trade? 6x= the number of years. = 7, A con 191. A person wishing to ascertain the area of a certain quadrilateral field, found that he could determine it the most readily by dividing it into two portions, one of which was of the form of a rectangular parallelogram, the shorter side of which measured 60 yards. The other was of the form of a right-angled triangle, whose shortest side was equal to the shorter side of the parellelogram, and the other side, containing the right angle, was equal to the diagonal of the parallelogram; and the area of the triangle was to the area of the parallelogram as 5 to 8. What was the area of the field? Ans. 7800 square yards. = Let the longer side of the parallelogram, .. x2: ✔√(x2+3600) = diagonal, and 60x the area of the parallelogram, and 30 (+3600) the area of the triangle,.. 60x 30 (x2+ 3600)::8 5, or x: (x2+3600)::45, .. x2: 3600 :: 16: 9,.. x 60::4: 3, and x: 20::4: 1, .. = 80, and the area 60X80+30/(6400+3600)=7800. 192. A Merchant laid out a certain sum upon a speculation, and found at the end of a year that he had gained $69. This he added to his stock, and at the end of another year found that he had gained exactly as much per cent. as in the year preceding. Proceeding in the same manner, and each year adding to his stock the gain of the year preceding, he found at the beginning of the fifth year that his stock was to the original stock as 81 to 16. What was the sum he first laid out? Ans. $138. Let x= = the sum, .. x+69 the stock at the beginning of the second year, and x: 69::x+69 the gain the 2d year = 69(x+69) 69 (x+69) +=(x + 69) x = the gain the third year. Hence (x+69)1 ̧(x+69)1 the stock at the beginning of the fifth year = x :: 81: 16, and x + 69 : x::3 : 2, .'. 69 : x::1 : 2, and x = 138. 193. There is a number consisting of two digits, which being multiplied by the digit on the left hand, the product is' 46; but if the sum of the digits be multiplied by the same digit, the product is only 10. Required the number. Ans. 23. Let 10x+y be the number, (10x+y=46)—(x2+xy=10)=9x2 -36, and x-§-2, and 2y-6 and y-3, the number is 23. 194. From two towns, C and D, which were at the distance of 396 miles, two persons, A and B, set out at the same time, and meet each other, after travelling as many days as are equal to the difference of the number of miles they travelled per day; when it appears that A has travelled 216 miles. How many miles did each travel per day? Ans. A went 36, and B 30. Let x and y be the number A and B each went, then by the question (x2-xy-216)(xy-y=180) = x2-2xy+y=36, or x2—2xy+y2—36, x-y-6, and x=36, and y=30. 195. There are two numbers, whose sum is to the greater as 40 is to the less, and whose sum is to the less as 90 is to the greater. What are the numbers? Ans. 36, and 24. Let x and y be the numbers, .". x+y: 40::x: y, and y x+y::x : 90, .. ex. æquali y: 40:.x2:: 90y, and 4x2= 40x 5x 9y,.. 2x-3y, and x+3x= or 60, and x-36, y=24. 3x 3 196. It is required to find 2 numbers such, that the product of the greater and the cube of the less may be to the product of the less and the cube of the greater as 4 to 9; and the sum of the cubes of the numbers may be 35. Ans. 3, and 2. Let x and y the numbers,.. xy: yx:: 49, or y2: :: 4 9, and y=x. Also x+y=35, or x+x=35, ... 35x3= 27×35, and x3-27, or x=3, and y=2. 197. The paving of two square court-yards cost £255; a yard of each costing one-fourth of as many shillings as there were yards in a side of the other. And a side of the greater and less together measure 41 yards. Required the length of a side of each. Ans. 25, and 16 yards. Let x and y = the lengths, .. y+xy2 = 205 × 20 × 4, and x+y=41, ... 41xy=16400, and xy-400. Hence (x-y=9)+ (x+y=41)=2x=50, or x=25, and 2y=32, and y=16. 198. A person bought a number of apples and pears, amounting together to 80. Now the apples cost twice as much as the pears; but had he bought as many apples as he did pears, and as many pears as he did apples, his apples would have cost 10d., and his pears 3s. 9d. How many did he buy of each? Let x and y be the number of apples and pears, then I have 45y 10x x y=-20, and x=60. and x2—9y3, ... x=3y, whence (3y+y)=80, and Ans. 60 apples, and 20 pears. 199. A person exchanged a quantity of brandy for a quantity of rum and £11.5s. ; the brandy and rum being each valued at as many shillings per gallon as there were gallons of that liquor. Now had the rum been worth as many shillings per gallon as the brandy was, the whole value of the rum and brandy would have been £56 5s. How many gallons were there of each? Let x and y be the number of gallons of brandy and rum, then and y will equal the prices of the brandy and rum, .. x2-y3 =225, and 2+xy=1125, hence x-25, and y=20. 200. There are two rectangular vats, the greater of which contains 20 solid feet more than the other. Their capacities are in the ratio of 4 to 5; and their bases are squares, a side of each of which is equal to the depth of the other. What are the depths ? -- Let x, and y denote the depths, .. xy — xy=20, and x'y : xy: 5 4, or x: y :: 5 : 4, .. y=x. Hence 23 -18x8 = 20, or 2-20, .. x=5, and y=4. 201. Bought two square carpets for £62 1s., for each of which I paid as many shillings per yard as there were yards in its side. Now had each of them cost as many shillings per yard as there were yards in a side of the other, I should have paid 17s. less. What was the size of each? = Let x and y be the lengths of the sides, then z3+y3=1241, and xy+xy 1224; adding 3 times the second equation to the first, x2+3x2y + 3xy+y3 4913, and .. x+y= 17, and xy= 1224 = =72, whence x2-2xy+y2=1, and x-y=±1, but x+y= x+y 17, and x-9 or 8, and y=8 or 9. 202. The number of men in both fronts of two columns of troops, A and B, when each consisted of as many ranks as it had men in front, was 84; but when the columns changed ground, and A was drawn up with the front B had, and B with the front A had, the number of ranks in both columns was 91. Required the number of men in each column. Ans. 2304, and 1296. x2, y2 Let 2 and y2 = the numbers, x+y=84,+91; whence y x {x3y3)=91xy, but x3+y3+3xy(x+y)=843,} .. 91xy+3(84xy) -843, or 343xy-843, and xy = 12-1728, and since x+y=84, ..x—y=±12, and x=48 or 36, y=36 or 48, Ans. 203. A field in the form of a rectangular parallelogram was planted with trees placed at such distances as to have four on every square yard. The expense of planting was such, that every 40 trees cost one-third of as many shillings as there were yards in the diagonal of the parallelogram. But had they been planted at such a price as that every hundred should have cost as many shillings as there were yards in the shorter side of the parallelogram, the would have been less by £224. Now a square described upon the diagonal of the parallelogram would be equal expense to of the square described on the less side, together with the square described on a line which is equal to the difference of the sides. Required the dimensions of the parallelogram. Let x, y, and 4xy denote the less and greater sides, and the number of trees, As 40: (x2 + y2):: 4xy: the price of xyNo (x2 + y2) x2y planting xy√(x2 + y2) 30 30 25 =4480, and 5xy/(x2+y2)—6x'y-150×4480, but x2+y2 = §x2+(y—x)2=—x2 y2+§x2—2xy, .'. 2xy—§x2, and y=4x... 20x2 × §x — 6x2 × ·ƒx=150×4480, ... 28x3=9×150×4480, and 23=9×150×160 216000, .. x-60 and y=80, Ans. Questions for practice without their solutions. App. p. 576 3x-3 204. Given 4 3x-4 27+4x to find x, x=9 |