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Lattefah84
- Senior | Next Rank: 100 Posts
- Posts: 93
- Joined: Mon Dec 14, 2009 4:15 am
I will write two solved tasks, and I just ask for an explanation (because, I would not do the 2nd task as it is done.
1st task: Sue can paint a room in 4 hours, and Sam can paint the same room in 5 hours. How long would it take them to paint the room together?
Solution:
Sue paints the room in 4 hours, so she does 1/4 of the room per hour, and Sam does 1/5 of the room per hour If w=t*r
(w-work, r- rate of work, t-time; rate of fork is constant person's work)
t/4 + t/5 =1
5t+4t=1
9t= 20
4= 29/9
2nd task: Abe and ben , working together, can build a cabinet in 6 days Abe works as twice as fast as Ben. How long would it take each of them to build an identical cabinet on his own?
Solution:
t=time for Abe, 2t= time for Ben.
6/t + 6/2t= 1
12+6= 2t
18=2t
9=t
Why is there 1/4*t in first task, and 6/t in second task? How can one explain this? I would do in second task like 1/6*t...?
1st task: Sue can paint a room in 4 hours, and Sam can paint the same room in 5 hours. How long would it take them to paint the room together?
Solution:
Sue paints the room in 4 hours, so she does 1/4 of the room per hour, and Sam does 1/5 of the room per hour If w=t*r
(w-work, r- rate of work, t-time; rate of fork is constant person's work)
t/4 + t/5 =1
5t+4t=1
9t= 20
4= 29/9
2nd task: Abe and ben , working together, can build a cabinet in 6 days Abe works as twice as fast as Ben. How long would it take each of them to build an identical cabinet on his own?
Solution:
t=time for Abe, 2t= time for Ben.
6/t + 6/2t= 1
12+6= 2t
18=2t
9=t
Why is there 1/4*t in first task, and 6/t in second task? How can one explain this? I would do in second task like 1/6*t...?
