3. A gentleman put out $49,103 on interest at 7% for the benefit of his three sons, aged 16, 17, and 18 years respectively, dividing it in such a manner that each, as he became of age, should receive the same amount; what were the shares of each? Ans. $15,488; $16,335; $17,280. CASE V. 1. It is between 10 and 11 o'clock, and the minute-hand of the clock is į as far after 12 as the hour-band is before it. What is the time of day? SOLUTION.-At 10 o'clock the hour-hand was at 10 and the minutehand at 12. Since that time, the minute-hand has moved of the hourhand's distance from 12, and the hour-hand its of } of that distance, or 14 of that distance. Then 14+2=1 of that distance=the distance from 10 to 12=10 minute-spaces, and as of 10 minute-spaces, and 3} or ļof the hour-hand's distance, which is the minute-hand's distance =}} of 10 minute-spaces=4spaces, and the time is 4 min. 48 seconds after 10 o'clock. 2. It is between 1 and 2 o'clock, and the minute-hand is as far past 2 as the hour-hand is before it; required the time. Ans. 13 min. 501% sec. after 1. 3. It is between 3 and 4 o'clock, and the minute-band is * as far before 12 as the hour-band is after it; required the time. Ans. 53 min. 3031 sec. past 3. CASE VI. 1. Suppose the hour-hand, minute-hand, and second-hand of a clock all turn upon the same centre; they will be together at 12 o'clock; how long before each band respectively will be half-way between the other two? 1 2 3 SOLUTION.–Case 1. Let A be 12 on the dial, and B, C, and D, the positions of the hour-hand, minute-hand, and second-hand respectively, the second-hand being equally distant from the other two. Then, while the hour-hand passes over a certain space, the minute-hand passes over 12 times that space, and the second-hand 720 times the space. Hence AC =12 AB, and ACD=720AB. But BD=CD=720AB-12AB=708AB, and BC=AC-AB=11AB; hence the whole circumference=CD+BD+ BC=708AB+708 AB+11AB=1427AB; hence AC, the space passed over by the minute-hand, is 143, of 3600 seconds, or 304297 seconds. CASE 2. The hour-hand being between the other two hands, we have BD=BC=11AB; hence AD=10AB, and AD+ACD=10AB+720AB= 730AB, the whole circumference; therefore AC=130 of 3600 seconds, or 5941 seconds. CASE 3. When the minute-hand is half-way between the other two, the second-hand will have gone once round the face of the clock; hence, since CD=BC=11AB, AD=AC+CD=23AB, and the circumference+ 23AB=720AB; hence the circumference=697 AB, and 12AB=ity of 3600 seconds, or 6168 seconds. 2. The three hands turning upon the same centre, how long will it be after 12 o'clock before the bour and second hands, the minute and second hands, and the hour and minute bands, will be together again? Ans. 60-415 seconds; 613 seconds ; 655 minutes. 3. How long will it be after 12 o'clock before the hour and second hands, the minute and second hands, and the hour and minute hands, are at right angles with each other? Ans. 1571 seconds; 1515 seconds; 161 minutes. 4. How long will it be after 12 o'clock before the hour and second hands, the minute and second hands, and the hour and minute bands, are exactly opposite each other? Ans. 30-315 seconds; 3038 seconds; 324 minutes. CASE VII. 1. If 6 acres of grass, together with what grows on the 6 acres during the time of grazing, keep 16 oxen 12 weeks, and 9 acres keep 26 oxen 9 weeks, how many oxen will 15 acres keep 10 weeks, the grass growing uniformly all the time? SOLUTION.-If 6 acres+the growth of 6 acres for 12 weeks, keep 16 oxen 12 weeks, one acre+the growth of 1 acre for 12 weeks, will support 16 oxen of 12 weeks, or 2 weeks, or 32 oxen 1 week. In the same manner, 1 acre+the growth of 1 acre for 9 weeks, will keep 26 oxen 1 week; subtracting these, we have the growth of 1 acre for 3 weeks will keep 6 oxen for 1 week, and the growth of 1 acre for 9 weeks will keep 18 oxen for 1 week; subtracting this latter expression from 1 acre W +the growth of 1 acre for 9 weeks, we have 1 acre, without the growth, will keep 8 oxen 1 week; hence, 15 acres will keep 120 oxen 1 week, or to of 120, or 12 oxen 10 weeks; again, since the growth of one acre for 3 weeks will keep 6 oxen 1 week, the growth of 1 acre for 1 eek will keep 2 oxen 1 week, hence the growth of 15 acres for 10 weeks will keep 30 oxen for 10 weeks; and adding, we have 15 acres + the growth of 15 acres for 10 weeks, will keep 42 oxen 10 weeks. 12 9 1 1 1 1 1 10 10 10 2. If 5 acres of grass, together with what grows on them during the time of grazing, keep 20 oxen 10 weeks, and 8 acres keep 29 oxen 16 weeks, how many weeks will 15 acres keep 70 oxen? Ans. 6 weeks. 3. If 10 acres of grass keep 48 oxen 15 weeks, and 7 acres keep 34 oxen 14 weeks, how many acres will keep 38 oxen 16 weeks, the grass growing uniformly all the time? Ans. 8 acres. CASE VIII. 1. What number divided by 13, leaves 12 for a remainder, by 7 leaves 3, by 6 leaves 5, and by 5 leaves 2 ? SOLUTION.