Let x= number of pounds A had at first. .. 40- -x= number of pounds B had at first. This must be so because they had £40 between them, and if A's money is subtracted the remainder must be B's. But B gives A £10. Hence A will have £10 more and B £10 less than at first. ..x+10= A's money after receiving £10, and 40-x-10=30-x-B's money after giving £10. But by the question A has now more money than B by £6. And 40-x-40-13=£27= B's money at first. (2) A bill of £1 10s. 6d. was paid in half-crowns and florins, and the total number of coins was 13. How many were there of each? .. 13-x= number of florins. Now the value of the half-crowns and florins must together = the amount of the bill, £1 10s. 6d. 1 5x 2 But the value of the half-crowns = x × 2== shillings. And the value of the florins (13-x)×2=26-2x shillings. 5x 2 1 = +26—2x=30—s. 5x+52-4x=61 30= = 61-52 X= =9= No. of half-crowns. 13-x=4= No. of florins. From these examples it will be seen that to work a problem by Algebra, we 1st. Put x for the unknown number or number sought. 2nd. Express in symbols the relations which hold among the quantities mentioned in the question, whether known or unknown; which will give rise to an equation. 3rd. Solve the equation thus formed according to the usual rules. The second process here named is the most difficult, and nothing but practice in working a large number of examples will enable the pupil to do it with certainty. When once mastered however, the use and power of Algebra will be to a great extent realised, and no pains ought to be spared to attain this end. EXERCISE XXIII. 1. What is that number whose double exceeds its half by 18? 2. Find a number such that the sum of its fifth and its fourth shall be less than the sum of its third and its half by 23. 3. Find a number which shall exceed the difference of its half and its third by 4. 4. Required the number whose half diminished by shall equal the sum of its third and fourth diminished by. 5. Find a number such that the difference of its seventh and its ninth shall exceed the difference of its ninth and its tenth by 13. 6. Find two numbers whose difference shall be 6, and their sum 16. 7. What number is that which when trebled is as much greater than a hundred as the number itself is below a hundred? 8. Divide 100 into two parts such that a third of the one part may exceed four times the other by 3. 9. Divide 80 into two parts such that 4 times the less may exceed three times the greater by 2. 10. Divide 60 into three parts such that the first may exceed the second by 10, and the third by 20. 11. Divide the number 500 into three parts such that the second may exceed one half of the first by 30, and one fourth of the third by 70. 12. A father is four times as old as his son; but 3 years ago he was seven times as old as his son then was. What is the age of each? 13. The ages of father and son amount to 60 years, and 10 years ago the father's age was exactly 7 times the son's. Required the age of each. 14. There are four brothers each of whom is 3 years younger than his next eldest brother, and the eldest brother is four times as old as the youngest. What is the age of each? 15. Divide £100 between A and B, so that A may have 3 times as much as B. 16. Divide £520 among A, B and C; so that A may have twice as much as B, and C half as much again as A and B together. 17. Two trains start at the same time from the same place and travel in opposite directions; the former travels at the rate of 30 miles per hour and the latter at the rate of 40. long must they travel before they are 350 miles apart? How 18. If the trains in the last example set out at the same time, the former from Edinburgh and the latter from London, when will they meet? The distance from London to Edinburgh is 400 miles. 19. Suppose the slower train in Ex. 17 to start at a certain time and the quicker to follow it after 24 hours, at what distance from the starting point would the latter overtake the former? 20. At what time are the hands of a watch together between 3 and 4 o'clock? 21. A bill of £1 6s. 6d. was paid in florins and half-crowns. The total number of coins was 12. Required the number of each sort. 22. One half of a certain number of poor people received half-a-crown each, one third received a shilling each, and the rest sixpence each; the whole sum distributed being £2. What was the number of persons relieved? 23. How much wine at 12 shillings a gallon must be mixed with 30 gallons at 15 shillings, that the mixture may be worth 14s. a gallon? 24. A grocer has tea at 3s. 6d. a lb. and also tea worth 58. a lb. From these he wishes to make a mixture of 50 lbs. worth 4s. 6d. a lb. How much must he take of each sort? 25. A person put £500 out to interest, part of it at 5 per cent. and the remainder at 3 per cent. The total interest received being £18, how much of it was put out at 3 per cent? 26. A post is of its length in the mud, of its length in the water, and 22 feet above the water. Required its length. 27. After cutting off of a rod from one end and 6 inches more than at the other, there remains 15 inches. Required the length of the rod. 28. A cistern which holds 2080 gallons is filled by 3 pipes in 20 minutes. The first conveys 12 gallons more and the second 8 gallons less, than the third, per minute. How much water does each pipe supply? 29. A and B together reap a field in 8 days, which A alone could reap in 10 days. How long would it take B alone to reap it? 30. If A could reap a field of 10 acres in 5 days of 12 hours each, and B could reap the same field in 8 days of 10 hours each; in how many days of 12 hours each could A and B together reap it? GREATEST COMMON MEASURE. (The proof of the following rules is not suited to an elementary work of this kind. The examples however afford excellent practice in the previous rules, and a selection is therefore appended.) To find the greatest common measure of two simple algebraical quantities: RULE. Write down the letters common to both quantities, assigning to each letter the lowest index which it has in either of the terms, and prefix the g.c.m. of their numerical coefficients. To find the G.C.M. of two compound expressions : (Generally the expressions may be resolved into their factors, and the highest factor common to both then becomes evident by inspection. When this cannot be done the G.c.M. is found in the following way.) Let a and b stand for the two expressions arranged according to the descending powers of some letter common to both, and let the highest power of this letter in a be not greater than the highest power of the letter in b. Divide b by a and obtain the remainder. Next make the remainder thus obtained a new divisor and the previous divisor a new dividend, and repeat the process until there is no remainder. The last divisor will be the greatest common measure. In order to avoid the introduction of fractional quotients into the above process: When the first term of the dividend at any stage of the working is not exactly divisible by the first term of the divisor, 1. Any factor may be struck out from all the terms of the divisor which is not also common to all the terms of the dividend. 2. All the terms of the dividend may be multiplied by any factor which is not contained in the divisor. (9) 10(a) and 12(a1-a2x2). (13) (+1) and (x+x+x+1)x. (20) x3+x2y—xy3 — y3 and x3-3x2y+3xy2 — y3. LEAST COMMON MULTIPLE. Rule to find the least common multiple of two quantities:Divide the product of the quantities by their greatest common measure; or Divide one of the quantities by their greatest common measure, and multiply the quotient by the other quantity. When there are more than two quantities the following rule will be found useful, especially when (as is usually the case in |