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Working alone at a constant rate, Alan can paint a house in a hours.
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26 Aug 2011, 02:10
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77% (02:07) correct 23% (02:34) wrong based on 324 sessions
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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) (4a12c)/(3ac) c) (3ac)/(4a12c) d) (ac)/(a+2c) e) (ac)/(a+c)
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Re: Working alone at a constant rate, Alan can paint a house in a hours.
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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 /(4a12c)
Answer is C.



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Re: Working alone at a constant rate, Alan can paint a house in a hours.
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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) (4a12c)/(3ac) c) (3ac)/(4a12c) 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: \(b=\frac{3ac}{4a12c}\) Ans: "C"
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Re: Working alone at a constant rate, Alan can paint a house in a hours.
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27 Aug 2011, 12:07
C only



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Re: Working alone at a constant rate, Alan can paint a house in a hours.
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30 Oct 2011, 10:19
Alan's time to complete the task: a Bob's time to complete the task: 4b Together: 3c
\(\frac{1}{a}+\frac{1}{4b}=\frac{1}{3c}\)
\(b=\frac{3ac}{4a12c}\)
Answer C



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Re: Working alone at a constant rate, Alan can paint a house in a hours.
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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
\(\frac{1}{a}+\frac{1}{4b}=\frac{1}{3c}\)
\(b=\frac{3ac}{4a12c}\)
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.
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09 Dec 2015, 00:11
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: GMAT assassins aren't born, they're made, Rich
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Working alone at a constant rate, Alan can paint a house in a hours.
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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: 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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Re: Working alone at a constant rate, Alan can paint a house in a hours.
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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.
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16 Jan 2017, 18:27
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) (4a12c)/(3ac) c) (3ac)/(4a12c) d) (ac)/(a+2c) e) (ac)/(a+c) We are given that Alan can paint a house in a hours and Bob can paint 1/4 of the same hours in b hours. Thus, we can say the following: rate of Alan = 1/a rate of Bob = (1/4)/b = 1/(4b) Rate together = 1/a + 1/(4b). However, since we are also given that working together Alan and Bob can paint 1/3 of the house in c hours, we can express their combined rate as (1/3)/c = 1/(3c). Since the combined rate has to be the sum of their individual rates, we can say 1/a + 1/(4b) = 1/(3c), and we can solve for b in terms of a and c. Multiplying the entire equation by 12abc, we obtain: 12bc + 3ac = 4ab 3ac = 4ab  12bc 3ac = b(4a  12c) 3ac/(4a  12c) = b Answer: C
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Re: Working alone at a constant rate, Alan can paint a house in a hours.
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30 Jan 2017, 04:35
1) First of all we need to find the rates of Alan and Bob: A: \(r*a=1; r=\frac{1}{a}, B:r*b=\frac{1}{4}, r=\frac{1}{4}/b=\frac{1}{4b}\) 2) Working together Alan and Bob's rate is \(\frac{1}{a}+\frac{1}{4b}\) 3) Let's set up the equation using the information in the third sentence. \((\frac{1}{a}+\frac{1}{4b})*c=\frac{1}{3}, \frac{c}{a}+\frac{c}{4b}=\frac{1}{3}\).........b=\(\frac{3ac}{4a12c}\)
The correct answer is C



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