Assume a number of widgets that needs to be produced. Say 120. So...
A+B=40/hr
B+C=30/hr
A+C=20/hr
Adding...
2A + 2B + 2C = 90/hr
Ergo...
A+B+C=45/hr
120/45=2-2/3 hr= 2 hr 40 min
Assume a number of widgets that needs to be produced. Say 120. So...
A+B=40/hr
B+C=30/hr
A+C=20/hr
Adding...
2A + 2B + 2C = 90/hr
Ergo...
A+B+C=45/hr
120/45=2-2/3 hr= 2 hr 40 min
To figure out the individual rates, just write each worker in terms of C. So B=30-C and A=20-C. So if A+B+C=45, then
(20-C) + (30-C) + C = 45
Solve for C and you get C=5 per hour.
So A = 15 per hour and B = 25 per hour
To figure out the individual rates, just write each worker in terms of C. So B=30-C and A=20-C. So if A+B+C=45, then
(20-C) + (30-C) + C = 45
Solve for C and you get C=5 per hour.
So A = 15 per hour and B = 25 per hour
Diophantus (father of Algebra) had the following riddle written about his life:
Diophantus’s youth lasted 1/6 of his life. He had his first beard in the next 1/12 of his life. At the end of the following 1/7 of his life, he got married. Five years from then his son was born. His son lived exactly 1/2 of Diophantus’s life. Diophantus died 4 years after the death of his son. How long did Diophantus live?
Diophantus (father of Algebra) had the following riddle written about his life:
Diophantus’s youth lasted 1/6 of his life. He had his first beard in the next 1/12 of his life. At the end of the following 1/7 of his life, he got married. Five years from then his son was born. His son lived exactly 1/2 of Diophantus’s life. Diophantus died 4 years after the death of his son. How long did Diophantus live?
Ok now this is hard but doable. Thanks! Let’s see...
Hidden:
I think 68/34 but it’s an estimate
Last edited by ExtraCredit on Tue, Oct 13 2020, 11:15 am; edited 1 time in total
