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# Working alone at a constant rate

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Manager
Joined: 18 Aug 2011
Posts: 60
Followers: 1

Kudos [?]: 19 [0], given: 6

Working alone at a constant rate [#permalink]  26 Aug 2011, 01:10
00:00

Difficulty:

(N/A)

Question Stats:

57% (02:20) correct 43% (02:07) wrong based on 25 sessions
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: 759
Followers: 14

Kudos [?]: 77 [0], given: 42

Re: Working alone at a constant rate [#permalink]  26 Aug 2011, 16:38
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 /(4a-12c)

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2026
Followers: 145

Kudos [?]: 1258 [0], given: 376

Re: Working alone at a constant rate [#permalink]  27 Aug 2011, 06:42
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) (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"
_________________
Intern
Joined: 31 May 2010
Posts: 38
Followers: 0

Kudos [?]: 2 [0], given: 2

Re: Working alone at a constant rate [#permalink]  27 Aug 2011, 11:07
C only
Manager
Joined: 25 May 2011
Posts: 158
Followers: 2

Kudos [?]: 47 [1] , given: 71

Re: Working alone at a constant rate [#permalink]  30 Oct 2011, 09:19
1
KUDOS
Alan's time to complete the task: a
Bob's time to complete the task: 4b
Together: 3c

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

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

Re: Working alone at a constant rate   [#permalink] 30 Oct 2011, 09:19
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# Working alone at a constant rate

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