rookie1ja 11 Report post Posted March 30, 2007 Filling the Pool - Back to the Cool Math Games A swimming pool has four faucets. The first can fill the entire pool with water in two days, the second – in three days, the third – in four days, and the last one can fill the pool in 6 hours. How long will it take to fill the pool using all 4 faucets together? This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Reservoir - solution Because there are 24 hours in one day, in one hour fills the first tap 1/48, the second tap 1/72, the third tap 1/96 and the fourth tap fills 1/6 of the reservoir. That is all together (6+4+3+48) / 288 = 61/288. The reservoir will be full in 288/61 hours, which is 4 hours 43 minutes and about 17 seconds. One reservoir has four taps. Using the first takes two days to saturate the reservoir, the second tap three days, the third four days and the last one 6 hours. How long will it take to fill the reservoir using all 4 taps at once? Share this post Link to post Share on other sites

Guest Report post Posted April 28, 2007 i divided them all by 4 and added them up to get 2 days and 7 hours Share this post Link to post Share on other sites

Guest Report post Posted June 3, 2007 I did it in the same way as rookie1ja, and got approximately 4 hrs and 44 minutes.. little bit less, actually. How did you get that solution with more than 2 days? Share this post Link to post Share on other sites

Guest Report post Posted June 5, 2007 natterbox, if all 4 taps are working together, you need to get a time less than 6 hours >< tap 4 can't do the whole job faster than all 4 working together. as a decimal, the time is approximately 4.72 hours. Share this post Link to post Share on other sites

Guest Report post Posted June 27, 2007 i came up with 4.73 hrs or 4 and 44 minutes Share this post Link to post Share on other sites

Guest Report post Posted July 3, 2007 I got 4.72 hours as well Share this post Link to post Share on other sites

Guest Report post Posted July 7, 2007 The answer is precisely 12/61 days. A = 1/2 (reservoirs/day) B = 1/3 C = 1/4 D = 4 (A + B + C + D) * Time = 1 (Time is in days. A, B, C, and D are in reservoirs per day.) (1/2 + 1/3 + 1/4 + 4) * Time = 1 (61 / 12) * Time = 1 Time = 12 / 61 days (~ 4.72 hours) Share this post Link to post Share on other sites

Guest Report post Posted August 29, 2007 2 days = 48 hours = x 3 days = 72 hours = y 4 days = 96 hours = z 6 hours = a (x + y)/2= 60 (60+ z)/2=78 (78+6)/2=42 Answer= 42 hours Share this post Link to post Share on other sites

Guest Report post Posted September 3, 2007 First of all we will calculate per hour work done by each tap like- 1st tap will fill the reservior by 1/48 in an hour. 2nd tap will fill it 1/72 3rd will fill 1/96 part of the reservior in an hour. and 4th will fill 1/6 now sum of all four taps if all are open simultaneously will be- 61/288 which gives the answer as- all four taps if open simultaneously will put the reservior in eq state in 4 hours 43 minutes and 2** seconds... ( Is that okay ) Share this post Link to post Share on other sites

Guest Report post Posted September 9, 2007 2 days = 48 hours = x 3 days = 72 hours = y 4 days = 96 hours = z 6 hours = a (x + y)/2= 60 (60+ z)/2=78 (78+6)/2=42 Answer= 42 hours you've calculated a weighted average ... i think. this is more like finding the factorial of the pipes (wrong), not the sum of the pipes and their "volume/time" flow rate. by your calculations, it would vary depending on the order calculated: (a + x) / 2 = (6 + 48) / 2 = 54 / 2 = 27 (27 + y) / 2 = (27 + 72) / 2 = 99 / 2 = 49.5 (49.5 + z) / 2 = (49.5 + 96) / 2 = 145.5 / 2 = 72.75 hours???? why would your answer change depending on plumbing order? step back from the problem and as mentioned earlier; if all four are ON, then they can't be slower than the fastest tap at 6 hours. then reverse determine a flow-rate ... say the reservoir is 1 litre. you should get calculations like: relative flowrates: 6hpl, 48hpl, 72hpl, 96hpl (hpl = hours per litre) invert to get lph from hpl (and find lowest common denominator for easy math) 6 hpl ==> 1/6 lph = 16/96 lph = 48/288 lph (lph = litres per hour) 2 days ==> 48 hpl ==> 1/48 lph = 2/96 lph = 6/288 lph 3 days ==> 72 hpl ==> 1/72 lph = 4/288 lph 4 days ==> 96 hpl ==> 1/96 lph = 3/288 lph combined flowrates (48+6+4+3)/288 lph = 61/288 lph ===> 288/61 hpl = 4.72 hours per litre Share this post Link to post Share on other sites

bonanova 83 Report post Posted September 9, 2007 The answer is precisely 12/61 days. A = 1/2 (reservoirs/day) B = 1/3 C = 1/4 D = 4 (A + B + C + D) * Time = 1 (Time is in days. A, B, C, and D are in reservoirs per day.) (1/2 + 1/3 + 1/4 + 4) * Time = 1 (61 / 12) * Time = 1 Time = 12 / 61 days (~ 4.72 hours) Lucid's analysis is the clearest. What you get when you turn on more taps is more flow. Find the flow rates and add them. The taps flow, respectively at rates of 1/2, 1/3, 1/4 and 4 reservoirs per day. Flowing together, [adding their flow rates] the rate is 61/12 reservoirs per day. They will fill 1 reservoir in exactly 12/61 days. That comes to 4 hours, 43 minutes and .. about 16.7213114 seconds. Share this post Link to post Share on other sites

