## rate issue with box method and combines rates

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### rate issue with box method and combines rates

by jzw » Sun Mar 18, 2012 1:11 pm
It takes 6 beavers 10 hours to build a certain dam, working at a uniform rate. If six beavers start to build the same dam at noon, and one beaver per hour is added beginning at 6:00 PM, at what time will the dam be complete?

(a) 7:40 PM

(b) 8:00 PM

(c) 8:20 PM

(d) 8:40 PM

(e) 9:00 PM

i got e, but i did it a very long way. does anyone have a quick way to do a rate problem like this?

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by pemdas » Sun Mar 18, 2012 1:59 pm
from six-beaver combined rate we calculate one-beaver average rate, (1/6)*(1/10). Through 6:00 PM, 0.6 part of work is completed; after this time the combined rate will increase by 1/60 each hour. That is the remaining 0.4 part of work will be completed at different rates each hour - by 7:00 PM the remaining part of work will be 4/10 - (1/10 + 1/60)= (24-6-1)/60 = 17/60. By 8:00 PM the remaining part of work will be 17/60 - (1/10 + 2/60)= (17-6-2)/60 = 9/60. Finally, by 9:00 PM the work left will be 9/60 - (1/10 + 3/60)= (9-6-3)/60 or 0. Answer must be [spoiler]9:00 PM[/spoiler]
jzw wrote:It takes 6 beavers 10 hours to build a certain dam, working at a uniform rate. If six beavers start to build the same dam at noon, and one beaver per hour is added beginning at 6:00 PM, at what time will the dam be complete?

(a) 7:40 PM

(b) 8:00 PM

(c) 8:20 PM

(d) 8:40 PM

(e) 9:00 PM

i got e, but i did it a very long way. does anyone have a quick way to do a rate problem like this?
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by jzw » Sun Mar 18, 2012 4:08 pm
I was asked how I did it by someone privately so I'm posting it here. Attached you'll see how originally I tried to use my trusty box method, but for this type of rate problem I couldn't figure out how to take the box method to the next step. So, I took the information I used by using the box method to draw it out. See the second picture for that.

The way we calculate the rate of the 6 beavers is, since they take 10 hours to do it, the rate is always the inverse of the time AS LONG AS the job they are completing it "one job". Meaning, 1/10 x 10/1 = 1. So when any problem says "to complete a job" or something of that nature, the job being (don't pick any answer choices in the verbal section that have the word "being") done means it's one job. So as long as you know any two parts of the equation, you can always get the third.

The way we calculate that the rate of one beaver is 1/60 is because we took the 1/10 (which is the rate of all six of them) and divided it by six. Had it been seven beavers we would have divided it by 7. etc...

So - since we know the rate of each beaver and we want to know how long it will take each beaver to complete one job, the time has to be the inverse of the rate, which comes to 60/1 (otherwise known as 60).

So now what... - I had been hoping that someone would be able to explain how to continue the box method here since it's awesome... but.... in the absence of that I did it the long way.

Ok - so we know that 6 beavers started the job at noon. By 1pm, 1/10 of the job was done, because they are working at a rate of 1/10 per hour, since over 10 hours they would complete the one full job. By 2pm they had completed 2/10 of the job (1/10+1/10=2/10), and so on until 6pm. Now at 6pm they added one additional beaver, so there were 7 beavers working from 6pm to 7pm. Now, you'll notice that I switched 6/10 to 42/60. Where did I get that from - so since we're adding one beaver using the format of 1/60, we need to have the 6 beavers represented with a common denominator. Easiest way to do that is to multiply both parts of the fraction by 1 in the form of 6/6, so instead of 7/10 we get 42/60. Means the same thing. Ah, but now we can add 1/60 to it - remember that the 1/60 represents the work of the additional beaver that started working at 6pm and it's 1/60 of the job was completed one hour later, at 7pm. So we take 42/60 + 1/60 and we get 43/60.

