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26 May 2017, 04:43
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Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
26 May 2017, 04:46
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Official Solution:
Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
If machine A produces toys for "a" hours and machine B produces toys for "b" hours, and both machine A and B produce toys for c hours at the same time, you get \(a + b + c = 20, 2a + 5b + (2 + 5)c = 88, 2a + 5b + 7c = 88\). If you multiply \(a + b + c = 20\) by 2, you get \(2a + 2b + 2c=40\). If you take away two equations, \(3b + 5c = 48\) becomes \((a,b,c) = (10,1,9), (8,6,6), (6,11,3).\) Among these, the possible maximum value of c is 9. Therefore, the answer is E.
Answer: E
Machine A can produce toys at a constant rate of 2 units per hour and machine B can produce toys at a constant rate of 5 units per hour. If at least one of either machine A or machine B produces toys, what is the greatest possible hours when machine A and machine B work together at their constant rates so that two machines, A and B, can produce 88 units of toys in 20 hours?
A. 5hrs
B. 6hrs
C. 7hrs
D. 8hrs
E. 9hrs
If machine A produces toys for "a" hours and machine B produces toys for "b" hours, and both machine A and B produce toys for c hours at the same time, you get \(a + b + c = 20, 2a + 5b + (2 + 5)c = 88, 2a + 5b + 7c = 88\). If you multiply \(a + b + c = 20\) by 2, you get \(2a + 2b + 2c=40\). If you take away two equations, \(3b + 5c = 48\) becomes \((a,b,c) = (10,1,9), (8,6,6), (6,11,3).\) Among these, the possible maximum value of c is 9. Therefore, the answer is E.
Answer: E
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