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R and S can complete a certain job in 6 and 4 days

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R and S can complete a certain job in 6 and 4 days

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e-GMAT

R and S can complete a certain job in 6 and 4 days respectively, while they work individually. What will be the least number of days they will take to complete the same job, if they work on alternate days?

A. 2.2 days
B. 2.67 days
C. 4.4 days
D. 4.67 days
E. 5 days

OA is D.

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AAPL wrote:
e-GMAT

R and S can complete a certain job in 6 and 4 days respectively, while they work individually. What will be the least number of days they will take to complete the same job, if they work on alternate days?

A. 2.2 days
B. 2.67 days
C. 4.4 days
D. 4.67 days
E. 5 days

OA is D.
R can complete 1/6 of the job in a day and S can complete 1/4

Try R performing first day work, then alternate:

Cumulative Progress
1/6 of the job 1st day 1/6
1/4 2nd day 1/6 + 1/4 = 5/12
1/6 3rd day 5/12 + 1/6 = 7/12
1/4 4th day 7/12 + 1/4 = 10/12 = 5/6
1/6 5th day 5/6 + 1/6 = 1

So 5 days works, but what if S goes first ? Well the first four days add up to the same, 5/6. S would work the 5th day, how much time is necessary ?

1 - 5/6 =1/6 of the job remaining. 1/6 = T(1/4) so T = 2/3

So total time is 4 days plus 2/3 of the 5th day, which is less than the first try at 5 days, so the minimum time is 4.67, D

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[quote="regor60"]
AAPL wrote:
e-GMAT

R and S can complete a certain job in 6 and 4 days respectively, while they work individually. What will be the least number of days they will take to complete the same job, if they work on alternate days?

A. 2.2 days
B. 2.67 days
C. 4.4 days
D. 4.67 days
E. 5 days
Let´s imagine the job is defined by exactly 12 identical tasks (LCM(6,4) = 12).

R does 2 tasks/day, while S does 3 tasks/day.

FOCUS: minimize the number of days to do the 12 tasks... S is more efficient, let´s make HIM/HER start as soon as possible! (*)

In four days, we have 3+2+3+2 = 10 tasks done.

At the beginning of the 5th day, it is S who works (*) and using UNITS CONTROL, one of the most powerful tools of our course, we have:
$$2\,\,\,{\rm{tasks}}\,\,\,\left( {{{1\,\,{\rm{day}}} \over {3\,\,{\rm{tasks}}}}\,\,\,\matrix{
\nearrow \cr
\nearrow \cr

} } \right)\,\,\,\,\, = \,\,\,{2 \over 3}\,\,{\rm{day}}$$
Obs.: arrows indicate licit converter.

$${\rm{?}}\,\,{\rm{ = }}\,\,{\rm{4}}{2 \over 3}\,\,{\rm{days}}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: www.GMATH.net (Math for the GMAT)
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AAPL wrote:
e-GMAT

R and S can complete a certain job in 6 and 4 days respectively, while they work individually. What will be the least number of days they will take to complete the same job, if they work on alternate days?

A. 2.2 days
B. 2.67 days
C. 4.4 days
D. 4.67 days
E. 5 days
OA is D.
Let's assign a "nice" value to the job, a value that works well with the given values (6 days and 4 days ).
So, let's say the ENTIRE job is to make 24 widgets

R can complete a certain job in 6 days
In other words, R can make 24 widgets in 6 days
So, R can make 4 widgets per day

S can complete a certain job in 4 days
In other words, S can make 24 widgets in 4 days
So, S can make 6 widgets per day

What will be the least number of days they will take to complete the same job, if they work on alternate days?
To MINIMIZE the time, the fastest worker (worker S) should go first.

DAY 1: Worker S makes 6 widgets (running total of widgets made at the end of day 1 = 6)
DAY 2: Worker R makes 4 widgets (running total of widgets made at the end of day 2 = 10)
DAY 3: Worker S makes 6 widgets (running total of widgets made at the end of day 3 = 16)
ASIDE: at this point, we can see that it will take more than 3 days to complete the job (ELIMINATE answer choices A and B)
DAY 4: Worker R makes 4 widgets (running total of widgets made at the end of day 4 = 20)

At this point, R and S have made 20 of the 24 needed widgets
So, on day 5, worker S need only make 4 widgets.

