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Bruce and Anne can clean their house in 4 hours working...

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Bruce and Anne can clean their house in 4 hours working...

by BTGmoderatorLU » Wed Feb 21, 2018 5:05 am
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

The OA is D.

Can I say,
$$R_A+R_B=\frac{1}{4}\left(1\right)$$
$$2\cdot R_A+R_B=\frac{1}{3}\left(2\right)$$
Then, subtracting (1) from (2),
$$R_A=\frac{1}{12}$$
That's mean that Anne takes 12 hours to clean the house working alone.

Experts, any suggestion about this PS question? Thanks in advance.
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by Jeff@TargetTestPrep » Mon Feb 26, 2018 10:11 am
LUANDATO 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
The normal combined rate of Bruce and Anne is:

1/B + 1/A = 1/4

With Ann's rate doubled, we have:

1/B + 2/A = 1/3

Subtracting the first equation from the second, we have:

1/A = 1/3 - 1/4

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

So it takes Anne 12 hours to clean the house on her own.

Answer: D

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Head of GMAT Instruction
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RE

by deloitte247 » Sun Mar 04, 2018 12:05 pm
Say, Anne can finish 'a' amount of work in 1 hour, and bruce can finish 'b' amount of work in 1 hour. If they worked together at their constant rates, they will do 'a+b' amount of work in 1 hour .
Therefore, let W be the total amount of work done in cleaning the house.
If the W amount of work done in 4 hours, w= 4(a+b)
if Annes rate is doubled, she can do (2a) amount of work in 1 hour.
for three hours, $$w=3\left(2a+b\right)\ \left(anne\ doubled\right)$$
$$i.e\ 4\left(a+b\right)=3\left(2a+b\right)\ $$
$$4b-3b=6a-4a$$
$$or\ b=2a$$
$$i.e\ w=4\left(a+2a\right)=12a$$
$$If\ anne\ do\ the\ whole\ work\ \left(i.e\ 12a\right)\ at\ her\ normal\ rate\ \left('a'\ amount\ of\ work\ in\ 1\ hour\right)$$
$$Then\ the\ number\ of\ hours\ used=\frac{\left('12a'work\right)}{'a'\ \frac{work}{hour}}=12hours$$
$$Hence,\ option\D\ is\ very\ correct$$
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by Brent@GMATPrepNow » Sun Mar 04, 2018 2:37 pm
LUANDATO 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
Brent Hanneson - Creator of GMATPrepNow.com
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