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Intern  Joined: 18 Aug 2011
Posts: 49
Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

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10 00:00

Difficulty:   35% (medium)

Question Stats: 78% (02:34) correct 22% (03:07) wrong based on 361 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) (4a-12c)/(3ac)
c) (3ac)/(4a-12c)
d) (ac)/(a+2c)
e) (ac)/(a+c)
Director  Joined: 01 Feb 2011
Posts: 653
Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

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1
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)

Retired Moderator Joined: 20 Dec 2010
Posts: 1790
Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

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2
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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Manager  Joined: 25 May 2011
Posts: 115
Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

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Alan's time to complete the task: a
Bob's time to complete the task: 4b
Together: 3c

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

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

Manager  Joined: 22 Sep 2015
Posts: 91
Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

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1
shahideh wrote:
Alan's time to complete the task: a
Bob's time to complete the task: 4b
Together: 3c

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

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

how do you isolate B on its own to figure out the other side of the equation?
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

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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)

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

Co-Founder & GMAT Assassin Follow
Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Manager  Joined: 22 Sep 2015 Posts: 91 Working alone at a constant rate, Alan can paint a house in a hours. [#permalink] ### Show Tags 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: $$\frac{1}{A} + \frac{1}{4B} =3C$$ EMPOWERgmat Instructor V Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13755 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Working alone at a constant rate, Alan can paint a house in a hours. [#permalink] ### Show Tags 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Follow Special Offer: Save$75 + GMAT Club Tests Free
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Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

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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)

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

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

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

The correct answer is 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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_________________ Re: Working alone at a constant rate, Alan can paint a house in a hours.   [#permalink] 08 Apr 2018, 07:38
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# Working alone at a constant rate, Alan can paint a house in a hours.

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