1. Machines A & B

Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten 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?

My Effort:

The rate of Machine A is 660/(x + 10).

The rate of Machine B is 660/x.

Let x = the amount of sprockets per hour produced by Machine A.

I am stuck here.

2. Re: Machines A & B

Your expressions are rates. That would indicate that x would be time in hours, correct?

The ratio of of B’s rate of production to A’s rate of production of sprockets per hour is 1.1

continue?

3. Re: Machines A & B

Originally Posted by Cervesa
Your expressions are rates. That would indicate that x would be time in hours, correct?

The ratio of of B’s rate of production to A’s rate of production of sprockets per hour is 1.1

continue?
Can you set the equations? I can then take it from there.

5. Re: Machines A & B

Originally Posted by harpazo
Can you set the equations? I can then take it from there.
“Setting the equations” is the whole point of the exercise. I have little doubt you can manipulate the algebra. Re-read the given statement regarding the ratios and make an attempt on your own.

6. Re: Machines A & B

To point out a rather obvious mistake, you have
$\displaystyle \frac{660}{x+ 10}$ as the "rate" of machine A and
$\displaystyle \frac{660}{x}$ as the rate of machine B

but then say "Let x = the amount of sprockets per hour produced by Machine A."

The numerator, 660, is a number of sprockets so "sprockets" divided by "sprockets per hour" has units of hours, which is NOT a "rate"!

In order that $\displaystyle \frac{660}{x}$ be the rate of machine B, in sprockets per hour, x must be the time in hours that it takes machine B to make 660 sprockets. Then, since "It takes machine A ten hours longer to produce 660 sprockets than machine B" it is true that A's rate, again in "sprockets per hour" is $\displaystyle \frac{660}{x+ 10}$. So in one hour, A will make $\displaystyle \frac{660}{x+ 10}$ and B will make $\displaystyle \frac{660}{x}$ sprockets. We are also told that the second of those is 10% more than the first. Saying "P is 10% more than Q" means P= Q+ 0.10Q= 1.1Q.

7. Re: Machines A & B

Originally Posted by HallsofIvy
To point out a rather obvious mistake, you have
$\displaystyle \frac{660}{x+ 10}$ as the "rate" of machine A and
$\displaystyle \frac{660}{x}$ as the rate of machine B

but then say "Let x = the amount of sprockets per hour produced by Machine A."

The numerator, 660, is a number of sprockets so "sprockets" divided by "sprockets per hour" has units of hours, which is NOT a "rate"!

In order that $\displaystyle \frac{660}{x}$ be the rate of machine B, in sprockets per hour, x must be the time in hours that it takes machine B to make 660 sprockets. Then, since "It takes machine A ten hours longer to produce 660 sprockets than machine B" it is true that A's rate, again in "sprockets per hour" is $\displaystyle \frac{660}{x+ 10}$. So in one hour, A will make $\displaystyle \frac{660}{x+ 10}$ and B will make $\displaystyle \frac{660}{x}$ sprockets. We are also told that the second of those is 10% more than the first. Saying "P is 10% more than Q" means P= Q+ 0.10Q= 1.1Q.
I was not able to set up the right equation(s) here. So, a friend solved it differently.

Friend's Work:

Since B produces 10 percent more sprockets per hour than A, for every 10 sprockets produced by A, B produces 11 sprockets. So, the RATE RATIO for A and B = 10:11.
Since rate and time are reciprocals, the TIME RATIO for A and B = 11:10. Since A's time is 10 hours longer than B's time, and 11:10 = 110:100, A's time = 110 hours, while B's time = 100 hours. He concluded that, A's rate = w/t = 660/110 = 6 sprockets per hour. Is this right?