1804: prob. 71, p. 140. Lincoln is 100 miles from London. Travellers set out at the same time and meet after seven hours, when they find that A has gone 1½ mph faster than B. What are their rates?
Bonnycastle. Algebra. 1782.
P. 84, no. 5. MR-(8, 7; 150). P. 85, no. 19 (1815: p. 107, no. 31). Same as Euler's no. 25.
Pike. Arithmetic. 1788. P. 350, no. 17. Circle 268 in circumference. Two men start at ends of a diameter and go in the same direction at rates 22/2 and 34/3. When and where do they meet? I. e. O-(34/3, 22/6), D = 134.
Eadon. Repository. 1794.
P. 80, no. 36. O-(30, 42), T = 4. P. 80, no. 37. MR-(23, 31; 162). P. 235, ex. 3. 5 + 9 + 13 + ... + (5 + 13x4). Find total travel.
John King, ed. John King 1795 Arithmetical Book. Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995.
P. 109. A meeting problem, but the times to meeting and the meeting point are given, so it reduces to: a + a+2 + a+4 + a+6 + a+8 = 50 = b + b+3 + b+6 + b+9. Pp. 117-118. Hare is 144 hare leaps ahead of a grayhound. Hare makes 4 leaps while grayhound makes 3, but grayhound leaps are 3/2 as big.
Thomas Manning. An Introduction to Arithmetic and Algebra. 2 vols., Nicholson, Lunn & Deighton, Cambridge, 1796 & 1798. Vol. 1, pp. 208-210. O-(5/2, 7/2), T = 2. Then considers what would happen if the second traveller went slower than the first [more simply, suppose the second traveller had the headstart, i.e. T is negative]. This gives a negative solution and he interprets this as that they must have met before the starting time. He gives a general solution and discussion of the problem. ??NX.
Hutton. A Course of Mathematics. 1798?
Prob. 5, 1833: 210-211; 1857: 214-215. Hare is 60 hare-leaps ahead of a greyhound. She makes 9 leaps while the hound does 6, but 7 hare-leaps are as long as 3 hound-leaps. = Lauremberger. Prob. 9, 1833: 213-214; 1857: 217-218. O-(31½/5, 22½/3), T = 8. Then does O (m/t, m'/t') with first having headstart of T. Prob. 20, 1833: 221; 1857: 225. M-(8, 7, 150). Remarks upon Equations of the First Degree, 1833: 224-231; 1857: 228-235, is an extensive discussion concerning possible negative roots and considers O-(m, n) with the second having a headstart of distance D. When m < n, he says the directions must be reversed.
D. Adams. Scholar's Arithmetic. 1801.
P. 134, no. 6. O-(20, 25), T = 5. P. 208, no. 64. Hare is 12 rods (= 198 ft) ahead of a hound and goes for 45 sec before the hound starts, running at 10 mph. Hound then starts at 16 mph. How long until the hound catches the hare and how far does the hound go? = O (44/3, 352/15), D = 858, using ft and sec.
Robert Goodacre. Op. cit. in 7.Y. 1804. Miscellaneous Questions, no. 125, p. 205 & Key p. 269. A goes 4 mi/hr for 7 hrs each day. B starts a day later at 5 mi/hr for 8 hours each day, both starting at the same time each morning. When does B overtake?
Silvestre François Lacroix. Élémens d'Algèbre, a l'Usage de l'École Centrale des Quatre-Nations. 14th ed., Bachelier, Paris, 1825. Sections 64-75, pp. 94-110 plus Addition, pp. 359-360. Discusses general problems MR-(a, b; D) and O-(a, b) and considers negative solutions and what happens when the divisor is zero!
Augustus De Morgan. Arithmetic and Algebra. (1831 ?). Reprinted as the second, separately paged, part of: Library of Useful Knowledge -- Mathematics I, Baldwin & Craddock, London, 1836. Art. 116, pp. 30-31. Hare is 80 hare-leaps ahead of a greyhound. Hare makes 3 leaps for every 2 of the hound, but a hound-leap is twice as long as a hare leap. Then considers a hound-leap as n/m of a hare-leap. Takes n/m = 4/3 and finds a negative solution which he discusses. Takes n/m = 3/2 and finds division by zero which he interprets as the hound never catching the hare.
Bourdon. Algèbre. 7th ed., 1834.
Art. 47, prob. 3, pp. 65-66. Same as Hutton, 1798?, pp. 210-211, with greyhound chasing a fox. = Lauremberger. Art. 190, question 6, p. 319. O-(10, 0; 3, 2).
D. Adams. New Arithmetic. 1835.
P. 243, no. 81. O-(6, 8) with headstart of distance 24. Pp. 243-244, no. 82. Hare is distance 12 and time 5/4 ahead of a hound. Hare runs 36 and hound runs 40. When and where does the hound catch the hare?
Augustus De Morgan. On The Study and Difficulties of Mathematics. First, separately paged, part of: Library of Useful Knowledge -- Mathematics I, Baldwin & Craddock, London, 1836. P. 30 mentions O-(2, 3), T = 4 and O-(a, b) with second delayed by time T. Pp. 37-39 discusses the general courier problems O-(m, n) with second having headstart of distance D and MR-(m, n; D). He considers different signs and sizes, getting six cases.
Unger. Arithmetische Unterhaltungen. 1838. Pp. 135 & 258, nos. 515 & 516. MR (5/4, 3/5; 57) with first delayed by time 2½. The second problem asks what delay of time for the first will make them meet at the half-way point?
Hutton-Rutherford. A Course of Mathematics. 1841?
Prob. 7, 1857: 81. Two persons on opposite sides of a wood of circumference 536, start to walk in the same direction at rates 11, 11⅓. How many times has the wood been gone round when they meet? = O-(11, 11⅓), D = 268. Answer is the number of times the faster has gone round. Prob. 23, 1857: 82. MR-(3, 4; 130), T = 8. Prob. 37, 1857: 83. Hare starts 40 yd and 40 sec in front of a hound. Hare goes 10 mph (= 44/9 yd/sec) and hound goes 18 mph (= 44/5 yd/sec). I.e. O-(44/9, 44/5), D = 40, T = 40.
Tate. Algebra Made Easy. Op. cit. in 6.BF.3. 1848. P. 46, no. 15. Man makes a journey at 4 mph and returns at 3 mph, taking 21 hours in total. How far did he go?
Family Friend 1 (1849) 122 & 150. Arithmetical problems -- 1. "A hare starts 40 yards before a greyhound, and is not perceived by him till she has been up 40 seconds: she gets away at the rate of 10 miles an hour and the dog pursues her at the rate of 18 miles an hour: how long will the course last, and what distance will the hare have run?" = Illustrated Boy's Own Treasury, 1860, Prob. 4, pp. 427 & 431. = Hutton, c1780?, prob. 68.
Anonymous. A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools. 3rd ed., 1850. Op. cit. in 7.H.
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