Or— (100 / 9·3)a.
Problem II.—Given the percentage of ‘solids, not fat’ (= a), also the percentage of fat (= b), in a specimen of sophisticated milk—required the number of grammes of
fat which have been removed by skimming from the genuine milk which was employed to form 100 grammes of it.
Answer.— (3·2 / 9·3)(a - b).
In translating fat into cream, the rule is that a removal of 0·2 gramme of fat equals a removal of 1·0 gramme of cream. This rule is directly founded on experiment. I do not, however, claim a high degree of accuracy for the measurement of the cream.
Finally, a slight refinement may be noticed. If a specimen of sophisticated milk has been produced by both skimming and watering, it will be obvious, on consideration, that the extraneous waters employed in manufacturing 100 grammes of it is equal to the difference between 100 and the quantity of genuine milk employed to make 100 grammes of sophisticated milk, together with a quantity of water equal to that of fat removed by skimming.
Extraneous water = 100(100 / 9·3)a + (3·2 / 9·3)(a - b)
= 100[(100 + 3·2) / 9·3](a - b)
Save for the purpose of finding out the presence of matters other than an excess of water in the milk (a contingency regarded as very improbable), the estimation of the casein and milk sugar is unnecessary. The determination of the ash is for the object of learning if foreign mineral matters, such as chalk or any other inorganic impurity, are present. Professor Wanklyn says he believes that such like extraneous bodies are never employed. The chief, if not the sole, form of dishonesty are watering and skimming.
The amount of ash, however, is a good criterion as to the extent of dilution that has been practised, a deficient amount being, of course, confirmatory of a watered milk.