QUEST. 7. To divide 20 into two such parts, that 3 times the one part added to 5 times the other may make 76. Let x and y denote the two parts. Then, by the question, x + y = 20, and 3x + 5y=76. From which x = 12, and y = = 8. QUEST. 8. A market woman bought in a certain number of eggs at 2 a penny, and as many more at 3 a penny, and sold them all out again at the rate of 5 for two-pence, and by so doing, contrary to expectation, found she lost three-pence; what number of eggs did she buy? Suppose she bought a eggs of each kind: then the cost of the first lot was x, and that of the second lot was. Also in selling 2x eggs at 5 for two-pence, she received. 2x pence: and by the question, this was three-pence less than she Hence + 3=and therefore x = 90, the number in gave for them. each lot, or 2x 2 3 4x 180, the whole number. QUEST. 9. Two persons, A and B, engage at play. Before they begin, A has 80 guineas, and B has 60: but after a certain number of games won and lost between them, A rises with three times as many guineas as B: how many guineas did A win of B? Denote by a the number of guineas won by A. Then they rise with 80 + x and 60- x respectively. But by the question 80 + x = 3(60 — x): hence x = 25, the guineas won by A. QUEST. 10. The sum of the three digits composing a certain number is 16; the sum of the left and middle digits is to the sum of the middle and right ones as 3 to 33; and if 198 be added to the number, the digits will be inverted in the expression of this sum. Let x, y, z denote the digits; then 100x + 10y + z will express the number itself, and 100% + 10y + z will express the number having the same digits in an inverted order. Whence the three conditions are x=2. x + y + z=16, x + y : y + z:: 3:33, and 100x10 y + z + 198 = 100z + 10y + x, or z— Whence the solution is a = 5, y = 4, and z = 7; and the number itself is 5. 1004. 10 + 7, or 547. 1 QUEST 11. If N men of a certain degree of skill can do a piece of work in n days, N, others of different skill in n, days, N1 others in n days, and so on for m sets of men in how many days would one of each set be able to do of the work, supposing they all worked together, without impeding each other's operations? Ρ Suppose that one of each of these men working together could execute the pth part of the work in a days: then they would execute the whole in pa days, or in one day. Hence equating the two expressions for their total work px in one day, we have It is recommended that the student should also solve these questions generally, by taking literal symbols instead of the given numbers. 1. DETERMINE two numbers such, that their difference may be 4, and the difference of their squares 64. Ans. 6 and 10. 2. Find two numbers with these conditions, viz. that half the first with a third part of the second may make 9, and that a fourth part of the first with a fifth part of the second may make 5. Ans. 8 and 15. 3. Divide the number 2 into two such parts, that a third of the one part added to a fifth of the other may make }. Ans. 1.5 and 5. 4. Find three numbers such, that the sum of the 1st and 2d shall be 7, the sum of the 1st and 3d 8, and the sum of the 2d and 3d 9. Ans. 3, 4, 5. 5. A father, dying, bequeathed his fortune, which was 28007, to his son and daughter, in this manner; that for every half-crown the son might have, the daughter was to have a shilling: what then were their two shares ? Ans. the son 20001, and the daughter 8007. 6. Three persons, A, B, C, make a joint contribution, which in the whole amounts to 4001: of which sum B contributes twice as much as A and 20/ more; and C as much as A and B together: what sum did each contribute? Ans. A 601, B 1407, and C 2007. 7. A person paid a bill of 1007 with half-guineas and crowns, using in all 202 pieces; how many pieces were there of each sort? Ans. 180 half-guineas and 22 crowns. 8. Says A to B, if you give me 10 guineas of your money, I shall then have twice as much as you will have left; but says B to A, give me 10 of your guineas, and then I shall have 3 times as many as you: how many had each ? Ans. A 22, B 26. 