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Cameron can paint a room in c hours. Cameron and Mackenzie, working to

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Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 05 Sep 2015, 13:41
2
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

72% (01:31) correct 28% (01:41) wrong based on 258 sessions

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Cameron can paint a room in c hours. Cameron and Mackenzie, working together, can paint the room in d hours. In terms of c and d, how long in hours would it take Mackenzie, working alone, to paint the room?

(A) 2d - c

(B) \(\frac{c+d}{cd}\)

(C) \(\frac{c-d}{cd}\)

(D) \(\frac{cd}{c+d}\)

(E) \(\frac{cd}{c-d}\)

Explanation: While it may be easier to work with actual numbers, this question is an explicit test of your knowledge of the work formula. If you know the time it takes for two individuals to do a job, you can plug those times into the equation \(\frac{xy}{x+y}\) to get the time it would take them, combined. Thus, plugging c in for x and setting the whole thing equal to d:

\(\frac{cy}{c + y}= d\)

\(cy= d(c + y)\)

\(cy = dc + dy\)

\(cy - dy = dc\)

\(y = \frac{dc}{c-d}\)

which is equivalent to \(\frac{cd}{c-d}\), choice (E).


Source: GMAT HACKS 1800 - Guide 1 - Rates, Ratios & Percents
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 05 Sep 2015, 13:42
I didn't fully understand the explanation--specifically the part about \(\frac{xy}{x+y}\). Anyone have a better approach?
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 05 Sep 2015, 15:22
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1
Hi gmatser1,

This prompt can be solved by TESTing VALUES and using the Work Formula:

Work = (A)(B)/(A+B) where A and B are the individual times that it takes to complete the 'job'

Here, we're told that one person can paint a room in C hours and that two people (when WORKING TOGETHER) can paint the room in D hours. We're asked for the individual amount of time that it would take the second person to paint the room when working along.

IF...
the 1st person can paint the room in 3 hours
the 2nd person can paint the room in 6 hours
Combined, it would take (3)(6)/(3+6) = 18/9 = 2 hours to paint the room when working together

Using the above example, we can TEST VALUES....
C = 3
D = 2

And we're looking for an answer that equals 6....There's only one answer that fits....

Final Answer:

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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 06 Sep 2015, 14:06
EMPOWERgmatRichC wrote:
Hi gmatser1,

This prompt can be solved by TESTing VALUES and using the Work Formula:

Work = (A)(B)/(A+B) where A and B are the individual times that it takes to complete the 'job'

Here, we're told that one person can paint a room in C hours and that two people (when WORKING TOGETHER) can paint the room in D hours. We're asked for the individual amount of time that it would take the second person to paint the room when working along.

IF...
the 1st person can paint the room in 3 hours
the 2nd person can paint the room in 6 hours
Combined, it would take (3)(6)/(3+6) = 18/9 = 2 hours to paint the room when working together

Using the above example, we can TEST VALUES....
C = 3
D = 2

And we're looking for an answer that equals 6....There's only one answer that fits....

Final Answer:

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


Rich,

Appreciate the response! However I still don't understand how the Work = (A)(B)/(A+B) combined rate equation is derived.

I understand that combined rates can be added or subtracted to determine the combined or individual rate, (i.e, Combined Rate = R1 + R2) and can be used in the overall rate formula (W = CR(T)) but not sure how that relates to time.
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 06 Sep 2015, 18:12
Hi gmatser1,

It sounds like you're asking about "why" the Work Formula provides the answer that it does. While I could explain the deeper math involved, that knowledge is really unnecessary. The Work Formula IS a formula that you can use to answer this type of question, whether you understand why it works or not. In that same way, do you need to know why the Pythagorean Theorem 'works' to answer 'right-triangle' questions? Or how the Combination Formula eliminates all of the duplicate entries?

I respect any 'quest' for knowledge that you might have, but if it's knowledge that isn't necessary to score at a high level on the GMAT, then I won't spend your valuable time on it.

GMAT assassins aren't born, they're made,
Rich
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Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 06 Sep 2015, 18:45
Cam's Time = C , Mack's Time = M ( Assume ) , Combined Time = D ( Given )

Let Work be = X

Then,

Cam's Speed = X / C, Mack's Speed = X / M.

Combined time should be ( D ) = Total Work/Their speed together( Addition of their speeds ).

i.e x / (x/c) + (x/m) = D

Solving the equation, M = dc/( c-d ).

Ans E.
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 06 Sep 2015, 20:15
3
gmatser1 wrote:
Cameron can paint a room in c hours. Cameron and Mackenzie, working together, can paint the room in d hours. In terms of c and d, how long in hours would it take Mackenzie, working alone, to paint the room?

(A) 2d - c

(B) \(\frac{c+d}{cd}\)

(C) \(\frac{c-d}{cd}\)

(D) \(\frac{cd}{c+d}\)

(E) \(\frac{cd}{c-d}\)

Explanation: While it may be easier to work with actual numbers, this question is an explicit test of your knowledge of the work formula. If you know the time it takes for two individuals to do a job, you can plug those times into the equation \(\frac{xy}{x+y}\) to get the time it would take them, combined. Thus, plugging c in for x and setting the whole thing equal to d:

\(\frac{cy}{c + y}= d\)

\(cy= d(c + y)\)

\(cy = dc + dy\)

\(cy - dy = dc\)

\(y = \frac{dc}{c-d}\)

which is equivalent to \(\frac{cd}{c-d}\), choice (E).


