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

Question

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

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

chetan2u · Answer

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

TheKingInTheNorth · Answer

OA is C ?
since the combined work done comes to be 500 
time taken = total /combined = 1000/500 = 2

generis · Answer

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

ydmuley · Answer

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 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

amanvermagmat · Answer

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

MathRevolution · Answer

Let 6 units  be the total work (500 feet fiber)

A complete 3 units in 1 hour, B complete 2 units in 1 hour, and C complete 1 unit in 1 hour.

Total 6 units combined work (500 feet fiber) by A + B + C in 1 hour.

For (1000 feet fiber = 500 * 2): Time taken = 2 * 1 = 2 hours

Answer C

sompant_98 · Answer

Rate of A= 500/2= 250 feet/hr
Rate of B= 500/3= 166.67 feet/hr
Rate of C = 500/6= 83.33 feet/hr

Combined rate = 250+166.67+83.33= 500 feet/hr

Therefore time taken by all 3 = 1000 feet/500 feet/hr= 2hr

bumpbot · Answer

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

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