This is sometimes called the method by Ratios. It is recommended for practice, in preference to any other, particularly with the following modifications. Let the third term, (or antecedent of the incomplete ratio,) be written down, with all the second terms, (or consequents of tho complete ratios,) as factors in the numerator of a Fraction, connected by the sign of Multiplication; and all the first terms, (or antecedents of the complete ratios,) in the same manner, for a denomina. tor, thus, Now observe whether any term in the Numerator has a common factor with one in the denominator, and, if so, divide both by that factor, and place the quotients directly under or over the terms divided. Thus, 27 and 54 may each be divided by 27. I accordingly divide and place the quotient, 1, under 27, and 2, over 54. In the same manner, I proceed with 24 and 48; 18 and 36; 16 and 64; and 10 and 30. I then see that the factor 6, in the denominator, and the quotient 3 over 30, may be divided by 3. I, therefore, divide and place the quotient 1, a step higher still, and the quotient 2, under 6, like the others. I then see a 2 over 54, and also a 2 under 6. I divide these, and put down the quotients, (1, in each case,) like. the 1 over 30. The whole Denominator is now exhausted. Each of its factors has become 1 by division, and consequently will not affect the Quo-. tient, or value of the Fraction. Taking, therefore, the last obtained factors in the Numerator, (which are of course, those which stand highest,) and multiplying them (rejecting 1s) with the terms unaltered by the process, (viz. in this case, the 13 and 8) we shall obtain the answer. Thus, 4×2×13×2×8=1,664. Ans. as before. We have been particular in explaining this mode, because it is usually much the shortest, and is never longer than any other. It also saves the trouble of a regular statement, for each ratio may be taken, (after the third term is written in the Numerator,) and written by itself, one term in the Numerator and the other in the Denominator, according to the direction given in Simple Proportion. ANALYTIC SOLUTION. If it take 64 men to dig 27 cellars, it will take as many to dig 1 cellar; that is, it will take 4 men. If it take 4 to dig one 24 ft. long, it will take as many to dig one 1 ft. long; that is, it will take & men, 64 If it take to dig one 18 ft. wide it will take as many to dig one 1 ft. wide; that is, it will take 17 men. If it take men to dig one 16 ft. deep, it will take as many to dig one 1 ft. deep; that is, it will take TT2016 men. If 916 648 81 will do it in 36 days, it will take 36 times as many to do it in one day; that is, it will take, 81 men. If will do it in a day of 8 hours long, it will take 8 times as many to do it in a day of 1 hour long; that is, it will take men. 27 6 Now if dig 1 cellar, it will take 48 times as many to dig 48 cellars; that is it will take 384128 men. If it take 12 when the length is 1 ft. it will take 54 times as many, when the length is 54 ft.; that is, it will take 912=256. If it take 256 men, when the width is 1 ft. it will take 30 times as many, when the width is 30 ft; that is, it will take 7,680 men. If it take 7,680, when the depth is 1 ft., it will take 13 times as many, when the depth is 13 ft.; that is, it will take 99,840 men. But if 99,840 do it in 1 day, it will require only the tenth as many for 10 days; that is, it will require 9,984 men. If it require 9,984 men, when the days are 1 hour long, it will require but the part as many, when the days are 6 hours long; that is it will require 1,664 men. Ans. as before. By the analytic method, the pupil will perceive that we take each term in the given or supposed case, and find what the answer would be, if that term were made 1. We then, consider what the answer would be, if we increase each term separately, from 1 to the num ber in the case required to be solved. The teacher should make this clear by illustration and he should require his pupil to find many answers by analysis, for the purpose of mental discipline. For common practice the mode recommended before is best. The pupil will observe, that, by the analytic method we sometimes obtain, in the course of the process, expressions for things which