digging for treasures, Un- less with the spade and the pends on his neigh-bour Will never work to be truth in the hoe. The bee has to work for the honey; The true, and to older, And man true, Keep have friends that will last: And do not put off till to - morrow The hat ing Your work, as to do it out. right. Know this, too, before you are saying," There al ways 15 room at the top." To con-science be f : Symph. sff mr drs, :m d:- sls fs:m:dr:m:rd: ms, s, s, ta ta ta 1, 1, .s, s, :f, :m, f,:f, :s, m:- ad: :- d:- -:m, f,:s, :f, m: mf:m:r m: :- S1: S d: The Practical Teacher. A MONTHLY EDUCATIONAL JOURNAL. To Subscribers.-The Practical Teacher is published on the 25th of every month. Price 6d. ; post free, 7d.; sent post free, three months for Is. iid.; six months, 3s. 9d.; a year, 7s. 6d. P. O. Orders should be made payable at Chief Office. Subscribers not receiving their copies regularly are respectfully requested to write to the Publisher. It would save time and expense if subscribers when remitting would state exactly what numbers they wish to be sent. To Correspondents.-All literary communications should be addressed, THE EDITOR OF THE PRAC TICAL TEACHER, Pilgrim Street, Ludgate Hill, London, E.C. Accepted contributions are paid for within twentyone days of publication. The Editor cannot return rejected MSS.; authors should therefore retain copies. Portfolios, fitted with elastic bands, for preserving twelve numbers of The Practical Teacher,' may now be had. Price 2s. 6d. Orders should be given through booksellers. 6. Write out clearly and concisely the rules for (a) Finding the G.C.M. of two numbers; (6) Finding mentally the product of 1616 by 625; (a) Divide the greater by the less; if there be a remainder, divide the first divisor by it; if there be still a remainder, the second divisor by it, and so on, always dividing the last preceding divisor by the last remainder till nothing remains. The last divisor is the G.C.M. (c) Reduce to a common denominator: subtract the lesser from the greater numerator, for a numerator to the common denominator. 7. If the larger wheel of a bicycle, whose circumference is 8 yds. o ft. 5 in., make 200 more revolutions than that of another bicycle in travelling 5 miles, find the circumference of the latter wheel. No. of revolutions of Ist wheel-5 miles÷8 yds. o ft. 53 in. =950400÷880=1080. No. of revolutions made by 2nd wheel 1080 - 200=880. = 8800 yds. ÷ 880= 10 yards. Ans. 8. 320 men begin a piece of work; it is completed in 6 days of 10 hours each, but on each day only half of those employed on the previous day are at work; in what time would 105 men working 6 hours a day have completed it? Average No. of men working (320 +160+80+40 +20 +10) ÷6=105. 105 men would do the work in 6 days of 10 hours .. 105 men would do the work in 10 days of 6 hours each. Ans. 9. Find the present value of £1363 due 5 years hence at 3 per cent. per annum simple interest. £100 would produce £34×5= £16} Present value of £116=100. 10. A sum of £8505 invested in the Three per Cents. produces an income of £252; what is the price of the stock? No. of cents.252÷3=84. For £8400 stock, £8505 was paid .. Price of stock=9505=£101}. Ans. II. Extract the square root of 892143 of 12 square feet. •892143 121 11·162224(3'332 ft. Ans. 9 63)210 189 663) 2122 1989 (662) 13324 12324 12. 800 yards of cloth are bought at 10s. 6d. per yard; half is sold for 10s. per yard, a fifth for 11s.; at what price must the remainder be sold to obtain a gain of 5 per cent. on the whole ? Selling price of remainder (800-560) 240=£155 Price per yard = £155 25.÷240=12s. 11d. Ans. Euclid, Algebra, and Mensuration. Three hours allowed for this Pater. Candidates who attempt either of the questions in Mensuration must omit questions II and 12. (Marks are given for portions of questions.) Euclid. In the Euclid questions all generally understood abbreviations for words may be used, but no symbols of operations (such as -,, x) are admissible. N.B.-Capital letters, not numbers, must be used in the diagrams. 1. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them of the other. On the base of an isosceles triangle, an equilateral triangle is described show that the line joining the vertices of the two triangles bisects their common base at right angles. Euclid I. 8. On the base BC of the isosceles triangle ABC, describe the equilateral triangle BDC. Join AD. AD or AD produced shall bisect the base BC in E. AB=AC; and AD is common to the two triangles BAD and CAD. .. BA, AD CA, AD, each to each, and base BD=base CD. .. angle BAD=angle CAD (Prop. 8.) Again BA = AC, and AE is common to the two triangles BAE, CAE; .. BA, AE=CA, AE; and angle BAE= angle CAE. .. base BE=base EC, and angle AEB= angle AEC, and they are adjacent angles. ... each is a right angle (Def. 10.) Wherefore AD or AD produced bisects BC at right angles. 2. What is the axiom on which Euclid bases his reasonings on parallel lines? Is Proposition 17 of the first book the converse of that axiom? If so, is there any objection to the axiom? If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles. x2 x-2 4 + 4 4) ÷ Sq. on DC=sqs. on DB, BC, (I. 47.) f. sq. on DC=sqs. on AD, BC. But sq. on AD=sqs. on AE ED (I. 47.) and the sq. on DC=sqs. on DE, EC (I. 47.) .. sqs. on DE, EC=sqs. on DE, EA, and BC. Take away the common sq. DE. .. sq. on EC=sqs. on EA and BC. i.e, the sq. on EC is greater than the sq. on EA by the sq. on BC. Q. E. D. 4. By what proposition of the first book is it proved that the area of a triangle, whose altitude is (a) units long, and whose base is (b) units long, is ab? If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. Euclid I. 41. Euclid II. 5. 5. Why cannot we satisfactorily demonstrate propositions of the second book by algebraical processes? In every triangle the square on the side subtending an acute angle is less than the squares on the sides containing that angle by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle and the acute angle. Because of the existence of incommensurable magnitudes. When the sides of rectangles can be expressed exactly in units of some measure of length, then the first 10 propositions of the second book can he expressed by algebraic symbols, but not otherwise. Euclid II. 13. Algebra. 1 4 Mensuration. 13. The area of a chess-board which contains 64 equal squares, and an outer rim an inch wide, is 134'56 square inches; find the side of each square. Find also the width of the outer rim of another board of the same size in which the area of each square is 1.361 square inches. √134.56=116 inches, the side of the board. Side of inner square = 116-2= 9'6 inches. 9'6÷8=1'2 in. the side of each small square. Ans. Candidates are not permitted to answer more than one question in each section. The solution must in every instance be given at such length as to be intelligible to the examiner, otherwise the answer will be considered of no value. SECTION I. 1. The first of 4 parcels of money contained two hundred and six pounds and twopence; the second, fifty sovereigns, seventeen half-sovereigns, and nine half-pence; the third, twentyseven half-guineas and eightpence; the fourth, nineteen halfsovereigns and three half-crowns. Distribute the amount equally among 29 societies. 2 K |