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15 Java programmers, working in a constant pace, finish a

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15 Java programmers, working in a constant pace, finish a [#permalink] New post 07 Sep 2004, 01:30
15 Java programmers, working in a constant pace, finish a web page in 3 days. If after one day, 9 programmers quit, how many more days are needed to finish the remainder of the job?

(a) 2.
(b) 4.
(c) 5.
(d) 6.
(e) 8.


will post the OA later
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 [#permalink] New post 07 Sep 2004, 02:33
Total work = 15*3 = 45 man days of work
In one day the programmers would have completed 15 mandays of work.

30 mandays is left over and only 6 programmers left. So, these 6 programmers will take 30/6 = 5days addl to complete.

Ans. is C 5.
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 [#permalink] New post 07 Sep 2004, 02:37
thanks venksune
i really mess with these kind of sums
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 [#permalink] New post 07 Sep 2004, 04:18
C for me

another way to look at it:

Men Work Days

15 1 3
15 2/3 2 (at end of 1st day, 1/3rd the work is done)
6 2/3 5 (using simple proportionality)

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 [#permalink] New post 07 Sep 2004, 11:34
I still find the most effective way to do these problems is to establish a total number of units that is required to finish the job. In this case I picked 135. (15*3 per day equals 45 units and 3 days equals 135 units). So if each programmer finishes 3 units per day that means that the group finished 45 after the first day. Therefore there are 90 units left to be completed and with 6 programmers finishing 3 units per day they are completing 18 per day. 90/18=5.
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 [#permalink] New post 07 Sep 2004, 12:04
I have C too.

I will give you my explanation.

15 persons = 3 days
1 person = 45 days

1 (day) * 15/45 (1/45+1/45+1/45, etc) + 6/45 (15-9) * x (days)==>
1 * 15/45 + 6/45 * x

1 * 15/45 + 6/45 * x = 1 (45/45)
15/45 + 6/45 * x = 45/45
6/45 * x = 30/45
x = 30/45 / 6/45
x = 30/45 * 45/6
x = 30/6
x = 5

I hope my explanation will help some of you.

Regards,

Alex
  [#permalink] 07 Sep 2004, 12:04
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15 Java programmers, working in a constant pace, finish a

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