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Machine A, operating alone at its constant rate, produces 500 feet of
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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
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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
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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
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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
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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 DoneMachine A = 500 in 2 Hrs
Machine B = 500 in 3 Hrs
Machine C = 500 in 6 HrsWork 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/HourTime 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
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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




Re: Machine A, operating alone at its constant rate, produces 500 feet of
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04 Jul 2017, 10:53






