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HKD1710
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) \(\frac{Pr}{S} - 1\)



\(P = \frac{S}{1+nr}\)
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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HKD1710
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) \(\frac{Pr}{S} - 1\)


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

A is the correct Answer
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HKD1710
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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HKD1710
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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