enigma123 wrote:

Machines X and Y run simultaneously at their respective constant rates. If machine X produces 400 bolts per hour, how many bolts do machines X and Y produce per hour?

(1) Machine X takes twice as long to produce 400 bolts as it does for machines X and Y, working together, to produce the same number of bolts.

(2) Machines X and Y produce bolts at the same rate.

Guys - I always struggle to do these questions. Can someone please let me know the step by step approach i.e. what should I be thinking as 1st step when I see this type of rate question.

Hello,

What worked for me: when talk about rates, the equation will be always similar to the following:

\(\frac{Px}{Tx}+ \frac{Py}{Ty} + ... + \frac{Pn}{Tn} = \frac{Pall}{Tall}\)

Where:

\(\frac{Px}{Tx}\) is production rate for a machine X alone

\(Px\) is how much it is producted by a machine X alone

\(Tx\) is the time spent by a machine x alone

and so goes for the other terms. Note that \(\frac{Pall}{Tall}\) is the variables when all machines are working together (and at different rates).

Resolution:The stem defines machine X rate, so:

\(\frac{Px}{Tx} = \frac{400 bolts}{1 hour}\)

Thus, the general equation is:

\(\frac{400}{1}+ \frac{Py}{Ty} = \frac{Pxy}{Txy}\)

Question: \(\frac{Pxy}{Txy}\) =?

(1) \(Tx = 2Txy => Txy = \frac{1}{2}hour\) = 0.5

and the production of machines X and Y is: \(Pxy = 400\)

So:

\(\frac{Pxy}{Txy} = \frac{400}{0.5} = 800\)

Note: The stem is describing the answer directly. We don't need to calculate it. The calculus is just to make it clearer.

Sufficient(2) \(\frac{400}{1}= \frac{Py}{Ty}\)

Thus:

\(\frac{400}{1}+ \frac{400}{1} = \frac{Pxy}{Txy}\)

=> \(\frac{Pxy}{Txy} = \frac{800}{1}\)

Note: We don't need to calculate it. The calculus is just to make it clearer.

SufficientBest,