9. A person goes to a tavern with a certain quantity of money in his pocket, where he spends 2 shillings; he then borrows as much money as he had left, and going to another tavern, he there spends 2 shillings also; then borrowing again as much money as was left, he went to a third tavern, where likewise he spent 2 shillings; and thus repeating the same at a fourth tavern, he then had nothing remaining: what sum had he at first, and what was he in debt? Ans. at first 3s 9d, and had borrowed 4s 3d. 10. A man with his wife and child dine together at an inn. The landlord charged 1 shilling for the child; for the woman as much as for the child, and as much as for the man; and for the man as much as for the woman and child together: how much was that for each ? Ans. the woman 20d, and the man 32d. 11. A cask, which held 60 gallons, was filled with a mixture of brandy, wine, and cyder. in this manner, viz. the cyder was 6 gallons more than the brandy, and the wine was as much as the cyder and of the brandy: how much was there of each ? Ans. brandy 15, cyder 21, wine 24. 12. A general, disposing his army into a square form, finds that he has 284 men more than a perfect square; but increasing the side by 1 man, he then wants 25 men to be a complete square : how many men had he under his command? Ans. 24000. 13. What number is that, to which if 3, 5, and 8 be severally added, the three sums shall be in geometrical progression? Ans. 1. 14. The stock of three traders amounted to 7607: the shares of the first and second exceeded that of the third by 2407; and the sum of the second and third exceeded the first by 3607: what was the share of each ? Ans. the 1st 2007, the 2d 3007, and the 3d 2607. 15. What two numbers are those, which, being in the ratio of 3 to 4, their product is equal to 12 times their sum ? Ans. 21 and 28. 16. A certain company at a tavern, when they came to settle their reckoning, found that had there been 4 more in company, they might have paid a shilling each less than they did; but that if there had been 3 fewer in company, they must have paid a shilling each more than they did: what then was the number of persons in company, what did each pay, and what was the whole reckoning? Ans. 24 persons, each paid 7s, and the whole reckoning was 8 guineas. 17. A jockey has two horses; and also two saddles, the one valued at 187, the other at 31. Now when he sets the better saddle on the 1st horse, and the worse on the 2d, it makes the 1st horse worth double the 2d; but when he places the better saddle on the 2d horse, and the worse on the 1st, it makes the 2d horse worth three times the 1st: what were the values of the two horses? Ans. the 1st 67, and the 2nd 97. 18. What two numbers are as 2 to 3, to each of which if 6 be added, the sums will be as 4 to 5? Ans. 6 and 9. 19. What are those two numbers, of which the greater is to the less as their sum is to 20, and as their difference is to 10 ? Ans. 15 and 45. 20. What two numbers are those, whose difference, sum, and product, are to each other, as the three numbers 2, 3, 5? Ans. 2 and 10. 21. Find three numbers in arithmetical progression, of which the first is to the third as 5 to 9, and the sum of all three is 63. Ans. 15, 21, 27. 22. It is required to divide the number 24 into two such parts, that the quotient of the greater part divided by the less, may be to the quotient of the less part divided by the greater, as 4 to 1. Ans. 16 and 8. 23. A gentleman being asked the age of his two sons, answered, that if to the sum of their ages 18 be added, the result will be double the age of the elder; but if 6 be taken from the difference of their ages, the remainder will be equal to the age of the younger: what then were their ages? Ans. 30 and 12. 24. Find four numbers such, that the sum of the 1st, 2d, and 3d shall be 13; the sum of the 1st, 2d, and 4th, 15; the sum of the 1st, 3d, and 4th, 18; and lastly, the sum of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9. 25. Divide 48 into 4 such parts, that the first increased by 3, the second diminished by 3, the third multiplied by 3, and the fourth divided by 3, may be all equal to each other. Ans. 6, 12, 3, 27. 