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R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 45% (01:55) correct 55% (01:44) wrong based on 262 sessions

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Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #2

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

To solve question 3: Question 3

To read the article: Solve Time and Work Problems Efficiently using Efficiency Method!

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Originally posted by EgmatQuantExpert on 23 May 2018, 04:13.
Last edited by EgmatQuantExpert on 13 Aug 2018, 05:58, edited 2 times in total.
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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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3
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EgmatQuantExpert wrote:
Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #2

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

Since R and S can complete a certain job in 6 and 4 days respectively,
let's assume the total work done to be done as 24 units(A number that works with individual rates of 6 and 4).

The individual rates are R - 4 units/day and S - 6 units/day.

Working alternatively, R and S do 4+6+4+6 = 20 units of the work in 4 days.

We are left with 4 units of work to do on the 5th day.
On the fifth day,
if R does the 4 units of work, it takes R 1 day to complete the work.
if S does the 4 units of work, it takes S $$\frac{4}{6}$$ or 0.67 day.
The minimum amount of time is taken if S does the work on the fifth day. Since they work alternatively,
S must have done work on the first and third day and R on the second and fourth day.

Therefore, the total time taken for R and S to complete the work is 4.67 days(Option D)
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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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1
LCM of 6 & 4 is 12 [Final answer does not change but this may be important for other questions as this amounts to a silly mistake]

Also in your solution you have to specifically assume who is doing the work first? Since the question stem is asking the "least" time, then the person who works faster will start the work first. The order mentioned by you assumes RSRS (4+6+4+6 = 20). If this is the assumption then on the fifth day R will work again and he will take entire day to complete 4 units. Which means the answer in this case comes to (E) 4+1= 5 days, which is the wrong answer as the stem asks you to find out the "least time". (Your final answer is right, I am just clarifying the logic behind your answer because someone could ignore the order and commit a silly mistake as the trap answer in one of the options)

Hence the correct order is S works on the first day because he does more units per day followed by R the next day. This gives us the following order of SRSR completing 20 units in first 4 days. On the 5th day S works for 2/3rd of the day as his individual rate is 6 units per day but he needs to complete only 4 units for which only 2/3 day or .67 day is required. Hence the answer is 4+.67=4.67 days.

pushpitkc wrote:
EgmatQuantExpert wrote:
Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #2

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

Since R and S can complete a certain job in 6 and 4 days respectively,
let's assume the total work done to be done as 24 units(LCM of 6 and 4).

The individual rates are R - 4 units/day and S - 6 units/day.

Working alternatively, R and S do 4+6+4+6 = 20 units of the work in 4 days.

We are left with 4 units of work to do on the 5th day.
On the fifth day,
if R does the 4 units of work, it takes R 1 day to complete the work.
if S does the 4 units of work, it takes S $$\frac{4}{6}$$ or 0.67 day.
The minimum amount of time is taken if S does the work on the fifth day.

Therefore, the total time taken for R and S to complete the work is 4.67 days(Option D)

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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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Since R and S can complete a certain job in 6 and 4 days respectively,
let's assume the total work done to be done as 12 units (LCM of 6 and 4).

The individual rates are R - 2 units/day and S - 3 units/day.

Since the question stem is asking the "least" time, then the person who works faster will start the work first. S works on the first day because he does more units per day followed by R the next day. This gives us the following order of SRSR completing 10 units in first 4 days. On the 5th day S works at his individual rate of 3 units per day but he needs to complete only 2 units for the entire work to be completed for which only 2/3 of the day or 0.67 day is required. Hence the answer is 4+.67=4.67 days.

PS: Someone could ignore the order and commit a silly mistake as the trap answer is E which is one of the options.
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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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CAMANISHPARMAR wrote:
Since R and S can complete a certain job in 6 and 4 days respectively,
let's assume the total work done to be done as 12 units (LCM of 6 and 4).

The individual rates are R - 2 units/day and S - 3 units/day.

