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

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swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

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

by swerve » Thu Mar 01, 2018 5:24 am

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Working alone at a constant rate, Alan can paint a house in an hour. 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)

The OA is C.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer and I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.

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Source: Beat The GMAT — Problem Solving | Thu Mar 01, 2018 5:24 am

mbawisdom: Senior | Next Rank: 100 Posts; Posts: 94; Joined: Tue Dec 16, 2014 9:50 am; Location: London, UK; Thanked: 2 times; Followed by:4 members; GMAT Score:770

by mbawisdom » Mon Mar 05, 2018 4:27 pm

swerve wrote:Working alone at a constant rate, Alan can paint a house in an hour. 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)

The OA is C.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer and I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.

PRESUMPTION: Alan can paint a house in A hours and not an hour. Otherwise we do not have an "a" value.

H = painted house
Ra = Rate of Alan
Rb - Rate of Bob

Work = Rate * Time
Equations:
(1) H = Ra * A --> Ra = H/A
(2) H/4 = Rb * B --> Rb = H/4B
(3) H/3 = (Ra + Rb)*C

Plug Ra and Rb into (3): H/3 = (H/A + H/4B)*C
Divide both sides by H: 1/3 = (1/A + 1/4B)*C
1/3 = C/A + C/4B
1/3 - C/A = C/4B
(A-3C)/3A = C/4B
Flip both sides: 3A/(A - 3C) = 4B/C
Multiply both sides by C/4: 3AC/(4A - 12C) = B

Answer is C.

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Mon Mar 05, 2018 8:30 pm

Hi swerve,

We're told that Alan can paint a house in 1 hour, Bob can point 1/4 of the same house in B hours and - working together - Alan and Bob can paint 1/3 of the house in C hours. We're asked for the value of B in terms of A and C. This is an example of a "Work Formula" question - when you have two entities working together on a task, you can use the following formula to determine how long it will take the two entities to complete the task:

(X)(Y)/(X+Y) = time to complete the task (where X and Y are the two times it takes the individuals to complete the task when working alone) .

Alan can paint a house in 1 hour, so X=1.
Bob can paint 1/4 of the house in B hours, so it takes him 4B hours to paint the full house (so Y = 4B)
Together, they can paint 1/3 of the house in C hours, so it takes them 3C hours to pail the full house.

(1)(4B)/(1 + 4B) = 3C

Now we can do Algebra and solve for B....

4B = (3C)(1+4B)
4B = 3C + 12BC
4B - 12BC = 3C
B(4 - 12C) = 3C
B = (3C)/(4 - 12C)

Remember that A=1, so we're looking for the above value of B when you plug A=1 into the answers...

Final Answer: C

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

Contact Rich at [email protected]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Mon Jun 10, 2019 6:11 pm

swerve wrote:Working alone at a constant rate, Alan can paint a house in an hour. 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)

The OA is C.

We are given that Alan can paint a house in a hours and Bob can paint 1/4 of the same house 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

Scott Woodbury-Stewart
Founder and CEO
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