« ZurückWeiter »
In the contracted process, as soon as the quotient figure 8 and the corresponding remainder 2856 have been obtained, no more ciphers are employed to extend the division, since the number of quotient figures remaining to be found (3) is less than the number of figures in the divisor. Accordingly, the right hand figure of the divisor is cut off, and 564 used as the next divisor. This goes into 2856 five times, and in multiplying by 5, the product of 5 x 4 is increased by the carriage from 5 x 3. Another figure being taken from the divisor, it is seen that 56 is not contained in the last remainder 35; consequently an 0 is placed in the quotient, and another figure cut off from the divisor. 6 times 5, with carriage of 4 from 6x6, is then subtracted, which closes the operation.
The vertical line drawn through part of the work of the ordinary process, cuts off all the figures omitted in the shorter method, and thus explains the principle of the contractions.
If the divisor contains more figures than the quotient is required to consist of, it will be unnecessary to find more than the first quotient figure by the ordinary process ; the contraction may begin as soon as the first remainder has been obtained ; but it is always advisable, as in the case of multiplication, to work for, at least, one place of decimals beyond that intended to be retained.
The method of contracted division is exceedingly useful in computing the areas of malt utensils, the amount of floor charges, &c., where a multiplier or factor is not used. Suppose it desired to ascertain the quotient of 1 divided by 2218 192 (the number of cubic inches in a bushel) to 8 decimal places. Here, as it is seen by the rule for valuing quotients by inspection (page 89), that there should be 3 ciphers prefixed to the first significant figure of the quotient, and consequently that there are only 5 figures to be found, we may adopt the contracted mode of division from the first step of the process. Thus,
22/118•111912) 1.000000 (-00045081
40 It must be obvious that the divisor itself may, in some instances, be abbreviated before beginning to operate with it, and a correct quotient to any limited number of decimal places attained, with much less labour than by the ordinary process.
Application of Decimals to Compound Quantities.—The mode of proceeding in all cases of this kind is so simple that rules may be advantageously dispensed with. Example (1.) To find the value in shillings, pence, &c., of £.615625.
3.00 farthings. 128. 33d. Answer. Here, as there are 20s. in a £, we first take the product of 98.62.86 x 20 which equals615625 x 10
12-3125 shillings. 1000000
Then, as there are 12 pence in a shilling, we multiply in a similar manner •3125 of a shilling by 12, which gives 3.75 pence; and, lastly, to find the number of farthings equivalent to •75d., we multiply that decimal by 4. It would lead to the same result, but in a less expeditious way, if we were to multiply the given decimal of a pound by 960, the number of farthings in £1, and reduce the product to the required denominations by dividing successively by 4 and 12.
Example (2.) What is the value of •56 of 8s. 72d.?
In working this question we may either multiply 8s. 7.d. by ,8% that is, multiply it by 56 and divide the product by 100, or, preferably, reduce the given sum to farthings—the lowest denomination present--and then multiply the number of farthings by •56; the integral part of the result divided by 4 and 12 will furnish the answer in shillings, pence, and farthings, to which should be annexed the fraction of a farthing, obtained in the multiplication by •56. d.
48. 9 d. 104
£1 198, 10d. 20
478d. The question now is, what decimal of 1063 is 478, that is, what is the decima corresponding to the vulgar fraction 476?
1063)478(.4497 Answer. Another way of obtaining this result would be, to reduce 8s. 7d., and 198. 10d. respectively to the decimal of a pound, prefix the whole pounds proper to each, and divide the second quantity by the first. (See the next example.)
Example : (4.) Reduce 138. 9 d. to the decimal of £1.
12)9: 75 pence.
20)13: 8125 shillings.
•690625 Answer. In this process we first find the decimal of the lowest denomination in the giren sum (farthings). To this we prefix the given number of pence, and divide the whole by 12, to obtain the corresponding decimal of a shilling. We then prefix the given number of shillings and divide by 20, which furnishes the decimal required.
A longer way of arriving at the same result would be to reduce 13s. 93d. to farthings, and divide them by 960, the number of farthings in a pound ; that is, to convert the fraction 6 6 3 into a decimal.
The following simple rule for determining at sight the first few places of the decimal fraction which answers to any given number of shillings, &c., will be found of use where entire accuracy is not essential. For every pair of shillings put 1 in the first place of decimals, and for the odd shilling, if any, put 5 in the second place, or, rather, 50 in the second and third. Reduce the pence and farthings to total farthings, increase these by 1 if the sum amount to sixpence, and write the number, with 50 added when there is an odd shilling, in the second and third places. The decimal thus obtained is strictly correct so far as it extends. Thus to decimalise 78. 7 d., we say
•3 for the pairs of shillings
031 for the pence and farthings
Total •381 It is easy, with a little practice, to go through these operations mentally, setting down nothing until the whole decimal is formed.
