34th. If for the carriage of 5 cwt. 14 lbs. a distance of 96 miles the charge is 1 £. 12 sh. 6 d., how far ought 3 cwt. I qr. to be carried for the same money? Ans. 151 mi. 3 fur. 3 po. 35th. What is the expense of keeping 11 horses for a year, at the rate of 11 d. per day for each horse? Ans. 192 £. 7 sh. 81 d. 36th. What cost 43 qrs. 5 bush. of corn at 1 £. 8 sh. 6d. per quarter?....... ......................Ans. 62 £. 3 sh. 33 d. 37th. If 3 cwt. of tea cost 40 £. 12 sh., at how much per lb. must it be retailed in order that 10 £. may be gained by the sale of the whole quantity?............. Ans. 334 sh. 38th. How many men are required to cut a trench of 135 yds. in length in 8 days, when 16 men complete 54 yds. of the same trench in 6 days?.... Ans. 30 men. 39th. A family of 9 persons expends 120 £. in 8 months; what sum ought to serve a family of 24 persons during 16 months, at the same rate of expenditure?...Ans. 640 £. 40th. 27 shillings being the wages of 4 men for 7 days; required the wages of 14 men for 10 days at the same rate? Ans. 6 £. 15 sh. 41st. A courier travels 130 miles in 3 days of 12 hours each; in how many days of 10 hours each can he travel 360 miles at the same hourly rate?.............. Ans. 9§3 days. 42d. If 120 bush. of corn serve 14 horses 56 days, how many days will 94 bush. serve 6 horses?...... Ans. 1021 days. 43d. If yds. cost £., what ought yds. to cost?.. Ans. 10 sh. 44th. If of a lottery ticket is worth 273 £. 2 sh. 6 d., what is of the same ticket worth?... Ans. 227 £. 12 sh. 1 d. 45th. If of a ship is worth 73 £. 1 sh. 3 d., what part of her is worth 250 £. 10 sh.?..................... 32 ....... Ans. 3. 46th. If the penny loaf weigh 7 oz. when wheat is at 5 sh. 6 d. the bushel, what is the bushel worth when the penny loaf weighs 21 oz. ?........ ......Ans. 15 sh. 4 d. 47th. If 2 lbs. of bread are bought for 21 d. when the price of wheat is 4 sh. per bushel, what ought the 20-penny loaf to weigh when wheat is 6 sh. per bush. ?......Ans. 12 lbs. 48th. The expense of building a wall 6 feet thick, 9 feet high, and 432 feet long, being 720 £., what must be the cost of building another wall 12 feet thick, 18 feet high, and 576 feet long, at the same rate?...............Ans. 3840 £. 49th. If 240 men in 5 days of 11 hours each dig a trench 230 yds. long, 3 yds. wide, and 2 yds. deep, in how many days of 9 hours long can 24 men dig a trench 420 yds. long, 5 yds. wide, and 3 yds. deep? Ans. 2786 days. 50th. If 24 pioneers in 21 days of 12 hours long can make a trench 144.75 yds. long, 44 yds. wide, and 2 yds. deep, what length of trench can 90 pioneers make in 4 days of 9 hours long, the trench being 47 yds. wide and 3 yds. deep?.. ...Ans. 5081123 yds. INTEREST. 347. If the sum of 5 £. is paid for the use of 100£. during 1 year, how much must be paid for the use of 1486 £. during 3 years, at the same rate. In this example of compound proportion, 5., the quantity of the same kind as the answer, is the third term. 100£. and 1486 £. are the terms of the first ratio. 1 year and 3 years are the terms of the second ratio. 100£. produce 5 £., 1486 £. must produce a greater sum; and 1 year giving 5., 3 years must give a greater sum. Whence, 5 x 1486 x 3 = =222·7 £.=222 £. 14 sh. 100 1 year: 3 years :: 5 £. : x £. — Questions, in which the compensation made for the loan of 100 £. for 1 year is given to find what must be paid for the use of any sum greater or less than 100£. during any given period of time, depend for their solutions on the rules of proportion. Some particular names are employed to describe the data in such questions, and the rule expressed in terms of these names is called the rule of Interest. The compensation paid yearly by the borrower to the lender for the use of 100£. is the interest of 100£. for 1 year. This (which in the preceding example is 5 £.) is described as the rate per cent. per annum. * The rate per cent. per annum is a matter of agreement between borrower and lender. See the acts of parliament 7 Will. 4. & 1 Vict. cap. 80., 2 & 3 Vict. cap. 37., 3 & 4 Vict. cap. 83., and 4 & 5 Vict. cap. 54. The money "lent is named the principal. The compensation for the loan is the interest. The sum of the principal and interest is the amount. Technically expressed, the preceding question runs thus: What is the interest of 1486 £. for 3 years at 5 per cent. per annum? The result already found is x= 5 × 3 × 1486 100 =222 £. 14 sh. If p is put for the principal or sum lent, r for the rate per cent. per annum, and for the time, it is found, as above, that 100: 1 prt 100° In words; to find the interest of any sum of money for any given rate and time, multiply the principal by the rate per cent. per annum, this product by the time during which the money is lent, and divide the result by 100. If the time is any number of years, t is a whole number; if months, or months and days (it being the custom in these calculations to reckon 12 months to the year, and 30 days to the month) the numerical value of tis fractional; if days only, the number of days may be made the numerator and 365 the denominator of the numerical value of t. The interest of 100 £. for 1 year being denoted by r, the interest of p£. for 1 year at rate r by i, and any period of time less or greater than a year by t; Cent. an abbreviation of centum, the Latin word for a hundred; per annum, or per an., also Latin for every year, yearly. a. The value of each of the quantities i, p, r, can be expressed in terms of the other quantities involved in the proportion 348. Case A. The principal, rate per cent., and time being given, to find the interest, Form the continual product of the numbers expressing the principal, rate per cent., and time, and divide this product by 100; the result is the interest required. Example. What is the interest of 170 £. for 13 years at 5 per cent. per annum? p=170 £., r=5, t=1}=1·5. Substituting these values of p, r, t in formula A, it= 170x5x1.5 1275 =12·75 £.