Guest Report post Posted September 20, 2007 Tap A=2 days=.5 reservoirs in 1 day Tap B=3 days=1/3 reservoirs in 1 day Tap C=4 days=.25 reservoirs in 1 day Tap D=6 hours=4 reservoirs in 1 day 4+.25+.33+.50=5.08 reservoirs/day with all 4 5.08 Res/1Day = 1 Res/x Days 5.08 Res * x = 1ResDay x = 1ResDay/5.08 Res x = .20 days = 4.8 hours = 4 hours 48 min. Share this post Link to post Share on other sites

Guest Report post Posted September 26, 2007 The answer is precisely 12/61 days. A = 1/2 (reservoirs/day) B = 1/3 C = 1/4 D = 4 (A + B + C + D) * Time = 1 (Time is in days. A, B, C, and D are in reservoirs per day.) (1/2 + 1/3 + 1/4 + 4) * Time = 1 (61 / 12) * Time = 1 Time = 12 / 61 days (~ 4.72 hours) Lucid's analysis is the clearest. What you get when you turn on more taps is more flow. Find the flow rates and add them. The taps flow, respectively at rates of 1/2, 1/3, 1/4 and 4 reservoirs per day. Flowing together, [adding their flow rates] the rate is 61/12 reservoirs per day. They will fill 1 reservoir in exactly 12/61 days. That comes to 4 hours, 43 minutes and .. about 16.7213114 seconds. 'Nuff said. Before even looking at what anyone was answering, I had already determined that it would be 12/61 (or less than 1/5) of a day. I didn't calculate any further, but since 1/5 of a day would be 4h48m, any answer approximating 4h43m16.72s sounds about right to me... BoilingOil Share this post Link to post Share on other sites

Guest Report post Posted February 13, 2008 4.7213114754098360655737704918041 Hours Share this post Link to post Share on other sites

Guest Report post Posted February 16, 2008 (edited) You add flow rates. Flow Rate = 1/Time 1/T = 1/t1 + 1/t2 + 1/ t3 + 1/t4 So, Total Rate (1/T) = 1/4days + 1/3days + 1/2days + 1/6hours = 5.083333333 Resevoirs/Day (also 61/12) Since Time = 1/Rate Then Total Time = 1/5.0833333 (also 12/61) = 0.19672 Days or 4.7213 Hours (also 288/61 or about 4 Hours, 43 minutes and 17 seconds) It's just like adding electrical resistance in parallel. Edited February 16, 2008 by Moodster Share this post Link to post Share on other sites

TwoaDay 0 Report post Posted February 16, 2008 fractions are more exact so simply saying 12/61days is the best/easyest answer Share this post Link to post Share on other sites

Guest Report post Posted February 26, 2008 You didn't saay the four tap... Share this post Link to post Share on other sites

Guest Report post Posted February 29, 2008 (edited) Seems to me that to reach 288 gallons, for example: Where A is the 4 day tap, B the 3 day tap, C the2 day tap, and D the wuarter day tap A= 3 gal/h (96 hrs*3=288 gal) B= 4 gal/h (72 hrs*4=288 gal) C= 6 gal/h (48 hrs*6=288 gal) D= 48 gal/h (6 hrs*48=288 gal) So 61 gal/h to fill 288 gal= 4.721311 hours. Edited February 29, 2008 by teej Share this post Link to post Share on other sites

Guest Report post Posted March 13, 2008 What I did was: a= how much water gets filled up in the tank in 6 hours 1a - first tap 0.125a - second tap 0.0833a - third tap 0.0625a - fourth tap thus 1+.125+.0833+.0625 = running four taps to fill up the reservoir in 6 hours 6 / 1.2708 = 4.7214 hours to fill up the reservoir Share this post Link to post Share on other sites

Guest Report post Posted March 18, 2008 haresh that's an interesting way of solving the puzzle. I suppose it's the same as asking how much each tap fills the reservoir in an hour and then finding the solution but it's a different perspective. nice. Share this post Link to post Share on other sites

Guest Report post Posted March 19, 2008 (edited) are'nt logical puzzles supposed to be logical not mathematical Edited March 19, 2008 by Talha Share this post Link to post Share on other sites

Guest Report post Posted March 19, 2008 The simplest answer is within the question. "the third takes four days" So, 4 Days. Share this post Link to post Share on other sites

Guest Report post Posted April 2, 2008 days hours % per hour (speed) a 2.00 48 hrs 0.020833333 = 1/48 b 3.00 72 hrs 0.013888889 = 1/72 c 4.00 96 hrs 0.010416667 = 1/96 d 0.25 6 hrs 0.166666667 = 1/6 Total Speed Per Hour 0.211805556 Time to Fill the 100% 4.721311475 hours = 1/0.211806 283.2786885 minutes (Done on an Excel spreadsheet) Share this post Link to post Share on other sites

Guest Report post Posted June 11, 2008 days hours % per hour (speed) a 2.00 48 hrs 0.020833333 = 1/48 b 3.00 72 hrs 0.013888889 = 1/72 c 4.00 96 hrs 0.010416667 = 1/96 d 0.25 6 hrs 0.166666667 = 1/6 Total Speed Per Hour 0.211805556 Time to Fill the 100% 4.721311475 hours = 1/0.211806 283.2786885 minutes (Done on an Excel spreadsheet) Answer is 12/61. Here is the solution. Let 1/2 be work A can do in a day/ Let 1/3 be the work B can do in a day. Let 1/4 be the work C can do in a day. Let 4 be the work D can do in a day. => X(1/2 + 1/3 + 1/4 + 4) = 1 => X=1/(1/2 + 1/3 + 1/4 + 4) => x=12/61 Share this post Link to post Share on other sites