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,