A college professor is grading 80 exam papers. He grades the papers at a certain rate for 3hours. Then fatigue sets in and he completes the grading in 2 more hours at a rate of 5 exam papers per hour slower. How fast was he grading the papers for the first 3 hours?
2006-10-06 10:55:04 · 9 answers · asked by primo026 1 in Science & Mathematics ➔ Mathematics
Let's denote the unknown rate of the professor as R. For the first 3 hours, the professor grades at a rate of R. Then for the next 2 hours, the professor grades at a rate of R-5
Set the sum of these relations to 80 and solve the equation for 'R'
3R + (R-5)*2 = 80
3R+2R-10 = 80
5R=90
Therefore, the professor was grading papers at a rate of 18 exam papers per hour.
Hope this helps
2006-10-06 10:57:07 · answer #1 · answered by JSAM 5 · 2⤊ 0⤋
the following is basic equation:
3x + 2(x-5) = 80
solving
x = 14 he was grading 18 papers per hour for the first 3 hour
2006-10-06 18:09:07 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Let the Starting rate be x papers / hour
.
x.(3hours) + (x-5). (2hours) = 80
.
3x +2x-10 = 80
.
x = 18 papers/hour
2006-10-06 21:40:38 · answer #3 · answered by Calculus 5 · 0⤊ 0⤋
Divide 80 by 3. Then add on per hour
2006-10-06 17:59:14 · answer #4 · answered by LadyOreo 4 · 0⤊ 0⤋
18 papers per hour
2006-10-06 18:02:31 · answer #5 · answered by sud 2 · 0⤊ 0⤋
let x= initial grading rate
The equation is
3x+2(x-5)=80
3x+2x-10=80
5x=90
x=18
check
3*18+2(18-5)
54+2*13
54+26=80
2006-10-07 14:41:03 · answer #6 · answered by yupchagee 7 · 0⤊ 0⤋
3x + 2y= 80
y=x-5
3x + 2(x-5)= 80
3x+2x -10=80
5x=90
x=18
so 18 papers per hr
2006-10-06 18:04:41 · answer #7 · answered by biracial 2 · 0⤊ 0⤋
18 papers pr hour
2006-10-06 18:46:17 · answer #8 · answered by curiousgirl? 2 · 0⤊ 0⤋
3R + 2(R-5) = 80
2006-10-06 18:01:51 · answer #9 · answered by Helmut 7 · 0⤊ 0⤋