I printed 25 images for someone and need to figure out how much that cost.
The product page says the printer comes with "Cyan, Magenta, Yellow: 1.000 pages; Black: 1.500 pages". But since we're talking about printers here I assume that this is is some way untrue.
If it were true then my calculation would be as follows:
I used 18% C, 24% M/Y and 7% K (why K and not B?), according to the printer.
1000 pages CMY is 91,40€ each. (from Xerox directly)
K is 117,48€ for 2500 pages. So let's just go with 60€ for 1500 (I realize that price is dumb but I already did all the calculations so we're going to have to live with it).
This would ammount to 334,20€ worth of toner.
18% of 91,40€ = 16,45
24% of 91,40€ = 21,94 * 2 = 43,88
7% of 60€ = 4,20€
Total: 64,45€
So 2,57€ per image? That seems insane, even if you slash 30% because there are probably cheaper shops.
The page amounts are at about 5% coverage, which is industry standard , you would get much less by adding a bar graph, and vastly less by printing pictures
CMYK refers to the four inks used in a standard "4-colour" printing process (cyan, magenta, yellow and black). Depending on the printing company or the press used, the inks are usually applied to the paper (or substrate) in the same order as per the CMYK abbreviation
The "K" in CMYK refers to "key". In 4-colour printing the black "key plate" is used to carefully align the other three colour plates, so that all the elements are in perfect registration. A popular notion is that if "B" was used to denote "black" it might be misconstrued as "blue", which was already taken by cyan. This is a handy explanation, but is technically and historically incorrect.
You may hear some of the older design fraternity referring to a thin black outline (such as around a photograph or a print ad) as a "keyline". Back in the "old" days (prior to using computers), a black keyline was used to show photolithographers where block colour boxes, ads and photographs were to appear in a page layout
I'm not sure if it's feasible for me to anaylze every single image and calculate the coverage for a one-off thing that isn't supposed to result in some sort of comparison with other printers.
I wouldn't even know how. I don't have the prints anymore and a software would need to know how large the image was actually printed.