can, actually, have no existence; as, for instance, of a man, in the last example. But, he should recollect that by this expression is only meant T of the labour of a man, and not of a man's person. Similar explanations may be given in all cases, where these quantities occur. As we were speaking just now, of supposed and required cases, contained in a question, it may be well to notice the distinction. It will be observed, that, in the last example, is contained a case complete in every term; viz. that "64 men can dig 27 cellars, 24 ft. long, 18 ft. wide and 16 ft. deep in 36 days, working 8 hours a day." This is complete in every term, because every number concern. ed is given. This is the supposed case, or supposition of the question. In the same question is contained another case, incomplete in one term, which term is required to be supplied; and hence, the case is called the required case, or the requisition, or the demand of the question. In the last example, the requisition or demand is, "how many men can dig 48 cellars, that are 54 feet long, and 30 feet wide, and 13 feet deep, in 10 days, working 6 hours a day?" Every term in this case is complete except the number of men, which is required, or de manded. The terms of the supposed case are called TERMS OF SUPPOSITION.. NOTE. The expression "required case," is not strictly proper, since but one term is required. Its conciseness, however, recommends it to use, and if it is understood, nothing more is necessary. From the preceding we derive the following rules: 1. BY PROPORTION. ARRANGE EACH COMPLETE RATIO, BY THE DIRECTIONS IN SIMPLE PROPORTION. COMPOUND THESE RATIOS, BY MULTIPLYING ANTECEDENTS TOGETHER, AND CONSEQUENTS TOGETHER. THE RESULT WILL BE THE RATIO OF THE REMAINING TERM TO THE ANSWER ; WHICH OBTAIN AS IN SIMPLE PROPORTION. Of this rule we have the following abbreviations. 1. EXPRESS EACH COMPLETE RATIO FRACTIONALLY, AND REDUCE IT FIND THE CONTINUED PRODUCT OF ALL THE REOr, TO ITS LOWEST TERMS. DUCED RATIOS AND THE REMAINING TERM. II. WRITE THE THIRD TERM, AND ALL THE CONSEQUENTS OF THE COMPLETE RATIOS AS FACTORS IN THE NUMERATOR OF A FRACTION; AND UNDER THEM ALL THE ANTECEDENTS, AS FACTORS IN THE DENOMINATOR. REJECT COMMON FACTORS FROM THE NUMERATOR AND DENOMINATOR UNTIL NO MORE CAN BE FOUND, WRITING THE QUOTIENTS, INSTEAD OF THEM. REDUCE THE FRACTION THUS OBTAINED, IF POSSIBLE, TO A WHOLE OR MIXED NUMBER. This method is best. II. BY ANALYSIS. CONSIDER WHAT THE ANSWER WOULD BE, SUPPOSING EACH TERM OF SUPPOSITION, SEPARATELY, CHANGED TO A UNIT. THEN SUPPOSE EACH TERM SEPARATELY CHANGED AGAIN, FROM A UNIT TO THE CORRESPONDING TERM OF DEMAND, AND CONSIDER WHAT THE ANSWER WOULD BE. THIS LAST ANSWER IS THE TERM REQUIRED. The above rule by proportion is sometimes called the DOUBLE RULE OF THREE. EXAMPLES FOR PRACTICE. A. 24. 12. If 20 bushels of wheat are sufficient for 8 persons, 5 months, how many will be sufficient for 4 persons, 12 months? RULE I. PROPORTION, 8 persons : 4 persons 5 months: 12 months 40 Compound ratio Or, : 48 :: 20 bu. : *** bu. ::20 20×48 : =24 40 1. Ratio of persons Ratio of mo.. Then 20xX1=24 Or, RULE II. ANALYTIC SOLUTION. If 20 bushels are enough for 8 persons, of 20 will answer for 1 person, that is, 20 bushels. If answer for 5 months, as much will answer for 1 month, that is bushel. If answer 1 person, 4 persons will need 4 times as much, that is 4=2 bushels. If they need 2 bu. in 1 month, they will need 24 bu. in 12 mo. Ans. 13. If 7 men build 36 rods of wall in 3 days, how many rods can men build in 14 days. A. 480 rods. 14. If $100 will gain $6 interest in 1 year. what will $400 gain 7 months? Ans. $14. 15. If 20 men perform a piece of work in 30 days, how many men will it take to accomplish a piece 4 times as large in 4 days? Ans. 600 men. 16. If 56 lb. of bread will last 7 men, 14 days, how much will last 3 men, 21 days? Ans. 36. 17. If 7 men can reap 84 acres of wheat in 12 days, how many men can reap 100 acres in 5 days? Ans. 20 men. 18. If a family of 9 persons spend $305 in 4 months, how much would they spend in 8 months if 5 persons were added? Ans. $948.888. 19. A hare is 50 leaps before a grey-hound, and takes 4 leaps, while the grey hound takes 3. But 2 of the grey-hound's leaps are equal 3 of the hare's. How many leaps must the grey-hound make to overtake the hare? Ans. 300. 