Bunuel 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.
For one day there are 15 workers. They finish a fraction of the work. Some people leave. How many [more] days are needed for the new number of workers to finish the remaining work? You could take out the word "more."
That word, though, signals you to find the amount of work already done. It takes 5 days for 6 people to finish what is left. "Five days to finish" is five days
more than one day.
Maybe trying individual rate will help. I think the work units and/or group efficiency methods might confuse. The group's numbers change, but the rate at which each person works does not. Fewer people to work equals more time to finish.
Find the rate of an individual programmer.
Just add (# of workers) to LHS in the standard (\(rt=W\)) formula. To find individual rate, you have to use t = 3 days
(# of workers * r * t) = W# of workers = 15, \(r\) = ??, \(t = 3\), \(W = 1\)
\((15 * r * 3) = 1\)
\(45r = 1\)
\(r = \frac{1}{45}\)
At that individual programmer's rate, in ONE day, the amount of work that 15 workers will finish, (# * r * t) = W:
\((15 *\frac{1}{45} * 1) =(\frac{15}{45}) =(\frac{1}{3})\) of work is finished.
\(\frac{2}{3}\) of work remains
But nine workers have gone. Use individual rate for the new number of workers.
How many [] days are needed for 6 workers to finish the remaining
\(\frac{2}{3}\) of W?
(# of workers * r * t) = W
# of workers = 6, \(r =\frac{1}{45}\),\(t\) = ??, \(W =\frac{2}{3}\)
Time needed: \((6 *\frac{1}{45} * t)\) = \(\frac{2}{3}\)
\((\frac{6}{45}t) = \frac{2}{3}\)
\(t = (\frac{2}{3}) * (\frac{45}{6})\)
\(t\) = 5 days
Total days:
ONE DAY for 15 people to finish 1/3, and
FIVE DAYS for 6 people to finish 2/3
1 + 5 = 6 days' total.
After the first day (subtract it from the total days) it takes: (6 - 1) = 5 MORE days
Hope it helps!