C.

In ALGEBRA the questions most frequently attempted were Q's. 13, 15, 16, 17, and on the whole they were well done.

find the numerical values of the three given expressions and of the remainder you have obtained, and show that the excess of the first value over the sum of the second and third values equals the fourth value.

There were very many failures in the numerical substitutions. The first part of the question was correct in most cases.

to its simplest form.
three factors.

(b) If we suppose that

Also show that it is the product of

stands for a positive whole number,

show that the three factors are either three consecutive even numbers or three consecutive odd numbers.

Was commonly avoided.

Q. 16. (a) Resolve the following expressions into their simplest factors:

(b) Find two consecutive numbers such that the difference of their squares is 49.

(a) (iii). The expression was very seldom put into four factors, and in (b) it was curious to notice in how many cases (x+1)" was treated as if it were less than x2.

Q. 18. A man has a certain number of horses, all of the same value ;

also he has seven more cows than he has horses, and each cow is worth two-thirds as much as a horse. If he had the same total number of horses and cows but seven more horses than cows, the whole value of the animals would have been 567. more in the latter case than in the former. Find the value of one of the horses.

Why cannot you determine the number of horses, as well as the value of each, from the data?

Is an easy question, and sometimes the value of one horse was found. The second part of the question proved difficult, e.g., in about four hundred consecutive papers the point was explained only twice.

STAGE 2.

Results 1st Class, 241; 2nd Class, 899; Failed, 843; Total, 1,983.

Each question was attempted fairly often; but Q's. 25, 26, 30, 32, 38, and in some schools Q. 36 were attempted markedly less often than the others. Still the work was, on the whole, good or fairly good; though, if estimated by the number of candidates who get 120 marks and upwards, it is not so good as it was last year. It may be added that several of the questions set this year are quite easy, e.g. Q's. 21, 22, 25, 27, 28, 31, not to mention others.

9291.

A 2

A.

In GEOMETRY the book work was well written out, with the exception given below:

Q. 22. Define a sector of a circle, a segment of a circle, and similar segments of two different circles :

(a) C is the middle point of the line joining two given points A and B ; a circle is drawn whose circumference passes through A and B ; show that its centre must be on the line drawn through C at right angles to AB.

(b) Two unequal circles are in the same plane and their centres coincide; show that the one must lie entirely within the other.

In Q. 22, the reasoning was in many cases given as if the writers had not made up their minds as to the point that they are required to prove. Q. 24. Show how to describe a circle about a given triangle.

Given the diameter of the circle circumscribing a triangle, an angle of the triangle, and one of the sides containing the angle, show how to construct the triangle.

Show also that there is an ambiguity in the construction exactly resembling that in the ambiguous case of the solution of triangles.

The "ambiguity" was seldom or never well explained.

Q. 25. ABCD is a parallelogram, and from B a line is drawn to cut CD or CD produced in E; from A a line is drawn at right angles to BE meeting it in F. Show that the area of ABCD equals that of the rectangle under BE and AF.

Is a very easy question. Thus, if AE is drawn, the rectangle AF, BC is twice the triangle ABE; also the parallelogram ABCD, is twice the same triangle; and therefore the rectangle is equal to the parallelogram. In fact the whole difficulty consists in drawing the line AE, yet in about 500 consecutive papers there were only 35 answers to the question.

In (a) scarcely anyone put the equation into the form

3√6+2√7

in its simplest surd form, and find the value of the expression as a decimal correct to four significant figures.

N.B.-√42 64807.

(c) Show that

√(13 + 3/13) + √ (13 − 3 √13) = √ (26 + 4√√13).

The surds were, on the whole, well handled.

Q. 29. Solve the following equations:

The equations (a) and (c) were solved fairly often; in (b) one root was sometimes found; both roots, seldom or never.

Q. 32. When are quantities said to be in continued proportion?

If five quantities a, b, c, d, e, are in continued proportion, show that the ratio of the first to the last is the fourth power of the ratio of any two consecutive quantities.

Show also that c is a mean proportional between a and e and that

ab+be+cd + de

is a mean proportional between

a2 + b2 + c2 + d2 and b2 + c2 + d2 + e2.

Was well answered in a few schools, but it was not often taken.

C.

In TRIGONOMETRY the work was on the whole fairly good.

Q. 33. Explain why the logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers.

By means of logarithms given below, find the fifth root, and the fifth power of 0'69889 correct to five decimal places.

The work requiring logarithms was often well done, and in particular, though there were many failures, x and y were calculated quite as often as could be fairly expected.

Q. 34. Draw an appropriate diagram, and from it find the numerical values of the sine, cosine, and tangent of an angle of 45o.

Find also the true logarithm and the tabular logarithm of sin 45° and of tan 45°.