=12 £. 15 sh. 349. Case B. The principal and its interest for any specified number of years being given, to find the rate per cent. per annum, Multiply 100£. by the interest for the given time; divide the product by the principal, and this quotient by the number of years; or multiply 100£. by the interest for 1 year, and divide the product by the principal; the result in either case is the rate per cent. per annum. Example. The interest of 25 £. for 3 years, being 3 £. 18 sh. 9 d.; what is the rate per cent. per annum? p=25, t=31=3.5, it=3 £. 18 sh. 9 d. 3.9375 £. Substituting these values of p, t, and it in formula B, 350. Case C. The rate per cent., time, and interest for that time being given, to find the principal. Multiply 100£. by the interest for the given time, and divide the result by the product of the rate per cent. and time; or multiply 100 £. by the interest for 1 year, and divide the result by the rate per cent.; the quotient is the principal. Example. What principal, being put to interest at 5 per cent. per annum, will in 7 years produce 86 £. 16 sh. of interest ? it=86 £. 16 sh.=86'8, r=5, t=7. Substituting these values of it, r, and t in formula C, p= 100 × 86.8 8680 5X7 =248 £. 351. The principal rate per cent. per annum and interest for an unknown time may be given to find the time. In this case let the interest of p for the unknown time be denoted by i'. Whence, rt= Then 100 100 xi and P Case D. The principal and its interest for an unknown time, at a given rate per cent., being given to find the time; Form first the product of 100 £. by the given interest; second, the product of the principal by the rate per cent. per annum; divide the first by the second product; the result is the time required. Example. In what time will the interest of 547 £. 15 sh. amount to 82 £. 3 sh. 3 d. at 5 per cent. per annum? p=547 £. 15 sh.=547·75 £., r=5, i'=82 £. 3 sh. 3 d.=82·1625 £. Substituting these values of p, r, i' in formula D, Now, if equal numbers are added to the same number, the sums are equal; therefore, it 1+100 100+rt_p+it = (Art. 171), or 100 100+rt::p : p+it. In this proportion rt being the interest of 100£. for t years at r per cent. per annum, 100+rt is the amount of 100£. for t years at rate r (Art. 347); and it being the interest of p £. for t years at rate r, p+it is the amount of p£. for time t at rate r. Denoting pit by a (the first letter of the word Amount), Whence, (100+rt)×p Ɑ= 100 352. Case E. The principal, rate per cent., and time being given, to find the amount, Multiply the amount of 100£. for the given rate and time by the principal, and divide this product by 100; the result is the amount required. Example. Find the amount of 450 £. in 2 years, at four per cent. per annum ? r=4, t=21=2·5, rt=4×2·5=10, 100+rt=110, p=450. Substituting these values of rt and p in formula E, The same result may be obtained by computing the interest of 450 £. for 2 years at four per cent. per annum, and adding that interest to 450£. (Art. 347). 353. Numbers expressing three of the four quantities a, p, r, t, being given, the fourth can be found as in Article 347 a. It seems unnecessary to repeat the detail of that article. Therefore let it further be only required to find p, having given r, t, a. To determine P, the proportion Case F. Having given the rate per cent., time, and amount, to find the principal. Multiply the given amount by 100, and divide the product by the amount of 100£. for the given rate and time; the result is the principal required. Example. The amount of an unknown principal in 3 years at 4 per cent. per annum is 289 £. 7 sh. 6 d. Required the principal? a=289 £. 7 sh. 6 d. = = 289·375 £. ; r=4}=4·5; t=3}=3·5; rt=4·5 × 3.5 15.75; 100+rt=100+15·75=115·75. Substituting these values of a, rt, in formula F, p= 289-375 x 100 28937-5 =250 £. 354. If a sum of money, a, due by one person to another, is not payable 100 x a 100+rt gives until after the expiration of a certain time, t, the expression that other sum of money, p, which, placed at interest for the time t, and at a certain rate of interest, r, will produce the sum a. When the rate per cent. per annum, time to come, and amount at the expiration of the given future time of an unknown principal are given to find that principal, the sum of money, a, is described as a £. due t years hence; p is termed the present worth, r the rate per cent. of discount; and the rule of interest, with these changes of terms and object, is called THE RULE OF DISCOUNT. The last example, expressed as a question of discount, is, What is the present worth of 289 £. 7 sh. 6 d. due 3 years hence, discount being allowed at the rate of 4 per cent. per annum? As already found, the present worth is, 289-375 × 100 28937.5 p= 115.75 115.75 =250 £. The difference between a and p, which is evidently the interest of p for the given time and rate, is termed in this rule the discount. Discount, then, is the difference between a sum of money payable at a future time and the value of that sum in ready money. Thus, in the last example, 289 £. 7 sh. 6 d. due 3 years hence at 44 per cent. discount is worth 250 £. payable now. And 289 £. 7 sh. 6 d.-250 £.39 £. 7 sh. 6 d. is the discount or compensation allowed for immediate payment. 355. The discount can be obtained otherwise than by subtraction of p from a. Since the interest of different principal sums at the same rate and for the same time are proportional to the principals, it follows that ..1+ =1+ |