COMPOUND SUBTRACTION. RULE. (1.) Place the less quantity below the greater, arranging that the quantities in each column may be of the same denomination. (2.) Then, beginning with the lowest denomination, subtract, if possible, each number in the lower line from that which stands above it. (3.) But when in any denomination this cannot be done, take the number in the lower line from the number which, in that denomination, is equivalent to a unit in the next higher; to the remainder add the upper number for the remainder to be set down in that denomination, and carry one to the number in the next higher denomination in the lower line. (4.) Proceed thus with all the denominations except the highest, in which the work is to be performed as in simple subtraction. The methods of proof, and the principles on which the operations depend, are the same as in simple subtraction. Exam. 2. Required the difference between £159 -9 - 41 and £86-17 -81. In this example, as a halfpenny is greater than a farthing, it is taken from a penny, and the remainder being added to the farthing, the sum, 3 farthings, is set down; £ s. d. 159 9 41 86 17 8 £72 11 73, answ. a penny is then carried to 8 pence, and the sum being taken from one shilling or 12 pence, the remainder 3 is added to the 4 pence, and the amount, 7 pence, is set down. We then proceed thus: carry 1, and 7 are 8; 8 from 9 and 1 remains, which we set down; then 1 from 2 (the tens' figure in the number of shillings in a pound), and 1 remains, and we set it down; carry 1, and 6 are 7; 7 from 9 and 2 remain, &c. 13. A person has to pay 8s. 2d. He has, in his purse, a halfcrown, two florins, a sixpence, a fourpenny piece, and a threepenny piece. He has also some copper pence. Paying all the silver money away from his purse, how much must he add in coppers to make up the required payment? 14. A person has to pay £2 - 14 - 7, and he has with him just two sovereigns, a crown piece, three half-crowns, a florin, and a shilling piece. Giving all this in payment, how much change ought e to get back? 15. The latitude of Rome (St. Peter's) is 41° 53′ 54′′ North; of Paris (Observatory of the Military School), 48° 51′ 6′′ N.; of London (St. Paul's), 51° 30′ 49′′ N.; of Dublin (Observatory), 53° 23′ 13′′ N.; of Edinburgh (Observatory), 55° 57′ 57′′ N.; and of St. Petersburgh, 59° 56′ 23′′ N. Required the difference of the latitudes of the first and second, the second and third, &c., of these capitals. 16. The three angles of a triangle are always equal to two right angles. A land surveyor has measured two angles and found them to be 43° 21', and 78° 54′. Find from these data what the third angle must be. 17. The latitude of the Cape of Good Hope is 33° 55′ 15′′ S., and that of Cape Horn 55° 58′ 30′′ S. Required their difference. 18. The latitudes of Belfast and Glasgow, are 54° 36′ N., and 55° 52′ N. respectively. Required their difference. 19. The following are the times in which the principal planets perform their revolutions round the sun; required the differences of the first and second, of the second and third, &c. :— 20. Reduce the difference between the Julian year of 365 days, 6 hours, and the true year of 365d. 5h. 48m. 50s. to seconds: reduce also one day to seconds; and divide the second result by the first. Answ. 670 seconds, and 86400 seconds; quotient, 12864.* * This exercise explains the old and new styles in the reckoning of time; showing that by using the Julian year, that is, by taking every fourth year as a leap year, an error of a day is accumulated in about 129 years, or a little more than three days in 400 years. Hence, in every four centuries, three years, that, by the Julian reckoning, would be leap years, are taken as common years.-(See pages 60 and 61.) 21. A person goes to the sea-side for three days, and on leaving home he has with him £4-9-7. On returning home he finds he has noted the following expenses paid:- Cab, 1s. 6d.; porter, 4d.; railway fare, 38. 7d.; newspaper, 1d.; time table, 4d.; boat hire, 7s. 6d.; fishing-hooks, 74d.: biscuits, 5d.; hotel bill, £1 - 15 - 4 ; railway fare, 38. 