8. A sets out from Oxford to London at the same time that B leaves London for Oxford, the former travels 5, and the latter 6 miles an hour; now supposing Oxford to be 58 miles from London, how far from the latter place will they meet if they travel the same road? If the whole distance be divided into two parts having the proportion of 5 to 6, it is evident those parts will be the respective distances travelled. (100) As 5+ 6: 58 :: m. 6 : 317 travelled by B. 5 + 6: 58 :: 5 : 264 travelled by A. 9. A detachment sets out at 6 in the morning, marching at the rate of 14 miles an hour; 3 hours after, another detachment from the same place follows them, but their march is 24 miles an hour. In what time will the latter overtake the former; and what distance will they have marched? m. 12 × 3 = 5 the first detachment is a-head when the other begins its march. m. m. m. The difference of 21 and 1 is, what the latter gains on the former per hour. h. (98) or, as 3: 1217 the time required. And 2 × 7 17 miles the distance required. 10. The hour and minute hands of a watch are together at 12 o'clock; at what time are they next together? The minute hand moves 1 circumference on the dial plate in 1 hour; but the hour hand moves only z the difference is which the minute hand gains per hour. But at setting off at 12 o'clock we may consider the hour hand as being 1 circumference before the minute hand; Therefore the minute hand has to gain I circumference: As : 1h.:: 1:43h. = 1h. the answer. 11. There is an Island 29 miles in circumference, and three travellers all start together to travel the same way about it; A goes 3 miles per hour, B 5, and C 7; when will they all be together again? B gains 2 miles an hour upon A; m. h. m. h. Therefore as 2: 1 :: 29: 14 the time from starting when Bovertakes A. C gains 4 miles an hour upon A; m. h. m. h. Hence 4 1 29 74 the time when C overtakes A. :: And since C will overtake A at the end of every 7 hours, they will be together at the end of twice 71 hours, or 144 hours: Therefore all three will be together again at the end of 14 hours from the time of starting. 12. Suppose a clock has 4 hands, A, B, C, D; and that A goes round once in 5d. 20h. B in 7d. 14h. C in 10d. 20h. and D in 18d. 23h. Now if the hands are all together at any particular time, how long will it be before they come in conjunc tion again? Now it is evident that at the end of any number of hours which is a common multiple of 140, 182, 260, and 455 (the times of I revolution) the hands will be together again: but the least common multiple is 13 x 5 X 7 X 4 or 1920 (46); therefore in 1820 hours, Consequently at the end of every 1820 hours the hands are together at the same place. Therefore since the hands come together at every like whole multiple of 13, 10, 7, 4 revolutions (as twice, thrice, four times, &c.), it follows, that if we can find like sub-multiples or aliquot parts of 13, 10, 7, and 4, having like fractions, the hands must have been in conjunction without performing entire revolutions: Thus, if we divide 13, 10, 7, and 4 by 3, we get 4, 3, 2, 1 revolutions for the elapsed time, or 12 6063 hours the time required. Or thus. A moves, BC, and D of the circumference in 1 hour, respectively. Now if we proceed according to Examp. 7. = we have 7 of the circumference which A gains on D in 1 hour: Therefore circumf. : 1h. :: 1 circumf.: 2023h. the time in which A is overtaking D. = 720 And circumf, which B gains on D in 1 hour: τίσ : 1h.:: 1: 303}h. the time in which B is overtaking D. As Also 25 = circumf. which C gains on D in 1 hour; And 18220 : 1h. :: 1 : 6063h. the time in which C is overtaking D. Now it is evident that the least common multiple of 2027, 303}, and 6063 will be the time when A, B, and C will first overtake D together; but 6063 is the least common multiple; for twice 303} is 6063, and three times 2023 is 6053; therefore 6063 hours is the time, as before. 13. What length must be cut off a rectangular board that is 71⁄2 inches broad, to make a foot or 144 square inches? In other words--What number is that which multiplied by 7 shall make 12 times 12, or 144? 14. A garrison of 488 men have provisions for 39 weeks, bow long will those provisions last if the garrison be increased to 732 men? It is evident that the provisions will last a less time, therefore the proportion must be wrought inversely: m. w. 172. As 488 39: 732: 488 x 39 26 weeks, answer. 15. If 1000 men besieged in a town with provisions for 28 days, allowing 18 ounces a day per man, be reinforced with 600 mcn, and supposing that they cannot be relieved till the end of 42 days; how many ounces a day must each man have that the provisions may last that time. 1000 × 18 × 28 ounces, the whole quantity of provisions. This quantity is to last 1600 men 42 days. Divide by 1600, and we have 1000 X 18 X 28 ounces the quantity which must last 1 man 42 days: this divided by 42 will give the allowance per day for 1 man: viz. for 40 miles; what will be the expence of 6ton 17cwt. for 94 miles at the same rate? 17. If a company of 160 men in six days of 11 hours each, can dig a trench 230 yards long, 5 wide, and 11⁄2 deep; in how many days of 8 hours long would another company consisting of 96 men dig a trench 220 yards long, 3 wide, and 1 deep; supposing the hardness of the ground in the former case is to that in the latter as 5 to 7, and that 4 men of the latter company can do as much work as 5 of the former in the same time? 230 × 5 × 118974 (by mensuration) the cubic yards in the first trench. 220 × 3 × 1 = 770 the cubic yards in the other. Now if we suppose the labour necessary to raise a like quantity of earth to be directly proportional to the hardness of the ground, it is evident that the strength required to dig the former trench, will be to that required for yds. yds. the latter, as 18971 × 5 to 770 x 7. m. m. m. m. And, as 4: 5 :: 96: 120, therefore the labour of 120 men of the first company is equal to that of the 96 men. Hence the question is reduced to the following. yds. If 160 men in 66 Hours (6 x 11) can dig 18974 × 5; in what time would 120 men men could dig in 66 hours. |