h.

When the result is not thus connected with the varying elements, the
Problem ceases to be Double Rule of Three and often becomes one of great
complexity.

To illustrate this, let us take two candidates for a prize, _A_ and _B_,
who are to compete in French, German, and Italian:

(_a_) Let it be laid down that the result is to depend on their
_relative_ knowledge of each subject, so that, whether their marks, for
French, be "1, 2" or "100, 200," the result will be the same: and let it
also be laid down that, if they get equal marks on 2 papers, the final
marks are to have the same ratio as those of the 3rd paper. This is a
case of ordinary Double Rule of Three. We multiply _A_'s 3 marks
together, and do the same for _B_. Note that, if _A_ gets a single "0,"
his final mark is "0," even if he gets full marks for 2 papers while _B_
gets only one mark for each paper. This of course would be very unfair
on _A_, though a correct solution under the given conditions.

(_b_) The result is to depend, as before, on _relative_ knowledge; but
French is to have twice as much weight as German or Italian. This is an
unusual form of question. I should be inclined to say "the resulting
ratio is to be nearer to the French ratio than if we multiplied as in
(_a_), and so much nearer that it would be necessary to use the other
multipliers _twice_ to produce the same result as in (_a_):" _e.g._ if
the French Ratio were 9/10, and the others 4/9, 1/9 so that the ultimate
ratio, by method (_a_), would be 2/45, I should multiply instead by 2/3,
1/3, giving the result, 1/3 which is nearer to 9/10 than if he had used
method (_a_).

(_c_) The result is to depend on _actual_ amount of knowledge of the 3
subjects collectively. Here we have to ask two questions. (1) What is
to be the "unit" (_i.e._ "standard to measure by") in each subject? (2)
Are these units to be of equal, or unequal value? The usual "unit" is
the knowledge shown by answering the whole paper correctly; calling this
"100," all lower amounts are represented by numbers between "0" and
"100." Then, if these units are to be of equal value, we simply add
_A_'s 3 marks together, and do the same for _B_.

(_d_) The conditions are the same as (_c_), but French is to have double
weight. Here we simply double the French marks, and add as before.

(_e_) French is to have such weight, that, if other marks be equal, the
ultimate ratio is to be that of the French paper, so that a "0" in