7d.; cab, 1s.: and, on counting the money he has remaining, he finds it to be £1 - 14 - 4. Find whether his notes account for all the money spent; and, if not, by how much they are deficient. COMPOUND MULTIPLICATION. RULE I. To multiply a quantity expressed in more denominations than one, by a number not exceeding 12: * Commencing with the lowest denomination, multiply, successively, the several numbers in the multiplicand by the multiplier, dividing, setting down, and carrying, as in compound addition. £ s. d. 1 14 Exam. 1. Multiply £1 - 14 - 7 by 9. In this example, the farthings, pence, and shillings are multiplied successively by 9; the numerical products,† as they are found, are respectively divided by 4, 12, and 20 (or the tens of the shillings by 2); the several remainders are written down, and the quotients carried. The pounds are multiplied as in simple multiplication; and the product is found to be £15 - 11 - 93. Exercises. 9 £15 11 93, answ. 10. 9 7 8 x 11 *Or not exceeding 19, if the learner have committed to memory the supplement to the multiplication table, page 25. The same circumstance will also modify the succeeding problems in a similar manner. + The reason for saying here, not that the products are divided, but that the numerical products are divided by 4, 12, and 20, is, that when we multiply the three farthings by 9, we get twenty-seven farthings as the product. Now, it is not this quantity of money that we are to divide by 4, but the number expressing it in farthings, viz. 27, that we are to divide by 4, to get a number of pence with a remainder of some farthings. Thus we get 6 pence with 3 farthings. If we were to divide the quantity, twenty-seven farthings, by 4, that is, to take a fourth part of it, we would get 6 farthings and a fourth of three farthings, for which, however, it happens we have no coin small enough. Like explanations might be given in respect to the pence and shillings. RULE II. To multiply by a number which exceeds 12, but is the product of two or more factors, each less than 13; (1.) By the preceding rule, multiply the given multiplicand by one of the factors. (2.) Multiply the result by another. (3.) Multiply this last result by another, if there be so many; and thus proceed, whatever is their number. Exam. 2. Multiply 18s. 34d. by 42. £ s. d. 0 18 3 6 In this example, the multiplicand is multiplied by 6, and the product is £59-71. This again is multiplied by 7, and the product is £38-74. The reason of the operation is sufficiently obvious, since 42 is the product of 6 and 7. The work might be proved by multiplying the multiplicand by 7, and the £38 7 41⁄2, answ. result by 6. 5 9 7 When the multiplicand contains one or more farthings, if one of the factors be even, it is better to use it first, as the farthings may thus disappear, in which case the rest of the work will be easier. But if the multiplicand end in pence, without farthings, and one of the factors be 3, 6, or 9, it is better to use that factor first, as the pence may thus disappear; and, in all cases in multiplication of money, when 12 is one of the factors, it should be used first, as part of the operation may be performed by inspection, by setting down 3 pence for each farthing, and carrying to the shillings 1 for every penny in the multiplicand. Thus, in multiplying £1-13-7 by 12, set down 3 pence, and carry 7 to the shillings, saying, 12 times 3 are 36, and 7 are 43, &c. £ s. d. 12 £20 3 3, answ. RULE III. To multiply by a number which exceeds 12, but is not produced by factors below 13: (1.) Use, as in the last rule, those factors whose product is nearly equal to the multiplier. (2.) Increase or diminish the result, as the case may require, by the product of the multiplicand and the difference between the multiplier and the product of the factors employed. Exam. 3. Multiply 14s. 103d. by 38. In this example, 38 not being the product of any two factors not exceeding 12, we multiply by 36, as before, and to the product we add twice the multiplicand, to find the product by 38. The answer would have been obtained with nearly the same facility, had we multiplied by 40 (4 x 10), and subtracted twice the multiplicand; and thus the operation might be proved. £ s. d. 0 14 10 8 18 12 3 6, prod. by 12 26 15 6, prod. by 36 1 9 9, prod. by 2 £28 5 3, prod. by 38 * In this and each of the six following exercises the multiplier is the product of three factors; since 75=3x5x5; 112=8x2x7, or 4x4x7; 128=8x8x2, or 4x4 x 8, &c. In like manner, 140=2×7 × 10; 192=12 × 4×4; 240=12×2×10; 1728 12x12 x 12, &c. |