We already know that S can make 6 widgets per day, so it will take LESS THAN ONE day to make the remaining 4 widgets (ELIMINATE answer choice E)
We can also conclude that, in 1/2 a day, S can make only 3 widgets. So, it will take MORE THAN 1/2 a day to make the remaining 4 widgets. (ELIMINATE answer choice C)

Answer: D

Cheers,
Brent

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Brent Hanneson – Creator of GMATPrepNow.com
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AAPL wrote:
e-GMAT

R and S can complete a certain job in 6 and 4 days respectively, while they work individually. What will be the least number of days they will take to complete the same job, if they work on alternate days?

A. 2.2 days
B. 2.67 days
C. 4.4 days
D. 4.67 days
E. 5 days
We see that the rate of R is 1/6 and the rate of S is 1/4. Since we want to determine the least number of days and since the rate of S is greater than that of R, we should begin with S first.

If S works the first day, then 1 - 1/4 = 3/4 of the job is left to be done after the first day.

If R works the second day, then 3/4 - 1/6 = 9/12 - 2/12 = 7/12 of the job is left to be done after the second day.

If S works the third day, then 7/12 - 1/4 = 7/12 - 3/12 = 4/12 = 1/3 of the job is left to be done after the third day.

If R works the fourth day, then 1/3 - 1/6 = 2/6 - 1/6 = 1/6 of the job is left to be done after the fourth day.

Since we have ⅙ of the job is left undone at the beginning of the fifth day and the rate of S is 1/4, then it only takes (1/6)/(1/4) = 4/6 = 2/3 of a day to finish the job. Therefore, in total it takes 4 + 2/3 = 4.67 days to complete the job.

Alternate Solution:

Since S does 1/4 of the job in one day and R does 1/6 of the job in one day, when they work on alternate days, 1/4 + 1/6 = 5/12 of the job is done on two days. Thus, in four days, 5/12 + 5/12 = 10/12 = 5/6 of the job is done no matter in what order they work.

If R started the job on the first day, then R will need to complete 1/6 of the job on the fifth day, which will take R exactly one day. In this scenario, it will take 5 days to complete the job.

If, on the other hand, S started the job on the first day, then S will need to complete 1/6 of the job on the fifth day. Since it takes 4 days for S to complete the whole job, it will take only 4 x 1/6 = 2/3 = 0.67 days to complete the job. Thus, the minimum required time to complete the job when R and S alternate is 4 + 0.67 = 4.67 days.

Answer: D

_________________
Scott Woodbury-Stewart Founder and CEO

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AAPL wrote:
e-GMAT

R and S can complete a certain job in 6 and 4 days respectively, while they work individually. What will be the least number of days they will take to complete the same job, if they work on alternate days?

A. 2.2 days
B. 2.67 days
C. 4.4 days
D. 4.67 days
E. 5 days
We see that the rate of R is 1/6 and the rate of S is 1/4. Since we want to determine the least number of days and since the rate of S is greater than that of R, we should begin with S first.

If S works the first day, then 1 - 1/4 = 3/4 of the job is left to be done after the first day.

If R works the second day, then 3/4 - 1/6 = 9/12 - 2/12 = 7/12 of the job is left to be done after the second day.

If S works the third day, then 7/12 - 1/4 = 7/12 - 3/12 = 4/12 = 1/3 of the job is left to be done after the third day.

If R works the fourth day, then 1/3 - 1/6 = 2/6 - 1/6 = 1/6 of the job is left to be done after the fourth day.

Since we have ⅙ of the job is left undone at the beginning of the fifth day and the rate of S is 1/4, then it only takes (1/6)/(1/4) = 4/6 = 2/3 of a day to finish the job. Therefore, in total it takes 4 + 2/3 = 4.67 days to complete the job.

Alternate Solution:

Since S does 1/4 of the job in one day and R does 1/6 of the job in one day, when they work on alternate days, 1/4 + 1/6 = 5/12 of the job is done on two days. Thus, in four days, 5/12 + 5/12 = 10/12 = 5/6 of the job is done no matter in what order they work.

If R started the job on the first day, then R will need to complete 1/6 of the job on the fifth day, which will take R exactly one day. In this scenario, it will take 5 days to complete the job.

If, on the other hand, S started the job on the first day, then S will need to complete 1/6 of the job on the fifth day. Since it takes 4 days for S to complete the whole job, it will take only 4 x 1/6 = 2/3 = 0.67 days to complete the job. Thus, the minimum required time to complete the job when R and S alternate is 4 + 0.67 = 4.67 days.

Answer: D

_________________
Scott Woodbury-Stewart Founder and CEO

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