Tag Archives: travel puzzles

Catching the Thief

This typical problem from the prolific H. E. Dudeney may be a bit tricky at first.

“104.—CATCHING THE THIEF.

“Now, constable,” said the defendant’s counsel in cross-examination,” you say that the prisoner was exactly twenty-seven steps ahead of you when you started to run after him?”

“Yes, sir.”

“And you swear that he takes eight steps to your five?”

“That is so.”

“Then I ask you, constable, as an intelligent man, to explain how you ever caught him, if that is the case?”

“Well, you see, I have got a longer stride. In fact, two of my steps are equal in length to five of the prisoner’s. If you work it out, you will find that the number of steps I required would bring me exactly to the spot where I captured him.”

Here the foreman of the jury asked for a few minutes to figure out the number of steps the constable must have taken. Can you also say how many steps the officer needed to catch the thief?”

Answer.

See Catching the Thief for solutions.

Three Runners Puzzle

Here is another problem from Presh Talwalkar which he says is adapted from India’s Civil Services Exam.

“There are three runners X, Y, and Z. Each runs with a different uniform speed in a 1000 meters race.  If X gives Y a start of 50 meters, they will finish the race at the same time.  If X gives Z a start of 69 meters, they will finish the race at the same time.  Suppose Y and Z are in a [1000 meter] race. How much of a start should Y give to Z so they would finish the race at the same time?”

Even though Talwalkar’s original graphic showed all the runners in a 1000 meter race, it was not immediately clear to me from the wording that the race between Y and Z was also 1000 meters.  But that was the case, so I made it explicit.

Answer.

See the Three Runners Puzzle for solutions.

Rufus Puzzle

Again we have a puzzle from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).  This one is quite a bit more challenging, at least for me.

“When Holmes and I met with Wiggins one afternoon, he was accompanied by a rather scrappy-looking mutt, who eyed me with evident suspicion.

‘This is Rufus,’ Wiggins said. ‘He’s a friend.’

‘Charmed,’ I said.

‘He’s very energetic,’ Wiggins told us. ‘Just this morning, he and I set out for a little walk.’

At the word ‘walk’, the dog barked happily.

‘When we set out, he immediately dashed off to the end of the road, then turned round and bounded back to me. He did this four times in total, in fact. After that, he settled down to match my speed, and we walked the remaining 81 feet to the end of the road at my pace. But it seems to me that if I tell you the distance from where we started to the end of the road, which is 625 feet, and that I was walking at four miles an hour, you ought to be able to work out how fast Rufus goes when he’s running.’

‘Indeed we should,’ said Holmes, and turned to look at me expectantly.

What’s the dog’s running speed?”

Answer.

See the Rufus Puzzle for solutions.

Tree Trunk Puzzle

Here is another problem (slightly edited) from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).

“Holmes and I were walking along a sleepy lane in Hookland, making our way back to the inn at which we had secured lodgings after scouting out the estates of the supposed major, C. L. Nolan. Up ahead, a team of horses were slowly pulling a chained tree trunk along the lane. Fortunately it had been trimmed of its branches, but it was still an imposing sight.

When we’d overtaken the thing, Holmes surprised me by turning sharply on his heel and walking back along the trunk. I stopped where I was to watch him. He continued at a steady pace until he’d passed the last of it, then reversed himself once more, and walked back to me.

‘Come along, old chap,’ he said as he walked past. Shaking my head, I duly followed.

‘It took me 140 paces to walk from the back of the tree to the front, and just twenty to walk from the front to the back,’ he declared.

‘Well of course,’ I said. ‘The tree was moving, after all.’

‘Precisely,’ he said. ‘My pace is one yard in length, so how long is that tree-trunk?’

Can you find the answer?”

Answer.

See the Tree Trunk Puzzle for solutions.

The Track Problem

Again we have a puzzle from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).