Since the question stem is asking the "least" time, then the person who works faster will start the work first. S works on the first day because he does more units per day followed by R the next day. This gives us the following order of SRSR completing 10 units in first 4 days. On the 5th day S works at his individual rate of 3 units per day but he needs to complete only 2 units for the entire work to be completed for which only 2/3 of the day or 0.67 day is required. Hence the answer is 4+.67=4.67 days.

PS: Someone could ignore the order and commit a silly mistake as the trap answer is E which is one of the options.

Thanks for noticing and notifying CAMANISHPARMAR

Have made the necessary changes in my solution.
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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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1
1

Solution

Given:
In this question, it is given that
• R and S can complete a certain job in 6 days and 4 days respectively, when they work individually.

To find:
• We need to find out the least number of days they will take to complete the job, if they work on alternate days.

Approach and Working:
As R and S take 6 days and 4 days respectively while working individually, we can assume the total work to be LCM (6, 4) = 12 units
• In 6 days, R can complete 12 units of work
• Hence, in 1-day R can complete $$\frac{12}{6}$$ = 2 units of work
• Similarly, in 4 days, S can complete 12 units of work
• Hence, in 1-day S can complete $$\frac{12}{4}$$ = 3 units of work
As they work on alternate days, the work that they will complete in a span of 2 days = (2 + 3) = 5 units
• Therefore, in the first 4 days, they will complete 5 * 2 = 10 units of the total work
Hence, remaining work = (12 – 10) units = 2 units

Now to ensure that the completion of work happens in the least possible day, the person, who is more efficient among the 2, should work on the 5th day
Among R and S, we know S is more efficient, and therefore, S should work on the 5th day to ensure the work gets completed at earliest
• As S can complete 3 units of work in 1 day, to complete 2 units of work, S will take = $$\frac{2}{3}$$ days = 0.67 days
Hence, the least total number of days needed to complete the work = (4 + 0.67) days = 4.67 days

Hence, the correct answer is option D.

Important Observation

• As they are working in alternate days, only one person is working in a day.
• Hence, to ensure the job gets completed in least number of days, we need to ensure the more efficient person(S) works on the last day.
• This also indicates that S should start the work on day 1, to ensure S works in the final day.

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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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EgmatQuantExpert 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

Let´s imagine the job is defined by exactly 12 identical tasks (LCM(6,4) = 12).

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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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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Top Contributor
EgmatQuantExpert 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

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)

Cheers,
Brent
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R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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The catch here is that you are not given the order. Although if any of them starts,after 2 days work completed will be 5/12.
Similarly after another 2 days work done will be 5/12 again. Now we are left with(1/6) work.
Here we have to decide the order. If we start with R on the first day, then it is his turn on the 5th day and he will take the whole day since his rate is 1/6.

Very good question. Not so difficult but tricky.Read question carefully.
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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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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

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Re: R and S can complete a certain job in 6 and 4 days respectively, while  [#permalink]

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[quote="EgmatQuantExpert"]Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #2

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

Rate of R = 1/6 (Slower)

Rate of S = 1/4 (Faster)

As we want the least time to complete, we will start with the faster one (S) on the first day.

Combined rate of R and S = 1/6 + 1/4 = 5/12

This means in 2 days the can do 5/12 of the job.

Work done on Day 1 and Day 2 = 5/12 ---- Remaining work = 7/12

Work done on Day 3 and Day 4 = 5/12 ----- Remaining work 7/12 - 5/12 = 2/12 Or 1/6 units

Now "S" will do the Day 5's work as we have started with S (faster one) on Day 1.

Time to finish remaining work = Work/Rate of S

Time = 1/6*4/1

Time = 2/3 Days or 0.666 days

Total Days = 4 + 0.666

Total Days =4.667 (D)

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"Do not watch clock; Do what it does. KEEP GOING." Re: R and S can complete a certain job in 6 and 4 days respectively, while   [#permalink] 09 Apr 2019, 23:49
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