Source: GMAT HACKS 1800 - Guide 1 - Rates, Ratios & Percents


Assume simple values.

If Cameron takes 2 hrs alone and Mackenzie takes 2 hrs alone, they both will take 1 hr working together. (If rate is same, for 2 people, time taken will be halved)
So put c = 2, d = 1.
The only option that gives you 2 (time taken by Mackenzie alone) is option (E).
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 11 Sep 2015, 21:34
EMPOWERgmatRichC wrote:
Hi gmatser1,

It sounds like you're asking about "why" the Work Formula provides the answer that it does. While I could explain the deeper math involved, that knowledge is really unnecessary. The Work Formula IS a formula that you can use to answer this type of question, whether you understand why it works or not. In that same way, do you need to know why the Pythagorean Theorem 'works' to answer 'right-triangle' questions? Or how the Combination Formula eliminates all of the duplicate entries?

I respect any 'quest' for knowledge that you might have, but if it's knowledge that isn't necessary to score at a high level on the GMAT, then I won't spend your valuable time on it.

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


Rich,

Thanks. So Work = (A)(B)/(A+B) is just another thing I have to memorize? Similar to how I just "know" Combined rate = R1 + R2?

Thanks so much for you help.
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 12 Sep 2015, 01:12
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gmatser1 wrote:
I still don't understand how the Work = (A)(B)/(A+B) combined rate equation is derived.

I understand that combined rates can be added or subtracted to determine the combined or individual rate, (i.e, Combined Rate = R1 + R2) and can be used in the overall rate formula (W = CR(T)) but not sure how that relates to time.


AB/(A+B) is the same thing as Combined Rate = R1 + R2.

Note that A and B is time taken in hours.

\(Rate = \frac{1}{Time}\) when work done is 1 (usually 1 job such as paint a room) because Work = Rate * Time

Combined Rate = R1 + R2

\(\frac{1}{Time Taken Together} = \frac{1}{A} + \frac{1}{B}\)

Time taken together = \(\frac{AB}{(A + B)}\)
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 24 Mar 2016, 23:49
Given : R(c) = 1/c
R(m) = 1/m (Say M alone can complete the job in 'm' hours)
Given: R(cm) = 1/d
therefor, 1/d = 1/c + 1/m
1/d-1/c=1/m
c-d/(cd) = 1/m ( this is the rate at which M can work)
therefore , Time Taken by M is... 1/Rate ie cd/(c-d) ie option E
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 27 Mar 2016, 09:34
Combined rate of work
1/c+1/M=1/d
1/M=1/d-1/c
1/M=c-d/dc
Time=reciprocal of rate=dc/c-d

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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 02 Feb 2017, 03:18
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1) Cameron's rate: \(\frac{1}{c}\)
2) Cameron and Mackenzie's rate when working together: \(\frac{1}{d}\)
3) Mackenzie's rate: \(\frac{1}{d}-\frac{1}{c}\)
4) T(Mackenzie)=\(\frac{1}{(1/d-1/c)}=\frac{1}{(c-d/cd)}=\frac{cd}{c-d}\)
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Re: Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 08 Feb 2017, 10:30
gmatser1 wrote:
Cameron can paint a room in c hours. Cameron and Mackenzie, working together, can paint the room in d hours. In terms of c and d, how long in hours would it take Mackenzie, working alone, to paint the room?

(A) 2d - c

(B) \(\frac{c+d}{cd}\)

(C) \(\frac{c-d}{cd}\)

(D) \(\frac{cd}{c+d}\)

(E) \(\frac{cd}{c-d}\)


We are given that Cameron can paint a room in c hours. Since rate = work/time, the rate of Cameron is 1/c. We are also given that, working together, Cameron and Mackenzie can paint the room in d hours. If we let m = the time it takes Mackenzie to paint the room alone, we can create the following equation:

1/c + 1/m = 1/d

We need to determine how long it would take Mackenzie to paint the room, so we need to isolate m.

We can start by multiplying the entire equation by cmd:

md + cd = cm

cd = cm - md

cd = m(c - d)

cd/(c-d) = m

Answer: E
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Cameron can paint a room in c hours. Cameron and Mackenzie, working to  [#permalink]

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New post 08 Feb 2017, 12:54
gmatser1 wrote:
Cameron can paint a room in c hours. Cameron and Mackenzie, working together, can paint the room in d hours. In terms of c and d, how long in hours would it take Mackenzie, working alone, to paint the room?

(A) 2d - c

(B) \(\frac{c+d}{cd}\)

(C) \(\frac{c-d}{cd}\)

(D) \(\frac{cd}{c+d}\)

(E) \(\frac{cd}{c-d}\)

Explanation: While it may be easier to work with actual numbers, this question is an explicit test of your knowledge of the work formula. If you know the time it takes for two individuals to do a job, you can plug those times into the equation \(\frac{xy}{x+y}\) to get the time it would take them, combined. Thus, plugging c in for x and setting the whole thing equal to d:

\(\frac{cy}{c + y}= d\)

\(cy= d(c + y)\)

\(cy = dc + dy\)

\(cy - dy = dc\)

\(y = \frac{dc}{c-d}\)

which is equivalent to \(\frac{cd}{c-d}\), choice (E).


Source: GMAT HACKS 1800 - Guide 1 - Rates, Ratios & Percents

1/d-1/c=(c-d)/cd=M's rate
inverting, cd/(c-d)=M's time to paint room alone
E
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