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Pumping alone at their respective constant rates, one inlet pipe fills
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16 Oct 2015, 06:31
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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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16 Oct 2015, 06:48
pipe 1 = 1/6 an hour pipe 2 = 1/9 an hour
Total time taken =1/6 + 1/9 =3.6 Hours



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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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16 Oct 2015, 07:06
Tank A does 1/2 in 3 hrs so we can fine for full capacity it would take 6 hrs. Similarly , Tank B takes 2/3rd of tank to be filled in 6 hrs so it would fill to full capacity in 6/(2/3) =>18/2 => 9
Now simple work problem states that it would fill the whole tank in :
1/9 + 1/6 = 5/18 unit =>18/5 hrs = 3.6 hrs
Hence Option B



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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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16 Oct 2015, 07:10
agree with the above solutions. Answer B is correct.



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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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17 Oct 2015, 03:48
Bunuel wrote: Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
(A) 3.25 (B) 3.6 (C) 4.2 (D) 4.4 (E) 5.5
Kudos for a correct solution. Total capacity = C Rate of 1 = C/6 Rate of 2 = C/9 Combined rate = 5C/18 Time = 18/5 = 3.6 (B)
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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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17 Oct 2015, 06:42
Bunuel wrote: Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
(A) 3.25 (B) 3.6 (C) 4.2 (D) 4.4 (E) 5.5
Kudos for a correct solution. Time taken by First tank Fill 1/2 the tank = 3 hours i.e. Time taken by First tank to Fill 1 complete the tank = 6 hours Time taken by Second tank Fill 2/3 the tank = 6 hours i.e. Time taken by First tank to Fill 1 complete the tank = (3/2)*6 = 9 hours in 1 Hour, Both tanks together Fill the tank = (1/6)+(1/9) = 5/18 tank i.e. Time taken by Both tank to Fill 1 complete the tank = 18/5 hours = 3.6 hours Answer: option B
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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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11 Dec 2015, 09:17
let c be tank capacity. Time taken by First pipe to fill 1/2 the tank = 3 hours. So work done in 1 hr= c*{1/(2*3)}= c/6 Time taken by Second pipe to Fill 2/3 the tank = 6 hours. So work done in 1hr= c*{2/(3*6)}=c/9 in 1 Hour, Both tanks together Fill the tank = (1c/6)+(c1/9) = 5c/18 i.e. Time taken by Both tank to Fill 1 complete the tank = 18/5 hours = 3.6 hours Answer: option B
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Pumping alone at their respective constant rates, one inlet pipe fills
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17 Feb 2016, 14:50
Bunuel wrote: Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
(A) 3.25 (B) 3.6 (C) 4.2 (D) 4.4 (E) 5.5
Kudos for a correct solution. An alternative, logical approach with minimal calculations and no variables: First pipe fills full capacity in 6 hours, therefore fills approx. 16.6% in 1 hr. Second pipe fills 2/3 in 6 hours, therefore fills approx. 11% in 1 hr. Combined, they'd fill 27.6% per hr and would need just a bit less than 4 hours to fill the tank completely. Only choice in that range is B.



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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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17 Feb 2016, 15:47
Time taken by First tank Fill 1/2 the tank = 3 hours i.e. Time taken by First tank to Fill 1 complete the tank = 6 hours Time taken by Second tank Fill 2/3 the tank = 6 hours i.e. Time taken by First tank to Fill 1 complete the tank = (3/2)*6 = 9 hours in 1 Hour, Both tanks together Fill the tank = (1/6)+(1/9) = 5/18 tank i.e. Time taken by Both tank to Fill 1 complete the tank = 18/5 hours = 3.6 hours correct answer .... b
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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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05 Jun 2017, 10:13
Bunuel wrote: Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
(A) 3.25 (B) 3.6 (C) 4.2 (D) 4.4 (E) 5.5 The inlet pipe fills the empty tank to full capacity in 6 hours The outlet pipe fills the empty tank to full capacity in 9 hours Let, the capacity of the tank be 18 units.. So, efficiency of the inlet pipe is 3 inits/hr And, efficiency of the outlet pipe is 2 inits/hr Combined efficiency of both the pipes is 5 units/hr Thus, time required by both the pipes to fill the tank will be \(\frac{18}{5}\) = 3.6 Hours Hence, answer will be (B) 3.6 hours...
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Re: Pumping alone at their respective constant rates, one inlet pipe fills
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18 Aug 2018, 19:12
Bunuel wrote: Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
(A) 3.25 (B) 3.6 (C) 4.2 (D) 4.4 (E) 5.5 We are given that the first inlet pipe fills an empty tank to 1/2 capacity in 3 hours. Since rate = work/time, the rate of the first inlet pipe is (1/2)/3 = 1/6. We are also given that the second inlet pipe fills the same empty tank to 2/3 capacity in 6 hours. Thus, the rate of the second inlet pipe is (2/3)/6 = 1/9. We need to determine how many hours it will take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity. If we let t = the time in hours the two inlet pipes are working together, then the work of the first inlet pipe = (1/6)t and the work of the second inlet pipe = (1/9)t. Since the tank is filled, we can set total work to 1 and create the following equation: (1/6)t + (1/9)t = 1 Multiplying the entire equation by 18, we obtain: 3t + 2t = 18 5t = 18 t = 18/5 = 3.6 Answer: B
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