John works twice as fast as Peter, but John takes a half hour break af

Let us assume the total work to be 1000 units

Since John completes the work in 5 hours(without any breaks), he must be doing 200 units/hour

As John works twice as fast as Peter, the rate at which Peter does the work must be 100 units/hour

Rate at which they take breaks are as follows:
John takes a half an hour break for an hour of work
Peter takes an hour break for two hours of work.

John does 200 units in \(1\frac{1}{2}\) hour, effectively doing 600 units in \(4\frac{1}{2}\) hours.

Peter does 200 units in 3 hours. In the additional \(1\frac{1}{2}\) hour of work, he would have completed 150 units of work.
Therefore, Peter does 350 units in \(4\frac{1}{2}\) hours

In \(4\frac{1}{2}\) hours John and Peter would have completed 950 units of work.
For the remaining 50 units, working at 300 units/hour, they would take an additional 10 minutes.

Therefore, the total time taken to complete the 1000 units of work is 4 hours and 40 minutes(Option C)

The rate of John is 1/5 and the rate of Peter is 1/10.

Their combined rate, with no breaks is, 1/5 + 1/10 = 2/5 + 1/10 = 3/10.

For the first 3 hours of working together, we see that each would have taken a one-hour break and thus each worked only 2 hours. Thus, they finished 2 x 3/10 = 6/10 of the job.

During the next hour, hour 4, they both worked for the full hour; thus, they finished another 3/10 of the job and so far they finished 6/10 + 3/10 = 9/10 of the job.

During the following hour, hour 5, John worked only half an hour (since he took a half-hour break) while Peter worked the full hour; thu,s they would have completed another ½(1/5) + 1/10 = 2/10 of the job. However, by the end of hour 5, we see that they would have completed 9/10 + 2/10 = 11/10 or more than 1 entire job. Thus we need to push the time back.

So let’s only focus on the first 30 minutes of hour 5; John would not be working since he’s on his half-hour break, while Peter worked the entire 30 minutes. Thus, Peter would have completed another ½(1/10) = 1/20 of the job. By the end of the first 30 minutes of hour 5, we see that they would have completed 9/10 + 1/20 = 19/20 of the job.

We see that it takes more than 4 hour 30 minutes but less than 5 hours to complete this job. We also see that there are two answer choices that are between these two times. Let’s analyze choice C, 4 hours and 40 minutes, first. In other words, let’s see how much more work they complete in the extra 10 minutes.

In the extra 10 minutes, or ⅙ of an hour, they were both working; thus, they completed ⅙(1/5) + ⅙(1/10) = 1/30 + 1/60 = 3/60 = 1/20 of the job. Add this to the 19/20 of the job they completed in 4 hours 30 minutes, and we see that they would have completed exactly one entire job.

Answer: C

Hi Bunuel , any suggestion on how to calculate this in a quicker way? Using some algebra or work rate formula?
I calculated the time it would took to do the job without any breaks which is 3 hours and 20 minutes , and then I tried to add the "break time" , but I got lost at some point.....I don't think is the right way to do it...

J=2P
J takes 5hrs so P ll take 10hrs
Let total work be 10units(lcm of 5 n 10)
J-5 hrs so 2units/hr
P-10 hrs so 1unit/hr
1st hr-3units
2nd hr-2units
3rd hr-1unit
4th hr-3 units
In next 30min
J- 0 units(rest)
P- 0.5 units
So 10-9.5=0.5 units they have to do together without rest.

Without rest they take 10/3hrs for completing 10 units, so for 0.5 units they lol take (10/3)*(1/10)*(1/2)=(1/6)hr= 10 min

So total 4hrs 40min(C)


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U can try hourly calculation,easiest method.



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Rj = 1/5 = 2/10
Rp = 1/10

It is known that John takes a 30' brake every one hour of work, hence his productivity (Rate of work) decreases. By taking into consideration the breaks, John needs 5 working hours plus 4*30' breaks (no need to take a break when the job is done). Having said that:

Rj' = 1/7
Rp' = 1/14 (Following the same logic)

R=W/t => 2/14 + 1/14 = 1/t => t = 14/3 = 4h and 2/3

D

please how did you get the 7hrs and 14hrs and why not 7.5hrs and 15hrs


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is this a fool proof way of solving this type of question?
Bunuel chetan2u

Because the last half hour (for John) and the last hour (for Peter) are break time when they already finished their work.

In case of John:
He works from 0h to 1h, takes break from 1h to 1.5h, then work from 1.5h to 2.5h, and so on.

0 -- work -- 1 -- break -- 1.5 -- work -- 2.5 -- break -- 3 -- work -- 4 -- break -- 4.5 -- work -- 5.5 -- break -- 6 -- work -- 7-- break -- 7.5

John complete the task in 7 hours. Break time from 7h to 7.5h should not be calculated in the time required for John to complete the task while maintaining his break habit.

The same logic is applied to the case of Peter.

Hope it helps.

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