20. If $2,000-will support a garrison of 150 men, 3 months, how long will $6,000 support 600 men? Ans. 24 mo. 21. If the transportation of 12 cwt. 3 qrs. 400 mls. cost $57.12, what will the transportation of 10 T. 75 mls. cost? Ans. $168. 22. If 8 men can build a wall 20 ft. long, 6 ft. high, and 4 ft. thick in 12 days; how long will it take 24 men to build one 200 ft. long, 8 ft. high and 6 ft. thick? Ans. 80 days. 23. If 18 men can build a wall 40 rds. long, 5 ft. high and 4 ft. thick in 15 days; in what time will 20 men build one 87 rds. long, 8 ft. high, and 5 ft. thick? Ans. 5829 days. 24. If 180 men, working 6 days, each day 10 hours, can dig a trench 200 yards long, 3 yds. wide, and 2 yds. deep; how many days will 100 men be occupied in digging a trench 360 yds. long, 4 wide, and 3 deep, working 8 hours a day? Ans. 483 days. 25. If 100 lb. weight English make 95 lb. Flemish, and 19 lb. Flemish, make 25 lb. at Bologna; how many lbs. English are equal to 50 lb. at Bolo na? Ans. 40 lb. English. 26. If 24 lb. at New York make 20 lb. at Amsterdam, and 50 lb. at Amsterdam, are equal to 60 lb. at Paris; how many lbs. at Paris, are equal to 100 lb. at New York? Ans. 100 lb. This comparison of the weights, measures and coin of two coun. tries, through the medium of those of other countries, is sometimes called CONJOINED PROPORTION. Merchants are accustomed to call it the CHAIN RULE. The process belongs to Compound Proportion. 27. If 70 braces at Venice are equal to 75 braces at Leghorn, and 7 braces at Leghorn be equal to 4 yards in Hartford; how many braces at Venice are equal to 64 yards in Hartford ? Ans. 104,5. 28. If 40 lbs. at Boston make 48 lb. at Antwerp, and 30 lb. at Antwerp make 36 lb. at Leghorn; how many lbs. at Leghorn are equal to 100 lb. at Boston. Ans. 144 lb. 29. A merchant in Petersburgh has to pay 1,000 ducats in Berlin, which he wishes to pay in rubles by the way of Holland; and he is acquainted with the following proportional values of moneys,→ viz. 2 rubles equal 95 stivers; 40 stivers make 2 florins; 5 florms make 2 rix-dollars, Hollandish; 100 rix-dollars Hollandish make 142 rix-dollars Prussian; and finally 3 ducats in Berlin equal 9 rixdollars Prussian. How many rubles must he pay? Ans. 2,223.87-rubles. § XCIV. COMPOUND FELLOWSHIP, by Analysis, has already been explained. (§ XLIX.) Questions in Compound Fellowship may be solved also, by Compound Proportion 1. A and B traded, A furnishing $215 for 6 months, and B $390 for 9 months. A's share of gain was then $53.75. What was B's, and what the whole gain? Ans. B's $146.25. Whole $200. PROPORTION, $215 $390}:: $53.75 : $*** 6 mo. : mo. 2. A and B traded, A funishing $1,000 for 12 mo. His gain was then $200, while B, whose capital was in trade only 8 mo. gained $300. What was B's capital? Ans. $2,250. PROPORTION, $200 $300 ( 8 mo.: 12 mo. ( :: $1,000: $*** 3. A furnished a common stock with $4,000 for 12 mo.; B, with $3,000 for 15 mo.; C, with $5,000 for 8 mo. C's gain is $200. What is the whole gain and what the gain of the other partners? Ans. Whole $665. A's $240. B's $225. 4. A has $500 capital for 12 mo. B $600 for 10 mo. and C $800 for 6 mo. B's gain is $250. What is the whole gain, and what the gain of the other partners? Ans. Whole gain $700. A's 250. B's 200. 5. A has $200 capital for 2 mo.; and B has $100 for 7 mo. Their joint gain is $55. What is that for each? Ans. A $20. B $35. 6. A, B and C trade in company, and gain $480. A's capital is $800 for 4 mo.; B's $600 for 2 mo.; and C's $720 for 5 mo.What is each man's share of the gain? Ans. A's $192, B's $72, and C's $216. ALLIGATION. MENTAL EXERCISES. § XCV. 1. A grocer mixed 8 gals. of water with 8 gals. of brandy worth $1 pr. gal. What was the mixture worth pr. gal. ? NOTE. The 8 gals. of brandy were worth $8; but, after the mixture, there were 16 gals. worth no more than before; that is, worth only $8. 2. A merchant mixed 6 lb. of tea at $1 pr. lb. with 4 lb. worth $31 pr. lb. What was the mixture worth pr. lb. ? NOTE. There were 6+4=10 lb., and the whole price was 6X1-$6+4X 3-$14, amounting to $20. The price of 1 Ib. is, therefore, 20=$2. Ans、 |