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Basic question regarding work rate | Experts plz comment

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by [email protected] » Fri Nov 11, 2016 12:37 am
Hi Experts,

Please help me with the understanding:

Is the rate derived for each correct in the table

A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. Then what will be the rate of each of the older machines.




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by DavidG@VeritasPrep » Fri Nov 11, 2016 7:50 am
[email protected] wrote:Hi Experts,

Please help me with the understanding:

Is the rate derived for each correct in the table

A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. Then what will be the rate of each of the older machines.




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Well, the labels in your table are a little ambiguous, but your basic reasoning looks okay. (I'm assuming the (1/4t) refers to the rate for each old machine, not the rate for 2 old machines together.)

Another way to think about it: If a new machine can do a job in half the time that 2 old machines can do a job, then the new machine has a rate that is double the rate of two old machines working together.

Rate for a new machine: N

Rate for an old machine D

If N's rate is twice the rate of two D's working together then N = 2(D+D) or N =2(2D) or N = 4D or N/4 = D
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by [email protected] » Fri Nov 11, 2016 9:27 am
Hi nishatfarhat87,

In these types of situations, instead of thinking 'algebraically', you might find it easier to come up with a simple example that you can use to define the relationships involved.

We're told that a newer machine can fill an order in HALF the time required by TWO older machines working together...

IF....
New Machine = fills order in 1 hour
TWO Old Machines = fill order in 2 hours

Since the two old machines are working together, each would fill HALF the job in 2 hours

ONE Old Machine = fills order in 4 hours

Now we have the relationship between one new machine and one old machine. One old machine takes 4 times as much time to complete the job as one new machine.

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by Matt@VeritasPrep » Fri Nov 11, 2016 2:05 pm
Let's call the amount of work to be done W, the rate of the new machine R, and the rate of each old machine S. This gives us

New machine: W = RT

Old machines: W = 2S * 2T

So RT = 4ST, or R = 4S, and the new machine is four times as fast as one of the old machines.
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