prove that the line joining the given points will be bisected by the given line at right angles.

(14) Describe a circle which shall pass through two given points, and have its centre in a given line.

The line must be of unlimited length, and even then there is one position in which this problem cannot be solved.

From E draw E F at right angles to A B, meeting C D, or C D produced, in F.

With centre F and distance F A describe the circle A B G.

Let the pupil now prove that F A is equal to F B, and that therefore the circle will pass through B.

(If the line joining A B be perpendicular to C D it is evident that the circle cannot be drawn as required.)

(15) A line which bisects the vertical angle of a triangle also bisects the base. Prove that the triangle is isosceles.

(16) Draw examples of the different kinds of four-sided figures, plane rectilineal angles, and triangles. Write a definition of each. What is a postulate? Mention those of Euclid.

(17) What is a rhombus? Show that the opposite angles of a rhombus are equal.

(18) Explain the following terms used in Euclid :-Proposition, problem, theorem, enunciation, hypothesis, axiom, postulate, and corollary.

(19) By the method of the first proposition describe on a given finite straight line an isosceles triangle, the sides of which shall be equal to twice the base.

(20) If two circles cut each other, the line joining their points of intersection is perpendicular to the line joining their centres. (21) If a triangle A B C be turned over about its side A B, show that the line joining the two positions of C is perpendicular to A B.

(22) Two straight lines are drawn to the base of a triangle from the vertex, one bisecting the vertical angle, the other bisecting the base. Prove that the latter is the greater of the two lines.

(23) Any two sides of a triangle are together greater than the third. How many triangles having two sides 5 feet and 6 feet long can be formed so that the third side shall contain a whole number of feet?

(24) If a point P be taken inside a quadrilateral A B C D, prove that the sum of the distances of P from the angular points is the least possible when P is situated at the intersection of the diagonals.

Solutions of geometrical problems, together with answers to questions in mathematics generally, as well as in history, grammar, etc., are given in Papers for Teachers, a monthly periodical, published by Messrs. Moffatt & Paige, and specially designed to assist pupil teachers in their studies.

ARITHMETIC.

(MALES.)

INTEREST.

Interest is the consideration paid for the use of money lent. The Rate of interest is the consideration paid for the use of a certain sum for a certain time.

Interest is usually calculated on the basis of so much per £100 for each year the money is lent; or at the rate of so much per cent. per annum.

Thus, if a person borrows money at the rate of £5 for each £100 for each year the loan continues, the interest is said to be at the rate of 5 per cent. per annum. So, "per cent." means, for each £100; “per annum means, for each year.

The money lent is called the Principal.

The Principal, together with the interest at any given period, is called the Amount. Thus, if £100 be lent for a year at 5 per cent., the interest at the end of the year would be £5, and the Amount at the end of that time would be £100 +£5 = £105. Interest may be either Simple or Compound.

Simple Interest is that which is received only on the original Principal.

Compound Interest is that which is received not only on the original principal, but on the interest which accrues from time to time, and which, being added to the sum originally lent, becomes new Principal. Thus, if £100 be lent for 5 years at Compound Interest, at the end of the first year the interest, £5, is not paid to the lender, but remains in the hands of the borrower, who therefore has to pay interest during the second year

on £105. Similarly, during the third year he has to pay interest on £105 + interest on £105 for one year, and so on till the end of the five years.

Thus, in Simple Interest the Principal always remains the same, but in Compound Interest the Principal is increased at the end of any given period by the Interest due for that given period.

SIMPLE INTEREST.

To find the Interest of a given sum of money at a given rate per cent., for a given number of years.

Rule. Multiply the Principal by the rate per cent., divide the product by 100, and multiply the result by the number of years. (1) Find the Interest on £300 for 3 years at 4 per cent. Here, Interest on £100 for 1 year at I per cent.=£1=£100.

100

£300

100

The reason of the rule will appear from the above example. (2) Find the Interest on £925 16s. 8d. at 7 per cent. for 5

Here the Principal is £925 16s. 8d. Following the rule, we multiply this by the rate per cent. (7) and divide by 100 (cutting off two figures to the right in each line). We thus get the Interest for one year, viz., £64 16s. 2d.; then multiplying by 5, we obtain the Interest for 5 years.

Note. It is sometimes convenient to multiply by both the rate per cent. and the number of years before dividing by 100. If the Amount were required in Example 2, it would be found by adding the Principal to the Interest, thus :

£925 16s. 8d. + £356 8s. 11d. = £1282 5s. 7d.

(3) Find the Interest and Amount of £1284 15s. 10d. at 2 per cent. for 3 yr. 5 m.

(4) What will £1126 amount to in 4 years 17 weeks, at 5 per