Working simultaneously at their respective constant rates, Machines \(A\) and \(B\) produce 800 nails in \(x\) hours.

[VJesus12]

Working simultaneously at their respective constant rates, Machines \(A\) and \(B\) produce 800 nails in \(x\) hours. Working alone at its constant rate, Machine \(A\) produces 800 nails in \(y\) hours. In terms of \(x\) and \(y,\) how many hours does it take Machine \(B,\) working alone at its constant rate, to produce 800 nails?

(A) \(\dfrac{x}{x+y}\)
(B) \(\dfrac{y}{x+y}\)
(C) \(\dfrac{xy}{x+y}\)
(D) \(\dfrac{xy}{x-y}\)
(E) \(\dfrac{xy}{y-x}\)

[spoiler]OA=E[/spoiler]

Source: Official Guide

[Scott@TargetTestPrep]

Working simultaneously at their respective constant rates, Machines \(A\) and \(B\) produce 800 nails in \(x\) hours. Working alone at its constant rate, Machine \(A\) produces 800 nails in \(y\) hours. In terms of \(x\) and \(y,\) how many hours does it take Machine \(B,\) working alone at its constant rate, to produce 800 nails?

(A) \(\dfrac{x}{x+y}\)
(B) \(\dfrac{y}{x+y}\)
(C) \(\dfrac{xy}{x+y}\)
(D) \(\dfrac{xy}{x-y}\)
(E) \(\dfrac{xy}{y-x}\)

[spoiler]OA=E[/spoiler]

Source: Official Guide

Solution:

Let r = the rate of B in producing 800 nails. Since the rate of A is 800/y and their combined rate is 800/x, we can create the equation:

800/y + r = 800/x

r = 800/x - 800/y

r = 800(1/x - 1/y) = 800(y/(xy) - x/(xy)) = 800(y - x)/(xy)

Therefore, the time it takes for B to produce 800 nails is 800/[800(y - x)/(xy)] = 1/[(y - x)/(xy)] = xy/(y - x).

Answer: E