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Bunuel
Will and Christine will work together to proofread a 420-page manuscript. Will takes 4 minutes to proofread one page and Christine takes 3 minutes to proofread one page. If each proofreader, at maximum, can proofread for 18 hours, what is the fewest number of hours that Will will have to spend proofreading this manuscript?

A. 0 hours
B. 2 hours
C. 4 hours
D. 8 hours
E. 18 hours
To minimize Will's time, maximize Christine's time, and thus, amount of work she completes.

Convert rates in minutes to rates in hours. For Christine, multiply 3 minutes by 20 to get 60 minutes = 1 hour. Multiply numerator by 20, too.

C's rate: \((\frac{1p}{3min}*\frac{20}{20})=\frac{20p}{60mins}=\frac{20p}{1hr}\)

C completes how many pages?
\(W=R*T\). Maximum T= 18 hrs
\(W\) completed: (20 p/hr * 18 hrs) = 360 pages

Work remaining for Will: (420-360)= 60 pages

Will's rate: \((\frac{1p}{4min}=\frac{15p}{60min})=\frac{15p}{1hr}\)

Will's time? 60 pages remain.\(T=\frac{W}{R}\)
Will needs only \(T =\frac{60p}{(\frac{15p}{1hr})}= 4\) hours to finish

Answer C
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Will will be reading 15 pages an hour --- Christine will be reading 20 pages an hour.

15x+20y=420
For x to be minimum, make Y maximum
x=28-4/3Y

Start with y=18
X=4 Hours.
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Christine reads 1 page in 3 minutes

Christine reads 20 pages in 60 minutes (1 hours)

Will reads 1 page in 4 minutes

Will reads 15 page in 60 minutes

To minimize the hours of Will, we have to maximize the hours of Christine

Christine reads 18*20=360 pages in 18 hours

Remaining pages = 420 - 360 = 60

Will time to proofread 60 pages = 60*4 = 240 min = 240/60= 4 hours
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