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Rate/Percent Problem

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Rate/Percent Problem

by bburton11 » Thu May 28, 2009 3:35 pm
Working alone at its constat rate a machine seal K cartons in 8 hours, and working alone at its own constat rate, a second machine seals K cartons in 4 hours. If the two machines , each working at it own contact rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate.

25%

33 1/3%

50%

60 2/3%

75%

answer is 60 2/3%
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by ssmiles08 » Thu May 28, 2009 3:44 pm
you first add the two rates:

rate of Machine1= k/8
rate of Machine2= k/4

k/8 + k/4 = 3k/8

faster rate is machine2: k/4

(k/4)/(3k/8) = 2/3 = 60.6667%
Last edited by ssmiles08 on Thu May 28, 2009 3:49 pm, edited 1 time in total.
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by truplayer256 » Thu May 28, 2009 3:48 pm
Let's call the first machine, Machine A and the second machine, Machine B.

Machine A can seal K cartons in 8 hours, which means that it can seal K/8 cartons per hour. Machine B can seal K cartons in 4 hours, which means that it can seal K/4 cartons per hour. Together, both machines can seal K/8+K/4 or 3K/8 cartons per hour. Let's say that they sealed a total of y cartons working together and that it took them 3K(t)/8=y t=8y/3k hours to do the job. In order to find out how how many cartons were sealed by Machine B (The machine working at the faster rate), all we have to do is multiply 8y/3k hours by K/4(since that's how many cartons Machine B seals in a hour).

K/4*8Y/3K=2Y/3

Now to find the percent of cartons sealed by Machine B, we do (2Y/3)/Y*100 to get 66.67% or 66 2/3%. I think there's an error or a typo in one of the answer choices, it should be 66.67%, not 60.67%.
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Re: Rate/Percent Problem

by sureshbala » Fri May 29, 2009 10:57 pm
bburton11 wrote:Working alone at its constat rate a machine seal K cartons in 8 hours, and working alone at its own constat rate, a second machine seals K cartons in 4 hours. If the two machines , each working at it own contact rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate.

25%

33 1/3%

50%

60 2/3%

75%

answer is 60 2/3%
Folks, this is very simple.

In 8 hrs first machine seals K cartons

In 8 hrs second machine seals 2K cartons.

So if total work is 3K, work done by the faster machine is 2K.

Hence its contribution is 2K/3K = 2/3 = 66.67%

Check your choices once again
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