NOTE. Some of these equations might have been solved in a more expeditious manner, but we have chosen to adhere to one uniform method, which is applicable to all cases, in order both to engrave it the more deeply into the minds of beginners, and to avoid perplexing them with particular methods, which practice alone will soon make them discover. PROBLEMS OF THE FIRST DEGREE, WITH ONE UNKNOWN QUANTITY. First. Divide 54 £. among three persons, so that the share of the second be double that of the first, and the third lot be equal to both the others; What will be each person's share? NOTE. The question seems to involve three unknown quantities, but a little reflection will shew, that the first person's share being known, the other two will of course be also known. This observation deserves particular attention, as it is applicable to a great number of problems, the enunciation of which seems to suppose more unknown quantities than there really are. Ber the share of the first person; then we will have the equation + 2 x + 3 x = 54 ... 6x54. ..x = 9. Second. Divide 112 £. among three persons, so that the second have three times the share of the third, and the first 3 times the share of the second and third together. What will be the share of each. Be a the share of the third person; we will have a 7. Third, Four Merchants have a sum of $.58000 to divide in the following manner: The second is to have the share of the first, the third of the second's, the ourth of the third's; Say the share of each. x the share of the first; we shall have the equation, Be Fourth. A father dying leaves an estate of 15000 £. to be distributed to his widow, a son and a daughter, as follows: the daughter to have only of the sou's share, and the mother of the sum of both the children's portions; what sum will each receive? Be a the share of the son, the equation will be, Fifth. Divide 720 in three parts, the greatest of which exceed the smallest by 80, and the mean one by 40. Ans. 280 ... 240. and 200. Sixth, How old are we, asks a son to his father; your age, replies his father, is now the third of mine, and six years ago it was but the fourth of it. Find the ages of both; Ans. The father's 54 years, and the son's 18. Seventh. Divide 25 in two parts, such that their difference divided by 3, be equal to 5. Ber one of the parts, 25 - will be the other; then we will have the equation, 25x15 X (25— a) = 5 3 ... x 20. Eighth. Divide 20 in two parts, so that the third of the greater be equal to the half of the smaller. Ans. 12 and 8. Ninth. Divide 45 in two parts, which be to one another as 5 is to 4. Ans. 25 and 20. Tenth. A courier sets off from Baltimore, and runs 9 miles an hour; another courier sets off 5 hours after, on the same road, and runs 12 miles an hour; At what distance from Baltimore will the second overtake the first. NOTE. The problems on couriers, which generally appear very perplexing to beginners, can all be solved in the same manner, by means of a proportion, which it is always very easy to establish; this proportion is founded upon the remark, that the spaces run by the two couriers in the same time, are evidently in the same ratio as their respective velocities; for instance, if one runs twice as fast as another, he will of course go twice the length of the other in the same time. Ꮓ Ber the distance demanded in the present problem, it is plain that during the time the second courier 45; these 45 representgoes, the first will go r ing the space performed by the first in 5 hours start, which he has over the second, at the rate of 9 miles an hour: The proportion will then be, 12:9::x: 2 ·45 and a 180 miles. Eleventh. A courier starts from Baltimore for New-York, and goes 12 miles an hour; another dispatched for the same city, starts from Philadelphia 3 hours after the first has left Baltimore, and goes 6 miles an hour. At what distance from Baltimore will they meet, supposing Philadelphia is 102 miles distant from Baltimore. Ber the distance demanded. Observe that the first courier has set off three hours before the second, and has consequently run 36 miles before the departure of this; that, therefore, the second has only æ— 102 to run in the same time the first will run - 36. The proportion will then be, —36: x- 102 12: 6, or 2:1 :: x — 168. and r = Twelfth, Two couriers set off at the same moment, the one from Baltimore, the other from Philadelphia, and direct their course towards each other. It is supposed that the one from Baltimore goes at the rate of 8 miles, and that from Philadelphia of 10 miles an hour. At what distance from Baltimore will they meet? 102 miles is the distance of these two cities from each other. Ans. 45 miles. |