MAGPIE (who makes the delightful remark "therefore 90 per cent. have
two of something," recalling to one's memory that fortunate monarch,
with whom Xerxes was so much pleased that "he gave him ten of
everything!"), S. S. G., and TOKIO.

BRADSHAW OF THE FUTURE and T. R. do the question in a piecemeal
fashion--on the principle that the 70 per cent. and the 75 per cent.,
though commenced at opposite ends of the 100, must overlap by _at least_
45 per cent.; and so on. This is quite correct working, but not, I
think, quite the best way of doing it.

The other five competitors will, I hope, feel themselves sufficiently
glorified by being placed in the first class, without my composing a
Triumphal Ode for each!

CLASS LIST.

I.

OLD CAT.
OLD HEN.
POLAR STAR.
SIMPLE SUSAN.
WHITE SUGAR.

II.

BRADSHAW OF THE FUTURE.
T. R.

III.

ALGERNON BRAY.
DINAH MITE.
G. S. C.
JANE E.
J. D. W.
MAGPIE.
S. S. G.
TOKIO.

Sec. 2. CHANGE OF DAY.

I must postpone, _sine die_, the geographical problem--partly because I
have not yet received the statistics I am hoping for, and partly because
I am myself so entirely puzzled by it; and when an examiner is himself
dimly hovering between a second class and a third how is he to decide
the position of others?

Sec. 3. THE SONS' AGES.

_Problem._--"At first, two of the ages are together equal to the third.
A few years afterwards, two of them are together double of the third.
When the number of years since the first occasion is two-thirds of the
sum of the ages on that occasion, one age is 21. What are the other two?

_Answer._--"15 and 18."

* * * * *

_Solution._--Let the ages at first be _x_, _y_, (_x_ + _y_). Now, if _a_
+ _b_ = 2_c_, then (_a_-_n_) + (_b_-_n_) = 2(_c_-_n_), whatever be the
value of _n_. Hence the second relationship, if _ever_ true, was
_always_ true. Hence it was true at first. But it cannot be true that
_x_ and _y_ are together double of (_x_ + _y_). Hence it must be true of
(_x_ + _y_), together with _x_ or _y_; and it does not matter which we
take. We assume, then, (_x_ + _y_) + _x_ = 2_y_; _i.e._ _y_ = 2_x_.
Hence the three ages were, at first, _x_, 2_x_, 3_x_; and the number of
years, since that time is two-thirds of 6_x_, _i.e._ is 4_x_. Hence the
present ages are 5_x_, 6_x_, 7_x_. The ages are clearly _integers_,
since this is only "the year when one of my sons co