Work...2

Is my set up right?


1. If Amy, Bianca and Carrie work together on a job, it will take one and one-third hours. If only Amy and Bianca work, it would take one and five-sevenths hours, but if Bianca and Carrie work, it would take two and two-fifths hours, how long would it take each girl working alone to complete the job?

This looks like three equations in three unknowns.


Amy = 1/x
Bianca = 1/y
Carrie = 1/z

1/x + 1/y + 1/z = 4/3
1/x + 1/y = 12/7
1/y + 1/z = 12/5

Yes?


2. John can paint a garage in 8 hours. Gary can do it in 6 hours. Fred can do it in 4 hours. How long will it take if they all paint together?

John = 1/8
Gary = 1/6
Fred = 1/4

1/8 + 1/6 + 1/4 = 1/x

Let x = length of time it will take all three to do the job.

 

NO! 1/x, 1/y, and 1/z are the rates at which Amy, Bianca, and Carrie work. The sum is the rate at which the three, together work in "jobs per hour". "One and one third hour", 4/3 hour is the TIME they take to do the job. Their rate is 3/4 job per hour.
1/x+ 1/y+ 1/z= 3/4, not 4/3.

Similarly, since "If only Amy and Bianca work, it would take one and five-sevenths hours"
1/x+ 1/y= 7/12, not 12/7, and since "if Bianca and Carrie work, it would take two and two-fifths hours" 1/y+ 1/z= 5/12, not 12/5.

IF you had completed the problem, which you apparently refuse to do, you would have found, subtracting the second equation from the first, that
1/z= 4/3- 12/7= 28/21- 36/21= -8/21 so z= -21/8, a negative number!

 

I don't understand the rate part of WORK problems. How does 1/(variable) represent rate? When I think of the word rate, DISTANCE word problems come to mind.

 

So it's more a problem of English than mathematics? No, "rate" does not refer only to distance. Anything that changes has a "rate" of change?

 

COUNTRY BOY:

GIVE IT UP! YOU HAVE BEEN PLACED ON MY IGNORE LIST NEVER TO BE REMOVED. MOVE ON. IF YOU REFUSE TO HEED MY WARNING, I WILL ASK THE MODERATOR TO BLOCK YOU FROM THIS SITE INDEFINITELY.