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Expert replies
by vaibhav101 » Sun Jun 03, 2018 10:02 am

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E

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Difficulty—

A can do a work in 10 days, B in 20 days and C in 40 days. They start working in turns with A on first day, B on second day and C on third day and so on till they finish the work. Find the time taken to complete the work.

A 15.5
B 16
C 17
D 15
E 16.5
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Source: — Problem Solving |

by Vincen » Tue Jun 05, 2018 12:28 am
vaibhav101 wrote:A can do a work in 10 days, B in 20 days and C in 40 days. They start working in turns with A on first day, B on second day and C on third day and so on till they finish the work. Find the time taken to complete the work.

A 15.5
B 16
C 17
D 15
E 16.5
Hello vaibhav101.

Here, we have to find the part of the work that each one does in one day, that is to say:

- A does a work in 10 days, then each day A does 1/10 of the work.
- B does a work in 20 days, then each day B does 1/20 of the work.
- C does a work in 40 days, then each day C does 1/40 of the work.

Since A starts, then B and finally C we can say that in three days they have done $$\frac{1}{10}+\frac{1}{20}+\frac{1}{40}=\frac{4}{40}+\frac{2}{40}+\frac{1}{40}=\frac{7}{40}\ of\ the\ work.$$ Then, in 6 days the have done $$2\left(\frac{7}{40}\right)=\frac{7}{20}\ of\ the\ work.$$ In 9 days they have done $$3\left(\frac{7}{40}\right)=\frac{21}{40}\ of\ the\ work.$$ In 12 days the have done $$4\left(\frac{7}{40}\right)=\frac{7}{10}\ of\ the\ work.$$ In 15 days the have done $$5\left(\frac{7}{40}\right)=\frac{7}{8}\ of\ the\ work.$$ Now, in 16 days the have done $$\frac{1}{10}+\frac{7}{8}=\frac{78}{80}=\frac{39}{40}\ of\ the\ work.$$ Now, we can see that left 1/40 of the work, but it is the turn of B, whose rate of work is 1/20 per day, therefore B just need to work half of the day to finish the job, because $$\frac{39}{40}+\frac{\frac{1}{20}}{2}=\frac{39}{40}+\frac{1}{40}=\frac{40}{40}=1.$$ Hence, the correct answer is the option E.

I hope it may help you.

Regards.
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by GMATGuruNY » Tue Jun 05, 2018 2:29 am
vaibhav101 wrote:A can do a work in 10 days, B in 20 days and C in 40 days. They start working in turns with A on first day, B on second day and C on third day and so on till they finish the work. Find the time taken to complete the work.

A 15.5
B 16
C 17
D 15
E 16.5
Let the job = 40 units.
Since A takes 10 days to complete the job, A's rate = w/t = 40/10 = 4 units per day.
Since B takes 20 days to complete the job, B's rate = w/t = 40/20 = 2 units per day.
Since C takes 40 days to complete the job, C's rate = w/t = 40/40 = 1 unit per day.
Work completed every 3 days by A+B+C = 4+2+1 = 7 units.
Since 7 units are produced every 3 days, the amount of work produced over 15 days = 5*7 = 35.
Remaining work = 40-35 = 5 units.
Since A's rate = 4 units per day, A produces on the 16th day 4 of the remaining 5 units, leaving 1 unit for B to produce on the last day.
Since B's rate = 2 units per day, B takes 1/2 day to produce the 1 remaining unit, bringing the total amount of time to 16.5 days.

The correct answer is E.
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by Scott@TargetTestPrep » Tue Jun 05, 2018 9:44 am
vaibhav101 wrote:A can do a work in 10 days, B in 20 days and C in 40 days. They start working in turns with A on first day, B on second day and C on third day and so on till they finish the work. Find the time taken to complete the work.

A 15.5
B 16
C 17
D 15
E 16.5
We see that in every 3 days, 1/10 + 1/20 + 1/40 = 4/40 + 2/20 + 1/40 = 7/40 of the work is completed. Therefore, in 15 days, 5 x 7/40 = 35/40 of the work will be completed. For the last 5/40 of the work, A will work on day 16, so 1/10 will be completed and 5/40 - 1/10 = 1/40 will be left undone. So B will work day 17, but he doesn't need the whole day to finish it. Since (1/40)/(1/20) = 1/2, B just needs half a day to finish the last 1/40 of the work. So the total time to complete the work is 16.5 days.

Answer: E

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