“Our pursuit of the dubious Alan Grey, whom we encountered during The Adventure of the Third Carriage, led Holmes and myself to a circular running track where, as the sun fell, we witnessed a race using bicycles. There was some sort of substantial wager involved in the matter, as I recall, and the track had been closed off specially for the occasion. This was insufficient to prevent our ingress, obviously.

One of the competitors was wearing red, and the other blue. We never did discover their names. As the race started, red immediately pulled ahead. A few moments later, Holmes observed that if they maintained their pace, red would complete a lap in four minutes, whilst blue would complete one in seven.

Having made that pronouncement, he turned to me. ‘How long would it be before red passed blue if they kept those rates up, old chap?’

Whilst I wrestled with the answer, Holmes went back to watching the proceedings. Can you find the solution?”

Answer.

See the Track Problem for a solution.

Calculating on the Way

In looking through some old files I came across a math magazine I had bought in 1998. It was called Quantum and was published by the National Science Teachers Association in collaboration with the Russian magazine Kvant during the period 1990 to 2001 (coinciding with the Russian thaw, which in the following age of Putin seems eons ago). Fortunately, they are all online now. Besides some fascinating math articles the magazine contains a column of “Brainteasers.” Here is one of them:

“Alice used to walk to school every morning, and it took 20 minutes for her from door to door. Once on her way she remembered she was going to show the latest issue of Quantum to her classmates but had forgotten it at home. She knew that if she continued walking to school at the same speed, she’d be there 8 minutes before the bell, and if she went back home for the magazine she’d arrive at school 10 minutes late. What fraction of the way to school had she walked at that moment in time? (S. Dvorianinov)”

This is fairly straight-forward, but other problems in the magazine are a bit more challenging.

Answer.

See Calculating on the Way for solutions.

The Bicycle Problem

A fun, relatively new, Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos) has puzzles couched in terms of the Holmes-Watson banter. The following problem is a variation on the Sam Loyd Tandem Bicycle Puzzle.

“ ‘Here’s something mostly unrelated for you to chew over, my dear Watson. Say you and I have a single bicycle between us, and no other transport options save walking. We want to get the both of us to a location eighteen miles distant as swiftly as possible. If my walking speed is five miles per hour compared to your four, but for some reason—perhaps a bad ligament—my cycling speed is eight miles per hour compared to your ten. How would you get us simultaneously to our destination with maximum rapidity?’

‘A cab,’ I suggested.

‘Without cheating,’ Holmes replied, and went back to tossing his toast in the air.”

Answer.

See the Bicycle Problem for solutions.

Perpetual Meetings Problem

The following problem from Five Hundred Mathematical Challenges was a challenge indeed, even though it appeared to be a standard travel puzzle.

Problem 118. Andy leaves at noon and drives at constant speed back and forth from town A to town B. Bob also leaves at noon, driving at 40 km per hour back and forth from town B to town A on the same highway as Andy. Andy arrives at town B twenty minutes after first passing Bob, whereas Bob arrives at town A forty-five minutes after first passing Andy. At what time do Any and Bob pass each other for the nth time?”

Answer.

See the Perpetual Meetings Problem for solutions.

Impossible Car Riddle

This is another intriguing problem from Presh Talwalkar.

“A car travels 75 miles per hour (mph) downhill, 60 mph on flat roads, and 50 mph uphill. It takes 3 hours to go from town A to B, and it takes 3 hours and 30 minutes for return journey by the same route. What is the distance in miles between towns A and B?”

Answer.

See the Impossible Car Riddle for solutions.

Tandem Bicycle Puzzle

A glutton for punishment I considered another Sam Loyd puzzle:

“Three men had a tandem and wished to go just forty miles. It could complete the journey with two passengers in one hour, but could not carry the three persons at one time. Well, one who was a good pedestrian, could walk at the rate of a mile in ten minutes; another could walk in fifteen minutes, and the other in twenty. What would be the best possible time in which all three could get to the end of their journey?”

Answer.

See the Tandem Bicycle Puzzle for a solution.