If A be any angle between 0° and 90°, find tan A in terms of sin A, and also in terms of cos A.

Q. 35. Show, in a carefully drawn diagram, an angle 234°, and explain, with reference to your diagram, why both the sine and the cosine of that angle are negative.

Assuming that sin 36° 56′ is 06, and that cos 36° 56′ is 0'8, find the angle whose sine is 06 and whose cosine is + 08.

Find from the annexed table the numerical value of sin 326° 42'. These two questions came in for many good answers, though, of course, there were failures as to one or other of the points involved, and in particular, very few found that the numerical value of sin 326° 42′ is -054902.

Q. 37. Two points A and B are 2,000 yards apart on a straight road, and P is a flagstaff off the road; it is found that the angles PAB and PBA are 33° 18′ and 105° 20' respectively.

Calculate the distance BP, and the number of square yards in the triangle ABP.

Was often well answered, but the first part more often than the second. Q. 38. Show that the area of a quadrilateral is equal to the area of a triangle having two sides equal to the diagonals of the quadrilateral, and the included angle equal to either of the angles between the diagonals.

Find the area of the quadrilateral in which the diagonals are 216.5 ft. and 447'5 ft. long respectively, and are inclined to each other at an angle of 116° 30′.

Was answered very occasionally.

STAGE 3.

Results 1st Class, 154; 2nd Class, 461; Failed, 253; Total, 868.

The results are distinctly better than those of last year. The work on the whole was satisfactory, and in many of the schools the teaching has been good.

A.

Q. 41. If a straight line cut two sides of a triangle proportionally show that it is parallel to the third side.

Show how, through a given point, to draw a straight line such that the part of it intercepted by two given straight lines is divided in a given ratio at the point.

Draw a triangle ABC whose sides AB, BC and CA are respectively 2, 2 and 3 inches long; find a point P in AB such that AP is to PB as 4 to 7, and construct a straight line PQ to meet AC produced in Q so that PQ is bisected by BC.

Q. 42. State the conditions under which two rectilineal figures are said to be similar.

Show that these conditions are satisfied in the case of two triangles which have the angles of the one respectively equal to the angles of the other.

Show also that polygons which are equiangular to one another are not necessarily similar, but that polygons which are similar to the same polygon must be similar to each other.

Q. 44. A quadrilateral is inscribed in a circle; show that the rectangle under the diagonals is equal to the sum of the rectangles under the opposite sides.

A, B, C, D are four points in order on the circumference of a circle, and A and C are fixed, but B and D move in such a way that the sum of the rectangle under AB and CD and that under BC and DA is constant ; find the locus of a point which divides BD into parts having a given ratio to each other.

These three questions were those usually selected; the propositions were nearly always well written out, and there were a good many solutions of the riders.

Q. 43. If A, B, P are hree given lines; show how, by a geometrical construction, to draw a line Q, such that 43 may be to B3 as P to Q.

Q. 45. AB is a chord of a circle, and tangents at A and B intersect at P; through P a straight line is drawn to cut the circumference at Q, AB at R, and the circumference again at S. If N is the middle point of QS, show that QN is a mean proportional to PN and NR.

Q. 46. If through a fixed point O a straight line be drawn to meet a given straight line in P, and on OP a point Q be taken so that the rectangle OP.OQ is constant, show that the locus of Q is a circle.

In a Peaucellier's cell, OA, OC are two equal rods, and AB, BC, CD, DA four other equal rods, the six rods being hinged together at the points O, A, B, C, D; show that in all positions of the rods, the rectangle OD. OB is constant. If O be fixed and B move in a straight line, what is the path of D?

Q. 43 and 45 were well answered several times, and a few of the best candidates answered Q. 46 correctly.

Q. 48. Resolve each of the following expressions into four factors:— (a) }{ (a + b + c)3 — a3-b5c5}

(b) a1 + b1 + ct - 262c2 - 2c2a2 - 2a2b2. ·

(c) (a2 + 2bc)3 + (b2 + 2ca)3 + (c2 + 2ab)3
- 3(a2 + 2bc) (b2 + 2ca) (c2 + 2ab).

A fair number of candidates used the remainder theorem successfully with (a); many made out (b); and a few succeeded with (c).

(c) x2-2xy + 3y2 = 3(x − y), 2x2 + xy-y2=9(xy).

--

The equations were fairly well solved, but in (b) the root x=2a+l, and

in (c) the solution x=0, y=0, were often left unnoticed.

Q. 50. If x1, x2 are the roots of the equation

(xb) (x−c) + (x −c) (x − a) + (x − a) (x —b) = 0

find the value of

--

(x − a) (x2-a) + (x − b) (x − b) + (x¡ — c) (x2 —c).