143. To reduce compound numbers to decimals of the highest denomination. RULE.-Divide the lowest denomination (annexing one or more cipher, as shall be found necessary) by the number which it takes of that to make one of the next higher denomination,(126) and write the quotient as a decimal of the higher; divide this higher | denomination by the number which it takes to make one still higher, and so continue to do till it is brought to the decimal required. 144. To find the value of a decimal in integers of a lower denomination. RULE.-Multiply the deci mal by that number which it takes of the next lower denomination to make one of the denomination in which the decimal is given, and point off as in the multiplication of decimals. (122) Multiply the decimal part of the product by the number it takes of the next lower denomination to make one of that, and so on; the several numbers at the left of the decimals will be the answer. QUESTIONS FOR PRACTICE. 1. Reduce 2 yards, 2 feet, and 9 inches to the decimal of a rod. 12)9.00(0.75 ft. 3)2.75(0.9166 yd. 5.5)2.9166(0.5303 rd. Ans. 2. Reduce 10s. 3d. to the decimal of a pound. 3d.=s.=0.25s. and 10.258.10.251.=0.51251. 3. Reduce 3qrs. to the decimal of a shilling. 4. Reduce 12s. 9d. 3qr. to the decimal of a pound. 1. Reduce 0.5303 rod to yards, feet and inches. 0.5303X5.5=2.91665yd. 0.91665X3=2.74995ft. 0.74995X12-8.9994in or 2yds. 2ft. 9in. nearly, Ans. 2. In 0.5125l. how many shillings and pence? 3. What is the value of 0.0625s.? 4. What is the value of 0.6406257. in integers? it is common to consider 30 Reduce Q.725 year months and days. 145. In computing interest, days one month, and 12 months a year. Reduce 8 months 21 days to the decimal of a year. 21d. m. 0.7m. and 8m. 2147-0.725yr. Ans. 0.725X12-8.7mo. and 0.8X30-21d 3. ADDITION. 146. 1. A person gave £2 17s. and 8d. for a load of hay, £1 5s. 3d. for 5 bushels of wheat, and 10s. 4d. for a load of wood; what did the whole cost? l. s. d. 2 17 8 1 53 10 4 4 13 3 Ans. As we may very evidently add pence to pence, shillings to shillings, &c. we write down the numbers so that pence shall stand under pence, shillings under shillings, and so on. We then add the pence, and find their sum to be 15d. but as 12d. 1s. 15=1s. 3d. We therefore write down 3d. under the column of pence, and reserve the 1s. to be joined with the shillings." We now add together the shillings, which, with the 1s. reserved, amount to 33s. 4 13 3 proof. =£1 13s. we therefore write 13s. under the column of. shillings, and reserve the £1 to be joined with the pounds. Lastly, we add the pounds, and joining the £1 reserved, write the amount, £4, under the column of pounds; and thus we find the whole cost to be £4 13s. 3d. The above process is called Compound Addition. COMPOUND ADDITION 147. Is the uniting together of several compound numbers into one sum.(48) RULE 148. Place the numbers to be added so that those of the same denomination may stand directly under each other. Add the numbers of the lowest denomination, and carry for that number which it takes of that denomination to make 1 of the next higher, writing the excess, if any, at the foot of the column. Proceed with each denomination in the same way, till you arrive at the last, whose amount is to be set down as in Simple Addition. PROOF. The same as in Simple Addition. QUESTIONS FOR PRACTICE. If a man purchase a yoke of oxen for £15 5s. 8d., four cows for £20 10s. 6d., and a horse for £26; what did they all cost? Ans. £61, 16s. 2d. The floors of 4 rooms in a certain house cover 5rd. 24in. of land; the remaining room 1rd. lyd. lft.; and the walls and chimney cover 2rd. 11in.; how much land does the whole house occupy? Ans. 8rd. lyd. 1ft. 35in. A certain field has four sides, whose lengths are as follows: 4ch. 27lin. 5ch. 19lin. 4ch. 50lin. and 6ch. 4lin.; what is the distance round it? Ans. 20 ch. What is the weight of 3hhd. of sugar, the first weighing 10cwt. 20lb.; the 2d, 9cwt. 1qr. 15oz.; and the 3d, 11cwt. 151b. 14dr.? Ans. 1 ton, 10 cwt. 2 qr. 7 lb. 15 oz. 14 dr. 3. Subtraction. 149. 1. A person bought a cow for £3 7s. 6s., and sold it for £4 12s. 3d., how much did he gain? 443 S. d. 4 12 3 7 6 Gain 1 4 9 We write the less number under the greater, so that pence shall stand under pence, shillings under shillings, and pounds under pounds; we then begin at the right hand, but as we cannot take 6d. from 3d., we borrow from the 12s. 1s. 12d., which we join with the 3d., making 15d., and then 6d. from 15d. leaves 9d., which we write under the pence. We now proceed to the shillings, but as we have borrowed 1s. from 12s. we call the 12s. 11s., and 7s. from 11s. leaves 4s., and lastly, £3 from £4 leaves £1. Thus we find that he gained £1 4s. 9d. The above process is called Compound Subtraction. Proof 4 12 3 COMPOUND SUBTRACTION 150. Is the taking of one compound number from another, so as to find the difference between them. (42) RULE. 151. Write the less number under the greater, so that the parts which are of the same name may stand directly under each other. Begin with the lowest denomination, and take the number in the lower line from the one standing over: proceed in the same way with all the denominations. Should the number in the upper line be less than the one standing under it, suppose as many units to be added to the upper number as will make a unit of the next higher denomination, remembering to diminish the number in the next place in the upper line by 1. PROOF. The same as in Simple Subtraction. |