ADVERTISEMENT. In the following pages are given, with complete solutions, the Mathematical problems set at the competitive examination for admission to the Royal Military College at Sandhurst, and for First Appointments in the Royal Marine Light Infantry, held under the direction of the Civil Service Commissioners in June and July 1883. 12, CARDIGAN ROAD, RICHMOND, AND 42, THE COMMON, WOOLWICH, August 1883. MILITARY ENTRANCE EXAMINATION. ROYAL MILITARY COLLEGE AND ROYAL MARINE LIGHT INFANTRY. MATHEMATICS (1). Saturday, 30th June, 1883. 10 A.M. to 1 P.M. (Great importance will be attached to accuracy in results.) 1.-Reduce to its simplest form Sol. This expression is equivalent to x2 + 2y2 x + y x2 + 2y2 x2 + 2y2 x2 + 2y2 xy ÷ = = 1. Ans. 2.-(1) Divide (ax+by)3 + (ax — by)3 + (bx — ay)3 + (bx + ay)3 (2) Find the Greatest Common Measure of 10x3 54x2+87x 45 and 5x4 (ax + by)3 + (ax — by)3 + (bx — ay)3 + (bx + ay)3 = 2 (a3 + b3) x3 + 6 (a + b) ab xy2 = 2(a + b) x {(a2 — ab + b2) x2 + 3ab y2}. Also (a+b)2x2 - 3ab (x2 — y2) = (a2 — ab + b2) x2 + 3ab y2. Hence the required quotient is 2 (a + b) x. Ans. B 2 36x387x2 6x3 29x2+45x-36 (y) 10.2 71x+123 (8) - 41x + 123 Every measure of two quantities is a measure both of the sum and difference of any multiples of the quantities. The Greatest Common Measure is not altered if either of the quantities be multiplied or divided by a factor which is not a measure of the other. Subtracting 28 from xa and dividing the result by 3 gives y Subtracting 10x (x-3) from 8 gives Dividing 41x123 by - 41 gives x 3. Hence x 3 is the Greatest Common Measure of (a) and (B). 3.-(1) Prove that a2 (a2 + ao)* + (aa — x2o) — {a^ − 1}' = 2. (a2 + x2)‡ − (a2 − x2)* (2) Find the square root of Sols. (1) Multiplying both numerator and denominator by the numerator, (a2 + x2)++ (a2x2) 2a2 + 2 (a* - 21)+ (a2 + a2)+ + (a2 − ∞2)* — { aa — 1}= a. Q.E.D. (a2 + x2)‡ — (a2 — x2)‡ (2) 2ab axbx -2 √ ab (ab x2 - axbx+x2) is equivalent to a (b − x) + b (a − x) − 2 √ a (b − x) √ b (a — x). Hence the required square root is √ a (b - x) - √b (a — x). Ans. 4.-(1) Prove that the cube of a number which has n digits cannot have less than 3 n 2 nor more than 3 n digits. (2) Find the cube root of 40 353607; (3) and of 1 3x to four terms in ascending powers of x. Sols. (1) 10" is represented by 1 and n zeros, and is therefore the least number containing n + 1 digits. 10-1, in like manner, is the least number containing n digits. Hence a number which has n digits must be less than 10", and not less than 10"-1. Therefore the cube of such a number cannot be less than 103-3, which has 3n 2 digits, and must be less than 103", the least number with 3n+ 1 digits, and therefore cannot have more than 3 n digits. 40.353607 9 27 18 13353 6 2700 12304 3 376 1049607 90 3076 1049607 4 392 94 346800 4 3069 98 349869 4 1020 3 3.43. Ans. | 1 x - x2 - 4x3, &c. Ans. · 3x2 + 3x3 + 2x1 3x2+6x3 + 3x1 2nd trial divr. 3 2nd divr. 3 3rd trial divr. 3 ენ 5x3 + 3x + 206 1 = a, (2) when x = a + √ a2 − 1 ; (1) when x + also 2-2 3+ x + 2→3 3 ̄x3 1 Sols. When x + = α, 1 10 (27x) when x = 64. x2+ 1⁄2 ï = (x + 1)3 −2=a2−2, x2+1-3 = (x+1)3 −3(x + 1) = a3 — 3a, 1 Hence, substituting 2a for a in the above expressions, 1 x2 + 1⁄2 = 4a3−2, x3+ - = 8a3— 6a, and a* + 1 = 16a1 16a2+2. 64, 2-23 x + 2-3 3-x3- 10 (27) x§ 2−3 x} 1 = = 1 √ 3 × 32 + 1 × √3 10 × 16 1 2 = 8√3+ 2√3 √3 √3 44, xy+xz = 36, x + y + z = 15. x2− 4 x + 4 + 3 x 1 2 = 0 х 3 (3) Sols. (1) |