snorkeler wrote:

Working simultaneously at their respective constant rates, machine A and B produces 20 widgets in c hours. Working alone at its constant rate, Machine A produces 20 widgets in 'a' hours.In terms of a and c, how many hours does it take Machine B, working alone at its constant rate, to produce 10 widgets

(A)ac/a+c

(B)2ac/a+c

(C)ac/2a+2c

(D)ac/2a-2c

(E)ac/2c-2a

Since machines A and B produce 20 widgets in c hours, their combined rate = 20/c. Since machine A produces 20 widgets in a hours, its rate = 20/a. We can let the time it takes machine B to produce 20 widgets = b; thus, its rate = 20/b and we can create the following equation:

20/a + 20/b = 20/c

Multiplying by abc, we have:

20bc + 20ac = 20ab

bc + ac = ab

ac = ab - bc

ac = b(a - c)

ac/(a - c) = b

Since the time it takes machine B to produce 20 widgets is ac/(a - c), the time it takes machine B to produce 10 widgets must be half as much, or:

ac/(a - c) x 1/2 = ac/(2a - 2c)

Answer: D

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