Project PS Butler : Question #06

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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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

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

Project PS Butler : Question #06

$$P = \frac{S}{1+nr}$$
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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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) $$\frac{Pr}{S} - 1$$

Project PS Butler : Question #06

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

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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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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

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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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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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