Machines X and Y run simultaneously at their respective

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Hi All,

This DS question doesn't actually require that much 'math' - but you do have to keep track of the information that you're given to get the correct answer.

We're told that Machine X and Machine Y each have a constant rate of production (Machine X's rate is 400 bolts/hour) and that the Machines run simultaneously. We're asked how many bolts are produced by the 2 machines 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.

We already know that Machine X produces 400 bolts in 1 hour. Fact 1 tells us that 1 hour is TWICE as long as the time it takes the two machines to produce those same 400 bolts. Thus, the two machines (working together) can produce 400 bolts in 1/2 hour. We know that Machine X would produce 200 of those bolts in 1/2 hour, so the other 200 bolts must be produced by Machine Y during that time. We now know Machine Y's rate (it's 400 bolts/hour).
Fact 1 is SUFFICIENT.

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

Since we already know that Machine X produces 400 bolts/hour, we now know that Machine Y also produces 400 bolts/hour.
Fact 2 is SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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.

Sufficient

Best,