MISCELLANEOUS PROBLEMS. 1. How many times will a wheel 11 ft. 6 in. in circumference revolve in going 50 miles? Ans. 2295623 times. 2. I have a watch which is 188 carats fine; how much pure gold is there in it? Ans. 3 of the whole. 3. My watch is į and my chain { pure gold; how many carats fine is each ? Ans. 18 carats; 16 carats. 4. How long will it take to count 6 millions at the rate of 80 a minute, working 10 hours a day? Ans. 125 days. 5. A young lady weighs 125 lb. Troy weight; how much does she weigh Avoirdupois weight? Ans. 1029 lb. 6. What is the weight of the silver in English silver coin worth $447.718? Ans. 25 lb. 9 oz. 8 pwt. 21.54 gr. 7. What is the difference in the weight of $960 in gold coin or in silver half-dollars? Ans. 59 lb. 8 oz. 8 pwt. 8. How many demijohns can be filled from 4 hhd. of wine, each demijohn holding 2 gal. 3 qt. 1 pt. ? Ans. 8725. 9. The Gregorian calendar adds 97 days in 400 years; how long will it require to gain a day? Ans. 3874,283 years. 10. What day of the week and what day of the year is the 4th of July in a common year which begins on Monday? 11. A captain of a vessel, taking the soundings, found the water to be 760 fathoms in depth; what part of a mile was it? Ans. 12. 12. How many copies of a duodecimo book can be printed on 66 reams 15 quires of paper, using 10 sheets per volume ? Ans. 3204. 13. A grocer bought 100 bushels of coarse salt (50 lb. to a bushel) at 62¢, and sold it at 1% of a cent a lb.; what was his gain? Ans. $31.75. 14. Find the number of minims in the following prescription: Tincture of digitalis Mxv, distilled vinegar f3j, syrup f3j, water fzjss. Ans. 855m. 15. The wheels of a locomotive are 15 ft. 3 in. in circumference and make 5 revolutions a second; in what time will the locomotive run 75 miles ? Ans. 1 h. 26 min. 3377 sec. 16. Ilow many quart, pint, and half-pint bottles, of each an equal number, may be filled from a vessel containing 54 gal. 1 qt. ? Ans. 124 of each. 17. A boy brought from the bank a bag of gold weighing 1 lb. 9 oz. 10 pwt. ; required its value. Ans. $400. 18. A man bought 50 cords of wood, 3 ft. long, and proposes to put it in a pile 12 ft. high ; how long will the pile be ? Ans. 152.8 ft. 19. William rises 40 minutes earlier and retires 30 minutes later than his companion; how much time will he gain in 4 school sessions of 26 weeks each? Ans. 35 da. 9 h. 20 min. 20. An apothecary bought 50 lb. 8 oz. of opium at 45 cents an ounce Avoirdupois, and sold it at 3 cents a scruple ; did he gain or lose, and how much? Ans. Gained $166.65. 21. A man bought 2 cwt. 87 lb. 10 oz. of sugar, at 6 cts. a pound, and retailed it at 6 cts. a pound, using by mistake Troy weights; how much did he make? Ans. $5.464 22. In the following prescription find the number of grains of the solid aud the number of minims of the liquid part: R. Mellis, confectionis rosæ caninæ aa zij; aceti distillati fziij; acidi hydrochlorici mxxx; aquæ rosæ fzj; aquæ puræ fzvj. Misce. Ans. 240 gr.; 4830 m. 23. A man having a piece of ground which he wished accurately surveyed, employed two surveyors, the first of wbom reported its contents to be 2 A. 54 P. 64 in.; the other reported 2 A. 53 P. 30 yd. 2 ft. 100 in. Wishing to know which was right, he employed a third, who gave the area as 2 A. 53 P. 272 ft. 100 in. How much did the three estimates differ from one another ? 24. The distance from Boston to Chicago being about 1040 miles, if one man should start from Boston and one from Chicago on Monday, November 3, 1873, the first traveling 2 miles 299 rd. 12 ft. per hour, and the second 3 miles 20 rd. 41 ft. per hour, both starting at 9 A. M., traveling 7 hours a day, and resting on the Sabbath, at what time will they meet, and how far will each have traveled ? Ans. Dec. 1st, 2 h. 20 min. P. M. OPERATION. S. 13 ADDITION OF COMPOUND NUMBERS. 430. Addition of Compound Numbers is the process of finding the sum of two or more similar compound numbers. 1. Find the sum of £9 13 s. 11 d.; £17 15 s. 9 d.; £15 12 s. 5 d.; and £23 11 s. 10 d. SOLUTION.–We write the numbers so that similar units shall stand in the same column, and begin at £ d. the right to add. 10d. plus 5 d., plus 9 d., plus 11 d. 9 11 are 35 d., which by reduction we find equals 2 s. and 17 15 9 11 d.; we write the 11 d. under the column of pence 15 12 5 and reserve the 2 s. to add to the column of shillings: 23 11 10 2 s. plus 11 s., plus 12 s., plus 15 s., plus 13 s. are 53 s., 66 13 11 which by reduction we find equals £2 and 13 s.: we write the 13 s. under the column of shillings and reserve the £2 to add to the column of pounds: £2 plus £23, plus £15, plus £17, plus £9, equals £66, which we write under the column of pounds. Rule.