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I was solving some mathematical questions and have come across the situation, where I need to divide 3900/139. Here is my question,

a. Can I assume 139 to 140 for the ease of division?

If so, how will I know what percentage of error I am introducing? How can I ensure that I am adding very less value to a number and the results will not be tremendously affected?

RobPratt
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dexterous
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2 Answers2

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N/139 = real answer

N/140 = your answer

your answer/real answer = (N/140)/(N/139) = 139/140.

Your answer will be 1/140 too small.

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in general if you replace p with (p + n) your result will by factor of n/(p+n)

Replace 487 with 500 and your be off by a factor of 13/500.

Basically your answer will be off be the same proportion as your rounding was off.

fleablood
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  • so, I can say 1/140 is kind of negligible and I don't have any deviation actually. But, this deviation matters if the multiplying factor is let's say 100000 to this fraction. – dexterous Oct 05 '15 at 22:34
  • total deviation will be larger but in proportional the same. 100,000/99 = 1010.1. But 100,000/100 = 1000 so you are off by 10. Which is 1%. But 2/99 = .020202020.... But 2/100 = .02 so you are off by .0002. Which is also 1%. – fleablood Oct 06 '15 at 00:08
  • So x/140 rather than x/139 will always give an answer that is about .7% too small. – fleablood Oct 06 '15 at 00:10
  • Thanks! That makes sense. So, can I add 0.7% to the value I calculated to get the close answer. i.e. (3900/140) * ( 0.7 /100) + (3900/140) – dexterous Oct 06 '15 at 01:45
  • Well at that point you might as well just divide by 139 in the first place. If you do this your dividing by 140/1.007 = 139.02681231380337636544190665343 which is darned close. – fleablood Oct 06 '15 at 01:57
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In general, if y = c/x (c is a constant) and you make a small change in x, say by h (> 0), then the following happens:

  1. The exact relative error is (-h/(c+h))
  2. This is approximately equal to (-h/c) [If you take the log and then the differential, that will be evident].
PTDS
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