Whence or .. (x2) (x-3) 3 (x − 1) (x − 3) + 2(x − 1) (x − 2) = 0; Hence - · x25x6+3x2+12x-9+2x2-6x+4=0. (2) From Ans. 1. 0, and .. x = 5х x + 1 = x + y x + y = we have 6 xy = = 6. (a) we have 5 x + y + 1 (x + y)2 + (x + y) = 30; :: (x + y)2 + (x + y) + † = 30 + 1 = 11 .. x 3 or 2 or y = 2 or 3 or 5 or 6 121 4 - 6. (b) 3 ± 3 Ans. 36 (3) From xy + xz = 36, we have y + z = (a) 36 Substituting this value in the 3rd equation we have x + = 15; whence 2-15x+36= 0, or (x − 3) (x - 12) = 0. ..∞ = 3 or 12; and, from (a), y + z = 12 or 3. Also from x2 + yz = 44, we have yz = 35, or y2+2yz + z2 = 144, or 9. 4yz 140, or 400. = .*. y2 — 2yz + 22 = y 4, or 409; z = ± 2, or ± √409 y + z = y = 7 or 5, or z = 5,, 7,,, 2 12 or 3 (3 ± √409) } Ans. 7.-A walks m miles in n hours; B walks 6 n miles in the difference of their rates of walking is mile per hour. rate at which each walks. m hours; Find the 8.-A bar of metal, of a certain weight, made of gold and silver mixed, is worth £111 38.; if it were all gold, it would be worth £162; and if the proportions of the metals were reversed, it would be worth £60 68. An ounce of gold is worth 278. more than a pound (troy) of silver. Find the prices of gold and silver per ounce, and the quantity of each metal in the bar. Sol. Let £x price of gold per oz., = Then from the given conditions we have the following equa tions: 12xy + (x-27) z = 111%. (1) 12x (y + z) = 162. (2) 12xz+(x-27) y = 60%. (3) (11x+27) (y + z) = 1521. 27 (6) 2x and substituting this value in (6) we have and x = £2 = £4 108. = price of gold per oz.; 63 = 58. 3d. = price of silver per oz. and (x-2)= £% and(x Substituting for x in (4) and (5), we obtain z = 1 lb. = y = 2 lb. = weight of silver in the bar, 9.-Prove that, if a, ẞ are the roots of the equation Sol. ax2 + (a + b) x + b = 0, then + b a + B = − (1 + 2), and as = 2 - aß a b b ...a+2aẞ+ẞ2=1+2+ 10.-Prove that the logarithm of a quotient is equal to the logarithm of the dividend diminished by the logarithm of the divisor. Given log, 269314718, log, 3 = 1.09861229; find the logarithms to base e of 8 3e2 (1) ; (2) (3) Sols. Let log. p = x, then p = = = =(log, 2-log, 3)+ log, 3 log. 2+ log, 2-log, 3 (+3) log, 2 − (1 − 3 + 2) log, 3 log, 2 log, 3 = .46209812 • 64089050 = 1.82320762. Ans. 11.-Find the logarithm of 3.375 to the base 2 25. What is the characteristic of log, 5463? Given log10 4 ·60206 and 8* 1252-2 = 2**+3 5*, find x to three places of decimals. Sols. Therefore 3.375 (1.5), and 2.25 = (1.5). = 3.375 (2.25), = and, therefore, log. 3.375 to base 2.25 Again, Hence 5463 is greater than 7a, and less than 75; and, therefore, log, 5463 is greater than 4 and less than 5; so that its characteristic or integral part is 4. Ans. -x Lastly, if 8 125*-* = 2**+*5*, Therefore (39) log 2 = 4x-6. (4-3 log 2) x = 6-9 log 2, Hence and .*. x = 6-9 log 2 = 12 9 log 4 3.29073 = 8 - 6 log 2 3.09691 = 1.062. Ans. 12.-A rectangular building, 256 feet long and 48 feet wide, has a roof running lengthwise, of which the height of the centre ridge above the top of the walls is 18 feet, and the slope of each end the same as at the sides; find the area of the roof, supposing there are no eaves. If a flat roof were substituted for the above, to what additional height must the walls be raised in order to have the same volume of air in the building? Sol. Let the accompanying figure represent the plan of the roof; then the portion A EH is equal to the portion A EG, and consequently the area of the roof is exactly the same as if the sloping end were removed, and the ridge ran the whole length, that is, it is equal to twice the parallelogram A B KH. Now, the height of the ridge above the walls being 18 feet, and half the breadth of the building being 24 feet, the breadth of the slope will be 182 + 242 or 30 feet. Therefore the required area is 2 × 30 × 256 = 15360 square feet. Ans. A B K H D Secondly. The volume of the space inclosed by the roof consists of two halves of a square pyramid 18 feet high, and having 48 feet in a side of the base, and a triangular prism 208 feet long, the triangular ends having a base of 48 feet, and a height of 18 feet; and if the walls were raised x feet, and a flat roof substituted, the volume included by this roof and the raised portion of the walls would be a parallelopiped 256 feet long, 48 feet wide, and x feet high. If this space is equal to that inclosed by the sloping roof, we have 256 x 48 x x = 6 × 482 + 9 x 48 x 208; |