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80 level." I suppose you were thinking of the relative rates, up hill and on the level; which we might express by saying that, if they went $$\scriptstyle x$$ miles up hill in a certain time, they would go $$\scriptstyle \frac{4x}{3}$$ miles on the level in the same time. You have, in fact, assumed that they took the same time on the level that they took in ascending the hill. assumes that, when the aged knight said they had gone "four miles in the hour" on the level, he meant that four miles was the distance gone, not merely the rate. This would have been—if  will excuse the slang expression—a "sell," ill-suited to the dignity of the hero.

And now "descend, ye classic Nine!" who have solved the whole problem, and let me sing your praises. Your names are, and. (These last two I count as one, as they send a joint answer.) and  and. do not actually state that the hill-top was reached some time between 6 and 7, but, as they have clearly grasped the fact that a mile, ascended and descended, took the same time as two level miles, I mark them as "right." and deserve honourable mention for their algebraical solutions being the only two who have perceived