SOLUTION: Ken and Ted can produce 1 chair in 3 hours working together at their respective constant rates. If Ken doubles his speed, then two of them could produce 1 chair in 2 hours working

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Ken and Ted can produce 1 chair in 3 hours working together at their respective constant rates. If Ken doubles his speed, then two of them could produce 1 chair in 2 hours working       Log On


   

Question 1090882: Ken and Ted can produce 1 chair in 3 hours working together at their respective constant rates. If Ken doubles his speed, then two of them could produce 1 chair in 2 hours working together at their respective rates. How many hours does it currently take Ken to produce 1 chair on his own?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +R%5B1%5D+ = Ken's rate in chairs/hr
Let +R%5B2%5D+ = Ted's rate in chairs/hr
-------------------------------------
Add their rate of making chairs to get their
rate working together
(1) +R%5B1%5D+%2B+R%5B2%5D+=+1%2F3+
(2) +2R%5B1%5D+%2B+R%5B2%5D+=+1%2F2+
----------------------------
(1) +R%5B1%5D+%2B+R%5B2%5D+=+2%2F6+
(2) +2R%5B1%5D+%2B+R%5B2%5D+=+3%2F6+
-----------------------------
Subtract (1) from (2)
+R%5B1%5D+=+1%2F6+
Ken can make 1 chair in 6 hrs
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check:
(1) +R%5B1%5D+%2B+R%5B2%5D+=+1%2F3+
(1) +1%2F6+%2B+R%5B2%5D+=+2%2F6+
(1) +R%5B2%5D+=+1%2F6+
and
(2) +2R%5B1%5D+%2B+R%5B2%5D+=+1%2F2+
(2) +2%2A%28+1%2F6+%29+%2B+R%5B2%5D+=+3%2F6+
(2) +R%5B2%5D+=+3%2F6+-+2%2F6+
(2) +R%5B2%5D+=+1%2F6+
OK