A complex fraction may also often be simplified with great readiness by applying the principle mentioned in (21) the numerator and denominator being both multiplied by a number which will make them both integral. Ex. (1) might, for instance have had its numerator and denominator multiplied by 5, and we should have had at 23 12 Here each term of the numerator and denominator may be made integral by multiplying it by 12, the L. C. M. of the denominators of the small fractions. Then 58-52+170-61+21_30_15 6+31-7 = 78+37-87 28 14 =I Ans. Questions in some of the preceding rules may sometimes be best worked by expressing the result as a complex fraction, and then simplifying it. (8) Express 1 hlf-crs. — 13 fl. + 13s. as the fraction o + 13 hlf-sov. (9) Simplify of of 1 ton 12 cwt as the fraction of I 5 of 54 cwt. MISCELLANEOUS EXAMPLES IN FRACTIONS. In questions involving the application of fractions, it may often be best for a beginner to determine the method of solution by considering in what way he would proceed, if the numbers involved were integral instead of fractional: as in Ex. (1). One class of questions must be noticed, the solution of which depends on the first principles of fractions, but might not be readily discovered. Exx. (2), (3), (4), are illustrations. In them the times are given which several agents require separately for completing some work, to determine the time in which they could do it when acting together; or conversely. For solving them, a change must be made in the principal unit of reference; so that instead of regarding the work as the principal unit, as the question suggests, a unit of time is made the principal, and it is considered what part of the work each agent would do in that. The sum of these fractions, if the agents assist one another, as in Exx. (2) and (3),—or their difference, if the agents oppose one another, as in Ex. (4),-will be the part of the work done in the unit of time when all are acting. From this may easily be determined the time they would take for doing the whole. Converse questions must be solved on the same principles. In some questions brackets may occur. Their meaning is that the total value of the quantities enclosed within them is affected by signs outside; and the solution of such questions must be begun by finding the value of any expression within brackets, and substituting this value for the expression itself. See Exx. (5) and (6). Ex. (1) A train travels 25 miles in its rate per hour. ΙΟ of an hour; find If the question had been :-a train travels 40 miles in 2 hours, find its rate per hour; we should obviously have divided the whole distance, 40 miles, by 2. Applying the same method here, we have Ex. (2) A can do some work in 6 days, B can do it in 8 days, and C in 12; how long would they take to complete it, all working together? Taking the whole work as the unit, supposes the whole work to be divided into 8 equr 1 parts, 3 of which are thus done every day; therefore the whole time is 83, i. e. 23 days. Ex. (3) A can do as much work in 3 days as B can in 4 days. If B in 12 days has completed half of some work, how long will it take them both to finish? B alone would take 24 days to do the whole: the time A would take is of 24 days, i.e. 18 days. Hence, as in the last question, they would take 10 days to do the whole, and therefore 54 days to do the half. The last part of the question might have been reasoned out on the same principle as Ex. (1) :— 7 of the work is to be done at the rate of every day; 72 therefore the time will be÷/2 I 7 I 72 5 days. Ans. 7 Ex. (4) A cistern can be filled by means of a tap in 45 minutes, and it can be emptied by a plug in 30 minutes. Supposing the cistern were full, and both the tap and plug opened, in what time would it be emptied ? Therefore the whole will be emptied in 90 minutes. Ex. (5) Find the value of 1÷÷÷(23—14)×8 I 2 7 ÷ ×60 Ans. 13 60 13 Ex. (6) Simplify { 31 of 51—124} × {3}−(4}−2})}. |