SOLUTION: solve. Round to the nearest tenth of necessary A certain computer can grade 2000 objective-answer exams in 10 sec. A new computer can grade the 2000 exams in less time. If the

Question 1034411: solve. Round to the nearest tenth of necessary

A certain computer can grade 2000 objective-answer exams in 10 sec. A new computer can grade the 2000 exams in less time. If the two computers together can grade 2000 exams in 3 sec, how long does it take the new computer alone to do the grading?
Found 2 solutions by Theo, robertb:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
r*t = q
r = rate
t = time
q = quantity

first computer can grade 2000 exams in 10 seconds.
formula r*t = q becomes r*10 = 2000
solve for r to get r = 200
first computer can grade 200 papers per second.

when they work together their rates are additive.
formula becomes:
(200 + r)*3 = 2000
simplify to get 600 + 3r = 2000
subtract 600 from both sides to get 3r = 1400.
divide both sides by 3 to get r = 1400/3
second computer can grade 1400/3 papers per second.
that's equivalent to 466 and 2/3 papers per second.

working together, the formula becomes (200 + 1400/3) * 3 = 2000
simplify to get 2000 = 2000 which confirms the solution is correct.

when the new computer works alone, the formula becomes (1400/3) * t = 2000.
solve for t to get t = 2000 / (1400/3) = 2000 * 3/1400 = 6000/1400 seconds.
this is the same as 60/14 = 30/7 seconds.
in decimal format, that's equal to 4.285714286 seconds.
round that to 4.3 seconds.

it would take the new computer 4.3 seconds to do the grading on its own.

Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
Let x be the number of seconds it takes for the new computer to grade the 2,000 exams. Then in 1 second, 1/x of the job is done.
The older computer grades the 2,000 exams in 10 seconds. Hence in one second 1/10 of the job is done.
The combined rate is .
==> ==> ==> ==> x = 30/7, or seconds.
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