--It is evident that 13+12, or 25, will satisfy the first condition. Dividing 25 by 7, the remainder is 4, which is greater than the required remainder; hence we must add to 25 a number which, divided by 7, will leave a remainder that increased by 4, will contain 7 once, with a remainder of 3. This remainder is 6. But the number added must also contain 13, and both conditions are fulfilled by 13 itself; hence 25+13, or 38, fulfills the first two conditions. Dividing 38 by 6, the remainder is 2, which is 3 less than the required remainder; hence we must add to 38 a number which, divided by 6, will leave a remainder of 3, and the number added must also be a multiple of 13 and 7. The least multiple of 13 and 7 is 91, which, divided by 6, leaves 1 ; hence to get a remainder of 3, we must add 3 times 91, or 273. 38+273 311, which satisfies the first three conditions. Continuing the operation in the same manner, we find that 857 is the least number that will satisfy all the conditions. Other numbers can be found by adding to 857 any common multiple of all the divisors. 2. What number divided by 11, leaves 10 remaining, by 9 leaves 7, by 7 leaves 6, by 4 leaves 3 ? Ans. 1231. 3. What number divided by 15, leaves 8 for a remainder, by 13 leaves 7, by 11 leaves 9, by 7 leaves 5 ? Ans. 7313. SECTION XVI. MISCELLANEOUS EXAMPLES. These problems are designed both as a review of the work and a test of the knowledge and arithmetical skill of the pupil. They are to be used in accordance with the needs of the pupil and the judgment of the teacher. 1. If Henry's capital is 20% less than William's, how many % is William's more than Henry's ? Ans. 25% 2. I wish to put 20 hogsheads of ale (54 gallons) into 10 empty wine pipes; what must be the capacity of a cask which shall contain what is left over ? Ans. 5844 wine gal. 3. A bought stock 5% below par value, and sold it 5% above par, and gained $550 ; what was the par value of the stock? Ans. $5500. 4. Mr. Russell bought stocks 21% above par, and was obliged to sell 24% below par, and lost $235; what did the stocks cost him ? Ans. $4817.50. 5. A asked at one time 33}% less for an article than cost, but afterwards sold it for 331% more than this price; required the loss per cent. Ans. 113%. 6. What must I ask for a house that cost me $7520, that after falling 6% on the price, I may gain 184% on the cost? Ans. $9500. 7. A grocer asked for flour 35% more than cost, but sold it for 66% of his asking price; required the loss per cent. Ans. 10%. 8. What must be asked for a farm which cost $8160, so that after raising the price 331%, I may gain 81% on the cost ? Ans. $6630. 9. A's gain at wholesale is 121%, and his retail price is 5% more than his wholesale price; required the gain per cent. at retail. Ans. 181% 10. B's gain at retail is 121%, and his wholesale price is 21 % less than his retail price; required the gain per cent. at wholesale. Ans. 94% 11. A barrel of molasses lost 20% by leakage, and the remainder was sold at a gain of 20% ; required the gain or loss per cent. Ans. 4% loss. 12. A log 19 in. thick is sawn into 15 boards, each izin. thick; what % of the board is wasted ? Ans. 1138% 13. A merchant lost 20% of his goods, and sold the remainder for 331% more than cost, and gained $250.75; what did his goods cost? Ans. $3761.25. 14. A man lost 25% of a purchase; what must he gain per cent. on the remainder, that he may gain 25% on the whole ? Ans. 663%. 15. A bought a house and barn, paying 3 times as much for the house as for the barn ; if he had paid 121% more for the bouse, it would have cost $4725}; what was the cost of each ? Ans. House, $4200.44; barn, $1400.148. 16. A borrowed of B a certain sum ; 374% of the debt is $45.84, which is 663% of what has been repaid ; how much does A still owe? Ans. $53.48. 17. If 7 horses or 6 cows eat 56 tons of bay in 17 days, how long will it take 6 horses and 7 cows to eat the same quantity ? Ans. 8} days. 18. If stock bought at 10% above par pays 8% on the investment, what per cent. will it pay if bought at 10% discount? Ans. 98%. 19, 12 men can do a piece of work in 85 days; how long may 3 men remain away, and the work be finished in the same time by their bringing 7 more with them? Ans. 671 days. 20. What will it cost to paper the walls of a room 25.5 ft. long, 14.5 ft. wide, and 9.25 ft. high, a roll of paper being 8 yards long and of a yard wide, and costing 75 cents a roll ? Ans. $12.33. 21. If an important vote in the English Parliament is taken at 1 h.30 min. A. M., Jan. 15th, and telegraphed immediately to San Francisco, 122° 26' 15' W., at what hour will it be received ? Ans. 5 h. 20 min. 15 sec. P. M., Jan. 14th. 22. At a certain time between 5 and 6 o'clock, the minute |