difficult rate question - kaplan

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

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

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.

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.

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.