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Machine A, operating alone at its constant rate, produces 500 feet of

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Machine A, operating alone at its constant rate, produces 500 feet of  [#permalink]

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New post 01 Jul 2017, 03:42
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Machine A, operating alone at its constant rate, produces 500 feet of a particular fiber in 2 hours. Machine B, operating alone at its constant rate, produces 500 feet of the same fiber in 3 hours. Machine C, operating alone at its constant rate, produces 500 feet of the same fiber in 6 hours. How many hours will it take machines A, B, and C, operating together at their respective constant rates, to produce 1,000 feet of the fiber?

(A) 1.0
(B) 1.5
(C) 2.0
(D) 2.5
(E) 3.0

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Re: Machine A, operating alone at its constant rate, produces 500 feet of  [#permalink]

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New post 01 Jul 2017, 03:49
Bunuel wrote:
Machine A, operating alone at its constant rate, produces 500 feet of a particular fiber in 2 hours. Machine B, operating alone at its constant rate, produces 500 feet of the same fiber in 3 hours. Machine C, operating alone at its constant rate, produces 500 feet of the same fiber in 6 hours. How many hours will it take machines A, B, and C, operating together at their respective constant rates, to produce 1,000 feet of the fiber?

(A) 1.0
(B) 1.5
(C) 2.0
(D) 2.5
(E) 3.0



Hi..
Lets find per hour work ..
1) 500' per 2 hr= 250fph
2) 500'per3h=\(\frac{500}{3}\) fph
3) 500'per 6hr=\(\frac{500}{6}\) fph

Add all to get one hour combined work
250+\(\frac{500}{3}+\frac{500}{6}=500\)
So 1000 in 1000/500=2hr
C
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Re: Machine A, operating alone at its constant rate, produces 500 feet of  [#permalink]

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New post 03 Jul 2017, 17:11
OA is C ?
since the combined work done comes to be 500
time taken = total /combined = 1000/500 = 2
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Machine A, operating alone at its constant rate, produces 500 feet of  [#permalink]

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New post 04 Jul 2017, 10:18
Bunuel wrote:
Machine A, operating alone at its constant rate, produces 500 feet of a particular fiber in 2 hours. Machine B, operating alone at its constant rate, produces 500 feet of the same fiber in 3 hours. Machine C, operating alone at its constant rate, produces 500 feet of the same fiber in 6 hours. How many hours will it take machines A, B, and C, operating together at their respective constant rates, to produce 1,000 feet of the fiber?

(A) 1.0
(B) 1.5
(C) 2.0
(D) 2.5
(E) 3.0

Rate of Machine A = \(\frac{500}{2}\)
Rate of Machine B = \(\frac{500}{3}\)
Rate of Machine C = \(\frac{500}{6}\)

Combined rate = 500 (\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{6}\)) -->

500(\(\frac{3}{6}\) + \(\frac{2}{6}\) + \(\frac{1}{6}\)) =

500 (\(\frac{6}{6}\)) or \(\frac{(500)(6)}{(6)}\) =

500 feet/1 hour

W/r = t: \(\frac{1,000}{500}\) = 2

Answer C
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Re: Machine A, operating alone at its constant rate, produces 500 feet of  [#permalink]

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New post 04 Jul 2017, 10:42
Machine A, operating alone at its constant rate, produces 500 feet of a particular fiber in 2 hours. Machine B, operating alone at its constant rate, produces 500 feet of the same fiber in 3 hours. Machine C, operating alone at its constant rate, produces 500 feet of the same fiber in 6 hours. How many hours will it take machines A, B, and C, operating together at their respective constant rates, to produce 1,000 feet of the fiber?

Rate x Time = Work Done

Machine A = 500 in 2 Hrs

Machine B = 500 in 3 Hrs

Machine C = 500 in 6 Hrs


Work Done by All three machines A, B & C

\(= \frac{500}{2} + \frac{500}{3} + \frac{500}{6}\)

\(= 500 (\frac{1}{2} +\frac{1}{3} + \frac{1}{6})\)

\(= 500 (\frac{6}{6})\)

\(= 500 * 1\)

\(= 500\)

Hence, Rate of all three machines together is 500 Feet/Hour

Time for making 1000 Feet of fiber \(= \frac{1000}{500}\)

= 2 Hours

Hence, Answer is C
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Re: Machine A, operating alone at its constant rate, produces 500 feet of  [#permalink]

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New post 04 Jul 2017, 10:53
Lets assume the 500 ft fibre work to be = 6 units (6 is LCM of 2, 3, 6)

Then, Machine A's per hour work = 3 units
Machine B's per hour work = 2 units
Machine C's per hour work = 1 unit

(A+B+C) combined per hour work = 3+2+1 = 6 units
Thus together they can do 500 ft fibre work in 1 hour, so to do double the work (1000 ft), they would need double time, i.e., 2 hours. Hence C answer
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Re: Machine A, operating alone at its constant rate, produces 500 feet of   [#permalink] 04 Jul 2017, 10:53
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