If a line is 65 char for Alpha, it takes about 75 "black characters" to - equal a 65 char line with spaces. Posted: Dec 17th, 2017 - 7:20 pm In Reply to: well - t
In other words, taking some sample paragraphs from different parts of reports (including lab sections), there are about 10 spaces per 65 character line so it takes 75 black characters to make up for the loss of spaces.
Now, if my math doesn't fail me, the way you approach this problem is to set up a proportion between the lines (75 vs 65) and the line rate (5 cents vs ?). The question mark is what we're trying to solve for and we'll let that be X.
So 75 is to 65 as 5 is to X. That's our proportion.
This looks like this: 75/65 :: 5/X
To solve a proportion, you cross-multiply the OPPOSITE members of the ratios in the proportion. In other words, you multiply the numerator of one ratio by the denominator of the other ratio, and vice versa, and express this as an equation.
So now we have 75 times X = 65 times 5.
Simplifying: 75X = 325
Now solve for X by dividing 325 by 75, which gives us 4.33.
In other words, 5 cents is the same proportion to 4.33 cents as 75 characters is to 65 characters. This might differ somewhat depending on your sample, but I think over a large sample the results would center somewhere pretty close to 4.3 cents, perhaps a bit more or less.
If you doubt this method, suppose I were to say that Company A pays 8 cents per 65 character line but Company B defines a line as 130 characters. What should company B pay to be equivalent to Company A?
Intuitively, you would say 16 cents, the "B" line being twice as many characters as the "A" line. And you'd be right. Your logical mind solves it easily, but let's solve it by formal mathematics instead.
65 char is to 130 char as 8 cents is to how many cents (X)?
Set up the proportion: 65/130 :: 8/X
Cross-multiplying numerators and denominators, we get:
65X = 1040
...and now we can solve for X by dividing 1040 by 65, which is 16 - exactly the number that simple logic gave us. We need math to solve the original problem because those numbers don't lend themselves so easily to an intuitive solution, but at least this simple exercise illlustrates the validity of the method that we used.
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