The writer’s insinuation that I care not how much annoyance I give to my readers I think it best to pass over in silence; but to his concluding remark I must entirely demur. I hold that to use language likely to annoy any of my correspondents would not be in the least justified by the plea that I was “quite certain of being correct.” I trust that the knot-untiers and I are not on such terms as those!
I beg to thank G. B. for the offer of a puzzle—which, however, is too like the old one “Make four 9’s into 100.”
Answers to Knot VIII
§1: The Pigs
Problem.—Place twenty-four pigs in four sties so that, as you go round and round, you may always find the number in each sty nearer to ten than the number in the last.
Answer.—Place 8 pigs in the first sty, 10 in the second, nothing in the third, and 6 in the fourth: 10 is nearer ten than 8; nothing is nearer ten than 10; 6 is nearer ten than nothing; and 8 is nearer ten than 6.
This problem is noticed by only two correspondents. Balbus says “it certainly cannot be solved mathematically, nor do I see how to solve it by any verbal quibble.” Nolens Volens makes Her Radiancy change the direction of going round; and even then is obliged to add “the pigs must be carried in front of her”!
§2: The Grurmstipths
Problem.—Omnibuses start from a certain point, both ways, every 15 minutes. A traveller, starting on foot along with one of them, meets one in 12½ minutes: when will he be overtaken by one?
Answer.—In 6¼ minutes.
Solution.—Let “a” be the distance an omnibus goes in 15 minutes, and “x” the distance from the starting-point to where the traveller is overtaken. Since the omnibus met is due at the starting-point in 2½ minutes, it goes in that time as far as the traveller walks in 12½; i.e. it goes 5 times as fast. Now the overtaking omnibus is “a” behind the traveller when he starts, and therefore goes “a + x” while he goes “x.” Hence a+x=5x; i.e. 4x=a, and x=a4. This distance would be traversed by an omnibus in 154 minutes, and therefore by the traveller in 5×154. Hence he is overtaken in 18¾ minutes after starting, i.e. in 6¼ minutes after meeting the omnibus.
Four answers have been received, of which two are wrong. Dinah Mite rightly states that the overtaking omnibus reached the point where they met the other omnibus 5 minutes after they left, but wrongly concludes that, going 5 times as fast, it would overtake them in another minute. The travellers are 5-minutes-walk ahead of the omnibus, and must walk ¼th of this distance farther before the omnibus overtakes them, which will be ⅕th of the distance traversed by the omnibus in the same time: this will require 1¼ minutes more. Nolens Volens tries it by a process like “Achilles and the Tortoise.” He rightly states that, when the overtaking omnibus leaves the gate, the travellers are ⅕th of “a” ahead, and that it will take the omnibus 3 minutes to traverse this distance; “during which time” the travellers, he tells us, go ¹⁄₁₅th of “a” (this should be ¹⁄₂₅th). The travellers being now ¹⁄₁₅th of “a” ahead, he concludes that the work remaining to be done is for the travellers to go ¹⁄₆₀th of “a,” while the omnibus goes ¹⁄₁₂th. The principle is correct, and might have been applied earlier.
Class List.
Balbus.
Delta.
Answers to Knot IX
§1: The Buckets
Problem.—Lardner states that a solid, immersed in a fluid, displaces an amount equal to itself in bulk. How can this be true of a small bucket floating in a larger one?
Solution.—Lardner means, by “displaces,” “occupies a space which might be filled with water without any change in the surroundings.” If the portion of the floating bucket, which is above the water, could be annihilated, and the rest of it transformed into water, the surrounding water would not change its position: which agrees with Lardner’s statement.
Five answers have been received, none of which explains the difficulty arising from the well-known fact that a floating body is the same weight as the displaced fluid. Hecla says that “only that portion of the smaller bucket which descends below the original level of the water can be properly said to be immersed, and only an equal bulk of water is displaced.” Hence, according to Hecla, a solid, whose weight was equal to that of an equal bulk of water, would not float till the whole of it was below “the original level” of the water: but, as a matter of fact, it would float as soon as it was all under water. Magpie says the fallacy is “the assumption that one body can displace another from a place where it isn’t,” and that Lardner’s assertion is incorrect, except when the containing vessel “was originally full to the brim.” But the question of floating depends on the present state of things, not on past history. Old King Cole takes the same view as Hecla. Tympanum and Vindex assume that “displaced” means “raised above its original level,” and merely explain how it comes to pass that the water, so raised, is less in bulk than the immersed portion of bucket, and thus land themselves—or rather set themselves floating—in the same boat as Hecla.
I regret that there is no Class-list to publish for this Problem.
§2: Balbus’ Essay
Problem.—Balbus states that if a certain solid be immersed in a certain vessel of water, the water will rise through a series of distances, two inches, one inch, half an inch, etc., which series has no end. He concludes that the water will rise without limit. Is this true?
Solution.—No. This series can never reach
