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Hoses A and B spout water at different constant rates, and hose A can

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

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New post 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  [#permalink]

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New post 07 Jan 2015, 06:57
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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.

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


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  [#permalink]

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New post 07 Jan 2015, 20:39
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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  [#permalink]

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New post 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  [#permalink]

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New post 07 Jan 2015, 08:01
noshtafoyza wrote:
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

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]


Your approach is correct:

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

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

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New post 07 Jan 2015, 08:39
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.
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Hoses A and B spout water at different constant rates, and hose A can  [#permalink]

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New post 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  [#permalink]

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New post 03 Nov 2016, 11:00
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 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  [#permalink]

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New post 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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Re: Hoses A and B spout water at different constant rates, and hose A can &nbs [#permalink] 31 May 2018, 16:19
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