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If \(N=\dfrac{K}{T+\frac{x}{y}},\) where \(T=\dfrac{K}5\) and \(x=5-T,\) which of the following expresses \(y\) in terms

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If \(N=\dfrac{K}{T+\frac{x}{y}},\) where \(T=\dfrac{K}5\) and \(x=5-T,\) which of the following expresses \(y\) in terms

by Vincen » Thu Sep 03, 2020 6:18 am

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If \(N=\dfrac{K}{T+\frac{x}{y}},\) where \(T=\dfrac{K}5\) and \(x=5-T,\) which of the following expresses \(y\) in terms of \(N\) and \(T ?\)


A. \(\dfrac{N(5-T)}{T(5-N)}\)

B. \(\dfrac{N(T-5)}{T(5-N)}\)

C. \(\dfrac{5-T}{T(5-N)}\)

D. \(\dfrac{5N(5-T)}{T(1-5N)}\)

E. \(\dfrac{N(5-T)}5\)

Answer: A

Source: Official Guide
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Re: If \(N=\dfrac{K}{T+\frac{x}{y}},\) where \(T=\dfrac{K}5\) and \(x=5-T,\) which of the following expresses \(y\) in t

by deloitte247 » Sat Sep 05, 2020 6:28 pm
$$If\ N=\frac{K}{T+\frac{x}{y}}\ ;\ T=\frac{k}{5};\ K=5T;\ and\ x=5-T$$
$$Therefore,\ N=\frac{5T}{T+\frac{5-T}{y}}$$
Cross multiply
$$N\left(T+\frac{5-T}{y}\right)=5T$$
$$NT+\frac{N\left(5-T\right)}{y}=5T$$
$$\frac{N\left(5-T\right)}{y}=\frac{5T-NT}{1}$$
$$N\left(5-T\right)=y\left(5T-NT\right)$$
$$y=\frac{N\left(5-T\right)}{5T-NT}$$
Factorising T
$$y=\frac{N\left(5-T\right)}{T\left(5-N\right)}$$
Answer = A
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Re: If \(N=\dfrac{K}{T+\frac{x}{y}},\) where \(T=\dfrac{K}5\) and \(x=5-T,\) which of the following expresses \(y\) in t

by Scott@TargetTestPrep » Tue Sep 08, 2020 9:46 am
Vincen wrote: ↑
Thu Sep 03, 2020 6:18 am
 If \(N=\dfrac{K}{T+\frac{x}{y}},\) where \(T=\dfrac{K}5\) and \(x=5-T,\) which of the following expresses \(y\) in terms of \(N\) and \(T ?\)


A. \(\dfrac{N(5-T)}{T(5-N)}\)

B. \(\dfrac{N(T-5)}{T(5-N)}\)

C. \(\dfrac{5-T}{T(5-N)}\)

D. \(\dfrac{5N(5-T)}{T(1-5N)}\)

E. \(\dfrac{N(5-T)}5\)

Answer: A

Source: Official Guide
Solution:

Since T = K/5, then K = 5T. If we replace K with 5T and x with 5 - T in the first given equation, we have:

N = 5T / (T + (5 - T)/y)

N[T + (5 - T)/y] = 5T

NT + N(5 - T)/y = 5T

N(5 - T) / y = 5T - NT

N(5 - T) / y = T(5 - N)


N(5 - T) / (T(5 - N)) = y

Answer: A

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