GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Mar 2019, 12:12

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

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

### Show Tags

26 Aug 2011, 02:10
10
00:00

Difficulty:

35% (medium)

Question Stats:

78% (02:34) correct 22% (03:07) wrong based on 361 sessions

### HideShow timer Statistics

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]

### Show Tags

26 Aug 2011, 17:38
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]

### Show Tags

27 Aug 2011, 07:42
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"
_________________
Manager
Joined: 25 May 2011
Posts: 115
Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

### Show Tags

30 Oct 2011, 10:19
Alan's time to complete the task: a
Bob's time to complete the task: 4b
Together: 3c

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

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

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

### Show Tags

08 Dec 2015, 08:57
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?
EMPOWERgmat Instructor
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

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)

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

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 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: $$\frac{1}{A} + \frac{1}{4B} =3C$$ EMPOWERgmat Instructor 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 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 _________________ 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 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!*****

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2826
Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

### Show Tags

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) (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

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

Current Student
Joined: 20 Jan 2017
Posts: 58
Location: United States (NY)
Schools: CBS '20 (A)
GMAT 1: 750 Q48 V44
GMAT 2: 610 Q34 V41
GPA: 3.92
Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

### Show Tags

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

Non-Human User
Joined: 09 Sep 2013
Posts: 10141
Re: Working alone at a constant rate, Alan can paint a house in a hours.  [#permalink]

### Show Tags

08 Apr 2018, 07:38
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Working alone at a constant rate, Alan can paint a house in a hours.   [#permalink] 08 Apr 2018, 07:38
Display posts from previous: Sort by