Working simultaneously at their respective constant rates, Machines A

Question

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

A. ac/(a + c)
B. 2ac/(a + c)
C. ac/(2a + 2c)
D. ac/(2a - 2c)
E. ac/(2c - 2a)

JeffTargetTestPrep · Accepted Answer

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

Klarbeinstreiter · Answer

Since there is no restriction given that a and b must be different, let's assume that Machine A and Machine B have the same rates. => a=b.

Let's take some small numbers and see how that works out:

Let a = b = 2, which means both Machines A and B need 2 hours each independently to produce 20 widgets. 
Combined, they'll need half the time, namely 1 hour to produce 20 widgets => c = 1

What happens if we plug-in these numbers now?  
Before we do that we have to keep in mind that we are looking for the time that b needs to produce 10 widgets. 
Therefore our target value = b/2 = 1.

First we see that we can immediately rule out answers A) and B) as we'd divide an even number ac or 2ac with an uneven number a+c. That would not result in an integer value.

(A)ac/a+c =>2/3, no 
(B)2ac/a+c =>4/3, no 
(C)ac/2a+2c => 2/6, no 
(D)ac/2a-2c => 2/2, yes!
(E)ac/2c-2a => 2/-2, no

Answer (D)

KSBGC · Answer

(A+B)"s 1 hr work = 1/c hrs.

A's 1 hr work = 1/a

B's 1 hr work = 1/c - 1/a = a-c/ca.

Total time needed for B to finish this work = ca/c-a.

ca/a-c is basically the time for 20 widgets. Now we have to know time for 10 widgets.

ca / c-a *1/2 = ca / 2a - 2c.

D is the correct answer.

Bunuel · Answer

Kinshook · Answer

Given:  Working simultaneously at their respective constant rates, Machines A and B produce 20 widgets in c hours. Working alone at its constant rate, Machine A produces 20 widgets in a hours.

Asked:  In terms of a and c, how many hours does it take Machine B, working alone at its constant rate, to produce 10 widgets?

For producing 20 widgets: -

\frac{1}{a} + \frac{1}{b} = \frac{1}{c} 
 \frac{1}{b} = \frac{1}{c} - \frac{1}{a} = \frac{a-c}{ac} 
 b = \frac{ac}{(a-c)}

For producing 10 widgets :- 
 \frac{b}{2}= \frac{ac}{(2a-2c)}

IMO D

Abhishek009 · Answer

Efficiency of Machine A + B is  \frac{20}{c} 
Efficiency of Machine A is  \frac{20}{a}

Thus, Efficiency of Machine B is  \frac{20}{c} - \frac{20}{a} = \frac{20a - 20c}{ac}

Hence time required to produce 10 widgets is  10 ÷ \frac{20a - 20c}{ac}= ac/(2a - 2c) , Answer must be (D)

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Working simultaneously at their respective constant rates, Machines A

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