SOLUTION: A machine caps a 1000 bottles in 10 minutes, another machine caps 1000 bottles in 8 minutes. a) If these machines were together how much time will it take to cap 1000 bottles?

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A machine caps a 1000 bottles in 10 minutes, another machine caps 1000 bottles in 8 minutes. a) If these machines were together how much time will it take to cap 1000 bottles?       Log On


   

Question 728564: A machine caps a 1000 bottles in 10 minutes, another machine caps 1000 bottles in 8 minutes.
a) If these machines were together how much time will it take to cap 1000 bottles?
b) if t1 and t2 are the time in the above problem. Determine the time taken by both machines working together?
Can you please help me? Thanks so much in advance:)
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
>>...A machine caps a 1000 bottles in 10 minutes,...<< 
So the slower machine's bottle capping rate is 1000 bottles 
per 10 minutes or

1000_bottles%2F10_min = 1000%2F10bottles%2Fmin = 100bottles%2Fmin

another machine caps 1000 bottles in 8 minutes.
So the faster machine's bottle capping rate is 1000 bottles 
per 8 minutes or

1000_bottles%2F8_min = 1000%2F8bottles%2Fmin = 125bottles%2Fmin

>>...If these machines were together how much time will it take to cap 1000 bottles?...<<
Let that time be x minutes

So their combined bottle capping rate is 1000 bottles 
per x minutes or

1000_bottles%2Fx_min = 1000%2Fxbottles%2Fmin

The equation comes from:

            %28matrix%284%2C1%2C%0D%0A%0D%0ASLOWER%2C%22MACHINE%27S%22%2CBOTTLING%2CRATE%29%29%22%22%2B%22%22%28matrix%284%2C1%2C%0D%0A%0D%0AFASTER%2C%22MACHINE%27S%22%2CBOTTLING%2CRATE%29%29%22%22=%22%22%28matrix%284%2C1%2C%0D%0A%0D%0ATHEIR%2CCOMBINED%2CBOTTLING%2CRATE%29%29

                100bottles%2Fmin + 125bottles%2Fmin%22%22=%22%221000%2Fx

                           100 + 125 = 1000%2Fx
                                 225 = 1000%2Fx
                                225x = 1000
                                   x = 1000%2F225
                                   x = 40%2F9%7D%7D%7B%7D%0D%0A+++++++++++++++++++++++++++++++++++x+=+%7B%7B%7B4%264%2F9 minutes

----------------------------------------------------------------

b) if t1 and t2 are the time in the above problem. Determine the time taken by both machines working together? 

>>...A machine caps a 1000 bottles in t1 minutes,...<< 

So the slower machine's bottle capping rate is 1000 bottles 
per t1 minutes or

1000_bottles%2Ft%5B1%5Dmin = 1000%2Ft%5B1%5Dbottles%2Fmin



>>...another machine caps 1000 bottles in t2 minutes...<<


So the faster machine's bottle capping rate is 1000 bottles 
per t2 minutes or

1000_bottles%2Ft%5B2%5Dmin = 1000%2Ft%5B2%5Dbottles%2Fmin



>>...If these machines were together how much time will it take to cap 1000 bottles?...<<


Let that time be x minutes

So their combined bottle capping rate is 1000 bottles 
per x minutes or

1000_bottles%2Fx_min = 1000%2Fxbottles%2Fmin

The equation comes from:

            %28matrix%284%2C1%2C%0D%0A%0D%0ASLOWER%2C%22MACHINE%27S%22%2CBOTTLING%2CRATE%29%29%22%22%2B%22%22%28matrix%284%2C1%2C%0D%0A%0D%0AFASTER%2C%22MACHINE%27S%22%2CBOTTLING%2CRATE%29%29%22%22=%22%22%28matrix%284%2C1%2C%0D%0A%0D%0ATHEIR%2CCOMBINED%2CBOTTLING%2CRATE%29%29

                1000%2Ft%5B1%5Dbottles%2Fmin + 1000%2Ft%5B2%5Dbottles%2Fmin = 1000%2Fxbottles%2Fmin

                           1000%2Ft%5B1%5D + 1000%2Ft%5B2%5D = 1000%2Fx

The LCD is t1t2x

                           1000t2x + 1000t1x = 1000t1t2
                 
Divide through by 1000

                                   t2x + t1x = t1t2

Factor out x on left
              
                                  x(t2 + t1) = t1t2

Divide both sides by (t2 + t1)

                                    x%28t%5B2%5D%2Bt%5B1%5D%29%2F%28%28t%5B2%5D%2Bt%5B1%5D%29%29 = %28t%5B1%5Dt%5B2%5D%29%2F%28%28t%5B2%5D%2Bt%5B1%5D%29%29

                                    x%28cross%28t%5B2%5D%2Bt%5B1%5D%29%29%2F%28%28cross%28t%5B2%5D%2Bt%5B1%5D%29%29%29 = %28t%5B1%5Dt%5B2%5D%29%2F%28%28t%5B2%5D%2Bt%5B1%5D%29%29

                                     x = %28t%5B1%5Dt%5B2%5D%29%2F%28t%5B2%5D%2Bt%5B1%5D%29 minutes 
 
That's the answer.
-------------------------------------------------------------------------  
So the formula is "the product of the times over the sum of the times":

We could have used that formula to solve (a) if we had had it, with

t1 = 10 and t2 = 8

                                     x = 10%2A8%2F%2810%2B8%29
                                     x = 80%2F18
                                     x = 40%2F9
                                     x = 4%264%2F9minutes  

Edwin