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Work Problem

by mynewgator » Sun Aug 08, 2010 4:09 pm
Each of 10 machines works at the same constant rate doing a certain job. The amount of times needed by the 10 machines, working together, to complete the job is 16 hours. How many hours would be needed if only 8 of the machines, working together, were used to complete the job?

A. 18
B. 20
C. 22
D. 24
E. 26
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by Rahul@gurome » Sun Aug 08, 2010 4:18 pm
10 machines takes 16 hours.
1 machine will take 10 * 16 = 160 hours
8 machines will take 160/8 = 20 hours

The correct answer is [spoiler](B)[/spoiler].
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by HPengineer » Sun Aug 08, 2010 5:17 pm
Is there any other approach? would 16 divided by 10 give you the rate of each the machines is working?
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by Rahul@gurome » Sun Aug 08, 2010 5:34 pm
HPengineer wrote:Is there any other approach? would 16 divided by 10 give you the rate of each the machines is working?
This is the case of inverse variation (more will be the no. of people working, less will be the time taken), and 1 machine alone will take more time than 10 machines.

If 10 machines take 16 hours, then the same work can be done by 1 machine in more time, that is 16 * 10 = 160 hours

Does that help?
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by HPengineer » Sun Aug 08, 2010 5:46 pm
my issue is that i always start rate problems with the equation below...

R(T) = W from here is where i got messed up...

They tell us that there are 10 machines working at constant rate.. and for them together to complete 1 job it takes 16 hours...

So i create the equation R*(16) = 1 job which R = 1/16... i took this as the rate for each machine individually...

i guess i don't understand why you multiplied by 10 as the rate to 160 hours?? maybe a bit more details please?
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by Rahul@gurome » Sun Aug 08, 2010 6:20 pm
HPengineer wrote:my issue is that i always start rate problems with the equation below...

R(T) = W from here is where i got messed up...

They tell us that there are 10 machines working at constant rate.. and for them together to complete 1 job it takes 16 hours...

So i create the equation R*(16) = 1 job which R = 1/16... i took this as the rate for each machine individually...

i guess i don't understand why you multiplied by 10 as the rate to 160 hours?? maybe a bit more details please?
Here, we don't need to calculate the rate, it is constant. We need to find the time taken by 8 machines to complete the job, working at the same rate as 16 machines.
We can do this by ratio-proportion method also.

Machines : Time
10 : 16 hrs
8 : x hrs

Note: This is the case of inverse variation, as increasing the no. of workers would decrease the time taken.

So, 10/8 = x/16, which when solved gives x = 20 hours.

Does this help?
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by HPengineer » Sun Aug 08, 2010 6:24 pm
Excellent... makes clear sense now... I like the idea of rates approach as well very straight forward...
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by Rahul@gurome » Sun Aug 08, 2010 6:26 pm
HPengineer wrote:Excellent... makes clear sense now... I like the idea of rates approach as well very straight forward...
Great! :-)
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by likithae » Mon Aug 09, 2010 3:26 am
Rahul@gurome wrote:
HPengineer wrote:my issue is that i always start rate problems with the equation below...

R(T) = W from here is where i got messed up...

They tell us that there are 10 machines working at constant rate.. and for them together to complete 1 job it takes 16 hours...

So i create the equation R*(16) = 1 job which R = 1/16... i took this as the rate for each machine individually...

i guess i don't understand why you multiplied by 10 as the rate to 160 hours?? maybe a bit more details please?
Here, we don't need to calculate the rate, it is constant. We need to find the time taken by 8 machines to complete the job, working at the same rate as 16 machines.
We can do this by ratio-proportion method also.

Machines : Time
10 : 16 hrs
8 : x hrs

Note: This is the case of inverse variation, as increasing the no. of workers would decrease the time taken.

So, 10/8 = x/16, which when solved gives x = 20 hours.

Does this help?
your way of approach is great ............but is there any shortcut rather than this.........
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