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Hoses A and B spout water at different constant rates, and hose A can
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07 Jan 2015, 05:37
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Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool? A 18 B 15 C 12 D 6 E 3 Thanks! My approach is the follwing and I don't understand what my mistake is: A filled 1/3 of the pool, so 2/3 are left to be filled in 3 hours by both hoses.
\(\frac{2}{3} / 3 = \frac{1}{6} + \frac{1}{B}\)
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Re: Hoses A and B spout water at different constant rates, and hose A can
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07 Jan 2015, 06:57
noshtafoyza wrote: Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool? A 18 B 15 C 12 D 6 E 3 Thanks! My approach is the follwing and I don't understand what my mistake is: A filled 1/3 of the pool, so 2/3 are left to be filled in 3 hours by both hoses.
\(\frac{2}{3} / 3 = \frac{1}{6} + \frac{1}{B}\) Since hose A can fill the pool in 6 hours, then in 2 + 3 = 5 hours it will fill 5/6th of the pool. Thus the remaining 1/6th is filled by hose B in 3 hours. This means that hose B,working alone, to fill the entire pool will need 3*6 = 18 hours. Answer: A. Hope it's clear.
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Re: Hoses A and B spout water at different constant rates, and hose A can
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07 Jan 2015, 20:39
Rate of hose A \(= \frac{1}{6}\) Work done by hose A in 2 hrs \(= \frac{1}{6} * 2 = \frac{1}{3}\) Pending work \(= 1  \frac{1}{3} = \frac{2}{3}\) Let rate of hose B \(= \frac{1}{b}\) \((\frac{1}{b} + \frac{1}{6}) * 3 = \frac{2}{3}\) b = 18 Answer = A
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Re: Hoses A and B spout water at different constant rates, and hose A can
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07 Jan 2015, 07:57
Thanks for the reply Bunuel! Your solution makes sense.
Could you please comment on my approach which I posted in the spoiler? I don't get the flaw in my logic there



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Re: Hoses A and B spout water at different constant rates, and hose A can
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07 Jan 2015, 08:01



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Re: Hoses A and B spout water at different constant rates, and hose A can
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07 Jan 2015, 08:09
Thanks for verifying bunuel! Turns out I just didnt get my math straight solving the equation



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Re: Hoses A and B spout water at different constant rates, and hose A can
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07 Jan 2015, 08:39
noshtafoyza wrote: Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool? A 18 B 15 C 12 D 6 E 3 Thanks! My approach is the follwing and I don't understand what my mistake is: A filled 1/3 of the pool, so 2/3 are left to be filled in 3 hours by both hoses.
\(\frac{2}{3} / 3 = \frac{1}{6} + \frac{1}{B}\) Another approach. let work = 18 units therefore, amount of work done by A in 1 hour = 18/6 = 3 units now, since A work alone for two hours. therefore amount of work completed by A in two hours = 2*3 =6 units. thus we have 186 = 12 units remaining. the remaining 12 units are completed by both A and B in 3 hours. Amount of work completed by A in 3 hours = 3*3 = 9units . Thus remaining 3 units must have been contributed by B. i.e. B completed 1unit in 1 hour. Therefore total time taken by B to complete 18 units of work alone = 18/1 = 18 hrs.



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Hoses A and B spout water at different constant rates, and hose A can
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03 Nov 2016, 10:12
Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool?
A) 18 B) 15 C) 12 D) 6 E) 3



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Re: Hoses A and B spout water at different constant rates, and hose A can
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03 Nov 2016, 11:00



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Re: Hoses A and B spout water at different constant rates, and hose A can
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12 Nov 2016, 00:30
noshtafoyza wrote: Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool? A 18 B 15 C 12 D 6 E 3 Thanks! My approach is the follwing and I don't understand what my mistake is: A filled 1/3 of the pool, so 2/3 are left to be filled in 3 hours by both hoses.
\(\frac{2}{3} / 3 = \frac{1}{6} + \frac{1}{B}\) Answer: Option A Please check the solution in attachment
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Re: Hoses A and B spout water at different constant rates, and hose A can
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31 May 2018, 16:19
noshtafoyza wrote: Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool?
A 18 B 15 C 12 D 6 E 3 We are given that hose A can fill the pool in 6 hours alone. Thus, the rate of hose A is 1/6. Since we are not given the rate of hose B, we can let B = the number of hours it takes hose B to fill the pool alone. Thus, the rate of hose B is 1/B. We are given that hose A worked alone for the first 2 hours and then the two hoses worked together for another 3 hours to complete the job. Thus, the time worked by hose A is 5 hours and the time worked by hose B is 3 hours. Since work = rate x time, we can calculate the work done by hose A and hose B. Work done by hose A = (1/6) x 5 = 5/6 Work done by hose B = (1/B) x 3 = 3/B Finally, since the completed job = 1, we can sum the work values of hoses A and B and set the sum to 1. 5/6 + 3/B = 1 Multiplying the entire equation by 6B gives us: 5B + 18 = 6B 18 = B Thus, hose B, when working alone, can complete the job in 18 hours. Answer: A
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Hoses A and B spout water at different constant rates, and hose A can
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15 Sep 2018, 03:34
Quote: Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool?
A 18 B 15 C 12 D 6 E 3
A worked for a total of 5 hrs (2 hrs alone and remaining 3 hrs with B). A alone can complete the task in 6 hrs. amount of work 'A' can do in 1 hr = \(\frac{1}{6}\) amount of work done by 'A' in 5 hrs = \(\frac{5}{6}\) this means that the remaining \(\frac{1}{6}\) of the work was done by 'B' in 3 days. this means that 'B' can complete \(\frac{1}{6}\) of the work in 3 days. this means that 'B' can complete \(\frac{1}{18}\) of the work in 1 day. Hence 'B' can complete the entire work alone in 18 days.



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Re: Hoses A and B spout water at different constant rates, and hose A can
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25 Sep 2018, 17:16
Where am i going wrong: my approach is long but i wd like to know where am i going wrong in my thinking:
1. after 2 hours, the remaining task is: 4/6> 2/3 2. in the next 3 hours, A fills (3/6)(2/3): 1/3 3. which means 2/3rd of the remaining 2/3 (from 1) is filled by B 4. thus 4/9 is filled by B in 3 hours 5. thus the rate of B is (4/9)/3 > 4/27 6. which means the # of hours is: 27/4> approx 8




Re: Hoses A and B spout water at different constant rates, and hose A can &nbs
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25 Sep 2018, 17:16






