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# 2 Part question!

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Joined: 01 Jun 2012
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07 Sep 2012, 07:00
Hello Friends,

Pl. try his 2 part question.

A, B, & C can do same job in 8, 24, & 48 hours respectively working individually. It was decided that each of them will work on this job alternatively for an hour. They can start the work in any order. Select 'Minimum hours', &
'Maximum hours' required to complete the job in the table. Make only two selections, one in each column.
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07 Sep 2012, 14:08
ShalabhsQuants wrote:
Hello Friends,

Pl. try his 2 part question.

A, B, & C can do same job in 8, 24, & 48 hours respectively working individually. It was decided that each of them will work on this job alternatively for an hour. They can start the work in any order. Select 'Minimum hours', &
'Maximum hours' required to complete the job in the table. Make only two selections, one in each column.

If A, B, and C each work for an hour, in a total of 3 hours they accomplish $$\frac{1}{8}+\frac{1}{24}+\frac{1}{48}=\frac{9}{48}=\frac{3}{16}$$ of the entire job.
After 5 such cycles, a total of 5 x 3 = 15 hours, they accomplish $$\frac{15}{16}$$ of the entire job. For the remaining 1/16th of the job, if it is the slowest worker's turn to work (this is C), another 1/48th is accomplished, then B would do another 1/24th, which gives the missing $$\frac{1}{48}+\frac{1}{24}=\frac{3}{48}=\frac{1}{16}$$ of the job. So, the maximum time is 5 x 3 + 2 =17 hours. Choice (D,B) (row, column)

For the minimum time, after the 5 cycles of 3 hours, the fastest worker, which is A, needs just 1/2 an hour to do the remaining 1/16th of the job.
So, minimum time is 5 x 3 + 0.5 =15.5 hours. Choice (B,A) (row, column)
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08 Sep 2012, 08:56
2
KUDOS
EvaJager wrote:
ShalabhsQuants wrote:
Hello Friends,

Pl. try his 2 part question.

A, B, & C can do same job in 8, 24, & 48 hours respectively working individually. It was decided that each of them will work on this job alternatively for an hour. They can start the work in any order. Select 'Minimum hours', &
'Maximum hours' required to complete the job in the table. Make only two selections, one in each column.

If A, B, and C each work for an hour, in a total of 3 hours they accomplish $$\frac{1}{8}+\frac{1}{24}+\frac{1}{48}=\frac{9}{48}=\frac{3}{16}$$ of the entire job.
After 5 such cycles, a total of 5 x 3 = 15 hours, they accomplish $$\frac{15}{16}$$ of the entire job. For the remaining 1/16th of the job, if it is the slowest worker's turn to work (this is C), another 1/48th is accomplished, then B would do another 1/24th, which gives the missing $$\frac{1}{48}+\frac{1}{24}=\frac{3}{48}=\frac{1}{16}$$ of the job. So, the maximum time is 5 x 3 + 2 =17 hours. Choice (D,B) (row, column)

For the minimum time, after the 5 cycles of 3 hours, the fastest worker, which is A, needs just 1/2 an hour to do the remaining 1/16th of the job.
So, minimum time is 5 x 3 + 0.5 =15.5 hours. Choice (B,A) (row, column)

An easier way to work will be take the amount of work to be 48 units this allows to do away with fractions.
in one hour A does 48/8 = 6 units
B does 48/24 = 2 units
C does 48/48 = 1 unit

In three hours no matter in which order they work they will complete 9 units of work
3 hr = 9 units
15 hrs = 45 units
now three units are left, for minimum time as A to do it he will do 6 units in 1 hr so 3 units in .5 hr answer = 15.5 units
For maximum time let a and b finish the job they will take 1unit + 2unit = 3unit that is 2 hrs to complete it.
Re: 2 Part question!   [#permalink] 08 Sep 2012, 08:56
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