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difficult rate question - kaplan

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by resilient » Tue Mar 18, 2008 11:33 am
Machine A and MAchine B are each used manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 % more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

a. 6
b.6.6
c.60
d.100
e.110


Backsolving is a great strategy here and I would like to learn how to tackle this question using the manhattan gmat variable chart approach. I am trying to use the CLICK chart (backsolve) approach but am ahving trouble setting this up! Any help? All input from proffesionalls welcome!

qa is A
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by 2008nv » Tue Mar 18, 2008 12:06 pm
Machine A produce X sprockets per hour
Machine B produce 10% more sprockets per hour than A - 1.1*X

660 / (1.1*X) + 10 = 660 / X
660 +11*X = 726
11*X=66
X=6
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by xilef » Tue Mar 18, 2008 12:12 pm
A - sprockets per hour machine A produces
T - Total time it takes machine B

2 equations:

660=A(T+10)
660=T(1.1A)

660=A(660/1.1A+10)
A=6

Answer A
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how

by resilient » Wed Mar 19, 2008 8:02 am
is there a formula for this. I am trying to set this up in a drt formula but am having trouble setting up the equation.
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by AleksandrM » Wed Mar 19, 2008 9:31 am
I'm with you, I used the rate,time, distance (or work, in this case) table to set up the equation, and the sheer amount of work I put in while trying to set up the equations took me about 5 minutes, and I still ended up without a result. If someone can show us both how to use the RTD/W table to solve this problem, it would be greatly appreciated.
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by xilef » Wed Mar 19, 2008 10:44 am
Just because it's a problem that talks about work, don't assume that you should be using the work formula. The formulas and charts will not be applicable to all types of problems, it will be more beneficial if you learn to understand what is being asked. You do that by following the logic, all the necessary info is provided:


It takes machine A 10 hours longer to produce 660 sprockets than machine B , which means that machine A produces 660 sprockets in (total time it takes machine B to produce 660 sprockets +10 hours), so if we have

T - Total time it takes machine B
A - sprockets per hour machine A produces - hourly rate

then

660=A(T+10)


we also know that Machine B produces 10 % more sprockets per hour than machine A, which means that machine B produces 660 sprockets in T * 110% of hourly rate of A.

so:


660=T(1.1A)

Now we solve for A.
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