DINAH MITE.
E. B. D. L.
JORAM.

II.

BALBUS.
THE ELDER TRAVELLER.

* * * * *

With regard to Knot V., I beg to express to VIS INERTIAE and to any
others who, like her, understood the condition to be that _every_ marked
picture must have _three_ marks, my sincere regret that the unfortunate
phrase "_fill_ the columns with oughts and crosses" should have caused
them to waste so much time and trouble. I can only repeat that a
_literal_ interpretation of "fill" would seem to _me_ to require that
_every_ picture in the gallery should be marked. VIS INERTIAE would have
been in the First Class if she had sent in the solution she now offers.

ANSWERS TO KNOT VII.

_Problem._--Given that one glass of lemonade, 3 sandwiches, and 7
biscuits, cost 1_s._ 2_d._; and that one glass of lemonade, 4
sandwiches, and 10 biscuits, cost 1_s._ 5_d._: find the cost of (1) a
glass of lemonade, a sandwich, and a biscuit; and (2) 2 glasses of
lemonade, 3 sandwiches, and 5 biscuits.

_Answer._--(1) 8_d._; (2) 1_s._ 7_d._

_Solution._--This is best treated algebraically. Let _x_ = the cost (in
pence) of a glass of lemonade, _y_ of a sandwich, and _z_ of a biscuit.
Then we have _x_ + 3_y_ + 7_z_ = 14, and _x_ + 4_y_ + 10_z_ = 17. And we
require the values of _x_ + _y_ + _z_, and of 2_x_ + 3_y_ + 5_z_. Now,
from _two_ equations only, we cannot find, _separately_, the values of
_three_ unknowns: certain _combinations_ of them may, however, be found.
Also we know that we can, by the help of the given equations, eliminate
2 of the 3 unknowns from the quantity whose value is required, which
will then contain one only. If, then, the required value is
ascertainable at all, it can only be by the 3rd unknown vanishing of
itself: otherwise the problem is impossible.

Let us then eliminate lemonade and sandwiches, and reduce everything to
biscuits--a state of things even more depressing than "if all the world
were apple-pie"--by subtracting the 1st equation from the 2nd, which
eliminates lemonade, and gives _y_ + 3_z_ = 3, or _y_ = 3-3_z_; and then
substituting this value of _y_ in the 1st, which gives _x_-2_z_ = 5,
_i.e._ _x_ = 5 + 2_z_. Now if we substitute these values of _x_, _y_, in
the quantities whose values are required, the first becomes (5 + 2_z_) +
(3-3_z_) + _z_, _i.e._ 8: and the second becomes 2(5 + 2_z_) + 3(3-3_z_)
+ 5_z_, _i.e._ 19. Hence the answers are (1) 8_d._, (2) 1_s._ 7_d._