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Working alone at a constant rate, Alan can paint a house in a hours.

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Working alone at a constant rate, Alan can paint a house in a hours. [#permalink]

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New post 26 Aug 2011, 02:10
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Working alone at a constant rate, Alan can paint a house in a hours. Working alone at a constant rate, Bob can point 1/4 of the same house in b hours. Working together, Alan and Bob can paint 1/3 of the house in c hours. What is the value of b in terms of a and c?

a) (3ac)/(a+c)
b) (4a-12c)/(3ac)
c) (3ac)/(4a-12c)
d) (ac)/(a+2c)
e) (ac)/(a+c)
[Reveal] Spoiler: OA
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New post 26 Aug 2011, 17:38
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work time

A 1 a
B 1/4 b

A+B 1/3 c


A's rate = 1/a

B's rate = 1/4b

when they are working together , their combined rate is 1/a + 1/(4b) = (1/3)/c

1/(4b) = 1/(3c) - 1/a

=> b = 3ac /(4a-12c)

Answer is C.
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New post 27 Aug 2011, 07:42
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Berbatov wrote:
Working alone at a constant rate, Alan can paint a house in a hours. Working alone at a constant rate, Bob can point 1/4 of the same house in b hours. Working together, Alan and Bob can paint 1/3 of the house in c hours. What is the value of b in terms of a and c?

a) (3ac)/(a+c)
b) (4a-12c)/(3ac)
c) (3ac)/(4a-12c)
d) (ac)/(a+2c)
e) (ac)/(a+c)


Alan's rate= 1/a
Bob's rate= 1/(4b)
Combined rate= 1/(3c)

\(\frac{1}{a}+\frac{1}{4b}=\frac{1}{3c}\)

Upon solving:
\(b=\frac{3ac}{4a-12c}\)

Ans: "C"
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New post 27 Aug 2011, 12:07
C only :)
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New post 30 Oct 2011, 10:19
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Alan's time to complete the task: a
Bob's time to complete the task: 4b
Together: 3c

\(\frac{1}{a}+\frac{1}{4b}=\frac{1}{3c}\)


\(b=\frac{3ac}{4a-12c}\)

Answer C
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Re: Working alone at a constant rate, Alan can paint a house in a hours. [#permalink]

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New post 08 Dec 2015, 08:57
shahideh wrote:
Alan's time to complete the task: a
Bob's time to complete the task: 4b
Together: 3c

\(\frac{1}{a}+\frac{1}{4b}=\frac{1}{3c}\)


\(b=\frac{3ac}{4a-12c}\)

Answer C


how do you isolate B on its own to figure out the other side of the equation?
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Re: Working alone at a constant rate, Alan can paint a house in a hours. [#permalink]

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New post 09 Dec 2015, 00:11
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Hi All,

This question is a variation of a 'Work Formula' question (it involves 2 'entities' working on the same task together), so we can use the Work Formula to solve it.

Work = (A)(B)/(A+B) where A and B are the individual times that it takes the 2 entities to complete the task on their own.

We're told that:
1) Alan can paint a house in A hours
2) Bob can paint the same house in 4B hours
3) Together, Alan and Bob can paint the house in 3C hours

Using the Work Formula, we would have....

(A)(4B) / (A + 4B) = 3C

We're asked to solve for B in terms of A and C....

(A)(4B) = (3C)(A + 4B)
4AB = 3AC + 12BC
4AB - 12BC = 3AC
B(4A - 12C) = 3AC
B = (3AC)/(4A - 12C)

Final Answer:
[Reveal] Spoiler:
C


GMAT assassins aren't born, they're made,
Rich
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Manager
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Kudos [?]: 15 [0], given: 131

Working alone at a constant rate, Alan can paint a house in a hours. [#permalink]

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New post 10 Dec 2015, 12:24
EMPOWERgmatRichC wrote:
Hi All,

This question is a variation of a 'Work Formula' question (it involves 2 'entities' working on the same task together), so we can use the Work Formula to solve it.

Work = (A)(B)/(A+B) where A and B are the individual times that it takes the 2 entities to complete the task on their own.

We're told that:
1) Alan can paint a house in A hours
2) Bob can paint the same house in 4B hours
3) Together, Alan and Bob can paint the house in 3C hours

Using the Work Formula, we would have....

(A)(4B) / (A + 4B) = 3C

We're asked to solve for B in terms of A and C....

(A)(4B) = (3C)(A + 4B)
4AB = 3AC + 12BC
4AB - 12BC = 3AC
B(4A - 12C) = 3AC
B = (3AC)/(4A - 12C)

Final Answer:
[Reveal] Spoiler:
C


GMAT assassins aren't born, they're made,
Rich


But isn't the equation:

\(\frac{1}{A} + \frac{1}{4B} =3C\)
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Followers: 311

Kudos [?]: 2122 [0], given: 161

Re: Working alone at a constant rate, Alan can paint a house in a hours. [#permalink]

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New post 10 Dec 2015, 18:26
Hi nycgirl212,

The approach I used involved the Work Formula. If you want to use the 'unit' approach, then you should review the explanations offered by Spidy001, shahideh and fluke.

GMAT assassins aren't born, they're made,
Rich
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Re: Working alone at a constant rate, Alan can paint a house in a hours.   [#permalink] 10 Dec 2015, 18:26
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Working alone at a constant rate, Alan can paint a house in a hours.

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