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# Bruce and Anne can clean their house in 4 hours working toge

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Joined: 05 Nov 2012
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Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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Updated on: 05 May 2013, 22:49
2
11
00:00

Difficulty:

35% (medium)

Question Stats:

73% (01:47) correct 27% (02:28) wrong based on 227 sessions

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Bruce and Anne can clean their house in 4 hours working together at their respective constant rates. If Anne’s speed were doubled, they could clean their house in 3 hours working at their respective rates. How many hours does it currently take Anne to clean the house on her own?

A. 6
B. 7
C. 8
D. 12
E. 14

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Originally posted by pikachu on 05 May 2013, 21:49.
Last edited by Bunuel on 05 May 2013, 22:49, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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06 May 2013, 00:44
4
2
pikachu wrote:
Bruce and Anne can clean their house in 4 hours working together at their respective constant rates. If Anne’s speed were doubled, they could clean their house in 3 hours working at their respective rates. How many hours does it currently take Anne to clean the house on her own?

A. 6
B. 7
C. 8
D. 12
E. 14

As rates are additive,

Ra+Rb = 1/4

2*Ra+Rb = 1/3

Subtract both the equations, we have

Ra = 1/12.

D.
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Re: Anne and Bruce  [#permalink]

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Updated on: 06 May 2013, 07:40
Lets suppose Anne and Bruce take A and B hrs working separately
So in 1 hour they can together finish 1/A + 1/B portion of the work which equals 1/4 (As the work is completed in 4 hours)
After anne doubles her rate of work the portion completed by the both is 1/A +2/B which is equal to 1/3 (As the work is completed in 3 hours)

Solving these 2 equations we can find B as 12
So, D
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Originally posted by aceacharya on 05 May 2013, 22:09.
Last edited by aceacharya on 06 May 2013, 07:40, edited 1 time in total.
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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06 May 2013, 02:03
1/a + 1/b =1/4.--(1)

now ann's rate is doubled.
2/a+1/b = 1/3 => 1/a+1/a+1/b = 1/4 using 1/a+1/b from from (1) 1/a+1/4 = 1/3 or a =12.
Option D
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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06 May 2013, 18:36
2
Another way to look at it is:

Rate of A and B combined = Rate A + Rate B = 1job/4hrs

If A doubles and works with B, we have:

Rate A + [Rate A + Rate B] = 1job/3hrs

or

Rate A +1job/4hrs = 1job/3hrs

so

Rate A = 1/3 - 1/4 = 4/12 - 3/12 = 1/12 = 1job/12hrs (so basically, one job takes 12 hours)

oop-- so answer D! Hope that helps!
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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07 May 2013, 07:09
1
1/A = Anna's rate; 1/B= Bruce's rate

1/A+1/B=1/4
&
2/A+1/B=1/3

Combining both equations:
1/A+(1/A+1/B)=1/3

1/A=1/3-1/4=1/12

R=W/T pick (D).
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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07 May 2013, 16:45
1
DaNG! All you people did some complex workout all i did was 3x4= 12 FINISHED! hahahahaaa
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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14 May 2013, 02:35
madzstar wrote:
DaNG! All you people did some complex workout all i did was 3x4= 12 FINISHED! hahahahaaa

I truly appreciate your answering quality. Rather having confusing and complex workout,solution must be precise and illustrative.
I agree to your answering skill.The answer is D.
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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14 May 2013, 07:48
2
Looking at it in a different .(I don't want to work with fractions)

a = amount of work done by anne in one hour
b = amount of work done by bruce in one hour

4a + 4b = W(work)

6a + 3b = W

which implies 2a = b

which gives W = 12a

So it takes anne 12 hours to complete the work
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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18 Dec 2013, 05:27
1
work = rate * time

set up equations.

Working together, normal rate: 1(job) = (A + B) *4 => 1/4 = A+B
Working together, double speed of A: 1(job) = (2A+ B) *3 => 1/3 = 2A+B

Substract first equation from second.
1/12 = A
This means she finishes 1(job) every 12(hours). Hence D.
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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04 Mar 2018, 15:40
Top Contributor
pikachu wrote:
Bruce and Anne can clean their house in 4 hours working together at their respective constant rates. If Anne’s speed were doubled, they could clean their house in 3 hours working at their respective rates. How many hours does it currently take Anne to clean the house on her own?

A. 6
B. 7
C. 8
D. 12
E. 14

Another approach is to assign a NICE value to the job.
We'll choose a value that works well with the given information (4 hours and 3 hours)
So let's say the cleaning job consists of cleaning 12 rooms

Let A = the number of rooms that Anne can clean in ONE hour
Let B = the number of rooms that Bruce can clean in ONE hour

Bruce and Anne can clean their house in 4 hours working together at their respective constant rates.
This tells us that their COMBINED rate = 12 rooms in 4 hours = 3 rooms per ONE HOUR
In other words, A + B = 3

Anne's speed were doubled, they could clean their house in 3 hours working at their respective rates.
This tells us that their NEW COMBINED rate = 12 rooms in 3 hours = 4 rooms per ONE HOUR

If we DOUBLE Anne's speed, we get 2A
So, we can write: 2A + B = 4

So, we have a system of two equations:
A + B = 3
2A + B = 4

When we solve the system, we get A = 1 and B = 2

If A = 1, then this means Anne can clean 1 room in ONE hour

How many hours does it currently take Anne to clean the house on her own?
Time = output/rate
= 12 rooms/1 room per hour
= 12 hours

Answer: D

Cheers,
Brent
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Re: Bruce and Anne can clean their house in 4 hours working toge  [#permalink]

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17 Mar 2019, 01:47
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Re: Bruce and Anne can clean their house in 4 hours working toge   [#permalink] 17 Mar 2019, 01:47
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