26. A cistern is to be filled with water from three different cocks: from the first it can be filled in 8 hours, from the second in 10, and from the third in 14: how soon would they all together fill it? Ans. in 3 h 22 min 24 sec. 27. Show at what periods the hands of a watch will be together during a complete revolution of the hour-hand. 28. A labourer engages to work for 3s 6d a day and his board, but to allow 9d for his board each day that he is unemployed. At the end of 24 days he has to receive 31 28 9d: how many days did he work? Ans. 19 days. 29. Three workmen are employed to dig a ditch of 191 yards in length. If A can dig 27 yards in 4 days, B 35 yards in 6 days, and C 40 yards in 12 days, in what time could they do it if they worked simultaneously? Ans. 12 days. QUADRATIC EQUATIONS. A QUADRATIC equation is that in which the unknown quantity is of the second degree, and is generally represented by ax2 + bx + c = 0, where a, b, c may any numbers positive or negative, integer, fractional or irrational. be When b = O, it takes the form ax2 + c = 0, and it is called a pure quadratic. It is treated as a simple equation, since in the solution no operation is required but the arithmetical one of extracting the square root of C a . When all the terms are present, the equation is called an adfected quadratic. There are two methods of solving such equations; one due to the Hindûs, the other to the early Italian algebraists. They are alike in principle, which is that of so modifying the first side as to render it a complete square; and by corresponding additions, subtractions, multiplications, or divisions, applied to the other side, to still retain the original truth of the equality. The operation is technically called completing the square. 1. The Italian or common method. Transpose c, and divide every term by a: then we have and the first side is a complete square. Extract the roots, and resolve the resulting simple equation. This gives successively This operation is often troublesome, on account of the reduction of the second b side of the complete square, in actual numbers. When, however, is an even a α number, positive or negative, and also integer, it is the most convenient : but as this is seldom the case in the quadratics that arise in practice, we shall give a preferable one, viz. : 2. The Hindú method. Let the equation be ax2 + bx — — c. Multiply by 4a, and add b2 to the product on each side. Then we have 4a2x2+4abx+ b2 = b2 — 4ac, and extracting 2ax +b=±√√/b2 — 4ac, or x = --b± √b2 4ac 2a which is precisely the same result as before obtained by the Italian method, though by a process which is arithmetically simpler *. It has already been explained, that the square root of any quantity a2 is either + a or- - a, and marked by writing the sign thus, a, signifying that the root has both values + a and a. Hence the answers above given are to be under stood as twofold in each case, viz.: x = -b+b2-4ac -b-b2-4ac 2a and " 2a either of which substituted for x in the given equation, will render all the terms on one side equivalent to those on the other, taken collectively. Also, since the square root of a negative quantity cannot be actually extracted in real numbers, positive or negative, when b2 - 4ac is negative, the equation does not admit of resolution in numbers, or it is only symbolical. The roots, or values of x, are said in this case to be imaginary; and wherever such a result appears, it is the indication of contradictory conditions involved in the conditions of the problem which gave rise to the equation. All equations, in which there are two terms including the unknown quantity, and which have the index of the one just double that of the other, are resolved like quadratics, by completing the square, as above. Thus, x2 + ax2 = b, or x2" + ax” = b, or x + ax3 =b, or (x2± ax)2±m (x2+ax) b, are analogous to quadratics, and the value of the unknown quantity may be determined accordingly. may = It also on some occasions be useful to remark, that any quadratic equation in which four times the product of the co-efficient of the first term by the third term is equal to the square of the coefficient of the second term, is already a complete square. For let ax2 + bx + c = 0 be the equation: then if 4ac = -b2, The same of course is true if ax2 + bx + c = k, where k designates any quantity or expression whatever. EXAMPLES. 1. Given 2+4x=60, to find the values of x. The Italian method applies to this example, and x2+4x+4= 64, or extracting, x+2= +8. 2. Resolve the equation of 3x The Hindú method is applicable in this case. Then 36x2 60x+25=144 + 25 = 169, and 6x 5: ±13: hence ∞ = 3, and x = 3 *When the co-efficient of the second term is an even number, it will be sufficient to multiply all the terms by the co-efficient of the first, and add the square of half that of the second to both the products. For let a,x2+26,xe, then the completed equation is a 2x2 + 2a,b,x + b,2 = b2 + a,c,, the first side of which is the square of ax + b,. |