Question

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) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

Answer 1

(E) xy/(y-x)

Explanation:

The sum of the individual rates is the total rate. Each machine's rate is nails per hour is the inverse of its rate in hours per nail.

total rate = (800 nails)/(x hours) = 800/x nails/hour

A rate = (800 nails)/(y hours) = 800/y nails/hour

We want to find the time "b" such that ...

B rate = (800 nails)/(b hours) = 800/b nails/hour

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As we said, the total rate is the sum of the individual rates:

800/x = 800/y + 800/b

Multiplying by xyb/800, we get

yb = xb + xy

Solving for b, we have ...

yb -xb = xy

b(y -x) = xy

b = xy/(y-x) . . . . . matches choice E

It takes Machine B xy/(y-x) hours to produce 800 nails.