As another example, let it be required to decimalise 14s. 22d. Here ,we may see at once that there should be 7-tenths and 11-thousandths, or in all - 711. There is no odd shilling, and the farthings do not amount to sixpence.
For many purposes the decimal of £l to three places will be sufficient. Should additional figures be wanted, a brief method of computing them may be learnt from De Morgan's Arithmetic, appendix VI.
In turning the decimal of a £ into shillings, &c., we double the tenths for pairs of shillings, and reckon 50, if there is one, in the second and third places, as a shilling more; the remainder after fifty is deducted, is counted as farthings, abating 1, should the thousandths so left exceed 24.
Thus to value £.508,- twice the tenths 10s., and the 8-thousandths =2d. Answer 10s. 2d.
Let the decimal be •874; twice 8 is 16, and 1 is 17 (as there is 50 in the 2nd and 3rd places); 74 less 50 = 24 which as farthings - 6d. Answer 178. 6d.
The first result should be 108. 18d.; the second 178. 52d.; from which it appears that this method cannot be depended on if correctness to a farthing be indispensable. In that case it will be necessary to employ the process of multiplying by 20, 12 and 4, as exemplified on page (91).
As to the principle of the rule for decimalising fractions of a pound on inspection, it is obvious that as two shillings equal one-tenth of a pound, every pair of shillings must stand for 1 in the place of tenths. One shilling is to £ and as zo
To = .050, an odd shilling is represented by 50 in the second and third places. For the pence and farthings, it is considered, that as d. is do
£=·00104 d, no error will be committed up to the thousandths' figure, if each farthing be reckoned as 1 in the third decimal place, until we come to sixpence. But sixpence= . £= .025 exactly ; hence it is proper to increase the number of farthings by 1 when the given sum amounts to sixpence. Example (5.) What are the values respectively of .0578 of a cwt., in qrs., Ibs., &c.,
•725 of a day, in hours, &c. ; and 9169 of a bushel, in pecks, gallons, &c. The proper multipliers in each case will be seen on reference to the Table of Weights and Measures in the Appendix. (1.) (2.)
(2.) 17 hours. 24 minutes.
(3.) 3 pecks. 1 gallon. 1 quart •6816 pints. Example (6.) Reduce 14 inches to the decimal of a mile ; and 6s. 68d. to the decimal of a guinea. (1.)
() 12)14 inches
6s. 6id. 12
12 3) 1.16666 feet
78 1760) •38888 yards (-00022095 miles.
2520 Answers (1.) •00022095 miles.
6040 (2.) •3125 guineas.
The reduction of 6s. 6 d. to the decimal of 218., may be effected also by reducing both sums to decimals of £1, and dividing the smaller by the greater, thus :
£.328125 218. = £1.05.
1•05)•328125(-3125 Answer. N.B.--Officers will find it of advantage to commit to memory the decimals
equivalent to a few simple and commonly occurring fractions, such as tida
Miscellaneous Exercises in Decimals.
(1.) Express as a decimal fraction the sum of 2, 5, 106
Answer, 2.6057. (2.) Multiply 325.701428 by •7218393, retaining only 3 decimal places in the product.
Answer, 235.104. (3.) Divide •00128 by 8.192 (using the contracted method) to 6 places of decimals.
Answer, .000156. (4.) Add 1.7235 to .00456 ; subtract •99598 from the sum ; multiply the remainder by .001, and divide the product by .25 to four significant figures of decimals.
Answer, .002928. (5.) Multiply 85 millionths by 6 ten-thousandths, and divide the result by 264 hundred-thousandths to six places of decimals, expressing the result in words.
Answer, Nineteen-millionths. (6.) Find the value of £.90625 added to 2.296875s. added to 1.4375d.
Answer, £1 Os. 6.d. (7.) Reduce £2 38. 3 d. to the decimal of £4.
Answer, •54140625. 41 (8.) Reduce
to a decimal, and divide it to 5 places of decimals by of 9
Answer, •00336. 4. REDUCTION.-Reduction in arithmetic is the process by which any quantity is changed to another of equal value in a higher or lower denomination. To transform shillings into an equivalent number of pence, or pence into an equivalent number of shillings, for example, is an operation in reduction. The method of effecting every species of reduction is so simple and obvious when tables of the various equivalent denominations are given,* that it is much better to dispense with rules and to perform the work entirely by the aid of a common-sense perception of the steps that should be taken in each case to arrive at the result. It is extremely difficult, also, to frame rules that will be readily understood and brought to bear in questions of reduction, the language in which such rules can, at the best, be conveyed, tending to perplex rather than assist the learner. Indeed, it may be doubted whether a practical computer ever thinks of referring to a rule on these occasions, as the proper mode of proceeding always suggests itself after a moment's reflection. Several examples of processes which strictly are cases of reduction, have been given in the preceding articles on fractions. A few additional illustrations, therefore, will supply all the guidance that any person is likely to require in this part of arithmetic.
* See Table of Weights and Measures in the Appendix