And now by 8 pm, at the "top" we have to account for the original 6 beavers continuing their 1/10 of work (now represented as 6/10), but we're adding that to 43/60 to get 49/60. And since we were told that there is an additional beaver being (again, don't use "being" in the verbal section) added per hour, there are two additional beaver's work that is added represented by 2/60. So we add 49/60 to 2/60 to get 51/60. Now the same thing as we did before - we add the work done by the same top 6 beavers 6/60 to 51/60 to get 57/60, and now since there were three beavers working, cuz remember yet another additional beaver started working at 8pm, we add 3/60 to the 57/60 to get 60/60, which is the same thing as "1", which means they completed the job @ 9pm.

Hope this long megillah helped!
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by GMATGuruNY » Sun Mar 18, 2012 5:30 pm
jzw wrote:It takes 6 beavers 10 hours to build a certain dam, working at a uniform rate. If six beavers start to build the same dam at noon, and one beaver per hour is added beginning at 6:00 PM, at what time will the dam be complete?

(a) 7:40 PM

(b) 8:00 PM

(c) 8:20 PM

(d) 8:40 PM

(e) 9:00 PM

i got e, but i did it a very long way. does anyone have a quick way to do a rate problem like this?
Let the rate per beaver = 1 unit per hour.
The combined rate for 6 beavers = 6 units per hour.
In 10 hours, the number of units produced by 6 beavers = 60 units. This is the value of the job.

From 12-6pm, the number of units produced by 6 beavers = r*t = 6*6 = 36.
Remaining work to be produced = 60-36 = 24 units.
After 6pm, since one beaver is added each hour, the work produced will increase by one unit each hour:
7 units, 8 units, 9 units, 10 units...
Since 7+8+9 = 24, the job will be completed after 3 more hours.
6pm + 3 hours = 9pm.

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by jzw » Sun Mar 18, 2012 5:40 pm
Mitch - u always make it all look so easy

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by LalaB » Mon Mar 19, 2012 11:34 am
6 beavers worked till 6 p.m. and did 6/10=3/5 of work
the rest of work =2/5(i.e. (2/5)= 1-(3/5))

the rate of 6 beavers is 1/10
(1/10)+x*(1/10)=2/5
(1+x)/10=2/5
x=3

p.s. God bless that beavers. if only that beavers could have built my house with such a speed
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by [email protected] » Thu Dec 14, 2017 4:29 pm
jzw wrote:It takes 6 beavers 10 hours to build a certain dam, working at a uniform rate. If six beavers start to build the same dam at noon, and one beaver per hour is added beginning at 6:00 PM, at what time will the dam be complete?

(a) 7:40 PM

(b) 8:00 PM

(c) 8:20 PM

(d) 8:40 PM

(e) 9:00 PM
If 6 beavers build a dam in 10 hours, the rate of 6 beavers is 1/10 of a dam per hour and the rate of 1 beaver is (1/10)/6 = 1/60 of a dam per hour.

If 6 beavers start to build a dam at noon, by 6 PM, they will complete 6 x (1/10) = 6/10 = 3/5 of a dam. So 1 - 3/5 = 2/5 of a dam remains to be completed at 6 PM.

At this time, a new beaver joins in; thus, the rate of the 7 beavers is 7 x (1/60) = 7/60 of a dam per hour and they will complete 7/60 of a dam by 7 PM. So 2/5 - 7/60 = 24/60 - 7/60 = 17/60 of a dam remains to be completed at 7 PM.

At this time, another new beaver joins in; thus, the rate of the 8 beavers is 8 x (1/60) = 8/60 of a dam per hour and they will complete 8/60 of a dam by 8 PM. So 17/60 - 8/60 = 9/60 of a dam remains to be completed at 8 PM.

At this time, yet another new beaver joins in; thus, the rate of the 9 beavers is 9 x (1/60) = 9/60 of a dam per hour and they will complete 9/60 of a dam by 9 PM. This completes the 9/60 of a dam that remained to be completed at 8 PM. Therefore, they complete the dam at 9 PM.