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Word Problems/Expressed Fractionally



How can the following be expressed fractionally?

2 men can lay 1 meter of road in 1 day.

Are these correct?

(2 men)/(1 meter)/(1 day) and as 2 men/1 meter X 1/(1 day) or ?

I thank you for your reply.

A fraction like that is not really legible.  Let X=men, Y=meters, and Z=days.
We have X/Y/Z, but is that (X/Y)/Z, which is X/(YZ), or X/(Y/Z), which is XZ/Y?

Since we have 2 men and 1 meter in a day, then it would be 4 men doing 2 meters in a day.
This gives X/Y is a constant value.

Since we have 2 men and 1 day for a meter of road, then 4 men would take 1/2 a day.
This gives XZ is a constant value.

Since we have 1 meter in 1 day, they could do 2 meters in 2 days.
This tells me there is a Y/Z.

Since there is an X/Y and XZ, the equation would be XZ/Y.
Looking at the 3rd requirement of having a Y/Z in the answer, it could also be Z/Y,
which we have, so the problem is T = KXZ/Y where T is time and K is a constant.
Since X=2, Y=1, and Z=1, this says that K = 0.5.

That gives the equation T = 0.5XZ/Y { or T = XZ/(2Y) }.
Thus, if we double the people, that days would be cuts in half, and that makes sense.
If we double the people and distance, the days would stay the same, and that makes sense.

If we doubled the distance, we would need to double the people to keep the same rate,
and that makes sense.

If we went to 2 days, we would only need 1 person, and that makes sense.

Another way to view this is to think about the affect of doubling each.  If people are doubled (X is doubled), either meters done in a day (Y) are doubled or the amount of time to get the job done (Z) is halved.  This means we have XZ/Y in the equation.

If days (Y) are doubled, the number of people hired (X) could be halved or the amount of time (Z) to get it done could be halved.  This would means we have XZ/Y.

If meters are double, the amount of time would be doubled or the number of workers would be doubled to get the same job done.  This would mean we have XZ/Y.

Any way the problem is looked at, the same result is gotten.  

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Scott A Wilson


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