119. 1. How much butter in 3 boxes, each containing 4 pounds and 75 hundredths of a pound? By Addition 4.75 4.75 Ans. 14.25 lb. The method of solving this question 4.75 by Addition, must be sufficiently obvi- By Multiplication. are hundredths of a pound, the product Ans. 14.25 lb. 0.1.and 0.05, we therefore write 5 in the place of hundredths, and reserve the 1 to be joined with the tenths. We then say, times 7 are 21, which are so many tenths, because the 7 are tenths, and to these we join the 1 tenth reserved, making 22 tenths; but 22 tenths of a pound are equal to 2 pounds and 2 tenths of a pound. We therefore write the 2 tenths in the place of tens, and reserve the 2 lbs. to be united with the pounds. Lastly, we say, 3 times 4 lbs. are 12 lbs. to which we join the 2 lbs. reserved, making 14 pounds, which we write as whole numbers on the left hand of the separatrix. From this example it appears, that when one of the factors contains decimals, there will be an equal number of decimal places in the product. 120. 2. If a person travel 4.3 miles per hour, how far will he travel in 2.5 hours? 4.3 2.5 2.15 Having written the numbers as at the left hand, we 5 times 3 are 15. say Now as the 3, which is multiplied, is tenths, it is evident, that if the 5, by which it is multiplied, were units, the product, 15, would be tenths,(119) But since the 5 is only tenths of units, the product, 15, can be only 10ths of 10ths, or 100ths of units; but as 0.15 are 0.1 and 0.05, we write 5 in the place of hundredths, reserving the 1 to be joined We then say 5 times 4 are 20, which are tenths, because joining the 0.1 reserved, we have 21 tenths, equal to 2.1 miles; we therefore write 1 in the place of tenths, and 2 in the place of Ans. 10.75 miles. with the tenths. the 5 is tenths; units. We then multiply by 2, as illustrated in article 119, and write the product, 8.6, under the corresponding parts of the first product, and, adding the two partial products together, we have 10.75 miles for the distance travelled in 2.5 hours. 0.5 1 foot. 121. 3. What is the product of 0.5 ft. multiplied by 0.5 ft.? 1 foot, multiplied by itself, gives a square, measuring 1 foot on each side. 0.5 ft. by 0.5 gives a square, measuring 0.5 ft. equal to foot on each side. But the latter square, as shown by the diagram, is only 0.25, or of the former; hence 0.25 is evidently the product of 0.5 by Ans. 0.25 ft. 0.5 ft. Here we perceive that multiplication by a decimal diminishes the multiplicand, or, in other words, gives a product which is less than the multiplicand. 4. If you multiply 0.25 ft. by 0.25 ft. what will be the product? 0.25 0.25 Ans. .0625 ft. Here the operation is performed as above; but since tenths multiplied by tenths, give hundredths,(120) the 5 at the left hand of the second partial product is evidently hundredths; it is therefore necessary to supply the place of tenths with a cipher. Or the necessity of a cipher at the left of the 6, in the answer, may be shown by a diagram. A square foot being the area of a square which measures 1 foot on each side, a square 0.25, or quarter, of a foot, is a square measuring 0.25 of a foot on each side; but such a square, as 18 evident from the diagram, is only one sixteenth part of a square foot. Hence to prove that the decimal 0.0625 ft. is equal in value to one sixteenth part of a square foot, we have only to multiply it by 16 (0.0625X16=1 ft.) and the product is 1 foot. In like manner it may be shown that every product will have as many decimal places as there are decimal places in both the factors. RULE. 1 foot. 0.25 1 foot. 0.25 122. Write the multiplier under the multiplicand, and proceed in all respects as in the multiplication of whole numbers. In the product, point off as many figures for decimals as there are decimal places in both the factors counted together. Note. If there be not so many figures in the product as there are decimal places in the factors, make up the deficiency by prefixing ciphers. QUESTIONS FOR PRACTICE. 5. If a box of sugar weigh 7. What will be the weight 87.64 lb. what will 9 such box-of 13 loads of hay,each weighes weigh? ing 1108.124 lb.? 87.64 Ans. 788.76 lb. 6. What is the product of Ans. 14405.612 lb. 8. Multiply 0.026 by 0.003. Prod. 0.000078. 9. Multiply 125 by 0.008. Prod. 1. ! 10. Multiply 25.238 by 12.17. Prod. 307.14646. 11. Multiply 5 thousand by 5 thousandths. Prod. 25. 12. Twenty-five X 0.25 are how many? 13. Seven +117 X 1.024= how many? 14. 128.75+144.25 X 0.06 16.38 Ans. 15. 0.004+0.0004X0.00002 =0.000000088 Ans. 123SUBTRACTION Of decimals. ANALYSIS. 