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02 Jun 2016, 15:32
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:27) correct
24%
(02:31)
wrong
based on 382
sessions
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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. 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)
09 Aug 2017, 12:47
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snorkeler
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
General Discussion
03 Aug 2017, 06:59
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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)
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)
