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

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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)

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

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27 Aug 2011, 07:42

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This post received KUDOS

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)

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....

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

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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....

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.

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

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11 Jan 2017, 18:45

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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)

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. [#permalink]

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