Question 289919: It takes two laser printers working together 12 minutes to complete a 580 page print job ,if each printer is printing this job alone, the first laser printer takes 10 minutes longer than the second printer . How long does it take each printer to complete the print job alone? What is the speed of each printer.
I don't even know where to begin with this. Someone please help me step by step?
Thanks!
Found 2 solutions by CharlesG2, Grinnell:
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! It takes two laser printers working together 12 minutes to complete a 580 page print job ,if each printer is printing this job alone, the first laser printer takes 10 minutes longer than the second printer . How long does it take each printer to complete the print job alone? What is the speed of each printer.
I don't even know where to begin with this. Someone please help me step by step?
Thanks!
[see: http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/HOW-TO-Solve-Rate-of-Work-Problems.lesson]
printer A, printer B
A in X minutes, B in Y minutes
time = 1/[1/X + 1/Y]
A takes X minutes to do a job, so 1/X jobs per minute
B takes Y minutes to do a job, so 1/Y jobs per minute
working together: 1/X + 1/Y jobs per minute
one job: 1/[1/X + 1/Y] minutes
1/[1/X + 1/Y] = 12 minutes to do 580 pages (48 1/3 pages/minute)
A would take 10 minutes longer than B
X = Y + 10 --> X - 10 = Y
1/[1/X + 1/(X - 10)] = 12
1/X + 1/(X - 10) = 1/12
(X - 10)/[X(X - 10)] + X/[X(X - 10)] = 1/12 (made common denominators)
[(X - 10) + X]/(X^2 - 10X) = 1/12
(2X - 10)/(X^2 - 10X) = 1/12
(24X - 120)/(X^2 - 10X) = 1 (multiplied numerators both sides by 12)
24X - 120 = X^2 - 10X
0 = X^2 - 34X + 120
0 = (X - 4)(X - 30) (by FOIL this equals above)
X = 4 minutes or X = 30 minutes
since X - 10 = Y, and Y can not be negative minutes,
X = 30 minutes for A, and Y = 20 minutes for B to do the 580 page print job alone
A's speed is 580 pages/30 minutes = 19 1/3 pages/minute
B's speed is 580 pages/20 minutes = 29 pages/minute
Answer by Grinnell(63) (Show Source):
You can put this solution on YOUR website! Pack a Lunch!
This is basically a CLASSIC RATE WORK PROBLEM.
Then at the end we have DIVISION to deal with.
Breathe slowly...we WILL get thru this!!!
Let x be the faster machine.
Let x+10 be the slower machine.
STOP--UNDERSTAND THAT WITH RATE WORK PROBLEMS FAST AND SLOW PARADIGM MEANS ADDING TO SLOWER. E.G. IT TAKES ME 20 MINUTES TO DO SOMETHING IT TAKES YOU 20 + 10 MINUTES TO DO (30 MINUTES) WE ADDED MINUTES TO THE SLOWER TIME. YOU MUST UNDERSTAND THIS IN WORK PROBLEMS.
Now we are concerned about the Job. NOT THE NATURE OF THE JOB OR THE NUMBER OF PAGES INVOLVED--JUST THAT 'A' JOB IS BEING COMPLETED IN 12 MINUTES. THIS IS WHY WE CAN NOW SAY...
1/X+1/X+10=1/12 STOP HERE AND REVIEW YOUR SECTION IN YOUR TEXT ABOUT CLASSIC RATE WORK PROBLEMS!
MULTIPLY THRU BY (X)(X+10)(12) WE GET 12X+120+12X=X^2+10X DO THE 'LIKE TERMS' WE GET...
X^2-14x-120=0
(x-20)(x+6)=0 (x cannot be negative because of the nature of the problem, therefore x is 20!
The faster machine takes 20 minutes to complete the job alone.
The slower machine takes 30 minutes to complete the job alone. (1/20+1/30=1/12--You do the check.)
N O W WE GET TO THE ACTUAL JOB. 580 PAGES!
580/20=29 SO THE FASTER MACHINE COMPLETES 29 PAGES PER MINUTE
WHILE THE SLOWER MACHINE (580/30) COMPLETES 19.3 PAGES PER MINUTE. Thanks for the question. Have wonderful afternoon:)
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