I searched for this one before I posted but didn't find it.
Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed b the machine working at the faster rate?
The answer is 66 2/3% but how do I get it?
Work Question
This topic has expert replies
-
- Junior | Next Rank: 30 Posts
- Posts: 12
- Joined: Mon Nov 12, 2007 6:34 pm
-
- Senior | Next Rank: 100 Posts
- Posts: 41
- Joined: Thu Nov 08, 2007 3:22 pm
The way I looked at it, especially when its asking for a simplistic calculation / proportion is just compare the rates.
Make sure you're basing it on the same denominators (LCD) would be helpful.
1/4 + 1/8 = 4/16 + 2/16
Faster machine is 4 : 2 in terms of speed.
4/6 is 2/3 which is ~ 66.67%
hope that helps!
Make sure you're basing it on the same denominators (LCD) would be helpful.
1/4 + 1/8 = 4/16 + 2/16
Faster machine is 4 : 2 in terms of speed.
4/6 is 2/3 which is ~ 66.67%
hope that helps!
Work Hard, Play Harder!