-I. Write the compound numbers so that similar units stand in the same column. II. Begin with the lowest denomination and add each column separately, placing the sum, when less than a unit of the next higher denomination, under the column added. III. When the sum equals one or more units of the next higher denomination, reduce it to this denomination, write the remainder under the column added, and add the quotient obtained by reduction to the next column. IV. Proceed in the same manner with all the columns to the last, under which write the entire sum. Proof.—The same as in addition of simple numbers. NOTE.- Addition of compound numbers is the same in principle as the addition of simple numbers. In each we carry for the number of units in the lower denomination which makes a unit of the next higher. The apparent difference is in their scales. In simple numbers the expression shows how much to carry ; in denominate numbers we must reduce to see what to carry. (2) (3) (4) mi. rd. yd. A. P. sq. yd deg. mi. rd. ft. in. 187 319 4 789 109 27 27 56 148 15 9 269 227 2 891 143 19 32 43 223 12 6 387 158 3 134 79 17 45 57 316 16 10 578 269 1 234 108 27 56 65 267 14 11 465 217 3 678 157 18 34 68 318 12 11 (6) Ib. 3 3 9 gr. 28 8 5 2 16 37 7 6 1 12 42 10 3 0 15 96 11 7 2 13 78 10 2 11 C. 216 135 738 217 392 (6) T. cwt. lb. oz. 25 16 68 11 43 12 40 14 67 15 23 12 85 17 92 15 61 19 14 13 8. Find the sum of 25 lb. 7 oz. 15 pwt. 20 gr.; 78 lb. 11 oz. 19 pwt. 23 gr.; 34 lb. 9 oz. 12 pwt. 15 gr.; 60 lb. 10 oz. 3 pwt. 4 gr.; 17 lb. 6 oz. 18 pwt. 22 gr. Ans. 217 lb. 10 oz. 10 pwt. 12 gr. 9. Find the sum of 21 mi. 67 ch. 3 rd. 21 li.; 28 mi. 78 ch. 2 rd. 23 li.; 47 mi. 6 cb. 2 rd. 18 li.; 56 mi. 59 ch. 2 rd. 16 li.; 25 mi. 38 ch. 3 rd. 23 li.; 46 mi. 75 ch. 2 rd. 21 li. Ans. 227 mi. 7 ch. 2 rd. 22 li. 10. Find the sum of 145 sq. yd. 7 sq. ft. 116 sq. in. ; 218 sq.yd. 3 sq. ft. 141 sq. in.; 317 sq. yd. 6 sq. ft. 108 sq. in.; 419 sq. yd. 5 sq. ft. 132 sq. in. ; 381 sq. yd. 4 sq. ft. 136 sq. in. Ans. 1483 sq. yd. 2 sq. ft. 57 sq. in. 11. Find the sum of 37 mi. 275 rd. 3 yd. 2 ft. 10 in.; 42 mi. 228 rd. 2 yd. 1 ft. 8 in.; 56 mi. 317 rd. 1 yd. 2 ft. 7 in.; 76 mi. 141 rd. 5 yd. 2 ft. 11 in. ; 85 mi. 272 rd. 4 yd. 1 ft. 10 in. Ans. 299 mi. 276 rd. 2 yd. 1 ft. 4 in, OPERATION. SUBTRACTION OF COMPOUND NUMBERS. 431. Subtraction of Compound Numbers is the process of finding the difference between two similar compound numbers. 1. From 33 oz. 14 pwt. 23 gr., take 17 oz. 16 pwt. 11 gr. SOLUTION.-We write the subtrahend under the minuend, placing similar units in the same column, oz. pwt. gr. and begin at the lowest denomination to subtract. 33 14 23 11 gr. from 23 gr. leaves 12 gr., which we write under 17 16 11 the grains: 16 pwt. from 14 pwt. we cannot take, we will therefore take 1 oz. from the 33 oz., leaving 15 18 12 32 oz.; 1 oz. equals 20 pwt., which added to 14 pwt., equals 34 pwt.; 16 pwt. from 34 pwt. leaves 18 pwt., which we write under the pwt.: 17 oz. from 32 oz. (or, since it will give the same result, we may add 1 oz. to the 17 oz. and say 18 oz. from 33 oz.) leaves 15 oz. Hence the following Rule.-I. Write the subtrahend under the minuend so that similar units stand in the same column. II. Begin with the lowest denomination and subtract each term of the subtrahend from the corresponding term of the minuend. III. If any term of the subtrahend exceeds the corresponding term of the minuend, add to the latter as many units of that denomination as make one of the next higher, and then subtract; add 1 also to the next term of the subtrahend before subtracting. IV. Proceed in the same manner with each term to the last. Proof.—The same as in subtraction of simple numbers. NOTE.—The pupil will notice that the general principle of subtraction is the same as in simple numbers, the difference being in the irregularity of the scale, the units themselves being expressed in the decimal scale. (3) (4) £ lb. oz. pwt. gr. mi. rd. 56 18 5 3 48 10 18 13 72 45 2 1 22 18 7 1 27 11 12 18 48 272 4 2 d. qr. yd. ft. mi. 48 23 (5) A. (6) P. sq. yd. sq. ft sq. in. 147 00 00 00 155 30 3 71 7. From 28 deg. 160 rd. 1 ft., subtract 16 deg. 69 mi. 232 rd. 5 yd. 2 ft. 7 in. Ans. 10 deg. 69 mi. 29 rd. 2 yd. 6 in. 8. From 1 circumference subtract 358 deg. 69 mi. 159 rd. 5 yd. 1 ft. 5 in. Ans. 68 mi. 262 rd. 2 yd. 8 in. 9. A has a field 15 rd. 5 yd. 2 ft. 11 in. long, B has one 16 rd. 1 ft. 4 in. long; which is the longer field and how much? Ans. A's, 1 inch. 10. From 56 A. 97 P. 8 sq. ft. 112 sq. in., take 49 A. 159 P. 30 sq. yd. 8 sq. ft. 120 sq. in. Ans. 6 A. 97 P. 2 sq. ft. 28 sq. in. |