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Intern  Joined: 27 Nov 2014
Posts: 4
Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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3
11 00:00

Difficulty:   35% (medium)

Question Stats: 72% (02:01) correct 28% (02:02) wrong based on 390 sessions

### HideShow timer Statistics 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.

$$\frac{2}{3} / 3 = \frac{1}{6} + \frac{1}{B}$$
Math Expert V
Joined: 02 Sep 2009
Posts: 56307
Re: Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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6
1
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.

Hope it's clear.
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Re: Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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9
2
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

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Intern  Joined: 27 Nov 2014
Posts: 4
Re: Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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Could you please comment on my approach which I posted in the spoiler? I don't get the flaw in my logic there
Math Expert V
Joined: 02 Sep 2009
Posts: 56307
Re: Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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noshtafoyza wrote:

Could you please comment on my approach which I posted in the spoiler? I don't get the flaw in my logic there

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.

$$\frac{2}{3} / 3 = \frac{1}{6} + \frac{1}{B}$$[/spoiler]

(2/3)/3=1/6 + 1/x
2/3 = 1/2 + 3/x
1/6 = 3/x
x = 18.

The same answer as in my solution.
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Intern  Joined: 27 Nov 2014
Posts: 4
Re: Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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1
Thanks for verifying bunuel! Turns out I just didnt get my math straight solving the equation Senior Manager  Joined: 13 Jun 2013
Posts: 271
Re: Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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2
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 18-6 = 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.
Manager  Joined: 23 Sep 2016
Posts: 92
Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 56307
Re: Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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SW4 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

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Re: Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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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.

$$\frac{2}{3} / 3 = \frac{1}{6} + \frac{1}{B}$$

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  [#permalink]

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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.

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Manager  B
Joined: 20 Jun 2017
Posts: 92
GMAT 1: 570 Q49 V19 Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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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  [#permalink]

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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   [#permalink] 25 Sep 2018, 17:16
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