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by gmatmachoman » Mon Apr 05, 2010 7:37 am
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) 5.
(b) 2.
(c) 8.
(d) 4.
(e) 6.
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by thephoenix » Mon Apr 05, 2010 8:20 am
eqn is
M1*d1/w1=M2*d2/w2
w1=1
w2=1-1/3( as in one day 1/3 of web page is finished)=2/3
m1=15
m2=6
d1=3
solving for d2=5
hence a
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by gmatmachoman » Mon Apr 05, 2010 8:24 am
thephoenix wrote:eqn is
M1*d1/w1=M2*d2/w2
w1=1
w2=1-1/3( as in one day 1/3 of web page is finished)=2/3
m1=15
m2=6
d1=3
solving for d2=5
hence a
Thx bro..
I have devised a small tic for this kind of problems:

Change in Men : Original Men :: Change in days : required days

Now try a example:
(15-9):15::(3-1):required days

6:15::2:X

X=5(Easy isnt it??)

Feedbacks are requested
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by pops » Mon Apr 05, 2010 9:13 pm
gmatmachoman wrote: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) 5.
(b) 2.
(c) 8.
(d) 4.
(e) 6.
Another way of thinking it is consider number of "Java programmers days" (JPD) required to complete whole work = 15*3 = 45
1 day 15 JP already worked hence 15 JPD is done left 45-15 = 30 JPD
JP left = 6
hence, D required = 30 JPD/ 6 JP = 5 Days :-)
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by harshavardhanc » Tue Apr 06, 2010 12:44 am
gmatmachoman wrote: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) 5.
(b) 2.
(c) 8.
(d) 4.
(e) 6.
Product constancy ! :) Man-days (man * days ) is constant.

the programmers have decreased by 60% or have become .4 of the original.

So, the days have to become 2.5 times of the original. OR 2.5 * 2 = 5
Regards,
Harsha
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