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

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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 D

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by Brent@GMATPrepNow » Fri Nov 23, 2018 7:37 am
BTGmoderatorDC wrote: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 D

Source: e-GMAT
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

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by [email protected] » Fri Nov 23, 2018 10:04 am
Hi All,

We're told that R and S can complete a certain job in 6 and 4 days respectively, while they work individually. We're asked for the LEAST number of days they will take to complete the same job, if they work on alternate days. Although the wording of this question is a bit 'clunky', the intent is that the machines will work on opposite days (for example, Machine R on the 1st day, Machine S on the 2nd day, Machine R on the 3rd day, etc.). This question can be approached in a number of different ways, including by determining what fraction of the job is completed each day.

Machine R requires 6 days to complete a job, so it completes 1/6 of the job each day it works.
Machine S requires 4 days to complete a job, so it completes 1/4 of the job each day it works.

To complete this job in the fastest way possible, we should use the faster machine on Day 1...

Day 1 Machine S --> 1/4 of the job done = 6/24 done
Day 2 Machine R --> 1/6 of the job done = 4/24 done
Day 3 Machine S --> 1/4 of the job done = 6/24 done
Day 4 Machine R --> 1/6 of the job done = 4/24 done
Etc.

Every TWO days, 6/24 + 4/24 = 10/24 of the job is done
After the 4th day, 20/24 is done.
On the 5th day, Machine S is working. That machine will complete 6/24 of the job, but we only need 4/24 to complete the job. Thus, Machine S will need to work for only 2/3 of that last day....

Total work days = 4 + 2/3 = 4 2/3

Final Answer: D

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by Scott@TargetTestPrep » Fri Mar 29, 2019 6:45 am
BTGmoderatorDC wrote: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 D

Source: e-GMAT
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 â…™ of the job remaining 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

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