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Working alone at a constant rate

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Manager
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Working alone at a constant rate [#permalink] 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)
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Re: Working alone at a constant rate [#permalink] New post 26 Aug 2011, 17:38
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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Re: Working alone at a constant rate [#permalink] New post 27 Aug 2011, 07:42
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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Re: Working alone at a constant rate [#permalink] New post 27 Aug 2011, 12:07
C only :)
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Re: Working alone at a constant rate [#permalink] New post 30 Oct 2011, 10:19
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
Re: Working alone at a constant rate   [#permalink] 30 Oct 2011, 10:19
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