GROUP I.—PURE AND APPLIED MATHEMATICS. BOARD OF EXAMINERS. V.-Pure Mathematics Rev. J. F. Twisden, M.A., Chairman. P. T. Wrigley, M.A. H. B. Goodwin, M.A., late R.N. Report on the Examinations in Pure Mathematics. EVENING EXAMINATIONS. STAGE 1. Results : 1st Class, 632 ; 2nd Class, 1,138 ; Failed, 1,423 ; Total, 3,193. A. The questions in ARITHMETIC were, perhaps, a little harder than usual. Those most frequently attempted were Questions 1 and 3. The latter contains two questions on the metrical system of measures, which were often correctly answered. Q. 2. Find by contracted multiplication the product of 32'5467 and 2.4918 so as to obtain the product true to four decimal places. 1 1 1 1 + 22 33 44 and find the square root of the expression true to four decima places. From this question it would appear that more attention is paid to contracted work than was formerly the case, so that several good answers were sent up to the first part of the question. Very few treated the second part properly. Q. 4. A rectangular box measures externally (when the lid is down) 4 ft. long, 2} ft. wide, 14 ft. high ; it is made of wood an inch thick. Taking account of the wood only, find the ratio of the weight of the lid to the weight of the whole box. By how much does the ratio exceed 1 ? Is an easy question in the mensuration of rectangles ; it requires care and a correct appreciation of the data ; these were often wanting, and so there were many failures, but the right result (81/308) was obtained fairly often. Q. 6. An alloy of silver is mixed with an alloy of gold in the ratio of 57 to 13; the percentage of lead in the silver alloy is 13.75 and that in the gold alloy 16-25 ; what is the percentage of lead in the mixture ? Was answered correctly in a good many cases; but only by those who otherwise did well. On the whole the work in Arithmetic was well done. 9291. 4,001-Wt. 7762. 12/03. Wy. & S. 4971r. A B. The book work in GEOMETRY was, on the whole, fairly well done. The deductions and problems came in for a good deal of attention and were done well by one and another. Q. 7. BC is the base of an isosceles triangle ABC, and D is the middle point of BC; show that the straight line AD is perpendicular to BC. Define a triangle, an isosceles triangle, and a right angle. Was, of course, very often taken. The work, however, was marked by some very prevalent faults, viz. : (a) It was very often said that in the triangles ABD and ACD, the sides AB, AC are equal, AD common, and the angle ABD equal to ACD, therefore the triangles are equal in all respects. (6) The definitions were often faulty, e.g, a triangle was defined as à figure having three sides ; an isosceles triangle as a triangle having two equal sides and two equal angles ; a right angle as an angle of 90°. It hardly admits of a doubt that the prevalence of such faults as (a) and (6) proves that the teaching in many schools was not good. Q. 8. If two straight lines cut one another, show that the vertically opposite angles are equal. A and B being two points on the same side of a straight line PQ, from A a perpendicular AC is drawn to P Q, and is produced to D so that CD is equal to AC ; show that, if DB cuts PQ in E, the straight lines AE, BE make equal angles with P Q. The second part of the question is given in most text books ; both the first and second parts were very often answered. Q. 9. Two triangles have equal bases, and the angles at the base of one triangle are equal to the angles at the base of the other triangle, each to each ; show that the triangles are equal in all respects. ABCD is a parallelogram, and in DC or DC produced a point E is taken such that B E is equal to BC; also CA, EA, and BD are joined ; show that the angle CAE equals the difference between the angles D CA and DBA. In reference to the first part of this question, it is worth noticing that the proof by superposition (in itself, of course, a good proof) did not seem to be as well understood as the proof by a reductio ad absurdum. The second part is, perhaps, harder than it looks ; it was very occasionally wered, e.g., there were four good answers in about 400 consecutive papers. Q. 10. Define parallelograms about a diagonal of a parallelogram, and the complements of those parallelograms. Show that the complements of the parallelograms about a diagonal of any parallelogram are equal to one another. In a given parallelogram construct complements each three sixteenths of the area of the parallelogram. Q. 11. Define a square. E, F, G, II are the middle points of the sides of a square ABCD; show that EFGH is a square. Hence, in the particular case when a right-angled triangle has two equal sides, show that the square on the hypotenuse equals the sum of the squares on the other two sides. Q. 12. ADE is a rhombus, having the angle at A a little greater than a right angle, and B and C are points in DE and EF respectively such that ABC is an equilateral triangle ; if ADEF be such that a side of it equals a side of the triangle ABC, show that the angle DEF is ten-ninths of a right angle. These three questions elicited a few good answers, particularly the first parts of Q’s. 10 and 11. The force of the word “Hence" in Q. 11 was not felt by more than a very few, but most of those few found the required answer easily. well answe C. In ALGEBRA the questions most frequently attempted were Q's. 13, 15, 16, 17, and on the whole they were well done. Q. 13. (a) From 5.13 6x2 + 7 - 8 8x2 + 6x + 4. 1 4' 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. Q. 14. (a) Reduce 7x + 12) (r 1) - (x2 4x + 3) to its simplest form. Also show that it is the product of three factors. (6) If we suppose that a 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 : (i) x2 + 9x 52. (ii) ar 40°r. (iii) (a” + b2 – 6%)2 - 4a6. (6) 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 (6) it was curious to notice in how many cases (x+1)% was treated as if it were less than r". 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 561. 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.y. Q’s. 21, 22, 25, 27, 28, 31, not to mention others. a A. segments of two different circles :- A and B ; a circle is drawn whose circumference passes line drawn through C at right angles to AB. (6) 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 Å BCD, is twice the same triangle ; and therefore the rectangle is equal to the parallelogram. In fact the whole difficulty consists in drawing the line A E, yet in about 500 consecutive papers there were only 35 answers to the question. B b b + c show that either ir = b, or a + b + c = 0. as – b) + ab(a - b) = m(a” – 1%), b. (a-b) (a+b+c)=0, and hence deduced the alternative conclusion, either a=b or a+b+c=0. So also in (6). Q. 28. (a) Extract the square root of 15 – 4914. (() Express the quotient of 10 J6 – 2.17 divided by 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.- 112=6*4807. (1 C ta' ( |