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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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

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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Source: Beat The GMAT — Problem Solving | Thu Sep 03, 2020 6:18 am

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

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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Scott@TargetTestPrep: GMAT Instructor; Posts: 8153; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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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