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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