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Basic Math/Units X Units Calculation

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Question
QUESTION: Hello:

15 workers can finish a job in 7 days.

1 worker can do the same job in 105 days.

Here's the calculation: 15 workers X 7 days = 105 worker-days

1/(1 worker) X 105 worker-days = 105 days

Does "105 worker-days represent the same as 105, 1 worker days?

I thank you for your reply.

ANSWER: yes, either way you calculate it.  It takes 105 days for one person to do the job.  If 15 ppl are working, it takes 7 days.  If 105 ppl were working it would take one day.

---------- FOLLOW-UP ----------

QUESTION: Hello:

I want to thank you for the reply.

If I want to find the number of days 21 workers can finish the job, the calculation becomes 1/(21 workers)  X 15 workers X 7 days.
If the 21 and 15 are reduced to 5 and 7 and the calculation becomes 1/7 X 5 X 7 days, what do these two reduced terms (7 and 5) indicate or represent?

I thank you for your follow-up reply.

ANSWER: Not sure what you're doing with your calculations, but you can't just "reduce" (ie divide by 3), 2 elements of the equation and not the third.  The easiest way to figure out how many days it would take 21 ppl to do the job is just divide 105/21 = 5 days.

---------- FOLLOW-UP ----------

QUESTION: Hello:

I think you need to look at the calculation again!

1/21 X 15 X 7 = (15 X 7)/21 The 15 and 21 reduce to 5 and 7. The calculation becomes (5 X 7)/7 = 5

I want to know what these two reduced terms (7 and 5) indicate or represent concerning the original calculation when the two terms were not reduced?

I do not agree with your "you can't just "reduce" (ie divide by 3), 2 elements of the equation and not the third."

I thank you for your follow-up reply.

Answer
Ok, the only reason your "reduction" works is because you've already multiplied the 1/21 and 15 and you're reducing a fraction.

We are saying the same thing.  I say the easiest way to solve is to divide 105 by 21.  Which is exactly what you're doing 1/21 x (15 x 7).  The 5 and 7 represent nothing of the original equation.  You are simply reducing a fraction to make the equation easier to solve.  

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Lynn Houston

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