123., 1. What is difference between 43.25 rods and 22.5 roos X48.25 22.5 We write down the numbers as for Addition, with the largest uppermost. As there are no hundredths in the subtrahend, we bring down the 5 hundredths. ProceedAns. 20.75 rods. ing to the 10ths, we are unable to take 0.5 from 0.2; we therefore borrow a unit from the 3 units, which being 10 tenths, we join 10 to the 2, making 12 tenths; from which we take 5 tenths, and write the remainder, 7 tenths, in the place of tenths below the line The rest of the operation must be obvious. From 24 hours take 18.75 hours, what remains? 24. Ans. 5.25 h. Here, as we cannot take 5 from nothing, we borrow 0.10 from the 4 units, or 400 hundredths; then taking 5 (=0.05) from 0.10, the remainder is 0.05. The 400 hundredths bas now become 390 hundredths, or 39 tenths, or 3.9; then 0.7 from 0.9 leaves 0.2, and so on. RULE. 124. Write down the numbers as in Addition of Decimals, observing to place the largest number uppermost. Beginning at the right, subtract as in Simple Subtraction, (99) and place the decimal point in the remainder directly under those in the given numbers. NOTE 1.-When the numbers are all properly written, and the results correctly pointed, the decimal points will all fall in one vertical column, or directly under one another, both in Subtraction and Addition. NOTE 2.-In numbers given for Addition or Subtraction, the decimal places may all be made equal by annexing ciphers to a part of them,(116) without altering their value, and then all the decimals will express similar parts of a unit, or be of the same denomination. 125. 1. If 14.25 lb. of butter be divided into 3 equal shares, ho many pounds will there be in each? 8)14.25(4.75 12 22 21. 15 15 Here we wish to divide 14.25 into two factors, one of which shall be 3, and the other such a number as, multiplied by 3, (101) will produce 14.25. We first seek how many times 3 in 14, and find it 4 times, and 2 units over. The 2 units being 20 tenths, we join them to the 2 tenths, making 22 tenths, and, dividing these by 3, the quotient is 0.7, and 0.1 over: but 0.1 being 0.10,(116) we join the 1 to the 5 hundredths, making 0.15, and dividing by 3, the quotient is 5 hundredths. The whole quotient then is 4.75 lb. prove that this is the true quotient, we multiply it by the divisor, 3, (4.75X3=14.25) and reproduce the dividend. Since any dividend may be regarded as the product of the divisor and quotient taken as factors, (101) and since the product must have as many decimal places as are contained in both the factors,(121) follows, that the number of decimal places in the divisor and quotient, counted together, must be just equal to the number of decimal places in the dividend. To 126. 2. If 18 bushels of wheat be divided equally among 4 men, how much will each receive? 4)18(4.5 bu. 16 20 20 Here we find that 18 bushels will give each man 4 bushels, and that there will be 2 bushels left. We now add a cipher to the 2, which multiplying it by 10,(91) reduces it to tenths, and dividing 20 tenths by 4, the quotient is 0.5; each man will, therefore, receive 4.5 bushels. Hence by annexing ciphers to the remainder of a division, the operation may be continued, and in pointing the result, the ciphers annexed are to be regarded as decimals belonging to the dividend. 127. 3. What is the quotient of 0.0084 by 0.42? Omitting the ciphers, we find 42 in 84 just 2 times; 0.42 0.0084 0.02 but since there are 4 places of decimals in the divi84 Ans. dend, and only 2 in the divisor, there must be 2 places also in the quotient; we therefore place a cipher at the left of the 2 in the quotient, between it and the separatrix, to make up the deficiency. We see by this example, that if a quantity be divided by a decimal, the quotient will be larger than the dividend. RULE. 128. Write down the divisor and dividend, and divide as in whole numbers. Point off as many places for decimals from the right hand of the quotient, as the decimal places in the dividend exceed those in the divisor. NOTE 1.-If there are not so many figures in the quotient as the number of decimal places required, supply the deficiency by prefixing ciphers. 2. Should the decimal places in the divisor exceed those in the dividend, make them equal by annexing ciphers to the latter. 3.-Whenever there is a remainder after division, by annexing ciphers to it, one or more additional figures may be obtained in the quotient.(126.) QUESTIONS FOR PRACTICE. VULGAR FRACTIONS CHANGED TO DECIMALS. ANALYSIS. 129. If we divide an apple equally between 2 boys, the part which each will receive will be an apple, or the quotient of 1 divided by 2; if we divide 1 apple between 3 boys, each will receive, or the quotient of 1 divided by 3. In like manner, if 3 apples be divided between 4 boys, |